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<title>베지에 곡선 입문</title>
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<header>
<h1>
베지에 곡선 입문<a class="rss-link" href="news/rss.xml"><img src="images/rss.png" /></a>
</h1>
<h2>베지에 곡선이 꼭 필요할 때 읽기 좋은 무료 온라인 서적.</h2>
<div>
<span>다른 언어로 읽기:</span>
<ul class="lang-switcher">
<li><a href="./index.html">English</a> &nbsp;</li>
<li><a href="./ja-JP/index.html">日本語</a> <span class="localisation-progress">(24%)</span></li>
<li><a href="./zh-CN/index.html">中文</a> <span class="localisation-progress">(37%)</span></li>
<li><a href="./ru-RU/index.html">Русский</a> <span class="localisation-progress">(24%)</span></li>
<li><a href="./uk-UA/index.html">Українська</a> <span class="localisation-progress">(2%)</span></li>
<li><a href="./ko-KR/index.html">한국어</a> <span class="localisation-progress">(9%)</span></li>
</ul>
<p>
(원하는 언어가 없거나, 100%까지 올리고 싶다면
<a href="https://github.com/Pomax/BezierInfo-2/wiki/help-localize-the-primer-on-bezier-curves">내용 번역을 도와주세요!</a>)
</p>
</div>
<p>
베지에 곡선 입문서에 오신 것을 환영합니다. 이곳은 베지에 곡선의 수학과 프로그래밍을 모두 다루는 무료 웹사이트/eBook으로, 포토샵 패스부터 CSS
easing 함수와 글꼴 테두리까지 어느 곳에서든 튀어나오는 바로 그 곡선을 다루고 그리는 데 대한 다양한 주제를 다룹니다.
</p>
<p>
이 책은 처음인가요? 어서 오세요! 이 입문서에 없는 것 중 꼭 다루었으면 좋겠는 것이 있다면
<a href="https://github.com/Pomax/BezierInfo-2/issues">이슈 트래커</a>로 의견을 나누어 주세요!
</p>
<h2>기부와 후원</h2>
<p>
이 책을 연구나 작업 목적, 혹은 직접 소프트웨어를 작성하는 데 참고하고 계시다면,
<a href="https://www.paypal.com/donate/?cmd=_s-xclick&hosted_button_id=3BNHGHZAS3DP6&locale.x=en_CA">일회성 기부</a>
(소액 기부도 괜찮습니다)나 <a href="https://www.patreon.com/bezierinfo">Patreon에서 후원자 등록</a>을 고려해 주세요. 저는 이 책을 유료료
집필하고 있지 않으니, 혹여나 이 사이트에서 도움을 받았고 오랫동안 유지되기를 원하신다면 지금까지 몇 년 동안 수백 잔의 커피를 마셨고 앞으로도
더 많은 커피가 필요하다는 것만 기억해 주세요. 커피 한 잔을 사주시는 것만으로도 이 자료를 보존하는 데 도움을 주실 수 있습니다.
</p>
<p>
특히 기업에서 이 책을 업무용이나 온보딩 참고서로 사용하고 계신다면, 부디 이 사이트에 후원을 부탁드립니다! 재무부에서 후원 문의를 해주신다면
기쁘게 수락하겠습니다.
</p>
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<br style="clear: both;" />
<p>— <a href="https://mastodon.social/@TheRealPomax">Pomax</a></p>
<noscript>
<div class="note">
<header>
<h2>This site (obviously) works best with JS enabled</h2>
<h3>But it's not required.</h3>
</header>
<p>
If you're reading this text block, then you have scripts disabled: thankfully, that's perfectly fine, and this site is not going to punish
you for making smart choices around privacy and security in your browser. All the content will show just fine, you can still read the
text, navigate to sections, and see the graphics that are used to illustrate the concepts that individual sections talk about.
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<strong>However</strong>, a big part of this primer's experience is the fact that all graphics are interactive, and for that to work, HTML
Custom Elements need to work, which requires Javascript to be enabled. If anything, you'll probably want to allow scripts to run just for
this site, and keep blocking everything else. Although that does mean you won't see comments, which use Disqus's comment system, and you
won't get convenient "share a link to the section you're reading right now" buttons, if that's something you like to do.
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<nav aria-labelledby="toc">
<h1 id="toc">목차</h1>
<h4>서문</h4>
<ol class="preamble">
<li><a href="ko-KR/index.html#preface">머리말 </a></li>
<li><a href="ko-KR/index.html#changelog">업데이트 내역</a></li>
</ol>
<h4>본문</h4>
<ol>
<li><a href="ko-KR/index.html#introduction">초고속 소개</a></li>
<li><a href="ko-KR/index.html#whatis">베지에 곡선은 어떻게 만드나요?</a></li>
<li><a href="ko-KR/index.html#explanation">베지에 곡선을 이루는 수학</a></li>
<li><a href="ko-KR/index.html#control">Controlling Bézier curvatures</a></li>
<li><a href="ko-KR/index.html#weightcontrol">Controlling Bézier curvatures, part 2: Rational Béziers</a></li>
<li><a href="ko-KR/index.html#extended">The Bézier interval [0,1]</a></li>
<li><a href="ko-KR/index.html#matrix">Bézier curvatures as matrix operations</a></li>
<li><a href="ko-KR/index.html#decasteljau">de Casteljau's algorithm</a></li>
<li><a href="ko-KR/index.html#flattening">Simplified drawing</a></li>
<li><a href="ko-KR/index.html#splitting">Splitting curves</a></li>
<li><a href="ko-KR/index.html#matrixsplit">Splitting curves using matrices</a></li>
<li><a href="ko-KR/index.html#reordering">Lowering and elevating curve order</a></li>
<li><a href="ko-KR/index.html#derivatives">Derivatives</a></li>
<li><a href="ko-KR/index.html#pointvectors">Tangents and normals</a></li>
<li><a href="ko-KR/index.html#pointvectors3d">Working with 3D normals</a></li>
<li><a href="ko-KR/index.html#components">Component functions</a></li>
<li><a href="ko-KR/index.html#extremities">Finding extremities: root finding</a></li>
<li><a href="ko-KR/index.html#boundingbox">Bounding boxes</a></li>
<li><a href="ko-KR/index.html#aligning">Aligning curves</a></li>
<li><a href="ko-KR/index.html#tightbounds">Tight bounding boxes</a></li>
<li><a href="ko-KR/index.html#inflections">Curve inflections</a></li>
<li><a href="ko-KR/index.html#canonical">The canonical form (for cubic curves)</a></li>
<li><a href="ko-KR/index.html#yforx">Finding Y, given X</a></li>
<li><a href="ko-KR/index.html#arclength">Arc length</a></li>
<li><a href="ko-KR/index.html#arclengthapprox">Approximated arc length</a></li>
<li><a href="ko-KR/index.html#curvature">Curvature of a curve</a></li>
<li><a href="ko-KR/index.html#tracing">Tracing a curve at fixed distance intervals</a></li>
<li><a href="ko-KR/index.html#intersections">Intersections</a></li>
<li><a href="ko-KR/index.html#curveintersection">Curve/curve intersection</a></li>
<li><a href="ko-KR/index.html#abc">The projection identity</a></li>
<li><a href="ko-KR/index.html#pointcurves">Creating a curve from three points</a></li>
<li><a href="ko-KR/index.html#projections">Projecting a point onto a Bézier curve</a></li>
<li><a href="ko-KR/index.html#circleintersection">Intersections with a circle</a></li>
<li><a href="ko-KR/index.html#molding">Molding a curve</a></li>
<li><a href="ko-KR/index.html#curvefitting">Curve fitting</a></li>
<li><a href="ko-KR/index.html#catmullconv">Bézier curves and Catmull-Rom curves</a></li>
<li><a href="ko-KR/index.html#catmullfitting">Creating a Catmull-Rom curve from three points</a></li>
<li><a href="ko-KR/index.html#polybezier">Forming poly-Bézier curves</a></li>
<li><a href="ko-KR/index.html#offsetting">Curve offsetting</a></li>
<li><a href="ko-KR/index.html#graduatedoffset">Graduated curve offsetting</a></li>
<li><a href="ko-KR/index.html#circles">Circles and quadratic Bézier curves</a></li>
<li><a href="ko-KR/index.html#circles_cubic">Circular arcs and cubic Béziers</a></li>
<li><a href="ko-KR/index.html#arcapproximation">Approximating Bézier curves with circular arcs</a></li>
<li><a href="ko-KR/index.html#bsplines">B-Splines</a></li>
<li><a href="ko-KR/index.html#comments">Comments and questions</a></li>
</ol>
</nav>
</header>
<main>
<section id="preface">
<h1>머리말</h1>
<p>
2D에 그림을 그리려면 보통 선을 사용하고, 선이라고 하면 직선과 곡선의 두 종류로 나뉩니다. 전자는 컴퓨터에게 그리기를 시킬 줄 안다면 쉽게 그릴
수 있습니다. 시작점과 끝점만 입력하면 짜잔! 직선이 나왔습니다. 어렵지 않죠.
</p>
<p>
하지만 곡선은 훨씬 까다롭습니다. 손그림으로는 곡선 그리기가 그렇게 쉬울 수 없었겠지만, 컴퓨터는 곡선을 어떻게 그릴지 기술하는 수학적 함수
없이 곡선을 그릴 수 없다는 제한이 있습니다. 실은 직선을 그릴 때조차도 함수가 필요하지만, 어차피 직선의 함수는 간단하니 컴퓨터를 사용할 때는
보통 이 사실을 무시합니다. 직선이든 곡선이든 모든 선은 "함수"입니다. 그러나 선 그리기에 함수가 필요하기에, 빠르게 계산할 수 있고 컴퓨터로
그렸을 때 보기도 좋은 함수를 만들어야 할 필요성이 생깁니다. 이런 함수는 여러 종류가 있는데, 이 글에서 다룰 것은 이 중 많은 주목을 받고
곡선을 그릴 수 있으면 어디서나 쓰이는 것, 바로 베지에 곡선입니다.
</p>
<p>
베지에 곡선이라는 이름은 이 곡선이 설계 작업을 하기에 좋다고 세상에 알린
<a href="https://en.wikipedia.org/wiki/Pierre_B%C3%A9zier">피에르 베지에</a>의 이름을 땄지만(1962년에 르노 소속으로서 연구 결과를 발표함),
베지에가 이 곡선을 "발명"한 최초의, 혹은 유일한 사람은 아닙니다. 베지에 곡선을 처음 발견한 사람은 1959년에 시트로엥에서 일하면서 이 곡선의
성질을 연구하고 이 곡선을 그리는 정말 우아한 방법을 찾아낸 <a href="https://en.wikipedia.org/wiki/Paul_de_Casteljau">폴 드 카스텔죠</a>라고
하는 경우도 있습니다. 그러나 드 카스텔죠는 본인의 연구 결과를 발표하지 않았기 때문에 누가 먼저인지를 단언하기는 어렵습니다... 정말일까요?
사실 베지에 곡선은 근본적으로 "베른시테인 다항식"으로,
<a
href="https://ko.wikipedia.org/wiki/%EC%84%B8%EB%A5%B4%EA%B2%8C%EC%9D%B4_%EB%82%98%ED%83%80%EB%85%B8%EB%B9%84%EC%B9%98_%EB%B2%A0%EB%A5%B8%EC%8B%9C%ED%85%8C%EC%9D%B8"
>세르게이 나타노비치 베른시테인</a
>이 연구하여 자그마치 1912년에 발표한 수학적 함수의 집합입니다.
</p>
<p>
아무튼 위의 내용은 잡설이었고, 독자가 관심을 가질 만한 내용은 이 곡선이 편리하다는 것입니다. 여러 개의 베지에 곡선을 연결해서 하나의
곡선처럼 보이게 만들 수도 있습니다. 포토샵에서 패스를 한 번이라도 그려봤거나, 플래시, 일러스트레이터, 잉크스케이프 등 벡터 드로잉 프로그램을
써본 적이 있다면, 이 프로그램에서 그동안 그리던 곡선이 전부 베지에 곡선입니다.
</p>
<p>
그런데 이를 직접 프로그래밍하려면 어떻게 해야 할까요? 함정이 어디에 도사리고 있을까요? 곡선의 바운딩 박스는 어떻게 구하고, 교차 판정은
어떻게 하고, 밖으로 튀어나오게는 어떻게 하고... 간단히 말해서, 이 곡선으로 할 만한 온갖 것들을 어떻게 하는 것이 좋을까요? 그것이 바로 이
책의 존재 이유입니다. 수학 할 준비 되셨나요?
</p>
<div class="note">
<h2>거의 모든 베지에 그래픽이 인터랙티브입니다.</h2>
<p>
이 페이지에서는 <a href="https://pomax.github.io/bezierjs/">Bezier.js</a>를 사용하여 인터랙티브 예제를 제공하며, 수학 공식은
<a href="https://ctan.org/pkg/xetex">XeLaTeX</a> 조판 체계를 이용해 조판하고 <a href="https://cityinthesky.co.uk/">David Barton</a>님의
<a href="https://github.com/dawbarton/pdf2svg">pdf2svg</a>를 이용해 SVG로 변환하여 표시하고 있습니다.
</p>
<h2>이 책은 오픈 소스입니다.</h2>
<p>
이 책은 오픈 소스 소프트웨어 프로젝트의 일환으로, 소스 코드가 GitHub 레포지토리 두 곳에 보관되어 있습니다. 하나는
<a href="https://github.com/pomax/bezierinfo">https://github.com/pomax/bezierinfo</a>로, 지금 읽고 계시는 완전히 공개용인 버전입니다. 다른
하나는 개발 버전인 <a href="https://github.com/pomax/BezierInfo-2">https://github.com/pomax/BezierInfo-2</a>로
<em>웹 버전으로 변환되는</em> 코드가 보관되어 있습니다. 버그를 찾아서 제보하려고 하거나 입문서의 내용을 수정하거나 추가할 아이디어가
있다면 이곳으로 보내 주세요.
</p>
<h2>이 책의 수식은 얼마나 어렵나요?</h2>
<p>
입문서에 등장하는 대부분의 수식은 고등학교 수준입니다. 기본적인 산수 계산을 할 수 있고 영어를 읽을 줄 안다면 대부분 이해할 수 있을
정도입니다. <em>훨씬</em> 어려운 수식도 가끔 등장하지만, 너무 어려워 보이는 상자 안의 세부 내용을 건너뛰거나, 읽고 있는 장의 끝으로
건너뛰어도 좋습니다. 각 장의 끝에는 도출된 값만 활용할 수 있도록 결론을 단순 나열합니다.
</p>
<h2>예제 코드는 무슨 언어로 되어 있나요?</h2>
<p>
특정한 프로그래밍 언어 하나를 선호하기에는 프로그래밍 언어가 너무 많기 때문에, 입문서에 수록된 모든 예제는 JS나 Python 등 현대적인
스크립트 언어와 문법이 어느 정도 비슷하지만 같지는 않은 의사코드로 작성했습니다. 그렇기 때문에 예제 코드를 생각 없이 복사해서 쓸 수
없지만, 이는 의도적인 결정입니다. 이 입문서를 보고 계시는 것은 웬만하면 <em>배우기 위해서</em>이고, 배움은 복붙으로 이루어지지 않죠.
학습은 무언가를 직접 해 보고, <em>실수를 하며</em>, 그 실수를 고치면서 이루어집니다. 독자들이 실수를 하라고 일부러 예제 코드에 오류를
넣는다거나 하는 것은 당연히 아니지만(그건 끔찍하겠죠!), 어떤 한 프로그래밍 언어에 편중된 코드는 <em>의도적으로</em> 지양하고 있습니다.
절차적 프로그래밍 언어를 하나라도 알고 있다면 예제 코드를 읽는 데는 전혀 어려움이 없으니 걱정은 하지 말아주세요.
</p>
<h2>질문이나 의견이 있다면</h2>
<p>
새로운 장을 제안하고 싶다면, <a href="https://github.com/pomax/BezierInfo-2/issues">GitHub 이슈 트래커</a>를 찾아 주세요(오른쪽 위에
링크한 레포지토리에서도 확인할 수 있습니다). 코드 재작성이 끝날 때까지는 댓글창이 없을 예정이지만, 내용에 대한 질문이 있다면 역시 이슈
트래커를 이용할 수 있습니다. 재작성을 마치면 내용 전반에 대한 댓글창을 추가하고, 확실치는 않지만 아마 "이 부분을 마우스로 선택하고 질문
버튼을 눌러서 질문하기" 시스템을 만들 수도 있겠습니다.
</p>
<h2>이 책을 지원해 주세요!</h2>
<p>
이 책을 재미있게 읽었거나 하려고 했던 작업에 도움이 되었고, 제게 고마움을 표하고 싶다면 두 가지 방법이 있습니다. 이 책의
<a href="https://www.patreon.com/bezierinfo">Patreon 페이지</a>에서 정기 후원을 하거나, 일회성 후원을 하고 싶다면
<a href="https://www.paypal.com/donate/?cmd=_s-xclick&hosted_button_id=3BNHGHZAS3DP6&locale.x=en_CA">buy Pomax a coffee</a> 페이지를 찾아
주세요. 이 책은 지난 몇 년 동안 베지에 곡선에 관한 조그마한 입문서에서 종이책 100페이지짜리 교재로 커졌고, 책을 쓰면서 수백 잔의 커피가
들었습니다. 저는 이 책을 쓰게 된 것이 전혀 후회되지 않지만, 약간의 커피만 더 있으면 집필을 이어나갈 수 있겠죠!
</p>
</div>
</section>
<section id="changelog">
<h1>업데이트 내역</h1>
<p>
이 입문서는 "살아 숨쉬는" 문서로, 오랜만에 다시 찾아왔다면 새로운 내용이 추가되었을 수도 있습니다. 아래 링크를 눌러서 무엇이 언제
추가되었는지 확인할 수도 있고, <a href="./news">뉴스 포스트</a>(영문, <a href="./news/rss.xml">RSS 피드</a>도 있습니다)에서 자세한 내역을
확인할 수도 있습니다.
</p>
<!-- non-JS content reveals are nice -->
<label for="changelogtoggle">수정사항 확인</label>
<input type="checkbox" id="changelogtoggle" />
<section>
<h2>November 2020</h2>
<ul>
<li><p>Added a section on finding curve/circle intersections</p></li>
</ul>
<h2>October 2020</h2>
<ul>
<li><p>Added the Ukranian locale! Help out in getting its localization to 100%!</p></li>
</ul>
<h2>August-September 2020</h2>
<ul>
<li>
<p>
Completely overhauled the site: the Primer is now a normal web page that works fine with JS disabled, but obviously better with JS
turned on.
</p>
</li>
</ul>
<h2>June 2020</h2>
<ul>
<li><p>Added automatic CI/CD using Github Actions</p></li>
</ul>
<h2>January 2020</h2>
<ul>
<li><p>Added reset buttons to all graphics</p></li>
<li><p>Updated to preface to correctly describe the on-page maths</p></li>
<li><p>Fixed the Catmull-Rom section because it had glaring maths errors</p></li>
</ul>
<h2>August 2019</h2>
<ul>
<li><p>Added a section on (plain) rational Bezier curves</p></li>
<li><p>Improved the Graphic component to allow for sliders</p></li>
</ul>
<h2>December 2018</h2>
<ul>
<li><p>Added a section on curvature and calculating kappa.</p></li>
<li>
<p>
Added a Patreon page! Head on over to <a href="https://www.patreon.com/bezierinfo">patreon.com/bezierinfo</a> to help support this
site!
</p>
</li>
</ul>
<h2>August 2018</h2>
<ul>
<li><p>Added a section on finding a curve&#39;s y, if all you have is the x coordinate.</p></li>
</ul>
<h2>July 2018</h2>
<ul>
<li><p>Rewrote the 3D normals section, implementing and explaining Rotation Minimising Frames.</p></li>
<li><p>Updated the section on curve order raising/lowering, showing how to get a least-squares optimized lower order curve.</p></li>
<li>
<p>(Finally) updated &#39;npm test&#39; so that it automatically rebuilds when files are changed while the dev server is running.</p>
</li>
</ul>
<h2>June 2018</h2>
<ul>
<li><p>Added a section on direct curve fitting.</p></li>
<li><p>Added source links for all graphics.</p></li>
<li><p>Added this &quot;What&#39;s new?&quot; section.</p></li>
</ul>
<h2>April 2017</h2>
<ul>
<li><p>Added a section on 3d normals.</p></li>
<li><p>Added live-updating for the social link buttons, so they always link to the specific section you&#39;re reading.</p></li>
</ul>
<h2>February 2017</h2>
<ul>
<li><p>Finished rewriting the entire codebase for localization.</p></li>
</ul>
<h2>January 2016</h2>
<ul>
<li><p>Added a section to explain the Bezier interval.</p></li>
<li><p>Rewrote the Primer as a React application.</p></li>
</ul>
<h2>December 2015</h2>
<ul>
<li><p>Set up the split repository between BezierInfo-2 as development repository, and bezierinfo as live page.</p></li>
<li>
<p>
Removed the need for client-side LaTeX parsing entirely, so the site doesn&#39;t take a full minute or more to load all the graphics.
</p>
</li>
</ul>
<h2>May 2015</h2>
<ul>
<li><p>Switched over to pure JS rather than Processing-through-Processing.js</p></li>
<li><p>Added Cardano&#39;s algorithm for finding the roots of a cubic polynomial.</p></li>
</ul>
<h2>April 2015</h2>
<ul>
<li><p>Added a section on arc length approximations.</p></li>
</ul>
<h2>February 2015</h2>
<ul>
<li><p>Added a section on the canonical cubic Bezier form.</p></li>
</ul>
<h2>November 2014</h2>
<ul>
<li><p>Switched to HTTPS.</p></li>
</ul>
<h2>July 2014</h2>
<ul>
<li><p>Added the section on arc approximation.</p></li>
</ul>
<h2>April 2014</h2>
<ul>
<li><p>Added the section on Catmull-Rom fitting.</p></li>
</ul>
<h2>November 2013</h2>
<ul>
<li><p>Added the section on Catmull-Rom / Bezier conversion.</p></li>
<li><p>Added the section on Bezier cuves as matrices.</p></li>
</ul>
<h2>April 2013</h2>
<ul>
<li><p>Added a section on poly-Beziers.</p></li>
<li><p>Added a section on boolean shape operations.</p></li>
</ul>
<h2>March 2013</h2>
<ul>
<li><p>First drastic rewrite.</p></li>
<li><p>Added sections on circle approximations.</p></li>
<li><p>Added a section on projecting a point onto a curve.</p></li>
<li><p>Added a section on tangents and normals.</p></li>
<li><p>Added Legendre-Gauss numerical data tables.</p></li>
</ul>
<h2>October 2011</h2>
<ul>
<li>
<p>
First commit for the <a href="https://pomax.github.io/bezierinfo/">bezierinfo</a> site, based on the pre-Primer webpage that covered
the basics of Bezier curves in HTML with Processing.js examples.
</p>
</li>
</ul>
</section>
</section>
<section id="chapters">
<section id="introduction">
<h1>
<div class="nav"><a href="#toc">목차</a><a href="ko-KR/index.html#whatis">다음</a></div>
<a href="ko-KR/index.html#introduction">초고속 소개</a>
</h1>
<p>
간단하게 시작해 봅시다. "베지에 곡선"이라고 하면 아래 그림에서 볼 수 있는 형태의 것들을 말하는 것입니다. 어떤 시작점에서 시작해서 끝점에서
끝나고, 하나 이상의 "중간" 조절점이 그 사이의 모양에 관여합니다. 이 페이지에 있는 모든 그림은 인터랙티브이니 아래 곡선을 마음 가는 대로
바꿔 보세요! 조절점을 끌어 옮기면서 이에 따라 곡선의 모양이 어떻게 변하는지 살펴보세요.
</p>
<div class="figure">
<graphics-element
title="2차 베지에 곡선"
width="275"
height="275"
src="./chapters/introduction/quadratic.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/introduction/54e9ec0600ac436b0e6f0c6b5005cf03.png" loading="lazy" />
<label>2차 베지에 곡선</label>
</fallback-image></graphics-element
>
<graphics-element
title="3차 베지에 곡선"
width="275"
height="275"
src="./chapters/introduction/cubic.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/introduction/8d158a13e9a86969b99c64057644cbc6.png" loading="lazy" />
<label>3차 베지에 곡선</label>
</fallback-image></graphics-element
>
</div>
<p>
이런 곡선은 컴퓨터 지원 설계 및 제조(CAD/CAM) 분야뿐만 아니라 어도비 일러스트레이터, 어도비 포토샵, 잉크스케이프, GIMP 등의 그래픽 디자인
프로그램과 SVG, 오픈타입 글꼴(TTF/OTF) 등의 그래픽 기술에서도 자주 사용하고 있습니다. 정말 많은 곳에서 베지에 곡선을 쓰고 있으니, 베지에
곡선을 더 자세히 알아보고 싶다면... 이 책을 마저 읽어 봅시다!
</p>
</section>
<section id="whatis">
<h1>
<div class="nav">
<a href="ko-KR/index.html#introduction">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#explanation">다음</a>
</div>
<a href="ko-KR/index.html#whatis">베지에 곡선은 어떻게 만드나요?</a>
</h1>
<p>
곡선의 점들을 이리저리 옮겨보고 나면 베지에 곡선이 어떤 모양을 하는지 어느 정도 감이 올 것입니다. 그런데 베지에 곡선이라는 게 정말
<em>무엇</em>일까요? 베지에 곡선의 정체는 두 가지 방법으로 설명할 수 있고, 알고 보면 둘 다 완전히 동일하지만, 하나는 복잡한 수식을 쓰고
다른 하나는 정말 쉬운 수식을 씁니다. 그러니까... 일단 쉬운 설명부터 들어 봅시다.
</p>
<p>
베지에 곡선은 <a href="https://ko.wikipedia.org/wiki/%EC%84%A0%ED%98%95_%EB%B3%B4%EA%B0%84%EB%B2%95">선형 보간</a>의 결과물입니다. 말로만
들으면 어려워 보이지만 사실은 독자 여러분도 어릴 때부터 선형 보간을 한 적이 있을 것입니다. 무언가 둘 사이에 있는 것을 가리킬 때 했던 것이
바로 선형 보간, 즉 "두 점 사이에 점을 하나 찍는 것"입니다.
</p>
<p>
이 두 점 사이의 거리를 알고 있고, 예를 들어서 새로운 점을 한 점에서 20% 거리에(즉, 다른 점에서 80% 거리에) 찍으려고 한다면 이 점의 위치는
정말 쉽게 계산할 수 있습니다.
</p>
<!--
\setmainfont[Ligatures=TeX]TeX Gyre Pagella \setmathfontTeX Gyre Pagella Math
╭ p = 한 점 ╮
│ 1 │
│ p = 다른 점 │
│ 2 │
│ 거리= (p - p ) │ 이라 할 때, 새 점의 위치 = p + 거리 · 비율
│ 2 1 │ 1
│ 퍼센트 │
│ 비율= ──── │
╰ 100 ╯
-->
<img class="LaTeX SVG" src="./images/chapters/whatis/00a671609da73ff60d06f4a0e1989b15.svg" width="451px" height="105px" loading="lazy" />
<p>
눈으로도 직접 확인해 봅시다. 아래 그림은 상하 방향키로 보간 비율을 늘리거나 줄이면서 어떻게 바뀌는지 확인할 수 있는 인터랙티브
그래픽입니다. 우선 세 점으로 시작해서 선분을 두 개 그립니다. 이 두 선분에 선형 보간을 해서 두 점을 얻을 수 있고, 이 두 점 사이에 다시 선형
보간을 하면 한 점이 됩니다. 바로 이 점(과 모든 비율에 대해 같은 방법으로 취한 모든 점)이 베지에 곡선이 됩니다.
</p>
<graphics-element
title="선형 보간으로 베지에 곡선을 얻는 과정"
width="825"
height="275"
src="./chapters/whatis/interpolation.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/whatis/524dd296e96c0fe2281fb95146f8ea65.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="10" max="90" step="1" value="25" class="slide-control" />
</graphics-element>
<p>바로 여기서 복잡한 수학... 미적분이 등장합니다.</p>
<p>
전혀 그렇게 느껴지지는 않겠지만, 우리가 여기서 한 것이 바로 이차곡선을 (한 획이 아니라 여러 단계에 걸쳐) 그리는 작업입니다. 베지에 곡선의
신기한 점 중 하나가 같은 대상을 다항함수로도, 혹은 간단하게 보간의 보간의 ... 보간으로도 나타낼 수 있다는 것입니다. 다시 말하자면, 우리는
이 곡선의 성질을 "진짜 수학"을 통해서든(곡선의 함수와 도함수 따위를 이용해서), "기계적" 구성을 통해서든(예를 들어, 베지에 곡선은 곡선을
그리는 데 사용한 점 밖으로 빠져나가지 않음을 알 수 있습니다) 탐구할 수 있습니다.
</p>
<p>
이제부터는 베지에 곡선을 조금 더 자세히 ― 수식으로 어떻게 표현하는지, 어떤 성질을 가지고 있는지, 어떻게 변형하고 어떻게 쓸 수 있는지 ―
알아보도록 하겠습니다.
</p>
</section>
<section id="explanation">
<h1>
<div class="nav"><a href="ko-KR/index.html#whatis">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#control">다음</a></div>
<a href="ko-KR/index.html#explanation">베지에 곡선을 이루는 수학</a>
</h1>
<p>
베지에 곡선은 "매개변수" 함수의 일종입니다. 수학적으로 말하자면 매개변수 함수는 꼼수입니다. "함수"란 임의의 개수의 입력에서
<strong>하나의</strong> 출력으로 가는 대응을 의미하는, 엄연한 정의가 있는 용어입니다. 수를/수들을 넣으면 하나의 값이 나옵니다. 다른 수를
넣어도 값이 하나만 나오는 점은 변하지 않습니다.
</p>
<p>
매개변수 함수는 꼼수입니다. "저는 여러 값이 나오는 함수가 필요한데 그냥 함수를 여러 개 쓰면 안 돼요?"인 것이죠. 예컨대 어떤
값(<i>x</i>라고 합시다)을 어떤 연산을 통해 다른 값과 대응시키는 함수가 있다고 합시다.
</p>
<!--
f(x) = cos (x)
-->
<img class="LaTeX SVG" src="./images/chapters/explanation/0cc876c56200446c60114c1b0eeeb2cc.svg" width="96px" height="17px" loading="lazy" />
<p>
<i>f(x)</i>라는 표기는 <i>f</i>(함수가 하나만 있을 때는 관습적으로 <i>f</i>를 씁니다)가 함수이며, 그 값이 하나의 변수(여기서는 <i>x</i>)에
따라 바뀜을 나타내는 표준적인 방법입니다. <i>x</i>를 바꾸면 <i>f(x)</i>의 값도 바뀝니다.
</p>
<p>여기까지는 좋습니다. 이제 매개변수 함수를 보고, 어떤 꼼수를 쓰는지 확인해 봅시다. 아래의 두 함수를 생각해 봅시다.</p>
<!--
f(a) = cos (a)
f(b) = sin (b)
-->
<img class="LaTeX SVG" src="./images/chapters/explanation/a2891980850ddbb27d308ac112d69f74.svg" width="93px" height="36px" loading="lazy" />
<p>
별 특이할 점이 없는 사인과 코사인 함수인데, 자세히 보면 두 입력이 다른 이름을 가지는 것을 확인할 수 있습니다. <i>a</i>의 값을 바꾸더라도
<i>f(b)</i>에서는 <i>a</i>를 쓰지 않으므로 <i>f(b)</i>의 값이 바뀌지는 않습니다. 바로 이 부분을 조금 고치는 것이 매개변수 함수의
꼼수입니다. 매개변수 함수에서는 아래와 같이 여러 가지 함수가 같은 변수를 공유합니다.
</p>
<!--
╭ f (t) = cos (t)
╡ a
│ f (t) = sin (t)
╰ b
-->
<img
class="LaTeX SVG"
src="./images/chapters/explanation/7acc94ec70f053fd10dab69d424b02a6.svg"
width="100px"
height="40px"
loading="lazy"
/>
<p>
함수는 여러 개인데 변수는 하나뿐이네요. <i>t</i>의 값을 바꾸면 <i>f<sub>a</sub>(t)</i>와 <i>f<sub>b</sub>(t)</i>의 값이 모두 바뀝니다.
이게 어딜 봐서 유용한지 궁금해하는 독자가 있다면, 답은 의외로 간단합니다. <i>f<sub>a</sub>(t)</i>와 <i>f<sub>b</sub>(t)</i>라는 이름표를
매개변수 곡선을 만들 때 보통 의미하는 바를 나타내도록 바꾸면 이 식이 눈에 더 잘 들어올지도 모르겠습니다.
</p>
<!--
{ x = cos (t)
y = sin (t)
-->
<img class="LaTeX SVG" src="./images/chapters/explanation/6914ba615733c387251682db7a3db045.svg" width="77px" height="40px" loading="lazy" />
<p>짜잔! <i>x</i>/<i>y</i>좌표가 어떤 미지의 값 <i>t</i>를 통해 이어져 있었습니다.</p>
<p>
정리하자면, 매개변수 곡선은 보통의 함수처럼 <i>y</i>좌표를 <i>x</i>좌표값에 대해 정의하는 대신 두 값을 어떤 "조절" 변수로 연결합니다.
<i>t</i>의 값을 바꿀 때마다 매개변수 함수에서 <strong>두</strong> 값을 얻을 수 있고, 이 값을 그래프에 (<i>x</i>,<i>y</i>) 좌표로 나타낼 수
있습니다. 예를 들어 위에서 소개한 매개변수 함수는 원 위의 점을 생성합니다. <i>t</i>의 값을 음의 무한대에서 양의 무한대까지 잡을 수 있고,
여기서 나오는 (<i>x</i>,<i>y</i>) 좌표는 항상 중심이 원점 (0,0)이고 반지름이 1인 원 위에 있습니다. <i>t</i>의 값 0부터 5까지에 대해
그래프를 그리면 다음과 같습니다.
</p>
<graphics-element
title="원(의 일부): x=sin(t), y=cos(t)"
width="275"
height="275"
src="./chapters/explanation/circle.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/explanation/959762e39ae32407e914a687d804ff3a.png" loading="lazy" />
<label>원(의 일부): x=sin(t), y=cos(t)</label>
</fallback-image>
<input type="range" min="0" max="10" step="0.1" value="5" class="slide-control" />
</graphics-element>
<p>
베지에 곡선은 여러 종류의 매개변수 함수 중 한 종류이고, 모든 출력값을 같은 기초 함수로 구합니다. 위에서 살펴본 예제에서는 <i>x</i>와
<i>y</i> 값을 다른 함수로 구했지만(하나는 사인, 다른 하나는 코사인), 베지에 곡선은 <i>x</i>, <i>y</i> 값을 모두 "이항 다항식"으로
구합니다. 이항 다항식은 또 뭘까요?
</p>
<p>중·고등학교에서 다항식을 배운 기억이 나시나요? 이렇게 생긴 합을 다항식이라고 합니다.</p>
<!--
3 2
f(x) = a · x + b · x + c · x + d
-->
<img
class="LaTeX SVG"
src="./images/chapters/explanation/855a34c7f72733be6529c3fb33fa1a23.svg"
width="213px"
height="20px"
loading="lazy"
/>
<p>
여기서 가장 높은 차수의 항이 <i>x³</i>이면 3차( 다항)식, <i>x²</i>이면 2차식이라 하고, <i>x</i>라면 1차식, 즉 직선이 됩니다. (혹시
<i>x</i>가 있는 항이 하나도 없다면 그건 애초에 다항식이 아니죠!)
</p>
<p>
베지에 곡선은 <i>x</i>가 아니라 <i>t</i>에 대한 다항식이고, <i>t</i>의 값이 0과 1 사이로 고정되며 <i>a</i>, <i>b</i> 등의 계수가
"이항계수"의 형태를 띱니다. 이항계수라고 하면 어려워 보이지만 사실은 두 값을 섞는 것을 간단하게 나타낸 것입니다.
</p>
<!--
1차= (1-t) + t
2 2
2차= (1-t) + 2 · (1-t) · t + t
3 2 2 3
3차= (1-t) + 3 · (1-t) · t + 3 · (1-t) · t + t
-->
<img
class="LaTeX SVG"
src="./images/chapters/explanation/28d5dfe5c9275ee236560041d9107938.svg"
width="340px"
height="64px"
loading="lazy"
/>
<p>
제가 한 말이긴 하지만 저 식이 딱히 간단해 보이지는 않죠! 그런데 여기서 <i>t</i>를 없애고 대신 "곱하기 1"을 넣으면, 복잡한 수식이 갑자기
간단하게 바뀝니다. 밑에 늘어놓은 이항계수를 보세요.
</p>
<!--
1차= 1 + 1
2차= 1 + 2 + 1
3차= 1 + 3 + 3 + 1
4차= 1 + 4 + 6 + 4 + 1
-->
<img
class="LaTeX SVG"
src="./images/chapters/explanation/085719517aacb7e37072113fd63040b0.svg"
width="153px"
height="83px"
loading="lazy"
/>
<p>
식을 잘 보면 2는 1+1과 같고, 3은 2+1과 1+2이고, 6은 3+3...이라는 것을 알 수 있습니다. 보신 바와 같이 한 차수가 높아질 때마다 양 끝은
1이고, 그 사이의 모든 수는 "윗줄에 있는 두 수를 더한 것"과 같으며, 이렇게 만들어진 모양을
<a href="https://ko.wikipedia.org/wiki/%ED%8C%8C%EC%8A%A4%EC%B9%BC%EC%9D%98_%EC%82%BC%EA%B0%81%ED%98%95">파스칼의 삼각형</a>이라고 합니다.
<i>이건</i> 기억하기 쉽죠?
</p>
<p>
다항식의 문자 부분도 쉽게 외울 수 있는 방법이 있습니다. <i>(1-t)</i>를 <i>a</i>로, <i>t</i>를 <i>b</i>로 바꾸고 계수를 잠깐 지우면 아래의
식을 얻습니다.
</p>
<!--
1차= \colorblue a + \colorred b
2차= \colorblue a · \colorblue a + \colorblue a · \colorred b + \colorred b · \colorred b
3차= \colorblue a · \colorblue a · \colorblue a + \colorblue a · \colorblue a · \colorred b + \colorblue a · \colorred b · \colorred b + \colorr
ed b · \colorred b · \colorred b
-->
<img
class="LaTeX SVG"
src="./images/chapters/explanation/43ec2ac2931d7c620736b56afa69d37b.svg"
width="275px"
height="60px"
loading="lazy"
/>
<p>
이제 이 식은 "<i>a</i>와 <i>b</i>의 모든 조합의 합"인데, + 부호가 나올 때마다 <i>a</i>가 하나씩 <i>b</i>로 바뀌는 것으로 외울 수 있습니다.
이쪽도 그렇게 어렵지 않았네요. 이것으로 이항 다항식을 모두 배웠고, 혹시 필요할 경우를 위해 아래에 일반항의 식을 추가합니다.
</p>
<!--
__ n n-i i
Bézier(n,t) = ❯ \underset 이항계수 부분\underbrace\binomni · \ \underset 문자 부분\underbrace(1-t) · t
‾‾ i=0
-->
<img
class="LaTeX SVG"
src="./images/chapters/explanation/0a5182969272e9b5c229bdd6e8ce7e6f.svg"
width="261px"
height="52px"
loading="lazy"
/>
<p>
위 식으로 베지에 곡선을 완전히 기술할 수 있습니다. 이때 위 함수에서 Σ(시그마) 기호는 여러 개의 값을 더한다는 의미입니다(Σ 기호 밑에 적힌
문자를 ...=&lt;값&gt;에서 시작해서 Σ 위에 적힌 값까지 바꿔가면서 더합니다).
</p>
<div class="howtocode">
<h3>기저 함수를 구현하는 방법</h3>
<p>기저 함수를 구현할 때, 아래와 같이 위에서 본 수식을 그대로 함수로 옮겨서 구현할 수는 있습니다.</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="5">
<textarea disabled rows="5" role="doc-example">
function Bezier(n,t):
sum = 0
for(k=0; k<n; k++):
sum += n!/(k!*(n-k)!) * (1-t)^(n-k) * t^(k)
return sum</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
</table>
<p>
구현할 "수는" 있다고 한 것은 이렇게 구현하지 않을 것이기 때문입니다. 계승 함수는 <em>매우</em> 무겁고, 위에서 배운 것과 같이 계승을 쓰지
않고도 파스칼의 삼각형은 쉽게 만들 수 있습니다. [1]부터 시작하고, 그 다음에는 [1,1], [1,2,1], [1,3,3,1]과 같은 식으로, 줄이 바뀔 때마다
수 하나를 더 넣고 양쪽 끝은 1로, 그 사이의 모든 수는 지금 계산하고 있는 줄 바로 위의 양쪽 원소의 합으로 채우면 됩니다.
</p>
<p>위의 삼각형은 리스트의 리스트로 눈 깜짝할 새 구현할 수 있고, 이제 룩업 테이블을 만들었으니 이항계수는 다시 구할 필요도 없습니다.</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="18">
<textarea disabled rows="18" role="doc-example">
lut = [ [1], // n=0
[1,1], // n=1
[1,2,1], // n=2
[1,3,3,1], // n=3
[1,4,6,4,1], // n=4
[1,5,10,10,5,1], // n=5
[1,6,15,20,15,6,1]] // n=6
function binomial(n,k):
while(n >= lut.length):
s = lut.length
nextRow = new array(size=s+1)
nextRow[0] = 1
for(i=1, prev=s-1; i<s; i++):
nextRow[i] = lut[prev][i-1] + lut[prev][i]
nextRow[s] = 1
lut.add(nextRow)
return lut[n][k]</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
<tr>
<td>18</td>
</tr>
</table>
<p>
코드를 해석해 볼까요? 우선 웬만한 <i>n/k</i> 값을 모두 다룰 수 있는 적당한 크기로 룩업 테이블을 선언합니다. 그 다음에는 필요한 값을
읽어오는 함수를 선언하고, 인자로 전달된 <i>n/k</i> 값이 테이블에 없으면 우선 테이블을 키우도록 합니다. 이제 기저 함수를 아래와 같이
구현할 수 있습니다.
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="5">
<textarea disabled rows="5" role="doc-example">
function Bezier(n,t):
sum = 0
for(k=0; k<=n; k++):
sum += binomial(n,k) * (1-t)^(n-k) * t^(k)
return sum</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
</table>
<p>
좋습니다. 물론 여기에서도 최적화의 여지가 남아 있습니다. 대부분의 컴퓨터 그래픽 용도로는 임의 차수의 곡선이 필요 없습니다(물론
입문서에서는 임의 차수 곡선을 다루는 코드를 제공할 예정입니다). 보통 2차와 3차 곡선만 사용하기 때문에, 아래와 같이 훨씬 간단한 코드로
고칠 수 있습니다.
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="13">
<textarea disabled rows="13" role="doc-example">
function Bezier(2,t):
t2 = t * t
mt = 1-t
mt2 = mt * mt
return mt2 + 2*mt*t + t2
function Bezier(3,t):
t2 = t * t
t3 = t2 * t
mt = 1-t
mt2 = mt * mt
mt3 = mt2 * mt
return mt3 + 3*mt2*t + 3*mt*t2 + t3</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
</table>
<p>이제 기저 함수를 프로그래밍하는 법을 모두 배웠습니다. 와!</p>
</div>
<p>지금까지 베지에 곡선의 기저 함수를 살펴보았습니다. 이제 베지에 곡선을 그토록 특별하게 만드는 것, 조절점을 살펴볼 시간입니다.</p>
</section>
<section id="control">
<h1>
<div class="nav">
<a href="ko-KR/index.html#explanation">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#weightcontrol">다음</a>
</div>
<a href="ko-KR/index.html#control">Controlling Bézier curvatures</a>
</h1>
<p>
Bézier curves are, like all "splines", interpolation functions. This means that they take a set of points, and generate values somewhere
"between" those points. (One of the consequences of this is that you'll never be able to generate a point that lies outside the outline
for the control points, commonly called the "hull" for the curve. Useful information!). In fact, we can visualize how each point
contributes to the value generated by the function, so we can see which points are important, where, in the curve.
</p>
<p>
The following graphs show the interpolation functions for quadratic and cubic curves, with "S" being the strength of a point's
contribution to the total sum of the Bézier function. Click-and-drag to see the interpolation percentages for each curve-defining point at
a specific <i>t</i> value.
</p>
<div class="figure">
<graphics-element
title="Quadratic interpolations"
width="275"
height="275"
src="./chapters/control/lerp.js"
data-degree="3"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/control/8332e5d34b7344bbee2a2e1f4521ce46.png" loading="lazy" />
<label>Quadratic interpolations</label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0" class="slide-control" />
</graphics-element>
<graphics-element
title="Cubic interpolations"
width="275"
height="275"
src="./chapters/control/lerp.js"
data-degree="4"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/control/1b8c5e574dc67bfb0afc3fb0a8727378.png" loading="lazy" />
<label>Cubic interpolations</label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0" class="slide-control" />
</graphics-element>
<graphics-element
title="15th degree interpolations"
width="275"
height="275"
src="./chapters/control/lerp.js"
data-degree="15"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/control/c26d2655e8741ef7e2eeb4f6554fc7a5.png" loading="lazy" />
<label>15th degree interpolations</label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0" class="slide-control" />
</graphics-element>
</div>
<p>
Also shown is the interpolation function for a 15<sup>th</sup> order Bézier function. As you can see, the start and end point contribute
considerably more to the curve's shape than any other point in the control point set.
</p>
<p>
If we want to change the curve, we need to change the weights of each point, effectively changing the interpolations. The way to do this
is about as straightforward as possible: just multiply each point with a value that changes its strength. These values are conventionally
called "weights", and we can add them to our original Bézier function:
</p>
<!--
__ n n-i i
Bézier(n,t) = ❯ \underset binomial term\underbrace\binomni · \ \underset polynomial term\underbrace(1-t) · t · \ \underset
‾‾ i=0
weight\underbracew
i
-->
<img class="LaTeX SVG" src="./images/chapters/control/c2f2fe0ef5d0089d9dd8e5e3999405cb.svg" width="336px" height="55px" loading="lazy" />
<p>
That looks complicated, but as it so happens, the "weights" are actually just the coordinate values we want our curve to have: for an
<i>n<sup>th</sup></i> order curve, w<sub>0</sub> is our start coordinate, w<sub>n</sub> is our last coordinate, and everything in between
is a controlling coordinate. Say we want a cubic curve that starts at (110,150), is controlled by (25,190) and (210,250) and ends at
(210,30), we use this Bézier curve:
</p>
<!--
╭ 3 2 2 3
╡ x = \colordarkred 110 · (1-t) + \colordarkgreen 25 · 3 · (1-t) · t + \colordarkblue 210 · 3 · (1-t) · t + \coloramber 210 · t
│ 3 2 2 3
╰ y = \colordarkred 150 · (1-t) + \colordarkgreen 190 · 3 · (1-t) · t + \colordarkblue 250 · 3 · (1-t) · t + \coloramber 30 · t
-->
<img class="LaTeX SVG" src="./images/chapters/control/be73034ac382b54863c7a18c2932cbbc.svg" width="473px" height="40px" loading="lazy" />
<p>Which gives us the curve we saw at the top of the article:</p>
<graphics-element
title="Our cubic Bézier curve"
width="275"
height="275"
src="./chapters/control/../introduction/cubic.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/control/8d158a13e9a86969b99c64057644cbc6.png" loading="lazy" />
<label>Our cubic Bézier curve</label>
</fallback-image></graphics-element
>
<p>
What else can we do with Bézier curves? Quite a lot, actually. The rest of this article covers a multitude of possible operations and
algorithms that we can apply, and the tasks they achieve.
</p>
<div class="howtocode">
<h3>How to implement the weighted basis function</h3>
<p>Given that we already know how to implement basis function, adding in the control points is remarkably easy:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="5">
<textarea disabled rows="5" role="doc-example">
function Bezier(n,t,w[]):
sum = 0
for(k=0; k<=n; k++):
sum += w[k] * binomial(n,k) * (1-t)^(n-k) * t^(k)
return sum</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
</table>
<p>And now for the optimized versions:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="13">
<textarea disabled rows="13" role="doc-example">
function Bezier(2,t,w[]):
t2 = t * t
mt = 1-t
mt2 = mt * mt
return w[0]*mt2 + w[1]*2*mt*t + w[2]*t2
function Bezier(3,t,w[]):
t2 = t * t
t3 = t2 * t
mt = 1-t
mt2 = mt * mt
mt3 = mt2 * mt
return w[0]*mt3 + 3*w[1]*mt2*t + 3*w[2]*mt*t2 + w[3]*t3</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
</table>
<p>And now we know how to program the weighted basis function.</p>
</div>
</section>
<section id="weightcontrol">
<h1>
<div class="nav"><a href="ko-KR/index.html#control">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#extended">다음</a></div>
<a href="ko-KR/index.html#weightcontrol">Controlling Bézier curvatures, part 2: Rational Béziers</a>
</h1>
<p>
We can further control Bézier curves by "rationalising" them: that is, adding a "ratio" value in addition to the weight value discussed in
the previous section, thereby gaining control over "how strongly" each coordinate influences the curve.
</p>
<p>Adding these ratio values to the regular Bézier curve function is fairly easy. Where the regular function is the following:</p>
<!--
__ n n-i i
Bézier(n,t) = ❯ \binomni · (1-t) · t · w
‾‾ i=0 i
-->
<img
class="LaTeX SVG"
src="./images/chapters/weightcontrol/ceac4259d2aed0767c7765d2237ca1a3.svg"
width="276px"
height="41px"
loading="lazy"
/>
<p>The function for rational Bézier curves has two more terms:</p>
<!--
__ n n-i i
❯ \binomni · (1-t) · t · w · \colorblue ratio
‾‾ i=0 i i
Rational Bézier(n,t) = ────────────────────────────────────────────────────────────
__ n n-i i
\colorblue ❯ \binomni · (1-t) · t · ratio
‾‾ i=0 i
-->
<img
class="LaTeX SVG"
src="./images/chapters/weightcontrol/942e3b3cacc7f403ad95fcd4acce7d19.svg"
width="408px"
height="48px"
loading="lazy"
/>
<p>
In this, the first new term represents an additional weight for each coordinate. For example, if our ratio values are [1, 0.5, 0.5, 1]
then <code>ratio<sub>0</sub> = 1</code>, <code>ratio<sub>1</sub> = 0.5</code>, and so on, and is effectively identical as if we were just
using different weight. So far, nothing too special.
</p>
<p>
However, the second new term is what makes the difference: every point on the curve isn't just a "double weighted" point, it is a
<em>fraction</em> of the "doubly weighted" value we compute by introducing that ratio. When computing points on the curve, we compute the
"normal" Bézier value and then <em>divide</em> that by the Bézier value for the curve that only uses ratios, not weights.
</p>
<p>
This does something unexpected: it turns our polynomial into something that <em>isn't</em> a polynomial anymore. It is now a kind of curve
that is a super class of the polynomials, and can do some really cool things that Bézier curves can't do "on their own", such as perfectly
describing circles (which we'll see in a later section is literally impossible using standard Bézier curves).
</p>
<p>
But the best way to show what this does is to do literally that: let's look at the effect of "rationalising" our Bézier curves using an
interactive graphic for a rationalised curves. The following graphic shows the Bézier curve from the previous section, "enriched" with
ratio factors for each coordinate. The closer to zero we set one or more terms, the less relative influence the associated coordinate
exerts on the curve (and of course the higher we set them, the more influence they have). Try to change the values and see how it affects
what gets drawn:
</p>
<graphics-element
title="Our rational cubic Bézier curve"
width="275"
height="275"
src="./chapters/weightcontrol/rational.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/weightcontrol/3d71e2b9373684eebcb0dc8563f70b18.png" loading="lazy" />
<label>Our rational cubic Bézier curve</label>
</fallback-image>
<input type="range" min="0.01" max="2" value="1" step="0.01" class="ratio-1" />
<input type="range" min="0.01" max="2" value="1" step="0.01" class="ratio-2" />
<input type="range" min="0.01" max="2" value="1" step="0.01" class="ratio-3" />
<input type="range" min="0.01" max="2" value="1" step="0.01" class="ratio-4" />
</graphics-element>
<p>
You can think of the ratio values as each coordinate's "gravity": the higher the gravity, the closer to that coordinate the curve will
want to be. You'll also notice that if you simply increase or decrease all the ratios by the same amount, nothing changes... much like
with gravity, if the relative strengths stay the same, nothing really changes. The values define each coordinate's influence
<em>relative to all other points</em>.
</p>
<div class="howtocode">
<h3>How to implement rational curves</h3>
<p>Extending the code of the previous section to include ratios is almost trivial:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="26">
<textarea disabled rows="26" role="doc-example">
function RationalBezier(2,t,w[],r[]):
t2 = t * t
mt = 1-t
mt2 = mt * mt
f = [
r[0] * mt2,
2 * r[1] * mt * t,
r[2] * t2
]
basis = f[0] + f[1] + f[2]
return (f[0] * w[0] + f[1] * w[1] + f[2] * w[2])/basis
function RationalBezier(3,t,w[],r[]):
t2 = t * t
t3 = t2 * t
mt = 1-t
mt2 = mt * mt
mt3 = mt2 * mt
f = [
r[0] * mt3,
3 * r[1] * mt2 * t,
3 * r[2] * mt * t2,
r[3] * t3
]
basis = f[0] + f[1] + f[2] + f[3]
return (f[0] * w[0] + f[1] * w[1] + f[2] * w[2] + f[3] * w[3])/basis</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
<tr>
<td>18</td>
</tr>
<tr>
<td>19</td>
</tr>
<tr>
<td>20</td>
</tr>
<tr>
<td>21</td>
</tr>
<tr>
<td>22</td>
</tr>
<tr>
<td>23</td>
</tr>
<tr>
<td>24</td>
</tr>
<tr>
<td>25</td>
</tr>
<tr>
<td>26</td>
</tr>
</table>
<p>And that's all we have to do.</p>
</div>
</section>
<section id="extended">
<h1>
<div class="nav"><a href="ko-KR/index.html#weightcontrol">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#matrix">다음</a></div>
<a href="ko-KR/index.html#extended">The Bézier interval [0,1]</a>
</h1>
<p>
Now that we know the mathematics behind Bézier curves, there's one curious thing that you may have noticed: they always run from
<code>t=0</code> to <code>t=1</code>. Why that particular interval?
</p>
<p>
It all has to do with how we run from "the start" of our curve to "the end" of our curve. If we have a value that is a mixture of two
other values, then the general formula for this is:
</p>
<!--
mixture = a · value + b · value
1 2
-->
<img class="LaTeX SVG" src="./images/chapters/extended/dfd6ded3f0addcf43e0a1581627a2220.svg" width="212px" height="16px" loading="lazy" />
<p>
The obvious start and end values here need to be <code>a=1, b=0</code>, so that the mixed value is 100% value 1, and 0% value 2, and
<code>a=0, b=1</code>, so that the mixed value is 0% value 1 and 100% value 2. Additionally, we don't want "a" and "b" to be independent:
if they are, then we could just pick whatever values we like, and end up with a mixed value that is, for example, 100% value 1
<strong>and</strong> 100% value 2. In principle that's fine, but for Bézier curves we always want mixed values <em>between</em> the start
and end point, so we need to make sure we can never set "a" and "b" to some values that lead to a mix value that sums to more than 100%.
And that's easy:
</p>
<!--
m = a · value + (1 - a) · value
1 2
-->
<img class="LaTeX SVG" src="./images/chapters/extended/4e0fa763b173e3a683587acf83733353.svg" width="213px" height="16px" loading="lazy" />
<p>
With this we can guarantee that we never sum above 100%. By restricting <code>a</code> to values in the interval [0,1], we will always be
somewhere between our two values (inclusively), and we will always sum to a 100% mix.
</p>
<p>
But... what if we use this form, which is based on the assumption that we will only ever use values between 0 and 1, and instead use
values outside of that interval? Do things go horribly wrong? Well... not really, but we get to "see more".
</p>
<p>
In the case of Bézier curves, extending the interval simply makes our curve "keep going". Bézier curves are simply segments of some
polynomial curve, so if we pick a wider interval we simply get to see more of the curve. So what do they look like?
</p>
<p>
The following two graphics show you Bézier curves rendered "the usual way", as well as the curves they "lie on" if we were to extend the
<code>t</code> values much further. As you can see, there's a lot more "shape" hidden in the rest of the curve, and we can model those
parts by moving the curve points around.
</p>
<div class="figure">
<graphics-element
title="Quadratic infinite interval Bézier curve"
width="275"
height="275"
src="./chapters/extended/extended.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/extended/37948bde4bf0d25bde85f172bf55b9fb.png" loading="lazy" />
<label>Quadratic infinite interval Bézier curve</label>
</fallback-image></graphics-element
>
<graphics-element
title="Cubic infinite interval Bézier curve"
width="275"
height="275"
src="./chapters/extended/extended.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/extended/2d17acb381ebdd28f0ff43be00d723c4.png" loading="lazy" />
<label>Cubic infinite interval Bézier curve</label>
</fallback-image></graphics-element
>
</div>
<p>
In fact, there are curves used in graphics design and computer modelling that do the opposite of Bézier curves; rather than fixing the
interval, and giving you freedom to choose the coordinates, they fix the coordinates, but give you freedom over the interval. A great
example of this is the <a href="https://levien.com/phd/phd.html">"Spiro" curve</a>, which is a curve based on part of a
<a href="https://en.wikipedia.org/wiki/Euler_spiral">Cornu Spiral, also known as Euler's Spiral</a>. It's a very aesthetically pleasing
curve and you'll find it in quite a few graphics packages like <a href="https://fontforge.org/en-US/">FontForge</a> and
<a href="https://inkscape.org">Inkscape</a>. It has even been used in font design, for example for the Inconsolata typeface.
</p>
</section>
<section id="matrix">
<h1>
<div class="nav"><a href="ko-KR/index.html#extended">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#decasteljau">다음</a></div>
<a href="ko-KR/index.html#matrix">Bézier curvatures as matrix operations</a>
</h1>
<p>
We can also represent Bézier curves as matrix operations, by expressing the Bézier formula as a polynomial basis function and a
coefficients matrix, and the actual coordinates as a matrix. Let's look at what this means for the cubic curve, using P<sub>...</sub> to
refer to coordinate values "in one or more dimensions":
</p>
<!--
3 2 2 3
B(t) = P · (1-t) + P · 3 · (1-t) · t + P · 3 · (1-t) · t + P · t
1 2 3 4
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/9a9a55f5b0323d9ea88f82fc6be58ad3.svg" width="468px" height="20px" loading="lazy" />
<p>Disregarding our actual coordinates for a moment, we have:</p>
<!--
3 2 2 3
B(t) = (1-t) + 3 · (1-t) · t + 3 · (1-t) · t + t
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/87cfac83cb8a4b0bee68ef006effc611.svg" width="353px" height="19px" loading="lazy" />
<p>We can write this as a sum of four expressions:</p>
<!--
3
... = (1-t)
2
+ 3 · (1-t) · t
2
+ 3 · (1-t) · t
3
+ t
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/cdd88611833f3b178df91278359a4193.svg" width="140px" height="75px" loading="lazy" />
<p>And we can expand these expressions:</p>
<!--
3 2
... = (1-t) · (1-t) · (1-t) = -t + 3 · t - 3 · t + 1
3 2
+ 3 · (1-t) · (1-t) · t = 3 · t - 6 · t + 3 · t
3 2
+ 3 · (1-t) · t · t = -3 · t + 3 · t
3
+ t · t · t = t
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/ec118f296511c6e9ac8727be3703a7ce.svg" width="397px" height="75px" loading="lazy" />
<p>Furthermore, we can make all the 1 and 0 factors explicit:</p>
<!--
3 2
... = -1 · t + 3 · t - 3 · t + 1
3 2
+ +3 · t - 6 · t + 3 · t + 0
3 2
+ -3 · t + 3 · t + 0 · t + 0
3 2
+ +1 · t + 0 · t + 0 · t + 0
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/67a5ea33d6c6558f7d954b18226f4956.svg" width="217px" height="75px" loading="lazy" />
<p>And <em>that</em>, we can view as a series of four matrix operations:</p>
<!--
┌ -1 ┐ ┌ 3 ┐ ┌ -3 ┐ ┌ 1 ┐
┌ 3 2 ┐ · │ 3 │ + ┌ 3 2 ┐ · │ -6 │ + ┌ 3 2 ┐ · │ 3 │ + ┌ 3 2 ┐ · │ 0 │
└ t t t 1 ┘ │ -3 │ └ t t t 1 ┘ │ 3 │ └ t t t 1 ┘ │ 0 │ └ t t t 1 ┘ │ 0 │
└ 1 ┘ └ 0 ┘ └ 0 ┘ └ 0 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/8ecff6b8a37d60385d287ea2b26876db.svg" width="607px" height="72px" loading="lazy" />
<p>If we compact this into a single matrix operation, we get:</p>
<!--
┌ -1 3 -3 1 ┐
┌ 3 2 ┐ · │ 3 -6 3 0 │
└ t t t 1 ┘ │ -3 3 0 0 │
└ 1 0 0 0 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/b9527f7d5a0f5d2d737eac118d69243e.svg" width="227px" height="72px" loading="lazy" />
<p>
This kind of polynomial basis representation is generally written with the bases in increasing order, which means we need to flip our
<code>t</code> matrix horizontally, and our big "mixing" matrix upside down:
</p>
<!--
┌ 1 0 0 0 ┐
┌ 2 3 ┐ · │ -3 3 0 0 │
└ 1 t t t ┘ │ 3 -6 3 0 │
└ -1 3 -3 1 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/1a64ed455c6dd2f8cacca5e5e12bdcc1.svg" width="227px" height="72px" loading="lazy" />
<p>And then finally, we can add in our original coordinates as a single third matrix:</p>
<!--
┌ P ┐
│ 1 │
┌ 1 0 0 0 ┐ │ P │
B(t) = ┌ 2 3 ┐ · │ -3 3 0 0 │ · │ 2 │
└ 1 t t t ┘ │ 3 -6 3 0 │ │ P │
└ -1 3 -3 1 ┘ │ 3 │
│ P │
└ 4 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/b32cae2dfc47d5f36df0bc3defb7dfa8.svg" width="323px" height="73px" loading="lazy" />
<p>We can perform the same trick for the quadratic curve, in which case we end up with:</p>
<!--
┌ P ┐
┌ 1 0 0 ┐ │ 1 │
B(t) = ┌ 2 ┐ · │ -2 2 0 │ · │ P │
└ 1 t t ┘ └ 1 -2 1 ┘ │ 2 │
│ P │
└ 3 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/matrix/1bae50fefa43210b3a6259d1984f6cbc.svg" width="263px" height="55px" loading="lazy" />
<p>
If we plug in a <code>t</code> value, and then multiply the matrices, we will get exactly the same values as when we evaluate the original
polynomial function, or as when we evaluate the curve using progressive linear interpolation.
</p>
<p>
<strong>So: why would we bother with matrices?</strong> Matrix representations allow us to discover things about functions that would
otherwise be hard to tell. It turns out that the curves form
<a href="https://en.wikipedia.org/wiki/Triangular_matrix">triangular matrices</a>, and they have a determinant equal to the product of the
actual coordinates we use for our curve. It's also invertible, which means there's
<a href="https://en.wikipedia.org/wiki/Invertible_matrix#The_invertible_matrix_theorem">a ton of properties</a> that are all satisfied. Of
course, the main question is "why is this useful to us now?", and the answer to that is that it's not <em>immediately</em> useful, but
you'll be seeing some instances where certain curve properties can be either computed via function manipulation, or via clever use of
matrices, and sometimes the matrix approach can be (drastically) faster.
</p>
<p>So for now, just remember that we can represent curves this way, and let's move on.</p>
</section>
<section id="decasteljau">
<h1>
<div class="nav"><a href="ko-KR/index.html#matrix">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#flattening">다음</a></div>
<a href="ko-KR/index.html#decasteljau">de Casteljau's algorithm</a>
</h1>
<p>
If we want to draw Bézier curves, we can run through all values of <code>t</code> from 0 to 1 and then compute the weighted basis function
at each value, getting the <code>x/y</code> values we need to plot. Unfortunately, the more complex the curve gets, the more expensive
this computation becomes. Instead, we can use <em>de Casteljau's algorithm</em> to draw curves. This is a geometric approach to curve
drawing, and it's really easy to implement. So easy, in fact, you can do it by hand with a pencil and ruler.
</p>
<p>Rather than using our calculus function to find <code>x/y</code> values for <code>t</code>, let's do this instead:</p>
<ul>
<li>treat <code>t</code> as a ratio (which it is). t=0 is 0% along a line, t=1 is 100% along a line.</li>
<li>Take all lines between the curve's defining points. For an order <code>n</code> curve, that's <code>n</code> lines.</li>
<li>
Place markers along each of these line, at distance <code>t</code>. So if <code>t</code> is 0.2, place the mark at 20% from the start,
80% from the end.
</li>
<li>Now form lines between <code>those</code> points. This gives <code>n-1</code> lines.</li>
<li>Place markers along each of these line at distance <code>t</code>.</li>
<li>Form lines between <code>those</code> points. This'll be <code>n-2</code> lines.</li>
<li>Place markers, form lines, place markers, etc.</li>
<li>
Repeat this until you have only one line left. The point <code>t</code> on that line coincides with the original curve point at
<code>t</code>.
</li>
</ul>
<p>
To see this in action, move the slider for the following sketch to changes which curve point is explicitly evaluated using de Casteljau's
algorithm.
</p>
<graphics-element
title="Traversing a curve using de Casteljau's algorithm"
width="275"
height="275"
src="./chapters/decasteljau/decasteljau.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/decasteljau/df92f529841f39decf9ad62b0967855a.png" loading="lazy" />
<label>Traversing a curve using de Casteljau's algorithm</label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0" class="slide-control" />
</graphics-element>
<div class="howtocode">
<h3>How to implement de Casteljau's algorithm</h3>
<p>
Let's just use the algorithm we just specified, and implement that as a function that can take a list of curve-defining points, and a
<code>t</code> value, and draws the associated point on the curve for that <code>t</code> value:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="8">
<textarea disabled rows="8" role="doc-example">
function drawCurvePoint(points[], t):
if(points.length==1):
draw(points[0])
else:
newpoints=array(points.size-1)
for(i=0; i<newpoints.length; i++):
newpoints[i] = (1-t) * points[i] + t * points[i+1]
drawCurvePoint(newpoints, t)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
</table>
<p>
And done, that's the algorithm implemented. Although: usually you don't get the luxury of overloading the "+" operator, so let's also
give the code for when you need to work with <code>x</code> and <code>y</code> values separately:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="10">
<textarea disabled rows="10" role="doc-example">
function drawCurvePoint(points[], t):
if(points.length==1):
draw(points[0])
else:
newpoints=array(points.size-1)
for(i=0; i<newpoints.length; i++):
x = (1-t) * points[i].x + t * points[i+1].x
y = (1-t) * points[i].y + t * points[i+1].y
newpoints[i] = new point(x,y)
drawCurvePoint(newpoints, t)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
</table>
<p>
So what does this do? This draws a point, if the passed list of points is only 1 point long. Otherwise it will create a new list of
points that sit at the <i>t</i> ratios (i.e. the "markers" outlined in the above algorithm), and then call the draw function for this
new list.
</p>
</div>
</section>
<section id="flattening">
<h1>
<div class="nav"><a href="ko-KR/index.html#decasteljau">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#splitting">다음</a></div>
<a href="ko-KR/index.html#flattening">Simplified drawing</a>
</h1>
<p>
We can also simplify the drawing process by "sampling" the curve at certain points, and then joining those points up with straight lines,
a process known as "flattening", as we are reducing a curve to a simple sequence of straight, "flat" lines.
</p>
<p>
We can do this is by saying "we want X segments", and then sampling the curve at intervals that are spaced such that we end up with the
number of segments we wanted. The advantage of this method is that it's fast: instead of evaluating 100 or even 1000 curve coordinates, we
can sample a much lower number and still end up with a curve that sort-of-kind-of looks good enough. The disadvantage of course is that we
lose the precision of working with "the real curve", so we usually can't use the flattened for doing true intersection detection, or
curvature alignment.
</p>
<div class="figure">
<graphics-element
title="Flattening a quadratic curve"
width="275"
height="275"
src="./chapters/flattening/flatten.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/flattening/6813bfc608aea11df1dda444b9f18123.png" loading="lazy" />
<label>Flattening a quadratic curve</label>
</fallback-image>
<input type="range" min="1" max="16" step="1" value="4" class="slide-control" />
</graphics-element>
<graphics-element
title="Flattening a cubic curve"
width="275"
height="275"
src="./chapters/flattening/flatten.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/flattening/0e0e4a2ee46bd89bcfde9f75cfe43292.png" loading="lazy" />
<label>Flattening a cubic curve</label>
</fallback-image>
<input type="range" min="1" max="24" step="1" value="8" class="slide-control" />
</graphics-element>
</div>
<p>
Try clicking on the sketch and using your up and down arrow keys to lower the number of segments for both the quadratic and cubic curve.
You'll notice that for certain curvatures, a low number of segments works quite well, but for more complex curvatures (try this for the
cubic curve), a higher number is required to capture the curvature changes properly.
</p>
<div class="howtocode">
<h3>How to implement curve flattening</h3>
<p>Let's just use the algorithm we just specified, and implement that:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="7">
<textarea disabled rows="7" role="doc-example">
function flattenCurve(curve, segmentCount):
step = 1/segmentCount;
coordinates = [curve.getXValue(0), curve.getYValue(0)]
for(i=1; i <= segmentCount; i++):
t = i*step;
coordinates.push[curve.getXValue(t), curve.getYValue(t)]
return coordinates;</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
</table>
<p>And done, that's the algorithm implemented. That just leaves drawing the resulting "curve" as a sequence of lines:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="7">
<textarea disabled rows="7" role="doc-example">
function drawFlattenedCurve(curve, segmentCount):
coordinates = flattenCurve(curve, segmentCount)
coord = coordinates[0], _coord;
for(i=1; i < coordinates.length; i++):
_coord = coordinates[i]
line(coord, _coord)
coord = _coord</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
</table>
<p>We start with the first coordinate as reference point, and then just draw lines between each point and its next point.</p>
</div>
</section>
<section id="splitting">
<h1>
<div class="nav">
<a href="ko-KR/index.html#flattening">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#matrixsplit">다음</a>
</div>
<a href="ko-KR/index.html#splitting">Splitting curves</a>
</h1>
<p>
Using de Casteljau's algorithm, we can also find all the points we need to split up a Bézier curve into two, smaller curves, which taken
together form the original curve. When we construct de Casteljau's skeleton for some value <code>t</code>, the procedure gives us all the
points we need to split a curve at that <code>t</code> value: one curve is defined by all the inside skeleton points found prior to our
on-curve point, with the other curve being defined by all the inside skeleton points after our on-curve point.
</p>
<graphics-element
title="Splitting a curve"
width="825"
height="275"
src="./chapters/splitting/splitting.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/splitting/fce5eb16dfcd103797c5e17bd77f1437.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0.5" class="slide-control" />
</graphics-element>
<div class="howtocode">
<h3>implementing curve splitting</h3>
<p>We can implement curve splitting by bolting some extra logging onto the de Casteljau function:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="16">
<textarea disabled rows="16" role="doc-example">
left=[]
right=[]
function drawCurvePoint(points[], t):
if(points.length==1):
left.add(points[0])
right.add(points[0])
draw(points[0])
else:
newpoints=array(points.size-1)
for(i=0; i<newpoints.length; i++):
if(i==0):
left.add(points[i])
if(i==newpoints.length-1):
right.add(points[i+1])
newpoints[i] = (1-t) * points[i] + t * points[i+1]
drawCurvePoint(newpoints, t)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
</table>
<p>
After running this function for some value <code>t</code>, the <code>left</code> and <code>right</code> arrays will contain all the
coordinates for two new curves - one to the "left" of our <code>t</code> value, the other on the "right". These new curves will have the
same order as the original curve, and can be overlaid exactly on the original curve.
</p>
</div>
</section>
<section id="matrixsplit">
<h1>
<div class="nav"><a href="ko-KR/index.html#splitting">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#reordering">다음</a></div>
<a href="ko-KR/index.html#matrixsplit">Splitting curves using matrices</a>
</h1>
<p>
Another way to split curves is to exploit the matrix representation of a Bézier curve. In <a href="#matrix">the section on matrices</a>,
we saw that we can represent curves as matrix multiplications. Specifically, we saw these two forms for the quadratic and cubic curves
respectively: (we'll reverse the Bézier coefficients vector for legibility)
</p>
<!--
┌ P ┐
┌ 1 0 0 ┐ │ 1 │
B(t) = ┌ 2 ┐ · │ -2 2 0 │ · │ P │
└ 1 t t ┘ └ 1 -2 1 ┘ │ 2 │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/1bae50fefa43210b3a6259d1984f6cbc.svg"
width="263px"
height="55px"
loading="lazy"
/>
<p>and</p>
<!--
┌ P ┐
│ 1 │
┌ 1 0 0 0 ┐ │ P │
B(t) = ┌ 2 3 ┐ · │ -3 3 0 0 │ · │ 2 │
└ 1 t t t ┘ │ 3 -6 3 0 │ │ P │
└ -1 3 -3 1 ┘ │ 3 │
│ P │
└ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/f690ff0502d9fd7d4697cc43d98afd5d.svg"
width="323px"
height="73px"
loading="lazy"
/>
<p>
Let's say we want to split the curve at some point <code>t = z</code>, forming two new (obviously smaller) Bézier curves. To find the
coordinates for these two Bézier curves, we can use the matrix representation and some linear algebra. First, we separate out the actual
"point on the curve" information into a new matrix multiplication:
</p>
<!--
┌ P ┐ ┌ P ┐
┌ 1 0 0 ┐ │ 1 │ ┌ 1 0 0 ┐ ┌ 1 0 0 ┐ │ 1 │
B(t) = ┌ 2 ┐ · │ -2 2 0 │ · │ P │ = ┌ 2 ┐ · │ 0 z 0 │ · │ -2 2 0 │ · │ P │
└ 1 (z · t) (z · t) ┘ └ 1 -2 1 ┘ │ 2 │ └ 1 t t ┘ │ 2 │ └ 1 -2 1 ┘ │ 2 │
│ P │ └ 0 0 z ┘ │ P │
└ 3 ┘ └ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/f565e66677138927335535d009409c3d.svg"
width="648px"
height="55px"
loading="lazy"
/>
<p>and</p>
<!--
┌ P ┐ ┌ P ┐
│ 1 │ ┌ 1 0 0 0 ┐ │ 1 │
┌ 1 0 0 0 ┐ │ P │ │ 0 z 0 0 │ ┌ 1 0 0 0 ┐ │ P │
B(t) = ┌ 2 3 ┐ · │ -3 3 0 0 │ · │ 2 │ = ┌ 2 3 ┐ · │ 2 │ · │ -3 3 0 0 │ · │ 2 │
└ 1 (z · t) (z · t) (z · t) ┘ │ 3 -6 3 0 │ │ P │ └ 1 t t t ┘ │ 0 0 z 0 │ │ 3 -6 3 0 │ │ P │
└ -1 3 -3 1 ┘ │ 3 │ │ 3 │ └ -1 3 -3 1 ┘ │ 3 │
│ P │ └ 0 0 0 z ┘ │ P │
└ 4 ┘ └ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/ebf8d72c6056476172deeb89726b75c8.svg"
width="805px"
height="75px"
loading="lazy"
/>
<p>
If we could compact these matrices back to the form <strong>[t values] · [Bézier matrix] · [column matrix]</strong>, with the first two
staying the same, then that column matrix on the right would be the coordinates of a new Bézier curve that describes the first segment,
from <code>t = 0</code> to <code>t = z</code>. As it turns out, we can do this quite easily, by exploiting some simple rules of linear
algebra (and if you don't care about the derivations, just skip to the end of the box for the results!).
</p>
<div class="note">
<h2>Deriving new hull coordinates</h2>
<p>
Deriving the two segments upon splitting a curve takes a few steps, and the higher the curve order, the more work it is, so let's look
at the quadratic curve first:
</p>
<!--
┌ P ┐
┌ 1 0 0 ┐ ┌ 1 0 0 ┐ │ 1 │
B(t) = ┌ 2 ┐ · │ 0 z 0 │ · │ -2 2 0 │ · │ P │
└ 1 t t ┘ │ 2 │ └ 1 -2 1 ┘ │ 2 │
└ 0 0 z ┘ │ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/c32007be095224e0d157a8f71c62c90e.svg"
width="348px"
height="55px"
loading="lazy"
/>
<!--
┌ P ┐
│ 1 │
= ┌ 2 ┐ · \underset we turn this...\underbrace\kern 2.25em Z · M \kern 2.25em · │ P │
└ 1 t t ┘ │ 2 │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/aa17f7e82cf50498f90deb6a21a2489a.svg"
width="247px"
height="55px"
loading="lazy"
/>
<!--
┌ P ┐
-1 │ 1 │
= ┌ 2 ┐ · \underset into this...\underbrace M · M · Z · M · │ P │
└ 1 t t ┘ │ 2 │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/4549b95450db3c73479e8902e4939427.svg"
width="247px"
height="55px"
loading="lazy"
/>
<!--
┌ P ┐
│ 1 │
= ┌ 2 ┐ · M \underset ...to get this!\underbrace \kern 1.25em · \kern 1.25em Q \kern 1.25em · \kern 1.25em │ P │
└ 1 t t ┘ │ 2 │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/266b71339b55ad3a312a9f41e6bcf988.svg"
width="252px"
height="55px"
loading="lazy"
/>
<p>
We can do this because [<em>M · M<sup>-1</sup></em
>] is the identity matrix. It's a bit like multiplying something by x/x in calculus: it doesn't do anything to the function, but it does
allow you to rewrite it to something that may be easier to work with, or can be broken up differently. In the same way, multiplying our
matrix by [<em>M · M<sup>-1</sup></em
>] has no effect on the total formula, but it does allow us to change the matrix sequence [<em>something · M</em>] to a sequence [<em
>M · something</em
>], and that makes a world of difference: if we know what [<em>M<sup>-1</sup> · Z · M</em>] is, we can apply that to our coordinates,
and be left with a proper matrix representation of a quadratic Bézier curve (which is [<em>T · M · P</em>]), with a new set of
coordinates that represent the curve from <em>t = 0</em> to <em>t = z</em>. So let's get computing:
</p>
<!--
┌ 1 0 0 ┐
-1 │ 1 │ ┌ 1 0 0 ┐ ┌ 1 0 0 ┐ ┌ 1 0 0 ┐
Q = M · Z · M = │ 1 ─ 0 │ · │ 0 z 0 │ · │ -2 2 0 │ = │ -(z-1) z 0 │
│ 2 │ │ 2 │ └ 1 -2 1 ┘ │ 2 2 │
└ 1 1 1 ┘ └ 0 0 z ┘ └ (z - 1) -2 · (z-1) · z z ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/5e008143622c66bb5e9cc4d5d6a8ea62.svg"
width="627px"
height="56px"
loading="lazy"
/>
<p>Excellent! Now we can form our new quadratic curve:</p>
<!--
┌ P ┐ ╭ ┌ P ┐ ╮
│ 1 │ │ │ 1 │ │
B(t) = ┌ 2 ┐ · M · Q · │ P │ = ┌ 2 ┐ · M · │ Q · │ P │ │
└ 1 t t ┘ │ 2 │ └ 1 t t ┘ │ │ 2 │ │
│ P │ │ │ P │ │
└ 3 ┘ ╰ └ 3 ┘ ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/dceed84990aaf6878bcc67ddbaa8d8d9.svg"
width="417px"
height="55px"
loading="lazy"
/>
<!--
╭ ┌ P ┐ ╮
┌ 1 0 0 ┐ │ ┌ 1 0 0 ┐ │ 1 │ │
= ┌ 2 ┐ · │ -2 2 0 │ · │ │ -(z-1) z 0 │ · │ P │ │
└ 1 t t ┘ └ 1 -2 1 ┘ │ │ 2 2 │ │ 2 │ │
│ └ (z - 1) -2 · (z-1) · z z ┘ │ P │ │
╰ └ 3 ┘ ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/f63067c2c3042c374a58dfa7f692309e.svg"
width="479px"
height="55px"
loading="lazy"
/>
<!--
┌ P ┐
│ 1 │
┌ 1 0 0 ┐ │ z · P - (z-1) · P │
= ┌ 2 ┐ · │ -2 2 0 │ · │ 2 1 │
└ 1 t t ┘ └ 1 -2 1 ┘ │ 2 2 │
│ z · P - 2 · z · (z-1) · P + (z - 1) · P │
└ 3 2 1 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/e58196b82b78f584779208cce88137f5.svg"
width="492px"
height="55px"
loading="lazy"
/>
<p>
<strong>Brilliant</strong>: if we want a subcurve from <code>t = 0</code> to <code>t = z</code>, we can keep the first coordinate the
same (which makes sense), our control point becomes a z-ratio mixture of the original control point and the start point, and the new end
point is a mixture that looks oddly similar to a
<a href="https://en.wikipedia.org/wiki/Bernstein_polynomial">Bernstein polynomial</a> of degree two. These new coordinates are actually
really easy to compute directly!
</p>
<p>
Of course, that's only one of the two curves. Getting the section from <code>t = z</code> to <code>t = 1</code> requires doing this
again. We first observe that in the previous calculation, we actually evaluated the general interval [0,<code>z</code>]. We were able to
write it down in a more simple form because of the zero, but what we <em>actually</em> evaluated, making the zero explicit, was:
</p>
<!--
┌ P ┐
┌ 1 0 0 ┐ │ 1 │
B(t) = ┌ 2 ┐ · │ -2 2 0 │ · │ P │
└ 1 ( 0 + z · t) ( 0 + z · t) ┘ └ 1 -2 1 ┘ │ 2 │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/6a22184e6ca869d28f4a252b64f23eff.svg"
width="381px"
height="55px"
loading="lazy"
/>
<!--
┌ P ┐
┌ 1 0 0 ┐ ┌ 1 0 0 ┐ │ 1 │
= ┌ 2 ┐ · │ 0 z 0 │ · │ -2 2 0 │ · │ P │
└ 1 t t ┘ │ 2 │ └ 1 -2 1 ┘ │ 2 │
└ 0 0 z ┘ │ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/3b5e41808b6c3bc66f3da2f40651410e.svg"
width="313px"
height="55px"
loading="lazy"
/>
<p>If we want the interval [<em>z</em>,1], we will be evaluating this instead:</p>
<!--
┌ P ┐
┌ 1 0 0 ┐ │ 1 │
B(t) = ┌ 2 ┐ · │ -2 2 0 │ · │ P │
└ 1 ( z + (1-z) · t) ( z + (1-z) · t) ┘ └ 1 -2 1 ┘ │ 2 │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/e079f44b56e07c8d7f83c17c8ebf1ecf.svg"
width="461px"
height="55px"
loading="lazy"
/>
<!--
┌ 2 ┐ ┌ P ┐
│ 1 z z │ ┌ 1 0 0 ┐ │ 1 │
= ┌ 2 ┐ · │ 0 1-z 2 · z · (1-z) │ · │ -2 2 0 │ · │ P │
└ 1 t t ┘ │ 2 │ └ 1 -2 1 ┘ │ 2 │
└ 0 0 (1-z) ┘ │ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/4764868f43815e471bb1ea95a81e1633.svg"
width="412px"
height="57px"
loading="lazy"
/>
<p>
We're going to do the same trick of multiplying by the identity matrix, to turn <code>[something · M]</code> into
<code>[M · something]</code>:
</p>
<!--
┌ 1 0 0 ┐ ┌ 2 ┐
-1 │ 1 │ │ 1 z z │ ┌ 1 0 0 ┐ ┌ 2 2 ┐
Q' = M · Z' · M = │ 1 ─ 0 │ · │ 0 1-z 2 · z · (1-z) │ · │ -2 2 0 │ = │ (z-1) -2 · z · (z-1) z │
│ 2 │ │ 2 │ └ 1 -2 1 ┘ │ 0 -(z-1) z │
└ 1 1 1 ┘ └ 0 0 (1-z) ┘ └ 0 0 1 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/c341532f693c2c1adfd298597bbfb5b5.svg"
width="729px"
height="57px"
loading="lazy"
/>
<p>So, our final second curve looks like:</p>
<!--
┌ P ┐ ╭ ┌ P ┐ ╮
│ 1 │ │ │ 1 │ │
B(t) = ┌ 2 ┐ · M · Q · │ P │ = ┌ 2 ┐ · M · │ Q' · │ P │ │
└ 1 t t ┘ │ 2 │ └ 1 t t ┘ │ │ 2 │ │
│ P │ │ │ P │ │
└ 3 ┘ ╰ └ 3 ┘ ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/e2622175dadafecc015f15c79ddf3002.svg"
width="421px"
height="55px"
loading="lazy"
/>
<!--
╭ ┌ P ┐ ╮
┌ 1 0 0 ┐ │ ┌ 2 2 ┐ │ 1 │ │
= ┌ 2 ┐ · │ -2 2 0 │ · │ │ (z-1) -2 · z · (z-1) z │ · │ P │ │
└ 1 t t ┘ └ 1 -2 1 ┘ │ │ 0 -(z-1) z │ │ 2 │ │
│ └ 0 0 1 ┘ │ P │ │
╰ └ 3 ┘ ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/4ce218bc968cbd98da0ca6ab66d415ed.svg"
width="473px"
height="57px"
loading="lazy"
/>
<!--
┌ 2 2 ┐
│ z · P - 2 · z · (z-1) · P + (z-1) · P │
┌ 1 0 0 ┐ │ 3 2 1 │
= ┌ 2 ┐ · │ -2 2 0 │ · │ z · P - (z-1) · P │
└ 1 t t ┘ └ 1 -2 1 ┘ │ 3 2 │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/9a4899b69e03cd4ad02c5eedffaa6a2f.svg"
width="492px"
height="57px"
loading="lazy"
/>
<p>
<strong><em>Nice</em></strong
>. We see the same thing as before: we can keep the last coordinate the same (which makes sense); our control point becomes a z-ratio
mixture of the original control point and the end point, and the new start point is a mixture that looks oddly similar to a bernstein
polynomial of degree two, except this time it uses (z-1) rather than (1-z). These new coordinates are <em>also</em> really easy to
compute directly!
</p>
</div>
<p>
So, using linear algebra rather than de Casteljau's algorithm, we have determined that, for any quadratic curve split at some value
<code>t = z</code>, we get two subcurves that are described as Bézier curves with simple-to-derive coordinates:
</p>
<!--
┌ P ┐
┌ P ┐ │ 1 │
┌ 1 0 0 ┐ │ 1 │ │ z · P - (z-1) · P │
│ -(z-1) z 0 │ · │ P │ = │ 2 1 │
│ 2 2 │ │ 2 │ │ 2 2 │
└ (z - 1) -2 · (z-1) · z z ┘ │ P │ │ z · P - 2 · z · (z-1) · P + (z - 1) · P │
└ 3 ┘ └ 3 2 1 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/480ebd0234e2fe1adc94926e8ed4339c.svg"
width="576px"
height="55px"
loading="lazy"
/>
<p>and</p>
<!--
┌ 2 2 ┐
┌ P ┐ │ z · P - 2 · z · (z-1) · P + (z-1) · P │
┌ 2 2 ┐ │ 1 │ │ 3 2 1 │
│ (z-1) -2 · z · (z-1) z │ · │ P │ = │ z · P - (z-1) · P │
│ 0 -(z-1) z │ │ 2 │ │ 3 2 │
└ 0 0 1 ┘ │ P │ │ P │
└ 3 ┘ └ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/17e308aa6d459b1d06d3160cc8e2e786.svg"
width="571px"
height="57px"
loading="lazy"
/>
<p>
We can do the same for cubic curves. However, I'll spare you the actual derivation (don't let that stop you from writing that out
yourself, though) and simply show you the resulting new coordinate sets:
</p>
<!--
┌ P ┐
┌ P ┐ │ 1 │
┌ 1 0 0 0 ┐ │ 1 │ │ z · P - (z-1) · P │
│ -(z-1) z 0 0 │ │ P │ │ 2 1 │
│ 2 2 │ · │ 2 │ = │ 2 2 │
│ (z-1) -2 · (z-1) · z z 0 │ │ P │ │ z · P - 2 · z · (z-1) · P + (z-1) · P │
│ 3 2 2 3 │ │ 3 │ │ 3 2 1 │
└ -(z-1) 3 · (z-1) · z -3 · (z-1) · z z ┘ │ P │ │ 3 2 2 3 │
└ 4 ┘ │ z · P - 3 · z · (z-1) · P + 3 · z · (z-1) · P - (z-1) · P │
└ 4 3 2 1 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/11505e0215ef026f2e49383ebb4a1abb.svg"
width="841px"
height="75px"
loading="lazy"
/>
<p>and</p>
<!--
┌ 3 2 2 3 ┐
┌ P ┐ │ z · P - 3 · z · (z-1) · P + 3 · z · (z-1) · P - (z-1) · P │
┌ 3 2 2 3 ┐ │ 1 │ │ 4 3 2 1 │
│ -(z-1) 3 · (z-1) · z -3 · (z-1) · z z │ │ P │ │ 2 2 │
│ 2 2 │ · │ 2 │ = │ z · P - 2 · z · (z-1) · P + (z-1) · P │
│ 0 (z-1) -2 · (z-1) · z z │ │ P │ │ 4 3 2 │
│ 0 0 -(z-1) z │ │ 3 │ │ z · P - (z-1) · P │
└ 0 0 0 1 ┘ │ P │ │ 4 3 │
└ 4 ┘ │ P │
└ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/matrixsplit/a899891096d82b7fdb23a90e6106b6df.svg"
width="837px"
height="77px"
loading="lazy"
/>
<p>
So, looking at our matrices, did we really need to compute the second segment matrix? No, we didn't. Actually having one segment's matrix
means we implicitly have the other: push the values of each row in the matrix <strong>Q</strong> to the right, with zeroes getting pushed
off the right edge and appearing back on the left, and then flip the matrix vertically. Presto, you just "calculated" <strong>Q'</strong>.
</p>
<p>
Implementing curve splitting this way requires less recursion, and is just straight arithmetic with cached values, so can be cheaper on
systems where recursion is expensive. If you're doing computation with devices that are good at matrix multiplication, chopping up a
Bézier curve with this method will be a lot faster than applying de Casteljau.
</p>
</section>
<section id="reordering">
<h1>
<div class="nav">
<a href="ko-KR/index.html#matrixsplit">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#derivatives">다음</a>
</div>
<a href="ko-KR/index.html#reordering">Lowering and elevating curve order</a>
</h1>
<p>
One interesting property of Bézier curves is that an <em>n<sup>th</sup></em> order curve can always be perfectly represented by an
<em>(n+1)<sup>th</sup></em> order curve, by giving the higher-order curve specific control points.
</p>
<p>
If we have a curve with three points, then we can create a curve with four points that exactly reproduces the original curve. First, we
give it the same start and end points, and for its two control points we pick "1/3<sup>rd</sup> start + 2/3<sup>rd</sup> control" and
"2/3<sup>rd</sup> control + 1/3<sup>rd</sup> end". Now we have exactly the same curve as before, except represented as a cubic curve
rather than a quadratic curve.
</p>
<p>
The general rule for raising an <em>n<sup>th</sup></em> order curve to an <em>(n+1)<sup>th</sup></em> order curve is as follows (observing
that the start and end weights are the same as the start and end weights for the old curve):
</p>
<!--
__ k k-i i
Bézier(k,t) = ❯ \underset binomial term\underbrace\binomki · \ \underset polynomial term\underbrace(1-t) · t · \ \underset new
‾‾ i=0
╭ (k-i) · w + i · w ╮
│ i i-1 │
weights\underbrace│ ─────────────────────── │ , with k = n+1 and w =0 when i = 0
╰ k ╯ i-1
-->
<img class="LaTeX SVG" src="./images/chapters/reordering/74e038deabd9e240606fa3f07ba98269.svg" width="755px" height="61px" loading="lazy" />
<p>
However, this rule also has as direct consequence that you <strong>cannot</strong> generally safely lower a curve from
<em>n<sup>th</sup></em> order to <em>(n-1)<sup>th</sup></em> order, because the control points cannot be "pulled apart" cleanly. We can
try to, but the resulting curve will not be identical to the original, and may in fact look completely different.
</p>
<p>
However, there is a surprisingly good way to ensure that a lower order curve looks "as close as reasonably possible" to the original
curve: we can optimise the "least-squares distance" between the original curve and the lower order curve, in a single operation (also
explained over on <a href="https://www.sirver.net/blog/2011/08/23/degree-reduction-of-bezier-curves/">Sirver's Castle</a>). However, to
use it, we'll need to do some calculus work and then switch over to linear algebra. As mentioned in the section on matrix representations,
some things can be done much more easily with matrices than with calculus functions, and this is one of those things. So... let's go!
</p>
<p>We start by taking the standard Bézier function, and condensing it a little:</p>
<!--
__ n n n n-i i
Bézier(n,t) = ❯ w B (t) , where B (t) = \binomni · (1-t) · t
‾‾ i=0 i i i
-->
<img class="LaTeX SVG" src="./images/chapters/reordering/9fc4ecc087d389dd0111bcba165cd5d0.svg" width="408px" height="41px" loading="lazy" />
<p>
Then, we apply one of those silly (actually, super useful) calculus tricks: since our <code>t</code> value is always between zero and one
(inclusive), we know that <code>(1-t)</code> plus <code>t</code> always sums to 1. As such, we can express any value as a sum of
<code>t</code> and <code>1-t</code>:
</p>
<!--
x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
-->
<img class="LaTeX SVG" src="./images/chapters/reordering/ff224ded6bbbc94b43130f5f8eeb5d29.svg" width="379px" height="16px" loading="lazy" />
<p>
So, with that seemingly trivial observation, we rewrite that Bézier function by splitting it up into a sum of a <code>(1-t)</code> and
<code>t</code> component:
</p>
<!--
Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
__ n n __ n n
= ❯ w (1 - t) B (t) + ❯ w t B (t)
‾‾ i=0 i i ‾‾ i=0 i i
-->
<img class="LaTeX SVG" src="./images/chapters/reordering/0cf0d5f856ec204dc32e0e42691cc70a.svg" width="316px" height="67px" loading="lazy" />
<p>
So far so good. Now, to see why we did this, let's write out the <code>(1-t)</code> and <code>t</code> parts, and see what that gives us.
I promise, it's about to make sense. We start with <code>(1-t)</code>:
</p>
<!--
n n! n-i i
(1 - t) B (t)= (1-t) ──────── (1-t) t
i (n-i)!i!
n+1-i (n+1)! n+1-i i
= ───── ────────── (1-t) t
n+1 (n+1-i)!i!
k-i k! k-i i
= ─── ──────── (1-t) t , where k = n + 1
k (k-i)!i!
k-i k
= ─── B (t)
k i
-->
<img
class="LaTeX SVG"
src="./images/chapters/reordering/0f5698b31598b2390e966fc5e43ab53e.svg"
width="387px"
height="160px"
loading="lazy"
/>
<p>
So by using this seemingly silly trick, we can suddenly express part of our n<sup>th</sup> order Bézier function in terms of an (n+1)<sup
>th</sup
>
order Bézier function. And that sounds a lot like raising the curve order! Of course we need to be able to repeat that trick for the
<code>t</code> part, but that's not a problem:
</p>
<!--
n n! n-i i
t B (t)= t ──────── (1-t) t
i (n-i)!i!
i+1 (n+1)! (n+1)-(i+1) i+1
= ─── ──────────────────── (1-t) t
n+1 ((n+1)-(i+1))!(i+1)!
i+1 k! k-(i+1) i+1
= ─── ──────────────── (1-t) t , where k = n + 1
k (k-(i+1))!(i+1)!
i+1 k
= ─── B (t)
k i+1
-->
<img
class="LaTeX SVG"
src="./images/chapters/reordering/ab7c087f7c070d43a42f3f03010a7427.svg"
width="471px"
height="159px"
loading="lazy"
/>
<p>
So, with both of those changed from an order <code>n</code> expression to an order <code>(n+1)</code> expression, we can put them back
together again. Now, where the order <code>n</code> function had a summation from 0 to <code>n</code>, the order <code>n+1</code> function
uses a summation from 0 to <code>n+1</code>, but this shouldn't be a problem as long as we can add some new terms that "contribute
nothing". In the next section on derivatives, there is a discussion about why "higher terms than there is a binomial for" and "lower than
zero terms" both "contribute nothing". So as long as we can add terms that have the same form as the terms we need, we can just include
them in the summation, they'll sit there and do nothing, and the resulting function stays identical to the lower order curve.
</p>
<p>Let's do this:</p>
<!--
__ n+1 n __ n+1 n
Bézier(n,t)= ❯ w (1 - t) B (t) + ❯ w t B (t)
‾‾ i=0 i i ‾‾ i=0 i i
__ n+1 k-i k __ n+1 i+1 k
= ❯ w ─── B (t) + ❯ w ─── B (t), where k = n + 1
‾‾ i=0 i k i ‾‾ i=0 i k i+1
__ n+1 k-i k __ n+1 i k
= ❯ w ─── B (t) + ❯ p ─ B (t)
‾‾ i=0 i k i ‾‾ i=0 i-1 k i
__ n+1 ╭ k-i i ╮ k
= ❯ │ w ─── + p ─ │ B (t)
‾‾ i=0 ╰ i k i-1 k ╯ i
__ n+1 k i
= ❯ (w (1-s) + p s) B (t), where s = ─
‾‾ i=0 i i-1 i k
-->
<img
class="LaTeX SVG"
src="./images/chapters/reordering/4ff41e183d60d5fd10a5d3d30dd63358.svg"
width="465px"
height="257px"
loading="lazy"
/>
<p>
And this is where we switch over from calculus to linear algebra, and matrices: we can now express this relation between Bézier(n,t) and
Bézier(n+1,t) as a very simple matrix multiplication:
</p>
<!--
M B = B
n k
-->
<img class="LaTeX SVG" src="./images/chapters/reordering/1f5b60d190a1c7099b3411e4cc477291.svg" width="71px" height="16px" loading="lazy" />
<p>where the matrix <strong>M</strong> is an <code>n+1</code> by <code>n</code> matrix, and looks like:</p>
<!--
┌ 1 0 . . . . . . ┐
│ 1 k-1 │
│ ─ ─── 0 . . . 0 . │
│ k k │
│ 2 k-2 │
│ 0 ─ ─── 0 . . . . │
│ k k │
│ 3 k-3 │
│ . 0 ─ ─── 0 . . . │
M = │ k k │
│ . . 0 ... ... 0 . . │
│ . . . 0 ... ... 0 . │
│ n-1 k-n+1 │
│ . . . . 0 ─── ───── 0 │
│ k k │
│ n k-n │
│ . 0 . . . 0 ─ ─── │
│ k k │
└ . . . . . . 0 1 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/reordering/056e25c397c524d80f378ce3823c7e78.svg"
width="336px"
height="187px"
loading="lazy"
/>
<p>
That might look unwieldy, but it's really just a mostly-zeroes matrix, with a very simply fraction on the diagonal, and an even simpler
fraction to the left of it. Multiplying a list of coordinates with this matrix means we can plug the resulting transformed coordinates
into the one-order-higher function and get an identical looking curve.
</p>
<p>Not too bad!</p>
<p>
Equally interesting, though, is that with this matrix operation established, we can now use an incredibly powerful and ridiculously simple
way to find out a "best fit" way to reverse the operation, called
<a href="https://mathworld.wolfram.com/NormalEquation.html">the normal equation</a>. What it does is minimize the sum of the square
differences between one set of values and another set of values. Specifically, if we can express that as some function
<strong>A x = b</strong>, we can use it. And as it so happens, that's exactly what we're dealing with, so:
</p>
<!--
M B = B
n k
T T
(M M) B = M B
n k
T -1 T T -1 T
(M M) (M M) B = (M M) M B
n k
T -1 T
I B = (M M) M B
n k
T -1 T
B = (M M) M B
n k
-->
<img
class="LaTeX SVG"
src="./images/chapters/reordering/46e64dc07502e14217ec83d755f736ee.svg"
width="272px"
height="116px"
loading="lazy"
/>
<p>The steps taken here are:</p>
<ol>
<li>We have a function in a form that the normal equation can be used with, so</li>
<li>apply the normal equation!</li>
<li>
Then, we want to end up with just B<sub>n</sub> on the left, so we start by left-multiply both sides such that we'll end up with lots of
stuff on the left that simplified to "a factor 1", which in matrix maths is the
<a href="https://en.wikipedia.org/wiki/Identity_matrix">identity matrix</a>.
</li>
<li>
In fact, by left-multiplying with the inverse of what was already there, we've effectively "nullified" (but really, one-inified) that
big, unwieldy block into the identity matrix <strong>I</strong>. So we substitute the mess with <strong>I</strong>, and then
</li>
<li>
because multiplication with the identity matrix does nothing (like multiplying by 1 does nothing in regular algebra), we just drop it.
</li>
</ol>
<p>
And we're done: we now have an expression that lets us approximate an <code>n+1</code><sup>th</sup> order curve with a lower <code>n</code
><sup>th</sup> order curve. It won't be an exact fit, but it's definitely a best approximation. So, let's implement these rules for
raising and lowering curve order to a (semi) random curve, using the following graphic. Select the sketch, which has movable control
points, and press your up and down arrow keys to raise or lower the curve order.
</p>
<graphics-element
title="A variable-order Bézier curve"
width="275"
height="275"
src="./chapters/reordering/reorder.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/reordering/c4874e1205aabe624e5504abe154eae9.png" loading="lazy" />
<label>A variable-order Bézier curve</label>
</fallback-image>
<button class="raise">raise</button>
<button class="lower">lower</button>
</graphics-element>
</section>
<section id="derivatives">
<h1>
<div class="nav">
<a href="ko-KR/index.html#reordering">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#pointvectors">다음</a>
</div>
<a href="ko-KR/index.html#derivatives">Derivatives</a>
</h1>
<p>
There's a number of useful things that you can do with Bézier curves based on their derivative, and one of the more amusing observations
about Bézier curves is that their derivatives are, in fact, also Bézier curves. In fact, the differentiation of a Bézier curve is
relatively straightforward, although we do need a bit of math.
</p>
<p>First, let's look at the derivative rule for Bézier curves, which is:</p>
<!--
__ n-1
Bézier'(n,t) = n · ❯ (b -b ) · Bézier(n-1,t)
‾‾ i=0 i+1 i i
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/a7b79877822a8f60e45552dcafc0815d.svg"
width="333px"
height="44px"
loading="lazy"
/>
<p>
which we can also write (observing that <i>b</i> in this formula is the same as our <i>w</i> weights, and that <i>n</i> times a summation
is the same as a summation where each term is multiplied by <i>n</i>) as:
</p>
<!--
__ n-1
Bézier'(n,t) = ❯ Bézier(n-1,t) · n · (w -w )
‾‾ i=0 i i+1 i
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/8f78fdb9ef54b1bc4dbc00f07263cc97.svg"
width="343px"
height="44px"
loading="lazy"
/>
<p>
Or, in plain text: the derivative of an n<sup>th</sup> degree Bézier curve is an (n-1)<sup>th</sup> degree Bézier curve, with one fewer
term, and new weights w'<sub>0</sub>...w'<sub>n-1</sub> derived from the original weights as n(w<sub>i+1</sub> - w<sub>i</sub>). So for a
3<sup>rd</sup> degree curve, with four weights, the derivative has three new weights: w'<sub>0</sub> = 3(w<sub>1</sub>-w<sub>0</sub>),
w'<sub>1</sub> = 3(w<sub>2</sub>-w<sub>1</sub>) and w'<sub>2</sub> = 3(w<sub>3</sub>-w<sub>2</sub>).
</p>
<div class="note">
<h3>"Slow down, why is that true?"</h3>
<p>
Sometimes just being told "this is the derivative" is nice, but you might want to see why this is indeed the case. As such, let's have a
look at the proof for this derivative. First off, the weights are independent of the full Bézier function, so the derivative involves
only the derivative of the polynomial basis function. So, let's find that:
</p>
<!--
d ╭ n ╮ k n-k d
B (t) ── = │ │ t (1-t) ──
n,k dt ╰ k ╯ dt
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/a992185a346518b5ca159484019b6917.svg"
width="209px"
height="36px"
loading="lazy"
/>
<p>
Applying the <a href="https://en.wikipedia.org/wiki/Product_rule">product</a> and
<a href="https://en.wikipedia.org/wiki/Chain_rule">chain</a> rules gives us:
</p>
<!--
╭ n ╮ k-1 n-k k n-k-1
... = │ │ (k · t (1-t) + t · (1-t) · (n-k) · -1)
╰ k ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/c3ac18fe4ba0606a15bc111e52b17a9a.svg"
width="412px"
height="28px"
loading="lazy"
/>
<p>Which is hard to work with, so let's expand that properly:</p>
<!--
kn! k-1 n-k (n-k)n! k n-1-k
... = ──────── t (1-t) - ──────── t (1-t)
k!(n-k)! k!(n-k)!
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/12fa7f83f055ef2078cc9f04e1468663.svg"
width="344px"
height="27px"
loading="lazy"
/>
<p>
Now, the trick is to turn this expression into something that has binomial coefficients again, so we want to end up with things that
look like "x! over y!(x-y)!". If we can do that in a way that involves terms of <i>n-1</i> and <i>k-1</i>, we'll be on the right track.
</p>
<!--
n! k-1 n-k (n-k)n! k n-1-k
... = ──────────── t (1-t) - ──────── t (1-t)
(k-1)!(n-k)! k!(n-k)!
╭ (n-1)! k-1 n-k (n-k)(n-1)! k n-1-k ╮
... = n │ ──────────── t (1-t) - ─────────── t (1-t) │
╰ (k-1)!(n-k)! k!(n-k)! ╯
╭ (n-1)! (k-1) (n-1)-(k-1) (n-1)! k (n-1)-k ╮
... = n │ ──────────────────── t (1-t) - ──────────── t (1-t) │
╰ (k-1)!((n-1)-(k-1))! k!((n-1)-k)! ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/64c06c61727d0912a67c0f287a395e47.svg"
width="545px"
height="76px"
loading="lazy"
/>
<p>And that's the first part done: the two components inside the parentheses are actually regular, lower-order Bézier expressions:</p>
<!--
╭ x! y x-y x! k x-k ╮
... = n │ ──────── t (1-t) - ──────── t (1-t) │ , with x=n-1, y=k-1
╰ y!(x-y)! k!(x-k)! ╯
... = n (B (t) - B (t))
(n-1),(k-1) (n-1),k
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/6a3672344bb571eadb72669f60a93ff4.svg"
width="533px"
height="48px"
loading="lazy"
/>
<p>
Now to apply this to our weighted Bézier curves. We'll write out the plain curve formula that we saw earlier, and then work our way
through to its derivative:
</p>
<!--
Bézier (t) = B (t) · w + B (t) · w + B (t) · w + B (t) · w + ...
n,k n,0 0 n,1 1 n,2 2 n,3 3
d
Bézier (t) ── = n · (B (t) - B (t)) · w +
n,k dt n-1,-1 n-1,0 0
n · (B (t) - B (t)) · w +
n-1,0 n-1,1 1
n · (B (t) - B (t)) · w +
n-1,1 n-1,2 2
n · (B (t) - B (t)) · w +
n-1,2 n-1,3 3
...
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/02cecadc92b8ff681edc8edb0ace53ce.svg"
width="527px"
height="112px"
loading="lazy"
/>
<p>
If we expand this (with some color to show how terms line up), and reorder the terms by increasing values for <i>k</i> we see the
following:
</p>
<!--
n · B (t) · w +
n-1,-1 0
n · B (t) · w - n · B (t) · w +
n-1,\colorblue 0 1 n-1,\colorblue 0 0
n · B (t) · w - n · B (t) · w +
n-1,\colorred 1 2 n-1,\colorred 1 1
n · B (t) · w - n · B (t) · w +
n-1,\colormagenta 2 3 n-1,\colormagenta 2 2
... - n · B (t) · w +
n-1,3 3
...
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/87403e5b0da3dcc4ceca74fae058fe69.svg"
width="300px"
height="109px"
loading="lazy"
/>
<p>
Two of these terms fall way: the first term falls away because there is no -1<sup>st</sup> term in a summation. As such, it always
contributes "nothing", so we can safely completely ignore it for the purpose of finding the derivative function. The other term is the
very last term in this expansion: one involving <i>B<sub>n-1,n</sub></i
>. This term would have a binomial coefficient of [<i>i</i> choose <i>i+1</i>], which is a non-existent binomial coefficient. Again,
this term would contribute "nothing", so we can ignore it, too. This means we're left with:
</p>
<!--
n · B (t) · w - n · B (t) · w +
n-1,\colorblue 0 1 n-1,\colorblue 0 0
n · B (t) · w - n · B (t) · w +
n-1,\colorred 1 2 n-1,\colorred 1 1
n · B (t) · w - n · B (t) · w +
n-1,\colormagenta 2 3 n-1,\colormagenta 2 2
...
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/f00423aaceac6ade478aba2a761664d8.svg"
width="295px"
height="71px"
loading="lazy"
/>
<p>And that's just a summation of lower order curves:</p>
<!--
d
Bézier (t) ── = n · B (t) · (w - w ) + n · B (t) · (w - w ) + n · B (t) · (w - w )
n,k dt (n-1),\colorblue 0 1 0 (n-1),\colorred 1 2 1 (n-1),\colormagenta 2 3 2
+ ...
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/de486728b41299269cc990ed09d5d286.svg"
width="716px"
height="36px"
loading="lazy"
/>
<p>We can rewrite this as a normal summation, and we're done:</p>
<!--
d __ n-1 __ n-1
Bézier (t) ── = ❯ n · B (t) · (w - w ) = ❯ B (t) · \underset derivative weights \underbracen · (w - w )
n,k dt ‾‾ k=0 n-1,k k+1 k ‾‾ k=0 n-1,k k+1 k
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/4eeb75f5de2d13a39f894625d3222443.svg"
width="545px"
height="51px"
loading="lazy"
/>
</div>
<p>
Let's rewrite that in a form similar to our original formula, so we can see the difference. We will first list our original formula for
Bézier curves, and then the derivative:
</p>
<!--
__ n n-i i
Bézier(n,t) = ❯ \underset binomial term\underbrace\binomni · \ \underset polynomial term\underbrace(1-t) · t · \ \underset
‾‾ i=0
weight\underbracew
i
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/153d99ce571bd664945394a1203a9eba.svg"
width="336px"
height="55px"
loading="lazy"
/>
<!--
__ k k-i i
Bézier'(n,t) = ❯ \underset binomial term\underbrace\binomki · \ \underset polynomial term\underbrace(1-t) · t · \ \underset derivative
‾‾ i=0
weight\underbracen · (w - w ) , with k=n-1
i+1 i
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/2368534c6e964e6d4a54904cc99b8986.svg"
width="520px"
height="59px"
loading="lazy"
/>
<p>
What are the differences? In terms of the actual Bézier curve, virtually nothing! We lowered the order (rather than <i>n</i>, it's now
<i>n-1</i>), but it's still the same Bézier function. The only real difference is in how the weights change when we derive the curve's
function. If we have four points A, B, C, and D, then the derivative will have three points, the second derivative two, and the third
derivative one:
</p>
<!--
B(n,t), w = {A,B,C,D}
B'(n,t), n = 3, w' = {A',B',C'} = {3 · (B-A), 3 · (C-B), 3 · (D-C)}
B''(n,t), n = 2, w'' = {A'',B''} = {2 · (B'-A'), 2 · (C'-B')}
B'''(n,t), n = 1, w''' = {A'''} = {1 · (B''-A'')}
-->
<img
class="LaTeX SVG"
src="./images/chapters/derivatives/2d733684f81b65a42c4cdb3f1e589c8b.svg"
width="523px"
height="73px"
loading="lazy"
/>
<p>
We can keep performing this trick for as long as we have more than one weight. Once we have one weight left, the next step will see
<i>k = 0</i>, and the result of our "Bézier function" summation is zero, because we're not adding anything at all. As such, a quadratic
curve has no second derivative, a cubic curve has no third derivative, and generalized: an <i>n<sup>th</sup></i> order curve has
<i>n-1</i> (meaningful) derivatives, with any further derivative being zero.
</p>
</section>
<section id="pointvectors">
<h1>
<div class="nav">
<a href="ko-KR/index.html#derivatives">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#pointvectors3d">다음</a>
</div>
<a href="ko-KR/index.html#pointvectors">Tangents and normals</a>
</h1>
<p>
If you want to move objects along a curve, or "away from" a curve, the two vectors you're most interested in are the tangent vector and
normal vector for curve points. These are actually really easy to find. For moving and orienting along a curve, we use the tangent, which
indicates the direction of travel at specific points, and is literally just the first derivative of our curve:
</p>
<!--
tangent (t) = B' (t)
x x
tangent (t) = B' (t)
y y
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointvectors/da069c7d6cbdb516c5454371dae84e7f.svg"
width="132px"
height="61px"
loading="lazy"
/>
<p>
This gives us the directional vector we want. We can normalize it to give us uniform directional vectors (having a length of 1.0) at each
point, and then do whatever it is we want to do based on those directions:
</p>
<!--
┌─────────────────┐
│ 2 2
d = \| tangent(t)\| = │B' (t) + B' (t)
⟍│ x y
tangent (t) B' (t)
^ x x
x(t) = \| tangent (t)\| =─────────────── = ──────
x \| tangent(t)\| d
tangent (t) B' (t)
^ y y
y(t) = \| tangent (t)\| = ─────────────── = ──────
y \| tangent(t)\| d
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointvectors/33afd1a141ec444989c393b3e51ec9ca.svg"
width="273px"
height="121px"
loading="lazy"
/>
<p>
The tangent is very useful for moving along a line, but what if we want to move away from the curve instead, perpendicular to the curve at
some point <i>t</i>? In that case we want the <em>normal</em> vector. This vector runs at a right angle to the direction of the curve, and
is typically of length 1.0, so all we have to do is rotate the normalized directional vector and we're done:
</p>
<!--
^ \pi ^ \pi ^
normal (t) = x(t) · cos ─── - y(t) · sin ─── = - y(t)
x 2 2
^ \pi ^ \pi ^
normal (t) = \underset quarter circle rotation \underbrace x(t) · sin ─── + y(t) · cos ─── = x(t)
y 2 2
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointvectors/c3d5f3506b763b718e567f90dbb78324.svg"
width="324px"
height="79px"
loading="lazy"
/>
<div class="note">
<p>
Rotating coordinates is actually very easy, if you know the rule for it. You might find it explained as "applying a
<a href="https://en.wikipedia.org/wiki/Rotation_matrix">rotation matrix</a>, which is what we'll look at here, too. Essentially, the
idea is to take the circles over which we can rotate, and simply "sliding the coordinates" over these circles by the desired angle. If
we want a quarter circle turn, we take the coordinate, slide it along the circle by a quarter turn, and done.
</p>
<p>
To turn any point <i>(x,y)</i> into a rotated point <i>(x',y')</i> (over 0,0) by some angle φ, we apply this nice and easy computation:
</p>
<!--
x' = x · cos (\phi) - y · sin (\phi)
y' = x · sin (\phi) + y · cos (\phi)
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointvectors/1df6c055ae8e41a46bfdebc55a4f17c0.svg"
width="183px"
height="37px"
loading="lazy"
/>
<p>Which is the "long" version of the following matrix transformation:</p>
<!--
┌ x' ┐ = ┌ cos (\phi) -sin (\phi) ┐ ┌ x ┐
└ y' ┘ └ sin (\phi) cos (\phi) ┘ └ y ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointvectors/f02e359a5e47667919738fff69d2625b.svg"
width="211px"
height="40px"
loading="lazy"
/>
<p>
And that's all we need to rotate any coordinate. Note that for quarter, half, and three-quarter turns these functions become even
easier, since <em>sin</em> and <em>cos</em> for these angles are, respectively: 0 and 1, -1 and 0, and 0 and -1.
</p>
<p>
But <strong><em>why</em></strong> does this work? Why this matrix multiplication?
<a href="https://en.wikipedia.org/wiki/Rotation_matrix#Decomposition_into_shears">Wikipedia</a> (technically, Thomas Herter and Klaus
Lott) tells us that a rotation matrix can be treated as a sequence of three (elementary) shear operations. When we combine this into a
single matrix operation (because all matrix multiplications can be collapsed), we get the matrix that you see above.
<a href="https://datagenetics.com/blog/august32013/index.html">DataGenetics</a> have an excellent article about this very thing: it's
really quite cool, and I strongly recommend taking a quick break from this primer to read that article.
</p>
</div>
<p>
The following two graphics show the tangent and normal along a quadratic and cubic curve, with the direction vector coloured blue, and the
normal vector coloured red (the markers are spaced out evenly as <em>t</em>-intervals, not spaced equidistant).
</p>
<div class="figure">
<graphics-element
title="Quadratic Bézier tangents and normals"
width="275"
height="275"
src="./chapters/pointvectors/pointvectors.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/pointvectors/d0b09bb338bd42d4164ced871d1f77ba.png" loading="lazy" />
<label>Quadratic Bézier tangents and normals</label>
</fallback-image></graphics-element
>
<graphics-element
title="Cubic Bézier tangents and normals"
width="275"
height="275"
src="./chapters/pointvectors/pointvectors.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/pointvectors/b8285282b8a0ce28fc2cc0392e9b607a.png" loading="lazy" />
<label>Cubic Bézier tangents and normals</label>
</fallback-image></graphics-element
>
</div>
</section>
<section id="pointvectors3d">
<h1>
<div class="nav">
<a href="ko-KR/index.html#pointvectors">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#components">다음</a>
</div>
<a href="ko-KR/index.html#pointvectors3d">Working with 3D normals</a>
</h1>
<p>
Before we move on to the next section we need to spend a little bit of time on the difference between 2D and 3D. While for many things
this difference is irrelevant and the procedures are identical (for instance, getting the 3D tangent is just doing what we do for 2D, but
for x, y, and z, instead of just for x and y), when it comes to normals things are a little more complex, and thus more work. Mind you,
it's not "super hard", but there are more steps involved and we should have a look at those.
</p>
<p>
Getting normals in 3D is in principle the same as in 2D: we take the normalised tangent vector, and then rotate it by a quarter turn.
However, this is where things get that little more complex: we can turn in quite a few directions, since "the normal" in 3D is a plane,
not a single vector, so we basically need to define what "the" normal is in the 3D case.
</p>
<p>
The "naïve" approach is to construct what is known as the
<a href="https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas">Frenet normal</a>, where we follow a simple recipe that works in
many cases (but does super bizarre things in some others). The idea is that even though there are infinitely many vectors that are
perpendicular to the tangent (i.e. make a 90 degree angle with it), the tangent itself sort of lies on its own plane already: since each
point on the curve (no matter how closely spaced) has its own tangent vector, we can say that each point lies in the same plane as the
local tangent, as well as the tangents "right next to it".
</p>
<p>
Even if that difference in tangent vectors is minute, "any difference" is all we need to find out what that plane is - or rather, what the
vector perpendicular to that plane is. Which is what we need: if we can calculate that vector, and we have the tangent vector that we know
lies on a plane, then we can rotate the tangent vector over the perpendicular, and presto. We have computed the normal using the same
logic we used for the 2D case: "just rotate it 90 degrees".
</p>
<p>So let's do that! And in a twist surprise, we can do this in four lines:</p>
<ul>
<li><strong>a</strong> = normalize(B'(t))</li>
<li><strong>b</strong> = normalize(<strong>a</strong> + B''(t))</li>
<li><strong>r</strong> = normalize(<strong>b</strong> × <strong>a</strong>)</li>
<li><strong>normal</strong> = normalize(<strong>r</strong> × <strong>a</strong>)</li>
</ul>
<p>Let's unpack that a little:</p>
<ul>
<li>
We start by taking the <a href="https://en.wikipedia.org/wiki/Unit_vector">normalized vector</a> for the derivative at some point on the
curve. We normalize it so the maths is less work. Less work is good.
</li>
<li>
Then, we compute <strong>b</strong> which represents what a next point's tangent would be if the curve stopped changing at our point and
just had the same derivative and second derivative from that point on.
</li>
<li>
This lets us find two vectors (the derivative, and the second derivative added to the derivative) that lie on the same plane, which
means we can use them to compute a vector perpendicular to that plane, using an elementary vector operation called the
<a href="https://en.wikipedia.org/wiki/Cross_product">cross product</a>. (Note that while that operation uses the × operator, it's most
definitely not a multiplication!) The result of that gives us a vector that we can use as the "axis of rotation" for turning the tangent
a quarter circle to get our normal, just like we did in the 2D case.
</li>
<li>
Since the cross product lets us find a vector that is perpendicular to some plane defined by two other vectors, and since the normal
vector should be perpendicular to the plane that the tangent and the axis of rotation lie in, we can use the cross product a second
time, and immediately get our normal vector.
</li>
</ul>
<p>
And then we're done, we found "the" normal vector for a 3D curve. Let's see what that looks like for a sample curve, shall we? You can
move your cursor across the graphic from left to right, to show the normal at a point with a t value that is based on your cursor
position: all the way on the left is 0, all the way on the right = 1, midway is t=0.5, etc:
</p>
<graphics-element
title="Some known and unknown vectors"
width="350"
height="300"
src="./chapters/pointvectors3d/frenet.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="350px" height="300px" src="./images/chapters/pointvectors3d/11c1da2357004bb51cf0c591fc492115.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0" class="slide-control" />
</graphics-element>
<p>
However, if you've played with that graphic a bit, you might have noticed something odd. The normal seems to "suddenly twist around the
curve" between t=0.65 and t=0.75... Why is it doing that?
</p>
<p>
As it turns out, it's doing that because that's how the maths works, and that's the problem with Frenet normals: while they are
"mathematically correct", they are "practically problematic", and so for any kind of graphics work what we really want is a way to compute
normals that just... look good.
</p>
<p>Thankfully, Frenet normals are not our only option.</p>
<p>
Another option is to take a slightly more algorithmic approach and compute a form of
<a href="https://www.microsoft.com/en-us/research/wp-content/uploads/2016/12/Computation-of-rotation-minimizing-frames.pdf"
>Rotation Minimising Frame</a
>
(also known as "parallel transport frame" or "Bishop frame") instead, where a "frame" is a set made up of the tangent, the rotational
axis, and the normal vector, centered on an on-curve point.
</p>
<p>
These type of frames are computed based on "the previous frame", so we cannot simply compute these "on demand" for single points, as we
could for Frenet frames; we have to compute them for the entire curve. Thankfully, the procedure is pretty simple, and can be performed at
the same time that you're building lookup tables for your curve.
</p>
<p>
The idea is to take a starting "tangent/rotation axis/normal" frame at t=0, and then compute what the next frame "should" look like by
applying some rules that yield a good looking next frame. In the case of the RMF paper linked above, those rules are:
</p>
<ul>
<li>Take a point on the curve for which we know the RM frame already,</li>
<li>take a next point on the curve for which we don't know the RM frame yet, and</li>
<li>
reflect the known frame onto the next point, by treating the plane through the curve at the point exactly between the next and previous
points as a "mirror".
</li>
<li>
This gives the next point a tangent vector that's essentially pointing in the opposite direction of what it should be, and a normal
that's slightly off-kilter, so:
</li>
<li>
reflect the vectors of our "mirrored frame" a second time, but this time using the plane through the "next point" itself as "mirror".
</li>
<li>Done: the tangent and normal have been fixed, and we have a good looking frame to work with.</li>
</ul>
<p>So, let's write some code for that!</p>
<div class="howtocode">
<h3>Implementing Rotation Minimising Frames</h3>
<p>
We first assume we have a function for calculating the Frenet frame at a point, which we already discussed above, inn a way that it
yields a frame with properties:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="6">
<textarea disabled rows="6" role="doc-example">
{
o: origin of all vectors, i.e. the on-curve point,
t: tangent vector,
r: rotational axis vector,
n: normal vector
}</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
</table>
<p>Then, we can write a function that generates a sequence of RM frames in the following manner:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="36">
<textarea disabled rows="36" role="doc-example">
generateRMFrames(steps) -> frames:
step = 1.0/steps
// Start off with the standard tangent/axis/normal frame
// associated with the curve at t=0:
frames.add(getFrenetFrame(0))
// start constructing RM frames:
for t0 = 0, t0 < 1.0, t0 += step:
// start with the previous, known frame
x0 = frames.last
// get the next frame: we're going to keep its position and tangent,
// but we're going to recompute the axis and normal.
t1 = t0 + step
x1 = { o: getPoint(t1), t: getDerivative(t) }
// First we reflect x0's tangent and axis of rotation onto x1,
// through/ the plane of reflection at the point between x0 x1
v1 = x1.o - x0.o
c1 = v1 · v1
riL = x0.r - v1 * 2/c1 * v1 · x0.r
tiL = x0.t - v1 * 2/c1 * v1 · x0.t
// note that v1 is a vector, but 2/c1 and (v1 · ...) are just
// plain numbers, so we're just scaling v1 by some constant.
// Then we reflect a second time, over a plane at x1, so that
// the frame tangent is aligned with the curve tangent again:
v2 = x1.t - tiL
c2 = v2 · v2
// and we're done here:
x1.r = riL - v2 * 2/c2 * v2 · riL
x1.n = x1.r × x1.t
frames.add(x1)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
<tr>
<td>18</td>
</tr>
<tr>
<td>19</td>
</tr>
<tr>
<td>20</td>
</tr>
<tr>
<td>21</td>
</tr>
<tr>
<td>22</td>
</tr>
<tr>
<td>23</td>
</tr>
<tr>
<td>24</td>
</tr>
<tr>
<td>25</td>
</tr>
<tr>
<td>26</td>
</tr>
<tr>
<td>27</td>
</tr>
<tr>
<td>28</td>
</tr>
<tr>
<td>29</td>
</tr>
<tr>
<td>30</td>
</tr>
<tr>
<td>31</td>
</tr>
<tr>
<td>32</td>
</tr>
<tr>
<td>33</td>
</tr>
<tr>
<td>34</td>
</tr>
<tr>
<td>35</td>
</tr>
<tr>
<td>36</td>
</tr>
</table>
<p>
Ignoring comments, this is certainly more code than when we were just computing a single Frenet frame, but it's not a crazy amount more
code to get much better looking normals.
</p>
</div>
<p>
Speaking of better looking, what does this actually look like? Let's revisit that earlier curve, but this time use rotation minimising
frames rather than Frenet frames:
</p>
<graphics-element
title="Some known and unknown vectors"
width="350"
height="300"
src="./chapters/pointvectors3d/rotation-minimizing.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="350px" height="300px" src="./images/chapters/pointvectors3d/f4a2fa1e0204c890b2bff07228ba678d.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0" class="slide-control" />
</graphics-element>
<p>That looks so much better!</p>
<p>
For those reading along with the code: we don't even strictly speaking need a Frenet frame to start with: we could, for instance, treat
the z-axis as our initial axis of rotation, so that our initial normal is <strong>(0,0,1) × tangent</strong>, and then take things from
there, but having that initial "mathematically correct" frame so that the initial normal seems to line up based on the curve's orientation
in 3D space is just nice.
</p>
</section>
<section id="components">
<h1>
<div class="nav">
<a href="ko-KR/index.html#pointvectors3d">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#extremities">다음</a>
</div>
<a href="ko-KR/index.html#components">Component functions</a>
</h1>
<p>
One of the first things people run into when they start using Bézier curves in their own programs is "I know how to draw the curve, but
how do I determine the bounding box?". It's actually reasonably straightforward to do so, but it requires having some knowledge on
exploiting math to get the values we need. For bounding boxes, we aren't actually interested in the curve itself, but only in its
"extremities": the minimum and maximum values the curve has for its x- and y-axis values. If you remember your calculus (provided you ever
took calculus, otherwise it's going to be hard to remember) we can determine function extremities using the first derivative of that
function, but this poses a problem, since our function is parametric: every axis has its own function.
</p>
<p>
The solution: compute the derivative for each axis separately, and then fit them back together in the same way we do for the original.
</p>
<p>
Let's look at how a parametric Bézier curve "splits up" into two normal functions, one for the x-axis and one for the y-axis. Note the
leftmost figure is again an interactive curve, without labeled axes (you get coordinates in the graph instead). The center and rightmost
figures are the component functions for computing the x-axis value, given a value for <i>t</i> (between 0 and 1 inclusive), and the y-axis
value, respectively.
</p>
<p>
If you move points in a curve sideways, you should only see the middle graph change; likewise, moving points vertically should only show a
change in the right graph.
</p>
<graphics-element
title="Quadratic Bézier curve components"
width="825"
height="275"
src="./chapters/components/components.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/components/1e6e38f6403dbe4c8b80295a94fc6748.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>&nbsp;</p>
<graphics-element
title="Cubic Bézier curve components"
width="825"
height="275"
src="./chapters/components/components.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/components/348694339257428a260144da4bbf80fc.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
</section>
<section id="extremities">
<h1>
<div class="nav">
<a href="ko-KR/index.html#components">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#boundingbox">다음</a>
</div>
<a href="ko-KR/index.html#extremities">Finding extremities: root finding</a>
</h1>
<p>
Now that we understand (well, superficially anyway) the component functions, we can find the extremities of our Bézier curve by finding
maxima and minima on the component functions, by solving the equation B'(t) = 0. We've already seen that the derivative of a Bézier curve
is a simpler Bézier curve, but how do we solve the equality? Fairly easily, actually, until our derivatives are 4th order or higher...
then things get really hard. But let's start simple:
</p>
<h3>Quadratic curves: linear derivatives.</h3>
<p>
The derivative of a quadratic Bézier curve is a linear Bézier curve, interpolating between just two terms, which means finding the
solution for "where is this line 0" is effectively trivial by rewriting it to a function of <code>t</code> and solving. First we turn our
quadratic Bézier function into a linear one, by following the rule mentioned at the end of the
<a href="#derivatives">derivatives section</a>:
</p>
<!--
B'(t) = a(1-t) + b(t)= 0,
a - at + bt= 0,
(b-a)t + a= 0
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/55e16ef652d30face0f6586b675a6c7b.svg"
width="187px"
height="63px"
loading="lazy"
/>
<p>And then we turn this into our solution for <code>t</code> using basic arithmetics:</p>
<!--
(b-a)t + a= 0,
(b-a)t= -a,
-a
t= ───
b-a
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/1fab66c84e7df38a2edda147f939bd80.svg"
width="135px"
height="77px"
loading="lazy"
/>
<p>Done.</p>
<p>
Although with the <a href="https://en.wikipedia.org/wiki/Caveat_emptor#Caveat_lector">caveat</a> that if <code>b-a</code> is zero, there
is no solution and we probably shouldn't try to perform that division.
</p>
<h3>Cubic curves: the quadratic formula.</h3>
<p>
The derivative of a cubic Bézier curve is a quadratic Bézier curve, and finding the roots for a quadratic polynomial means we can apply
the <a href="https://en.wikipedia.org/wiki/Quadratic_formula">Quadratic formula</a>. If you've seen it before, you'll remember it, and if
you haven't, it looks like this:
</p>
<!--
┌────────┐
│ 2
2 -b ±⟍│b - 4ac
Given f(t) = at + bt + c, f(t)=0 when t = ───────────────
2a
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/3125ab785fb039994582552790a2674b.svg"
width="409px"
height="40px"
loading="lazy"
/>
<p>
So, if we can rewrite the Bézier component function as a plain polynomial, we're done: we just plug in the values into the quadratic
formula, check if that square root is negative or not (if it is, there are no roots) and then just compute the two values that come out
(because of that plus/minus sign we get two). Any value between 0 and 1 is a root that matters for Bézier curves, anything below or above
that is irrelevant (because Bézier curves are only defined over the interval [0,1]). So, how do we convert?
</p>
<p>
First we turn our cubic Bézier function into a quadratic one, by following the rule mentioned at the end of the
<a href="#derivatives">derivatives section</a>:
</p>
<!--
B(t) uses { p ,p ,p ,p }
1 2 3 4
B'(t) uses { v ,v ,v }, where v = 3(p -p ), v = 3(p -p ), v = 3(p -p )
1 2 3 1 2 1 2 3 2 3 4 3
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/bf0ad4611c47f8548396e40595c02b55.svg"
width="537px"
height="36px"
loading="lazy"
/>
<p>And then, using these <em>v</em> values, we can find out what our <em>a</em>, <em>b</em>, and <em>c</em> should be:</p>
<!--
2 2
B'(t)= v (1-t) + 2v (1-t)t + v t
1 2 3
2 2 2
...= v (t - 2t + 1) + 2v (t-t ) + v t
1 2 3
2 2 2
...= v t - 2v t + v + 2v t - 2v t + v t
1 1 1 2 2 3
2 2 2
...= v t - 2v t + v t - 2v t + v + 2v t
1 2 3 1 1 2
2
...= (v -2v +v )t + 2(v -v )t + v
1 2 3 2 1 1
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/d1c65d927825f20c3c358d1ff96ce881.svg"
width="315px"
height="119px"
loading="lazy"
/>
<p>
This gives us three coefficients {a, b, c} that are expressed in terms of <code>v</code> values, where the <code>v</code> values are
expressions of our original coordinate values, so we can do some substitution to get:
</p>
<!--
a= v -2v +v = 3(-p + 3p - 3p + p )
1 2 3 1 2 3 4
b= 2(v -v ) = 6(p - 2p + p )
2 1 1 2 3
c= v = 3(p -p )
1 2 1
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/a6acf08f43aa1f48c08a40e76bdd2a31.svg"
width="308px"
height="63px"
loading="lazy"
/>
<p>Easy-peasy. We can now almost trivially find the roots by plugging those values into the quadratic formula.</p>
<p>
And as a cubic curve, there is also a meaningful second derivative, which we can compute by simple taking the derivative of the
derivative.
</p>
<h3>Quartic curves: Cardano's algorithm.</h3>
<p>
We haven't really looked at them before now, but the next step up would be a Quartic curve, a fourth degree Bézier curve. As expected,
these have a derivative that is a cubic function, and now things get much harder. Cubic functions don't have a "simple" rule to find their
roots, like the quadratic formula, and instead require quite a bit of rewriting to a form that we can even start to try to solve.
</p>
<p>
Back in the 16<sup>th</sup> century, before Bézier curves were a thing, and even before <em>calculus itself</em> was a thing,
<a href="https://en.wikipedia.org/wiki/Gerolamo_Cardano">Gerolamo Cardano</a> figured out that even if the general cubic function is
really hard to solve, it can be rewritten to a form for which finding the roots is "easier" (even if not "easy"):
</p>
<!--
3 2
very hard: solve at + bt + ct + d = 0
3
easier: solve t + pt + q = 0
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/d31432533bd7940545d4a269eefbabf2.svg"
width="253px"
height="44px"
loading="lazy"
/>
<p>
We can see that the easier formula only has two constants, rather than four, and only two expressions involving <code>t</code>, rather
than three: this makes things considerably easier to solve because it lets us use
<a href="https://www.wolframalpha.com/input/?i=t%5E3+%2B+pt+%2B+q">regular calculus</a> to find the values that satisfy the equation.
</p>
<p>
Now, there is one small hitch: as a cubic function, the solutions may be
<a href="https://en.wikipedia.org/wiki/Complex_number">complex numbers</a> rather than plain numbers... And Cardano realised this,
centuries before complex numbers were a well-understood and established part of number theory. His interpretation of them was "these
numbers are impossible but that's okay because they disappear again in later steps", allowing him to not think about them too much, but we
have it even easier: as we're trying to find the roots for display purposes, we don't even <em>care</em> about complex numbers: we're
going to simplify Cardano's approach just that tiny bit further by throwing away any solution that's not a plain number.
</p>
<p>
So, how do we rewrite the hard formula into the easier formula? This is explained in detail over at
<a href="https://trans4mind.com/personal_development/mathematics/polynomials/cubicAlgebra.htm">Ken J. Ward's page</a> for solving the
cubic equation, so instead of showing the maths, I'm simply going to show the programming code for solving the cubic equation, with the
complex roots getting totally ignored, but if you're interested you should definitely head over to Ken's page and give the procedure a
read-through.
</p>
<div class="howtocode">
<h3>Implementing Cardano's algorithm for finding all real roots</h3>
<p>
The "real roots" part is fairly important, because while you cannot take a square, cube, etc. root of a negative number in the "real"
number space (denoted with ℝ), this is perfectly fine in the
<a href="https://en.wikipedia.org/wiki/Complex_number">"complex" number</a> space (denoted with ℂ). And, as it so happens, Cardano is
also attributed as the first mathematician in history to have made use of complex numbers in his calculations. For this very algorithm!
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="81">
<textarea disabled rows="81" role="doc-example">
// A helper function to filter for values in the [0,1] interval:
function accept(t) {
return 0<=t && t <=1;
}
// A real-cuberoots-only function:
function cuberoot(v) {
if(v<0) return -pow(-v,1/3);
return pow(v,1/3);
}
// Now then: given cubic coordinates {pa, pb, pc, pd} find all roots.
function getCubicRoots(pa, pb, pc, pd) {
var a = (3*pa - 6*pb + 3*pc),
b = (-3*pa + 3*pb),
c = pa,
d = (-pa + 3*pb - 3*pc + pd);
// do a check to see whether we even need cubic solving:
if (approximately(d,0)) {
// this is not a cubic curve.
if (approximately(a,0)) {
// in fact, this is not a quadratic curve either.
if (approximately(b,0)) {
// in fact in fact, there are no solutions.
return [];
}
// linear solution
return [-c / b].filter(accept);
}
// quadratic solution
var q = sqrt(b*b - 4*a*c), 2a = 2*a;
return [(q-b)/2a, (-b-q)/2a].filter(accept)
}
// at this point, we know we need a cubic solution.
a /= d;
b /= d;
c /= d;
var p = (3*b - a*a)/3,
p3 = p/3,
q = (2*a*a*a - 9*a*b + 27*c)/27,
q2 = q/2,
discriminant = q2*q2 + p3*p3*p3;
// and some variables we're going to use later on:
var u1, v1, root1, root2, root3;
// three possible real roots:
if (discriminant < 0) {
var mp3 = -p/3,
mp33 = mp3*mp3*mp3,
r = sqrt( mp33 ),
t = -q / (2*r),
cosphi = t<-1 ? -1 : t>1 ? 1 : t,
phi = acos(cosphi),
crtr = cuberoot(r),
t1 = 2*crtr;
root1 = t1 * cos(phi/3) - a/3;
root2 = t1 * cos((phi+2*pi)/3) - a/3;
root3 = t1 * cos((phi+4*pi)/3) - a/3;
return [root1, root2, root3].filter(accept);
}
// three real roots, but two of them are equal:
if(discriminant === 0) {
u1 = q2 < 0 ? cuberoot(-q2) : -cuberoot(q2);
root1 = 2*u1 - a/3;
root2 = -u1 - a/3;
return [root1, root2].filter(accept);
}
// one real root, two complex roots
var sd = sqrt(discriminant);
u1 = cuberoot(sd - q2);
v1 = cuberoot(sd + q2);
root1 = u1 - v1 - a/3;
return [root1].filter(accept);
}</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
<tr>
<td>18</td>
</tr>
<tr>
<td>19</td>
</tr>
<tr>
<td>20</td>
</tr>
<tr>
<td>21</td>
</tr>
<tr>
<td>22</td>
</tr>
<tr>
<td>23</td>
</tr>
<tr>
<td>24</td>
</tr>
<tr>
<td>25</td>
</tr>
<tr>
<td>26</td>
</tr>
<tr>
<td>27</td>
</tr>
<tr>
<td>28</td>
</tr>
<tr>
<td>29</td>
</tr>
<tr>
<td>30</td>
</tr>
<tr>
<td>31</td>
</tr>
<tr>
<td>32</td>
</tr>
<tr>
<td>33</td>
</tr>
<tr>
<td>34</td>
</tr>
<tr>
<td>35</td>
</tr>
<tr>
<td>36</td>
</tr>
<tr>
<td>37</td>
</tr>
<tr>
<td>38</td>
</tr>
<tr>
<td>39</td>
</tr>
<tr>
<td>40</td>
</tr>
<tr>
<td>41</td>
</tr>
<tr>
<td>42</td>
</tr>
<tr>
<td>43</td>
</tr>
<tr>
<td>44</td>
</tr>
<tr>
<td>45</td>
</tr>
<tr>
<td>46</td>
</tr>
<tr>
<td>47</td>
</tr>
<tr>
<td>48</td>
</tr>
<tr>
<td>49</td>
</tr>
<tr>
<td>50</td>
</tr>
<tr>
<td>51</td>
</tr>
<tr>
<td>52</td>
</tr>
<tr>
<td>53</td>
</tr>
<tr>
<td>54</td>
</tr>
<tr>
<td>55</td>
</tr>
<tr>
<td>56</td>
</tr>
<tr>
<td>57</td>
</tr>
<tr>
<td>58</td>
</tr>
<tr>
<td>59</td>
</tr>
<tr>
<td>60</td>
</tr>
<tr>
<td>61</td>
</tr>
<tr>
<td>62</td>
</tr>
<tr>
<td>63</td>
</tr>
<tr>
<td>64</td>
</tr>
<tr>
<td>65</td>
</tr>
<tr>
<td>66</td>
</tr>
<tr>
<td>67</td>
</tr>
<tr>
<td>68</td>
</tr>
<tr>
<td>69</td>
</tr>
<tr>
<td>70</td>
</tr>
<tr>
<td>71</td>
</tr>
<tr>
<td>72</td>
</tr>
<tr>
<td>73</td>
</tr>
<tr>
<td>74</td>
</tr>
<tr>
<td>75</td>
</tr>
<tr>
<td>76</td>
</tr>
<tr>
<td>77</td>
</tr>
<tr>
<td>78</td>
</tr>
<tr>
<td>79</td>
</tr>
<tr>
<td>80</td>
</tr>
<tr>
<td>81</td>
</tr>
</table>
</div>
<p>
And that's it. The maths is complicated, but the code is pretty much just "follow the maths, while caching as many values as we can to
prevent recomputing things as much as possible" and now we have a way to find all roots for a cubic function and can just move on with
using that to find extremities of our curves.
</p>
<p>
And of course, as a quartic curve also has meaningful second and third derivatives, we can quite easily compute those by using the
derivative of the derivative (of the derivative), just as for cubic curves.
</p>
<h3>Quintic and higher order curves: finding numerical solutions</h3>
<p>
And this is where thing stop, because we <em>cannot</em> find the roots for polynomials of degree 5 or higher using algebra (a fact known
as <a href="https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem">the Abel–Ruffini theorem</a>). Instead, for occasions like these,
where algebra simply cannot yield an answer, we turn to <a href="https://en.wikipedia.org/wiki/Numerical_analysis">numerical analysis</a>.
</p>
<p>
That's a fancy term for saying "rather than trying to find exact answers by manipulating symbols, find approximate answers by describing
the underlying process as a combination of steps, each of which <em>can</em> be assigned a number via symbolic manipulation". For example,
trying to mathematically compute how much water fits in a completely crazy three dimensional shape is very hard, even if it got you the
perfect, precise answer. A much easier approach, which would be less perfect but still entirely useful, would be to just grab a buck and
start filling the shape until it was full: just count the number of buckets of water you used. And if we want a more precise answer, we
can use smaller buckets.
</p>
<p>
So that's what we're going to do here, too: we're going to treat the problem as a sequence of steps, and the smaller we can make each
step, the closer we'll get to that "perfect, precise" answer. And as it turns out, there is a really nice numerical root-finding
algorithm, called the <a href="https://en.wikipedia.org/wiki/Newton-Raphson">Newton-Raphson</a> root finding method (yes, after
<em><a href="https://en.wikipedia.org/wiki/Isaac_Newton">that</a></em> Newton), which we can make use of. The Newton-Raphson approach
consists of taking our impossible-to-solve function <code>f(x)</code>, picking some initial value <code>x</code> (literally any value will
do), and calculating <code>f(x)</code>. We can think of that value as the "height" of the function at <code>x</code>. If that height is
zero, we're done, we have found a root. If it isn't, we calculate the tangent line at <code>f(x)</code> and calculate at which
<code>x</code> value <em>its</em> height is zero (which we've already seen is very easy). That will give us a new <code>x</code> and we
repeat the process until we find a root.
</p>
<p>
Mathematically, this means that for some <code>x</code>, at step <code>n=1</code>, we perform the following calculation until
<code>f<sub>y</sub>(x)</code> is zero, so that the next <code>t</code> is the same as the one we already have:
</p>
<!--
f (x )
y n
x = x - ───────
n+1 n f' (x )
y n
-->
<img
class="LaTeX SVG"
src="./images/chapters/extremities/53e67a29f134bd561aca550a2091a196.svg"
width="132px"
height="45px"
loading="lazy"
/>
<p>(The Wikipedia article has a decent animation for this process, so I will not add a graphic for that here)</p>
<p>
Now, this works well only if we can pick good starting points, and our curve is
<a href="https://en.wikipedia.org/wiki/Continuous_function">continuously differentiable</a> and doesn't have
<a href="https://en.wikipedia.org/wiki/Oscillation_(mathematics)">oscillations</a>. Glossing over the exact meaning of those terms, the
curves we're dealing with conform to those constraints, so as long as we pick good starting points, this will work. So the question is:
which starting points do we pick?
</p>
<p>
As it turns out, Newton-Raphson is so blindingly fast that we could get away with just not picking: we simply run the algorithm from
<em>t=0</em> to <em>t=1</em> at small steps (say, 1/200<sup>th</sup>) and the result will be all the roots we want. Of course, this may
pose problems for high order Bézier curves: 200 steps for a 200<sup>th</sup> order Bézier curve is going to go wrong, but that's okay:
there is no reason (at least, none that I know of) to <em>ever</em> use Bézier curves of crazy high orders. You might use a fifth order
curve to get the "nicest still remotely workable" approximation of a full circle with a single Bézier curve, but that's pretty much as
high as you'll ever need to go.
</p>
<h3>In conclusion:</h3>
<p>
So now that we know how to do root finding, we can determine the first and second derivative roots for our Bézier curves, and show those
roots overlaid on the previous graphics. For the quadratic curve, that means just the first derivative, in red:
</p>
<graphics-element
title="Quadratic Bézier curve extremities"
width="825"
height="275"
src="./chapters/extremities/extremities.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/extremities/fd68347a917c9b703ff8005287ac6ca4.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>And for cubic curves, that means first and second derivatives, in red and purple respectively:</p>
<graphics-element
title="Cubic Bézier curve extremities"
width="825"
height="275"
src="./chapters/extremities/extremities.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/extremities/fbfe9464c9653f5efcd04411e683faf9.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
</section>
<section id="boundingbox">
<h1>
<div class="nav"><a href="ko-KR/index.html#extremities">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#aligning">다음</a></div>
<a href="ko-KR/index.html#boundingbox">Bounding boxes</a>
</h1>
<p>
If we have the extremities, and the start/end points, a simple for-loop that tests for min/max values for x and y means we have the four
values we need to box in our curve:
</p>
<p><em>Computing the bounding box for a Bézier curve</em>:</p>
<ol>
<li>Find all <em>t</em> value(s) for the curve derivative's x- and y-roots.</li>
<li>Discard any <em>t</em> value that's lower than 0 or higher than 1, because Bézier curves only use the interval [0,1].</li>
<li>
Determine the lowest and highest value when plugging the values <em>t=0</em>, <em>t=1</em> and each of the found roots into the original
functions: the lowest value is the lower bound, and the highest value is the upper bound for the bounding box we want to construct.
</li>
</ol>
<p>
Applying this approach to our previous root finding, we get the following
<a href="https://en.wikipedia.org/wiki/Bounding_volume#Common_types">axis-aligned bounding boxes</a> (with all curve extremity points
shown on the curve):
</p>
<div class="figure">
<graphics-element
title="Quadratic Bézier bounding box"
width="275"
height="275"
src="./chapters/boundingbox/bbox.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/boundingbox/12ec4a5039de2e2cc06611db5e826282.png" loading="lazy" />
<label>Quadratic Bézier bounding box</label>
</fallback-image></graphics-element
>
<graphics-element
title="Cubic Bézier bounding box"
width="275"
height="275"
src="./chapters/boundingbox/bbox.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/boundingbox/daad01218ba430e2355d151811aa971b.png" loading="lazy" />
<label>Cubic Bézier bounding box</label>
</fallback-image></graphics-element
>
</div>
<p>
We can construct even nicer boxes by aligning them along our curve, rather than along the x- and y-axis, but in order to do so we first
need to look at how aligning works.
</p>
</section>
<section id="aligning">
<h1>
<div class="nav">
<a href="ko-KR/index.html#boundingbox">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#tightbounds">다음</a>
</div>
<a href="ko-KR/index.html#aligning">Aligning curves</a>
</h1>
<p>
While there are an incredible number of curves we can define by varying the x- and y-coordinates for the control points, not all curves
are actually distinct. For instance, if we define a curve, and then rotate it 90 degrees, it's still the same curve, and we'll find its
extremities in the same spots, just at different draw coordinates. As such, one way to make sure we're working with a "unique" curve is to
"axis-align" it.
</p>
<p>
Aligning also simplifies a curve's functions. We can translate (move) the curve so that the first point lies on (0,0), which turns our
<em>n</em> term polynomial functions into <em>n-1</em> term functions. The order stays the same, but we have less terms. Then, we can
rotate the curves so that the last point always lies on the x-axis, too, making its coordinate (...,0). This further simplifies the
function for the y-component to an <em>n-2</em> term function. For instance, if we have a cubic curve such as this:
</p>
<!--
╭ 3 2 2 3
╡ x = \colorblue 120 · (1-t) \colorblue + 35 · 3 · (1-t) · t \colorblue + 220 · 3 · (1-t) · t \colorblue + 220 · t
│ 3 2 2 3
╰ y = \colorblue 160 · (1-t) \colorblue + 200 · 3 · (1-t) · t \colorblue + 260 · 3 · (1-t) · t \colorblue + 40 · t
-->
<img class="LaTeX SVG" src="./images/chapters/aligning/e5db5e945606d3429e4475ff92283a9c.svg" width="497px" height="40px" loading="lazy" />
<p>
Then translating it so that the first coordinate lies on (0,0), moving all <em>x</em> coordinates by -120, and all <em>y</em> coordinates
by -160, gives us:
</p>
<!--
╭ 3 2 2 3
╡ x = \colorblue 0 · (1-t) \colorblue - 85 · 3 · (1-t) · t \colorblue + 100 · 3 · (1-t) · t \colorblue + 100 · t
│ 3 2 2 3
╰ y = \colorblue 0 · (1-t) \colorblue + 40 · 3 · (1-t) · t \colorblue + 100 · 3 · (1-t) · t \colorblue - 120 · t
-->
<img class="LaTeX SVG" src="./images/chapters/aligning/d412c72ed342c2036965ff251c8fb443.svg" width="481px" height="40px" loading="lazy" />
<p>
If we then rotate the curve so that its end point lies on the x-axis, the coordinates (integer-rounded for illustrative purposes here)
become:
</p>
<!--
╭ 3 2 2 3
╡ x = \colorblue 0 · (1-t) \colorblue - 85 · 3 · (1-t) · t \colorblue - 12 · 3 · (1-t) · t \colorblue + 156 · t
│ 3 2 2 3
╰ y = \colorblue 0 · (1-t) \colorblue - 40 · 3 · (1-t) · t \colorblue + 140 · 3 · (1-t) · t \colorblue + 0 · t
-->
<img class="LaTeX SVG" src="./images/chapters/aligning/e49d7bb45a18d14fbb12df4d91e2c67b.svg" width="473px" height="40px" loading="lazy" />
<p>If we drop all the zero-terms, this gives us:</p>
<!--
╭ 2 2 3
╡ x = \colorblue - 85 · 3 · (1-t) · t \colorblue - 12 · 3 · (1-t) · t \colorblue + 156 · t
│ 2 2
╰ y = \colorblue - 40 · 3 · (1-t) · t \colorblue + 140 · 3 · (1-t) · t
-->
<img class="LaTeX SVG" src="./images/chapters/aligning/016f37843b2af2ae34dc8b2975505f3d.svg" width="408px" height="40px" loading="lazy" />
<p>
We can see that our original curve definition has been simplified considerably. The following graphics illustrate the result of aligning
our example curves to the x-axis, with the cubic case using the coordinates that were just used in the example formulae:
</p>
<graphics-element
title="Aligning a quadratic curve"
width="550"
height="275"
src="./chapters/aligning/aligning.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="550px" height="275px" src="./images/chapters/aligning/28cc0f129fa0c028a1addd702e99f162.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>&nbsp;</p>
<graphics-element
title="Aligning a cubic curve"
width="550"
height="275"
src="./chapters/aligning/aligning.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="550px" height="275px" src="./images/chapters/aligning/9a6755a1e31a990e8f072a6da98f811a.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
</section>
<section id="tightbounds">
<h1>
<div class="nav"><a href="ko-KR/index.html#aligning">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#inflections">다음</a></div>
<a href="ko-KR/index.html#tightbounds">Tight bounding boxes</a>
</h1>
<p>
With our knowledge of bounding boxes, and curve alignment, We can now form the "tight" bounding box for curves. We first align our curve,
recording the translation we performed, "T", and the rotation angle we used, "R". We then determine the aligned curve's normal bounding
box. Once we have that, we can map that bounding box back to our original curve by rotating it by -R, and then translating it by -T.
</p>
<p>We now have nice tight bounding boxes for our curves:</p>
<div class="figure">
<graphics-element
title="Aligning a quadratic curve"
width="275"
height="275"
src="./chapters/tightbounds/tightbounds.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/tightbounds/ed91133976018ec032d9115344debb36.png" loading="lazy" />
<label>Aligning a quadratic curve</label>
</fallback-image></graphics-element
>
<graphics-element
title="Aligning a cubic curve"
width="275"
height="275"
src="./chapters/tightbounds/tightbounds.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/tightbounds/9ee5abc64b3fba71e284c70539279d74.png" loading="lazy" />
<label>Aligning a cubic curve</label>
</fallback-image></graphics-element
>
</div>
<p>
These are, strictly speaking, not necessarily the tightest possible bounding boxes. It is possible to compute the optimal bounding box by
determining which spanning lines we need to effect a minimal box area, but because of the parametric nature of Bézier curves this is
actually a rather costly operation, and the gain in bounding precision is often not worth it.
</p>
</section>
<section id="inflections">
<h1>
<div class="nav"><a href="ko-KR/index.html#tightbounds">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#canonical">다음</a></div>
<a href="ko-KR/index.html#inflections">Curve inflections</a>
</h1>
<p>
Now that we know how to align a curve, there's one more thing we can calculate: inflection points. Imagine we have a variable size circle
that we can slide up against our curve. We place it against the curve and adjust its radius so that where it touches the curve, the
curvatures of the curve and the circle are the same, and then we start to slide the circle along the curve - for quadratic curves, we can
always do this without the circle behaving oddly: we might have to change the radius of the circle as we slide it along, but it'll always
sit against the same side of the curve.
</p>
<p>
But what happens with cubic curves? Imagine we have an S curve and we place our circle at the start of the curve, and start sliding it
along. For a while we can simply adjust the radius and things will be fine, but once we get to the midpoint of that S, something odd
happens: the circle "flips" from one side of the curve to the other side, in order for the curvatures to keep matching. This is called an
inflection, and we can find out where those happen relatively easily.
</p>
<p>What we need to do is solve a simple equation:</p>
<!--
C(t) = 0
-->
<img class="LaTeX SVG" src="./images/chapters/inflections/ed68dcfb203517ca080fe48914769fb0.svg" width="59px" height="16px" loading="lazy" />
<p>
What we're saying here is that given the curvature function <em>C(t)</em>, we want to know for which values of <em>t</em> this function is
zero, meaning there is no "curvature", which will be exactly at the point between our circle being on one side of the curve, and our
circle being on the other side of the curve. So what does <em>C(t)</em> look like? Actually something that seems not too hard:
</p>
<!--
C(t) = Bézier \prime(t) · Bézier \prime\prime(t) - Bézier \prime(t) · Bézier \prime\prime(t)
x y y x
-->
<img
class="LaTeX SVG"
src="./images/chapters/inflections/852f0346f025c671b8a1ce6b628028aa.svg"
width="385px"
height="21px"
loading="lazy"
/>
<p>
The function <em>C(t)</em> is the cross product between the first and second derivative functions for the parametric dimensions of our
curve. And, as already shown, derivatives of Bézier curves are just simpler Bézier curves, with very easy to compute new coefficients, so
this should be pretty easy.
</p>
<p>
However as we've seen in the section on aligning, aligning lets us simplify things <em>a lot</em>, by completely removing the
contributions of the first coordinate from most mathematical evaluations, and removing the last <em>y</em> coordinate as well by virtue of
the last point lying on the x-axis. So, while we can evaluate <em>C(t) = 0</em> for our curve, it'll be much easier to first axis-align
the curve and <em>then</em> evaluating the curvature function.
</p>
<div class="note">
<h3>Let's derive the full formula anyway</h3>
<p>
Of course, before we do our aligned check, let's see what happens if we compute the curvature function without axis-aligning. We start
with the first and second derivatives, given our basis functions:
</p>
<!--
3 2 2 3
Bézier(t) = x (1-t) + 3x (1-t) t + 3x (1-t)t + x t
1 2 3 4
\prime 2 2
Bézier (t) = a(1-t) + 2b(1-t)t + ct { a=3(x -x ),b=3(x -x ),c=3(x -x ) }
2 1 3 2 4 3
\prime\prime
Bézier (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
-->
<img
class="LaTeX SVG"
src="./images/chapters/inflections/35299f4eb8e0bed76b68c7beb2038031.svg"
width="601px"
height="71px"
loading="lazy"
/>
<p>And of course the same functions for <em>y</em>:</p>
<!--
3 2 2 3
Bézier(t) = y (1-t) + 3y (1-t) t + 3y (1-t)t + y t
1 2 3 4
\prime 2 2
Bézier (t) = d(1-t) + 2e(1-t)t + ft
\prime\prime
Bézier (t) = w(1-t) + zt
-->
<img
class="LaTeX SVG"
src="./images/chapters/inflections/8278b9bec92ae49927283396692b51d5.svg"
width="399px"
height="69px"
loading="lazy"
/>
<p>
Asking a computer to now compose the <em>C(t)</em> function for us (and to expand it to a readable form of simple terms) gives us this
rather overly complicated set of arithmetic expressions:
</p>
<!--
2 2 2 2 2
-18 t x y + 36 t x y - 18 t x y + 18 t x y - 54 t x y
2 1 3 1 4 1 1 2 3 2
2 2 2 2 2
+36 t x y - 36 t x y + 54 t x y - 18 t x y + 18 t x y
4 2 1 3 2 3 4 3 1 4
2 2
-36 t x y + 18 t x y + 36 t x y - 54 t x y + 18 t x y - 36 t x y
2 4 3 4 2 1 3 1 4 1 1 2
+54 t x y - 18 t x y + 54 t x y - 54 t x y - 18 t x y + 18 t x y
3 2 4 2 1 3 2 3 1 4 2 4
-18 x y + 18 x y + 18 x y - 18 x y - 18 x y + 18 x y
2 1 3 1 1 2 3 2 1 3 2 3
-->
<img
class="LaTeX SVG"
src="./images/chapters/inflections/75fae2d0a94eae4addf074c294855fc7.svg"
width="552px"
height="97px"
loading="lazy"
/>
<p>
That is... unwieldy. So, we note that there are a lot of terms that involve multiplications involving x1, y1, and y4, which would all
disappear if we axis-align our curve, which is why aligning is a great idea.
</p>
</div>
<p>
Aligning our curve so that three of the eight coefficients become zero, and observing that scale does not affect finding
<code>t</code> values, we end up with the following simple term function for <em>C(t)</em>:
</p>
<!--
2
(3 x y +2 x y +3 x y -x y ) t + (3 x y -x y -3 x y ) t + (x y -x y )
3 2 4 2 2 3 4 3 3 2 4 2 2 3 2 3 3 2
-->
<img
class="LaTeX SVG"
src="./images/chapters/inflections/e50243eaa99b5acc08533dd2e9b71a74.svg"
width="533px"
height="20px"
loading="lazy"
/>
<p>
That's a lot easier to work with: we see a fair number of terms that we can compute and then cache, giving us the following
simplification:
</p>
<!--
a = x · y ╮
3 2 │
b = x · y │ 2
4 2 ╞ C(t) = (-3a + 2b + 3c - d)t + (3a - b - 3c)t + (c - a)
c = x · y │
2 3 │
d = x · y │
4 3 ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/inflections/2dbf3071d74e2ba37ab888aaa3c1a17c.svg"
width="467px"
height="73px"
loading="lazy"
/>
<p>This is a plain quadratic curve, and we know how to solve <em>C(t) = 0</em>; we use the quadratic formula:</p>
<!--
┌──────────┐
│ 2
x = -3a + 2b + 3c - d ╮ -y ±⟍│y - 4 x z
y = 3a - b - 3c ╞ C(t) = 0 ==> t = ─────────────────
z = c - a ╯ 2x
-->
<img
class="LaTeX SVG"
src="./images/chapters/inflections/d7d564126099bc0740058a7cdd744772.svg"
width="405px"
height="55px"
loading="lazy"
/>
<p>
We can easily compute this value <em>if</em> the discriminator isn't a negative number (because we only want real roots, not complex
roots), and <em>if</em> <em>x</em> is not zero, because divisions by zero are rather useless.
</p>
<p>
Taking that into account, we compute <em>t</em>, we disregard any <em>t</em> value that isn't in the Bézier interval [0,1], and we now
know at which <em>t</em> value(s) our curve will inflect.
</p>
<graphics-element
title="Finding cubic Bézier curve inflections"
width="275"
height="275"
src="./chapters/inflections/inflection.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/inflections/726ece45630c43be14589c51f1606bd7.png" loading="lazy" />
<label>Finding cubic Bézier curve inflections</label>
</fallback-image></graphics-element
>
</section>
<section id="canonical">
<h1>
<div class="nav"><a href="ko-KR/index.html#inflections">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#yforx">다음</a></div>
<a href="ko-KR/index.html#canonical">The canonical form (for cubic curves)</a>
</h1>
<p>
While quadratic curves are relatively simple curves to analyze, the same cannot be said of the cubic curve. As a curvature is controlled
by more than one control point, it exhibits all kinds of features like loops, cusps, odd colinear features, and as many as two inflection
points because the curvature can change direction up to three times. Now, knowing what kind of curve we're dealing with means that some
algorithms can be run more efficiently than if we have to implement them as generic solvers, so is there a way to determine the curve type
without lots of work?
</p>
<p>
As it so happens, the answer is yes, and the solution we're going to look at was presented by Maureen C. Stone from Xerox PARC and Tony D.
deRose from the University of Washington in their joint paper
<a href="https://graphics.pixar.com/people/derose/publications/CubicClassification/paper.pdf"
>"A Geometric Characterization of Parametric Cubic curves"</a
>. It was published in 1989, and defines curves as having a "canonical" form (i.e. a form that all curves can be reduced to) from which we
can immediately tell what features a curve will have. So how does it work?
</p>
<p>
The first observation that makes things work is that if we have a cubic curve with four points, we can apply a linear transformation to
these points such that three of the points end up on (0,0), (0,1) and (1,1), with the last point then being "somewhere". After applying
that transformation, the location of that last point can then tell us what kind of curve we're dealing with. Specifically, we see the
following breakdown:
</p>
<graphics-element
title="The canonical curve map"
width="400"
height="400"
src="./chapters/canonical/canonical.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/canonical/c086e72bd8aaeab37436515ab251b2df.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>This is a fairly funky image, so let's see what the various parts of it mean...</p>
<p>
We see the three fixed points at (0,0), (0,1) and (1,1). The various regions and boundaries indicate what property the original curve will
have, if the fourth point is in/on that region or boundary. Specifically, if the fourth point is...
</p>
<ol>
<li>
<p>
...anywhere inside the red zone, but not on its boundaries, the curve will be self-intersecting (yielding a loop). We won't know
<em>where</em> it self-intersects (in terms of <em>t</em> values), but we are guaranteed that it does.
</p>
</li>
<li>
<p>
...on the left (red) edge of the red zone, the curve will have a cusp. We again don't know <em>where</em>, but we know there is one.
This edge is described by the function:
</p>
<!--
2
-x + 2x + 3
y = ────────────, { x ≤1 }
4
-->
<img
class="LaTeX SVG"
src="./images/chapters/canonical/674d251590411398d06fb99cba7920f7.svg"
width="180px"
height="37px"
loading="lazy"
/>
</li>
<li>
<p>
...on the almost circular, lower right (pink) edge, the curve's end point touches the curve, forming a loop. This edge is described by
the function:
</p>
<!--
┌──────────┐
│ 2
⟍│3(4x - x ) - x
y = ─────────────────, { 0 ≤x ≤1 }
2
-->
<img
class="LaTeX SVG"
src="./images/chapters/canonical/6959a552f2c90a2bcaa787c23e19f488.svg"
width="231px"
height="39px"
loading="lazy"
/>
</li>
<li>
<p>...on the top (blue) edge, the curve's start point touches the curve, forming a loop. This edge is described by the function:</p>
<!--
2
-x + 3x
y = ────────, { x ≤0 }
3
-->
<img
class="LaTeX SVG"
src="./images/chapters/canonical/464b4ec0b67f248459792752be86d46d.svg"
width="153px"
height="37px"
loading="lazy"
/>
</li>
<li>
<p>...inside the lower (green) zone, past <code>y=1</code>, the curve will have a single inflection (switching concave/convex once).</p>
</li>
<li>
<p>
...between the left and lower boundaries (below the cusp line but above the single-inflection line), the curve will have two
inflections (switching from concave to convex and then back again, or from convex to concave and then back again).
</p>
</li>
<li><p>...anywhere on the right of self-intersection zone, the curve will have no inflections. It'll just be a simple arch.</p></li>
</ol>
<p>Of course, this map is fairly small, but the regions extend to infinity, with well defined boundaries.</p>
<div class="note">
<h3>Wait, where do those lines come from?</h3>
<p>
Without repeating the paper mentioned at the top of this section, the loop-boundaries come from rewriting the curve into canonical form,
and then solving the formulae for which constraints must hold for which possible curve properties. In the paper these functions yield
formulae for where you will find cusp points, or loops where we know t=0 or t=1, but those functions are derived for the full cubic
expression, meaning they apply to t=-∞ to t=∞... For Bézier curves we only care about the "clipped interval" t=0 to t=1, so some of the
properties that apply when you look at the curve over an infinite interval simply don't apply to the Bézier curve interval.
</p>
<p>
The right bound for the loop region, indicating where the curve switches from "having inflections" to "having a loop", for the general
cubic curve, is actually mirrored over x=1, but for Bézier curves this right half doesn't apply, so we don't need to pay attention to
it. Similarly, the boundaries for t=0 and t=1 loops are also nice clean curves but get "cut off" when we only look at what the general
curve does over the interval t=0 to t=1.
</p>
<p>
For the full details, head over to the paper and read through sections 3 and 4. If you still remember your high school pre-calculus, you
can probably follow along with this paper, although you might have to read it a few times before all the bits "click".
</p>
</div>
<p>
So now the question becomes: how do we manipulate our curve so that it fits this canonical form, with three fixed points, and one "free"
point? Enter linear algebra. Don't worry, I'll be doing all the math for you, as well as show you what the effect is on our curves, but
basically we're going to be using linear algebra, rather than calculus, because "it's way easier". Sometimes a calculus approach is very
hard to work with, when the equivalent geometrical solution is super obvious.
</p>
<p>
The approach is going to start with a curve that doesn't have all-colinear points (so we need to make sure the points don't all fall on a
straight line), and then applying three graphics operations that you will probably have heard of: translation (moving all points by some
fixed x- and y-distance), scaling (multiplying all points by some x and y scale factor), and shearing (an operation that turns rectangles
into parallelograms).
</p>
<p>
Step 1: we translate any curve by -p1.x and -p1.y, so that the curve starts at (0,0). We're going to make use of an interesting trick
here, by pretending our 2D coordinates are 3D, with the <em>z</em> coordinate simply always being 1. This is an old trick in graphics to
overcome the limitations of 2D transformations: without it, we can only turn (x,y) coordinates into new coordinates of the form (ax + by,
cx + dy), which means we can't do translation, since that requires we end up with some kind of (x + a, y + b). If we add a bogus
<em>z</em> coordinate that is always 1, then we can suddenly add arbitrary values. For example:
</p>
<!--
┌ 1 0 a ┐ ┌ x ┐ ┌ 1 · x + 0 · y + a · z ┐ ┌ x + a · 1 ┐ ┌ x + a ┐
│ 0 1 b │ · │ y │ = │ 0 · x + 1 · y + b · z │ = │ y + b · 1 │ = │ y + b │
└ 0 0 1 ┘ └ z=1 ┘ └ 0 · x + 0 · y + 1 · z ┘ └ 1 · z ┘ └ z=1 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/7ed8b53100737cbf7d87aa6267395d2b.svg" width="464px" height="55px" loading="lazy" />
<p>
Sweet! <em>z</em> stays 1, so we can effectively ignore it entirely, but we added some plain values to our x and y coordinates. So, if we
want to subtract p1.x and p1.y, we use:
</p>
<!--
┌ 1 0 -P ┐ ┌ 1 · x + 0 · y - P · 1 ┐ ┌ x - P ┐
│ 1x │ ┌ x ┐ │ 1x │ │ 1x │
T = │ 0 1 -P │ · │ y │ = │ 0 · x + 1 · y - P · 1 │ = │ y - P │
1 │ 1y │ └ 1 ┘ │ 1y │ │ 1y │
└ 0 0 1 ┘ └ 0 · x + 0 · y + 1 · 1 ┘ └ 1 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/f855cbf1d73e4bb7bccbbd4721d95f41.svg" width="447px" height="57px" loading="lazy" />
<p>
Running all our coordinates through this transformation gives a new set of coordinates, let's call those <strong>U</strong>, where the
first coordinate lies on (0,0), and the rest is still somewhat free. Our next job is to make sure point 2 ends up lying on the
<em>x=0</em> line, so what we want is a transformation matrix that, when we run it, subtracts <em>x</em> from whatever <em>x</em> we
currently have. This is called <a href="https://en.wikipedia.org/wiki/Shear_matrix">shearing</a>, and the typical x-shear matrix and its
transformation looks like this:
</p>
<!--
┌ 1 S 0 ┐ ┌ x ┐ ┌ x + S · y ┐
│ 0 1 0 │ · │ y │ = │ y │
└ 0 0 1 ┘ └ 1 ┘ └ 1 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/2b6478075f2f9f5e5973e01b3b3a0c8b.svg" width="195px" height="53px" loading="lazy" />
<p>
So we want some shearing value that, when multiplied by <em>y</em>, yields <em>-x</em>, so our x coordinate becomes zero. That value is
simply <em>-x/y</em>, because <em>-x/y \</em> y = -x*. Done:
</p>
<!--
┌ U ┐
│ 2x │
│ 1 -─── 0 │
T = │ U │
2 │ 2y │
│ 0 1 0 │
└ 0 0 1 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/ccbfd22cbccf633d182f7f451dee5164.svg" width="133px" height="67px" loading="lazy" />
<p>
Now, running this on all our points generates a new set of coordinates, let's call those <strong>V</strong>, which now have point 1 on
(0,0) and point 2 on (0, some-value), and we wanted it at (0,1), so we need to
<a href="https://en.wikipedia.org/wiki/Scaling_%28geometry%29">do some scaling</a> to make sure it ends up at (0,1). Additionally, we want
point 3 to end up on (1,1), so we can also scale x to make sure its x-coordinate will be 1 after we run the transform. That means we'll be
x-scaling by 1/point3<sub>x</sub>, and y-scaling by point2<sub>y</sub>. This is really easy:
</p>
<!--
┌ 1 ┐
│ ─── 0 0 │
│ V │
│ 3x │
T = │ 1 │
3 │ 0 ─── 0 │
│ V │
│ 2y │
└ 0 0 1 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/8e39a9e0c7469b4b45a260dd23bd4c6a.svg" width="137px" height="71px" loading="lazy" />
<p>
Then, finally, this generates a new set of coordinates, let's call those W, of which point 1 lies on (0,0), point 2 lies on (0,1), and
point three lies on (1, ...) so all that's left is to make sure point 3 ends up at (1,1) - but we can't scale! Point 2 is already in the
right place, and y-scaling would move it out of (0,1) again, so our only option is to y-shear point three, just like how we x-sheared
point 2 earlier. In this case, we do the same trick, but with <code>y/x</code> rather than <code>x/y</code> because we're not x-shearing
but y-shearing. Additionally, we don't actually want to end up at zero (which is what we did before) so we need to shear towards an
offset, in this case 1:
</p>
<!--
┌ 1 0 0 ┐
│ 1 - W │
│ 3y │
T = │ ─────── 1 0 │
4 │ W │
│ 3x │
└ 0 0 1 ┘
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/9420fd9d7a8de30714e23b8f31b3aa6d.svg" width="140px" height="65px" loading="lazy" />
<p>
And this generates our final set of four coordinates. Of these, we already know that points 1 through 3 are (0,0), (0,1) and (1,1), and
only the last coordinate is "free". In fact, given any four starting coordinates, the resulting "transformation mapped" coordinate will
be:
</p>
<!--
╭ (-x +x )(-y +y ) ╮
│ 1 2 1 4 │
│ -x + x - ──────────────── │
│ 1 4 -y +y │
│ 1 2 │
│ ─────────────────────────── │
│ (-x +x )(-y +y ) │
│ 1 2 1 3 │
│ -x +x -──────────────── │
│ 1 3 -y +y │
mapped = (x) = │ 1 2 │
4 y │ ╭ -y +y ╮ ╭ (-x +x )(-y +y ) ╮ │
│ │ 1 3 │ │ 1 2 1 4 │ │
│ │ 1 - ────── │ │ -x + x - ──────────────── │ │
│ (-y +y ) │ -y +y │ │ 1 4 -y +y │ │
│ 1 4 ╰ 1 2 ╯ ╰ 1 2 ╯ │
│ ──────── + ────────────────────────────────────────────── │
│ -y +y (-x +x )(-y +y ) │
│ 1 2 1 2 1 3 │
│ -x +x -──────────────── │
│ 1 3 -y +y │
╰ 1 2 ╯
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/fff37fa4275e43302f71cf052417a19f.svg" width="455px" height="91px" loading="lazy" />
<p>
Okay, well, that looks plain ridiculous, but: notice that every coordinate value is being offset by the initial translation, and also
notice that <em>a lot</em> of terms in that expression are repeated. Even though the maths looks crazy as a single expression, we can just
pull this apart a little and end up with an easy-to-calculate bit of code!
</p>
<p>
First, let's just do that translation step as a "preprocessing" operation so we don't have to subtract the values all the time. What does
that leave?
</p>
<!--
╭ x · y x · y ╮
│ 2 4 2 3 │
│ x - ──────── / x -──────── │ ╭ x ╮
│ 4 y 3 y │ │ 43 │ ╭ x · y x · y ╮
│ 2 2 │ │ │ │ 2 4 2 3 │
... = │ │ = │ y ╭ y ╮ │, where x = │ x - ──────── \middle / x -──────── │
│ y ╭ y ╮ ╭ x · y x · y ╮ │ │ 4 │ 3 │ │ 43 │ 4 y 3 y │
│ 4 │ 3 │ │ 2 4 2 3 │ │ │ ── + x │ 1 - ── │ │ ╰ 2 2 ╯
│ ── + │ 1 - ── │ · │ x - ──────── / x -──────── │ │ │ y 43 │ y │ │
│ y │ y │ │ 4 y 3 y │ │ ╰ 2 ╰ 2 ╯ ╯
╰ 2 ╰ 2 ╯ ╰ 2 2 ╯ ╯
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/5f174bc5019245f467ca63ae84b90a4b.svg" width="775px" height="67px" loading="lazy" />
<p>
Suddenly things look a lot simpler: the mapped x is fairly straight forward to compute, and we see that the mapped y actually contains the
mapped x in its entirety, so we'll have that part already available when we need to evaluate it. In fact, let's pull out all those common
factors to see just how simple this is:
</p>
<!--
╭ x ╮ ╭ x - x · y ╮ y y
│ 43 │ │ 4 2 42 │ 4 3
... = │ │, where x = │ ────────────── │, y = ──, and y = ──
│ y + x (1 - y ) │ 43 │ x - x · y │ 42 y 32 y
╰ 42 43 32 ╯ ╰ 3 2 32 ╯ 2 2
-->
<img class="LaTeX SVG" src="./images/chapters/canonical/88e3fae7aeef6d7614290587422542c9.svg" width="563px" height="55px" loading="lazy" />
<p>
That's kind of super-simple to write out in code, I think you'll agree. Coding math tends to be easier than the formulae initially make it
look!
</p>
<div class="note">
<h3>How do you track all that?</h3>
<p>
Doing maths can be a pain, so whenever possible, I like to make computers do the work for me. Especially for things like this, I simply
use <a href="https://www.wolfram.com/mathematica/">Mathematica</a>. Tracking all this math by hand is insane, and we invented computers,
literally, to do this for us. I have no reason to use pen and paper when I can write out what I want to do in a program, and have the
program do the math for me. And real math, too, with symbols, not with numbers. In fact,
<a href="https://pomax.github.io/gh-weblog-2/downloads/canonical-curve.nb">here's</a> the Mathematica notebook if you want to see how
this works for yourself.
</p>
<p>
Now, I know, you're thinking "but Mathematica is super expensive!" and that's true, it's
<a href="https://www.wolfram.com/mathematica-home-edition/">$344 for home use, up from $295 when I original wrote this</a>, but it's
<strong>also</strong> <a href="https://www.wolfram.com/raspberry-pi/">free when you buy a $35 raspberry pi</a>. Obviously, I bought a
raspberry pi, and I encourage you to do the same. With that, as long as you know what you want to <em>do</em>, Mathematica can just do
it for you. And we don't have to be geniuses to work out what the maths looks like. That's what we have computers for.
</p>
</div>
<p>
So, let's write up a sketch that'll show us the canonical form for any curve drawn in blue, overlaid on our canonical map, so that we can
immediately tell which features our curve must have, based on where the fourth coordinate is located on the map:
</p>
<graphics-element
title="A cubic curve mapped to canonical form"
width="800"
height="400"
src="./chapters/canonical/interactive.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="800px" height="400px" src="./images/chapters/canonical/83fe2473e20ea68b768765129ee44ae4.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
</section>
<section id="yforx">
<h1>
<div class="nav"><a href="ko-KR/index.html#canonical">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#arclength">다음</a></div>
<a href="ko-KR/index.html#yforx">Finding Y, given X</a>
</h1>
<p>
One common task that pops up in things like CSS work, or parametric equalizers, or image leveling, or any other number of applications
where Bézier curves are used as control curves in a way that there is really only ever one "y" value associated with one "x" value, you
might want to cut out the middle man, as it were, and compute "y" directly based on "x". After all, the function looks simple enough,
finding the "y" value should be simple too, right? Unfortunately, not really. However, it <em>is</em> possible and as long as you have
some code in place to help, it's not a lot of a work either.
</p>
<p>
We'll be tackling this problem in two stages: the first, which is the hard part, is figuring out which "t" value belongs to any given "x"
value. For instance, have a look at the following graphic. On the left we have a Bézier curve that looks for all intents and purposes like
it fits our criteria: every "x" has one and only one associated "y" value. On the right we see the function for just the "x" values:
that's a cubic curve, but not a really crazy cubic curve. If you move the graphic's slider, you will see a red line drawn that corresponds
to the <code>x</code> coordinate: this is a vertical line in the left graphic, and a horizontal line on the right.
</p>
<graphics-element
title="Finding t, given x=x(t). Left: our curve, right: the function x=f(t)"
width="550"
height="275"
src="./chapters/yforx/basics.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="550px" height="275px" src="./images/chapters/yforx/e469af5bf27a2c27d1dd6fc62a78ac27.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" class="slide-control" />
</graphics-element>
<p>
Now, if you look more closely at that right graphic, you'll notice something interesting: if we treat the red line as "the x axis", then
the point where the function crosses our line is really just a root for the cubic function x(t) through a shifted "x-axis"... and
<a href="#extremities">we've already seen</a> how to calculate roots, so let's just run cubic root finding - and not even the complicated
cubic case either: because of the kind of curve we're starting with, we <em>know</em> there is at most a single root in the interval
[0,1], simplifying the code we need!
</p>
<p>First, let's look at the function for x(t):</p>
<!--
3 2 2 3
x(t) = a(1-t) + 3b(1-t) t + 3c(1-t)t + dt
-->
<img class="LaTeX SVG" src="./images/chapters/yforx/378d0fd8cefa688d530ac38930d66844.svg" width="335px" height="19px" loading="lazy" />
<p>
We can rewrite this to a plain polynomial form, by just fully writing out the expansion and then collecting the polynomial factors, as:
</p>
<!--
3 2
x(t) = (-a + 3b- 3c + d)t + (3a - 6b + 3c)t + (-3a + 3b)t + a
-->
<img class="LaTeX SVG" src="./images/chapters/yforx/021718d3b46893b271f90083ccdceaf8.svg" width="445px" height="19px" loading="lazy" />
<p>
Nothing special here: that's a standard cubic polynomial in "power" form (i.e. all the terms are ordered by their power of
<code>t</code>). So, given that <code>a</code>, <code>b</code>, <code>c</code>, <code>d</code>, <em>and</em> <code>x(t)</code> are all
known constants, we can trivially rewrite this (by moving the <code>x(t)</code> across the equal sign) as:
</p>
<!--
3 2
(-a + 3b - 3c + d)t + (3a - 6b + 3c)t + (-3a + 3b)t + (a-x) = 0
-->
<img class="LaTeX SVG" src="./images/chapters/yforx/058a76e3e7d67c03f733e075829a6252.svg" width="465px" height="19px" loading="lazy" />
<p>
You might be wondering "where did all the other 'minus x' for all the other values a, b, c, and d go?" and the answer there is that they
all cancel out, so the only one we actually need to subtract is the one at the end. Handy! So now we just solve this equation using
Cardano's algorithm, and we're left with some rather short code:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="10">
<textarea disabled rows="10" role="doc-example">
// prepare our values for root finding:
x = a value we already know
xcoord = our set of Bézier curve's x coordinates
foreach p in xcoord: p.x -= x
// find our root, of which we know there is exactly one:
t = getRoots(p[0], p[1], p[2], p[3])[0]
// find our answer:
if t in [0,1] y = curve.get(t).y</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
</table>
<p>
So the procedure is fairly straight forward: pick an <code>x</code>, find the associated <code>t</code> value, evaluate our curve
<em>for</em> that <code>t</code> value, which gives us the curve's {x,y} coordinate, which means we know <code>y</code> for this
<code>x</code>. Move the slider for the following graphic to see this in action:
</p>
<graphics-element
title="Finding By(t), by finding t for a given x"
width="275"
height="275"
src="./chapters/yforx/yforx.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/yforx/2fc5c57e5d1ed0eaa1655edc31026252.png" loading="lazy" />
<label>Finding By(t), by finding t for a given x</label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" class="slide-control" />
</graphics-element>
</section>
<section id="arclength">
<h1>
<div class="nav"><a href="ko-KR/index.html#yforx">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#arclengthapprox">다음</a></div>
<a href="ko-KR/index.html#arclength">Arc length</a>
</h1>
<p>
How long is a Bézier curve? As it turns out, that's not actually an easy question, because the answer requires maths that —much like root
finding— cannot generally be solved the traditional way. If we have a parametric curve with <em>f<sub>x</sub>(t)</em> and
<em>f<sub>y</sub>(t)</em>, then the length of the curve, measured from start point to some point <em>t = z</em>, is computed using the
following seemingly straight forward (if a bit overwhelming) formula:
</p>
<!--
┌───────────────┐
╭ z │ 2 2
| │f '(t) +f '(t) dt
╯ 0⟍│ x y
-->
<img class="LaTeX SVG" src="./images/chapters/arclength/59ebc3a7c3547a50998d1ea3664fb688.svg" width="140px" height="33px" loading="lazy" />
<p>or, more commonly written using Leibnitz notation as:</p>
<!--
┌─────────────────┐
╭ z │ 2 2
length = | ⟍│(dx/dt) +(dy/dt) dt
╯ 0
-->
<img class="LaTeX SVG" src="./images/chapters/arclength/85620f0332fcf16f56c580794fd094c5.svg" width="245px" height="35px" loading="lazy" />
<p>
This formula says that the length of a parametric curve is in fact equal to the <strong>area</strong> underneath a function that looks a
remarkable amount like Pythagoras' rule for computing the diagonal of a straight angled triangle. This sounds pretty simple, right? Sadly,
it's far from simple... cutting straight to after the chase is over: for quadratic curves, this formula generates an
<a
href="https://www.wolframalpha.com/input/?i=antiderivative+for+sqrt((2*(1-t)*t*B+%2B+t%5E2*C)%27%5E2+%2B+(2*(1-t)*t*E)%27%5E2)&incParTime=true"
>unwieldy computation</a
>, and we're simply not going to implement things that way. For cubic Bézier curves, things get even more fun, because there is no "closed
form" solution, meaning that due to the way calculus works, there is no generic formula that allows you to calculate the arc length. Let
me just repeat this, because it's fairly crucial:
<strong
><em
>for cubic and higher Bézier curves, there is no way to solve this function if you want to use it "for all possible coordinates"</em
></strong
>.
</p>
<p>Seriously: <a href="https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem">It cannot be done</a>.</p>
<p>
So we turn to numerical approaches again. The method we'll look at here is the
<a href="https://www.youtube.com/watch?v=unWguclP-Ds&feature=BFa&list=PLC8FC40C714F5E60F&index=1">Gauss quadrature</a>. This approximation
is a really neat trick, because for any <em>n<sup>th</sup></em> degree polynomial it finds approximated values for an integral really
efficiently. Explaining this procedure in length is way beyond the scope of this page, so if you're interested in finding out why it
works, I can recommend the University of South Florida video lecture on the procedure, linked in this very paragraph. The general solution
we're looking for is the following:
</p>
<!--
┌─────────────────┐
╭ 1 │ 2 2 ╭ 1
| ⟍│(dx/dt) +(dy/dt) dt = | f(t) dt ≃ ┌ \underset strip 1 \underbrace C · f(t ) + ... + \underset strip n \underbrace C · f(t ) ┐ =
╯ -1 ╯ -1 └ 1 1 n n ┘
__ n
\underset strips 1 through n \underbrace ❯ C · f(t )
‾‾ i=1 i i
-->
<img class="LaTeX SVG" src="./images/chapters/arclength/b76753476ad6ecfe4b8f39bcf9432980.svg" width="623px" height="71px" loading="lazy" />
<p>
In plain text: an integral function can always be treated as the sum of an (infinite) number of (infinitely thin) rectangular strips
sitting "under" the function's plotted graph. To illustrate this idea, the following graph shows the integral for a sinusoid function. The
more strips we use (and of course the more we use, the thinner they get) the closer we get to the true area under the curve, and thus the
better the approximation:
</p>
<div class="figure">
<graphics-element
title="A function's approximated integral"
width="275"
height="275"
src="./chapters/arclength/draw-slices.js"
data-steps="10"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arclength/56533f47e73ad9fea08fa9bb3f597d49.png" loading="lazy" />
<label>A function's approximated integral</label>
</fallback-image></graphics-element
>
<graphics-element
title="A better approximation"
width="275"
height="275"
src="./chapters/arclength/draw-slices.js"
data-steps="24"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arclength/5ce02cbdbc47585c588f2656d5161a32.png" loading="lazy" />
<label>A better approximation</label>
</fallback-image></graphics-element
>
<graphics-element
title="An even better approximation"
width="275"
height="275"
src="./chapters/arclength/draw-slices.js"
data-steps="99"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arclength/fe2663b205d14c157a5a02bfbbd55987.png" loading="lazy" />
<label>An even better approximation</label>
</fallback-image></graphics-element
>
</div>
<p>
Now, infinitely many terms to sum and infinitely thin rectangles are not something that computers can work with, so instead we're going to
approximate the infinite summation by using a sum of a finite number of "just thin" rectangular strips. As long as we use a high enough
number of thin enough rectangular strips, this will give us an approximation that is pretty close to what the real value is.
</p>
<p>
So, the trick is to come up with useful rectangular strips. A naive way is to simply create <em>n</em> strips, all with the same width,
but there is a far better way using special values for <em>C</em> and <em>f(t)</em> depending on the value of <em>n</em>, which indicates
how many strips we'll use, and it's called the Legendre-Gauss quadrature.
</p>
<p>
This approach uses strips that are <em>not</em> spaced evenly, but instead spaces them in a special way based on describing the function
as a polynomial (the more strips, the more accurate the polynomial), and then computing the exact integral for that polynomial. We're
essentially performing arc length computation on a flattened curve, but flattening it based on the intervals dictated by the
Legendre-Gauss solution.
</p>
<div class="note">
<p>
Note that one requirement for the approach we'll use is that the integral must run from -1 to 1. That's no good, because we're dealing
with Bézier curves, and the length of a section of curve applies to values which run from 0 to "some value smaller than or equal to 1"
(let's call that value <em>z</em>). Thankfully, we can quite easily transform any integral interval to any other integral interval, by
shifting and scaling the inputs. Doing so, we get the following:
</p>
<!--
┌─────────────────┐
╭ z │ 2 2
| ⟍│(dx/dt) +(dy/dt) dt
╯ 0
z ┌ ╭ z z ╮ ╭ z z ╮ ┐
≃ \ ─ · │ C · f│ ─ · t + ─ │ + ... + C · f│ ─ · t + ─ │ │
2 └ 1 ╰ 2 1 2 ╯ n ╰ 2 n 2 ╯ ┘
z __ n ╭ z z ╮
= \ ─ · ❯ C · f│ ─ · t + ─ │
2 ‾‾ i=1 i ╰ 2 i 2 ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/arclength/0748ad25185548150b6c1c4c7039207e.svg"
width="341px"
height="72px"
loading="lazy"
/>
<p>
That may look a bit more complicated, but the fraction involving <em>z</em> is a fixed number, so the summation, and the evaluation of
the <em>f(t)</em> values are still pretty simple.
</p>
<p>
So, what do we need to perform this calculation? For one, we'll need an explicit formula for <em>f(t)</em>, because that derivative
notation is handy on paper, but not when we have to implement it. We'll also need to know what these <em>C<sub>i</sub></em> and
<em>t<sub>i</sub></em> values should be. Luckily, that's less work because there are actually many tables available that give these
values, for any <em>n</em>, so if we want to approximate our integral with only two terms (which is a bit low, really) then
<a href="./legendre-gauss.html">these tables</a> would tell us that for <em>n=2</em> we must use the following values:
</p>
<!--
C = 1
1
C = 1
2
1
t = - ────
1 ┌─┐
⟍│3
1
t = + ────
2 ┌─┐
⟍│3
-->
<img class="LaTeX SVG" src="./images/chapters/arclength/a91fbfb7abc38ff712ef660d85679f2e.svg" width="63px" height="93px" loading="lazy" />
<p>
Which means that in order for us to approximate the integral, we must plug these values into the approximate function, which gives us:
</p>
<!--
┌─────────────────┐
╭ z │ 2 2 z ┌ ╭ z -1 z ╮ ╭ z 1 z ╮ ┐
| ⟍│(dx/dt) +(dy/dt) dt ≃ ─ · │ f│ ─ · ──── + ─ │ + f│ ─ · ──── + ─ │ │
╯ 0 2 │ │ 2 ┌─┐ 2 │ │ 2 ┌─┐ 2 │ │
└ ╰ ⟍│3 ╯ ╰ ⟍│3 ╯ ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/arclength/046bbb52e8c8ed617fdf3a4fd18d62e1.svg"
width="476px"
height="44px"
loading="lazy"
/>
<p>
We can program that pretty easily, provided we have that <em>f(t)</em> available, which we do, as we know the full description for the
Bézier curve functions B<sub>x</sub>(t) and B<sub>y</sub>(t).
</p>
</div>
<p>
If we use the Legendre-Gauss values for our <em>C</em> values (thickness for each strip) and <em>t</em> values (location of each strip),
we can determine the approximate length of a Bézier curve by computing the Legendre-Gauss sum. The following graphic shows a cubic curve,
with its computed lengths; Go ahead and change the curve, to see how its length changes. One thing worth trying is to see if you can make
a straight line, and see if the length matches what you'd expect. What if you form a line with the control points on the outside, and the
start/end points on the inside?
</p>
<graphics-element
title="Arc length for a Bézier curve"
width="275"
height="275"
src="./chapters/arclength/arclength.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arclength/fa4c587126e8097206b88d9ea51974ca.png" loading="lazy" />
<label>Arc length for a Bézier curve</label>
</fallback-image></graphics-element
>
</section>
<section id="arclengthapprox">
<h1>
<div class="nav"><a href="ko-KR/index.html#arclength">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#curvature">다음</a></div>
<a href="ko-KR/index.html#arclengthapprox">Approximated arc length</a>
</h1>
<p>
Sometimes, we don't actually need the precision of a true arc length, and we can get away with simply computing the approximate arc length
instead. The by far fastest way to do this is to flatten the curve and then simply calculate the linear distance from point to point. This
will come with an error, but this can be made arbitrarily small by increasing the segment count.
</p>
<p>
If we combine the work done in the previous sections on curve flattening and arc length computation, we can implement these with minimal
effort:
</p>
<div class="figure">
<graphics-element
title="Approximate quadratic curve arc length"
width="275"
height="275"
src="./chapters/arclengthapprox/approximate.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arclengthapprox/3fc083ea7bdcc6b021560f2f2491f8aa.png" loading="lazy" />
<label>Approximate quadratic curve arc length</label>
</fallback-image>
<input type="range" min="2" max="24" step="1" value="4" class="slide-control" />
</graphics-element>
<graphics-element
title="Approximate cubic curve arc length"
width="275"
height="275"
src="./chapters/arclengthapprox/approximate.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arclengthapprox/537260c4aa9e98ffdea7c8120afbd427.png" loading="lazy" />
<label>Approximate cubic curve arc length</label>
</fallback-image>
<input type="range" min="2" max="32" step="1" value="8" class="slide-control" />
</graphics-element>
</div>
<p>
You may notice that even though the error in length is actually pretty significant in absolute terms, even at a low number of segments we
get a length that agrees with the true length when it comes to just the integer part of the arc length. Quite often, approximations can
drastically speed things up!
</p>
</section>
<section id="curvature">
<h1>
<div class="nav">
<a href="ko-KR/index.html#arclengthapprox">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#tracing">다음</a>
</div>
<a href="ko-KR/index.html#curvature">Curvature of a curve</a>
</h1>
<p>
If we have two curves, and we want to line them in up in a way that "looks right", what would we use as metric to let a computer decide
what "looks right" means?
</p>
<p>
For instance, we can start by ensuring that the two curves share an end coordinate, so that there is no "gap" between the end of one and
the start of the next curve, but that won't guarantee that things look right: both curves can be going in wildly different directions, and
the resulting joined geometry will have a corner in it, rather than a smooth transition from one curve to the next.
</p>
<p>
What we want is to ensure that the <a href="https://en.wikipedia.org/wiki/Curvature">curvature</a> at the transition from one curve to the
next "looks good". So, we start with a shared coordinate, and then also require that derivatives for both curves match at that coordinate.
That way, we're assured that their tangents line up, which must mean the curve transition is perfectly smooth. We can even make the
second, third, etc. derivatives match up for better and better transitions.
</p>
<p>Problem solved!</p>
<p>
However, there's a problem with this approach: if we think about this a little more, we realise that "what a curve looks like" and its
derivative values are pretty much entirely unrelated. After all, the section on <a href="#reordering">reordering curves</a> showed us that
the same looking curve can have an infinite number of curve expressions of arbitrarily high Bézier degree, and each of those will have
<em>wildly</em> different derivative values.
</p>
<p>
So what we really want is some kind of expression that's not based on any particular expression of <code>t</code>, but is based on
something that is invariant to the <em>kind</em> of function(s) we use to draw our curve. And the prime candidate for this is our curve
expression, reparameterised for distance: no matter what order of Bézier curve we use, if we were able to rewrite it as a function of
distance-along-the-curve, all those different degree Bézier functions would end up being <em>the same</em> function for "coordinate at
some distance D along the curve".
</p>
<p>We've seen this before... that's the arc length function.</p>
<p>
So you might think that in order to find the curvature of a curve, we now need to solve the arc length function itself, and that this
would be quite a problem because we just saw that there is no way to actually do that. Thankfully, we don't. We only need to know the
<em>form</em> of the arc length function, which we saw above and is fairly simple, rather than needing to <em>solve</em> the arc length
function. If we start with the arc length expression and the
<a href="https://mathworld.wolfram.com/Curvature.html">run through the steps necessary</a> to determine <em>its</em> derivative (with an
alternative, shorter demonstration of how to do this found
<a href="https://math.stackexchange.com/questions/275248/deriving-curvature-formula/275324#275324">over on Stackexchange</a>), then the
integral that was giving us so much problems in solving the arc length function disappears entirely (because of the
<a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus">fundamental theorem of calculus</a>), and what we're left with us
some surprisingly simple maths that relates curvature (denoted as κ, "kappa") to—and this is the truly surprising bit—a specific
combination of derivatives of our original function.
</p>
<p>Let me highlight what just happened, because it's pretty special:</p>
<ol>
<li>we wanted to make curves line up, and initially thought to match the curves' derivatives, but</li>
<li>that turned out to be a really bad choice, so instead</li>
<li>we picked a function that is basically impossible to work with, and then <em>worked with that</em>, which</li>
<li>gives us a simple formula that is <em>and expression using the curves' derivatives</em>.</li>
</ol>
<p><em>That's crazy!</em></p>
<p>
But that's also one of the things that makes maths so powerful: even if your initial ideas are off the mark, you might be much closer than
you thought you were, and the journey from "thinking we're completely wrong" to "actually being remarkably close to being right" is where
we can find a lot of insight.
</p>
<p>So, what does the function look like? This:</p>
<!--
x'y'' - x''y'
\kappa = ─────────────
3
─
2 2 2
(x' +y' )
-->
<img class="LaTeX SVG" src="./images/chapters/curvature/060acd6ff0a050fe4d98a7802a2b3a3f.svg" width="113px" height="47px" loading="lazy" />
<p>
Which is really just a "short form" that glosses over the fact that we're dealing with functions of <code>t</code>, so let's expand that a
tiny bit:
</p>
<!--
B '(t)B ''(t) - B ''(t)B '(t)
x y x y
\kappa(t) = ─────────────────────────────
3
─
2 2 2
(B '(t) +B '(t) )
x y
-->
<img class="LaTeX SVG" src="./images/chapters/curvature/afd8cb8b0fe291ff703752c1c9cc33d4.svg" width="239px" height="55px" loading="lazy" />
<p>
And while that's a little more verbose, it's still just as simple to work with as the first function: the curvature at some point on any
(and this cannot be overstated: <em>any</em>) curve is a ratio between the first and second derivative cross product, and something that
looks oddly similar to the standard Euclidean distance function. And nothing in these functions is hard to calculate either: for Bézier
curves, simply knowing our curve coordinates means <a href="#derivatives">we know what the first and second derivatives are</a>, and so
evaluating this function for any <strong>t</strong> value is just a matter of basic arithematics.
</p>
<p>In fact, let's just implement it right now:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="7">
<textarea disabled rows="7" role="doc-example">
function kappa(t, B):
d = B.getDerivative(t)
dd = B.getSecondDerivative(t)
numerator = d.x * dd.y - dd.x * d.y
denominator = pow(d.x*d.x + d.y*d.y, 3/2)
if denominator is 0: return NaN;
return numerator / denominator</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
</table>
<p>
That was easy! (Well okay, that "not a number" value will need to be taken into account by downstream code, but that's a reality of
programming anyway)
</p>
<p>
With all of that covered, let's line up some curves! The following graphic gives you two curves that look identical, but use quadratic and
cubic functions, respectively. As you can see, despite their derivatives being necessarily different, their curvature (thanks to being
derived based on maths that "ignores" specific function derivative, and instead gives a formula that smooths out any differences) is
exactly the same. And because of that, we can put them together such that the point where they overlap has the same curvature for both
curves, giving us the smoothest transition.
</p>
<graphics-element
title="Matching curvatures for a quadratic and cubic Bézier curve"
width="825"
height="275"
src="./chapters/curvature/curvature.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/curvature/4f2647446363ca5d93b11e414fd976df.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>
One thing you may have noticed in this sketch is that sometimes the curvature looks fine, but seems to be pointing in the wrong direction,
making it hard to line up the curves properly. A way around that, of course, is to show the curvature on both sides of the curve, so let's
just do that. But let's take it one step further: we can also compute the associated "radius of curvature", which gives us the implicit
circle that "fits" the curve's curvature at any point, using what is possibly the simplest bit of maths found in this entire primer:
</p>
<!--
1
R(t) = ─────────
\kappa(t)
-->
<img class="LaTeX SVG" src="./images/chapters/curvature/561ab3a938d655550de0abf458ac2494.svg" width="81px" height="37px" loading="lazy" />
<p>
So let's revisit the previous graphic with the curvature visualised on both sides of our curves, as well as showing the circle that "fits"
our curve at some point that we can control by using a slider:
</p>
<graphics-element
title="(Easier) curvature matching for a quadratic and cubic Bézier curve"
width="825"
height="275"
src="./chapters/curvature/curvature.js"
data-omni="true"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/curvature/392624cedf7c78aed6d4c6065a014b42.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="0" max="2" step="0.0005" value="0" class="slide-control" />
</graphics-element>
</section>
<section id="tracing">
<h1>
<div class="nav">
<a href="ko-KR/index.html#curvature">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#intersections">다음</a>
</div>
<a href="ko-KR/index.html#tracing">Tracing a curve at fixed distance intervals</a>
</h1>
<p>
Say you want to draw a curve with a dashed line, rather than a solid line, or you want to move something along the curve at fixed distance
intervals over time, like a train along a track, and you want to use Bézier curves.
</p>
<p>Now you have a problem.</p>
<p>
The reason you have a problem is that Bézier curves are parametric functions with non-linear behaviour, whereas moving a train along a
track is about as close to a practical example of linear behaviour as you can get. The problem we're faced with is that we can't just pick
<code>t</code> values at some fixed interval and expect the Bézier functions to generate points that are spaced a fixed distance apart. In
fact, let's look at the relation between "distance along a curve" and "<code>t</code> value", by plotting them against one another.
</p>
<p>
The following graphic shows a particularly illustrative curve, and its distance-for-t plot. For linear traversal, this line needs to be
straight, running from (0,0) to (length,1). That is, it's safe to say, not what we'll see: we'll see something very wobbly, instead. To
make matters even worse, the distance-for-t function is also of a much higher order than our curve is: while the curve we're using for
this exercise is a cubic curve, which can switch concave/convex form twice at best, the distance function is our old friend the arc length
function, which can have more inflection points.
</p>
<graphics-element
title="The t-for-distance function"
width="550"
height="275"
src="./chapters/tracing/distance-function.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="550px" height="275px" src="./images/chapters/tracing/d6239520389637a3c42e76ee44d86c41.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>
So, how do we "cut up" the arc length function at regular intervals, when we can't really work with it? We basically cheat: we run through
the curve using <code>t</code> values, determine the distance-for-this-<code>t</code>-value at each point we generate during the run, and
then we find "the closest <code>t</code> value that matches some required distance" using those values instead. If we have a low number of
points sampled, we can then even refine which <code>t</code> value "should" work for our desired distance by interpolating between two
points, but if we have a high enough number of samples, we don't even need to bother.
</p>
<p>
So let's do exactly that: the following graph is similar to the previous one, showing how we would have to "chop up" our distance-for-t
curve in order to get regularly spaced points on the curve. It also shows what using those <code>t</code> values on the real curve looks
like, by coloring each section of curve between two distance markers differently:
</p>
<graphics-element
title="Fixed-interval coloring a curve"
width="825"
height="275"
src="./chapters/tracing/tracing.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/tracing/1cd7304fb8d044835bfbc305ca5e5d10.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="2" max="24" step="1" value="8" class="slide-control" />
</graphics-element>
<p>Use the slider to increase or decrease the number of equidistant segments used to colour the curve.</p>
<p>
However, are there better ways? One such way is discussed in "<a
href="https://www.geometrictools.com/Documentation/MovingAlongCurveSpecifiedSpeed.pdf"
>Moving Along a Curve with Specified Speed</a
>" by David Eberly of Geometric Tools, LLC, but basically because we have no explicit length function (or rather, one we don't have to
constantly compute for different intervals), you may simply be better off with a traditional lookup table (LUT).
</p>
</section>
<section id="intersections">
<h1>
<div class="nav">
<a href="ko-KR/index.html#tracing">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#curveintersection">다음</a>
</div>
<a href="ko-KR/index.html#intersections">Intersections</a>
</h1>
<p>
Let's look at some more things we will want to do with Bézier curves. Almost immediately after figuring out how to get bounding boxes to
work, people tend to run into the problem that even though the minimal bounding box (based on rotation) is tight, it's not sufficient to
perform true collision detection. It's a good first step to make sure there <em>might</em> be a collision (if there is no bounding box
overlap, there can't be one), but in order to do real collision detection we need to know whether or not there's an intersection on the
actual curve.
</p>
<p>
We'll do this in steps, because it's a bit of a journey to get to curve/curve intersection checking. First, let's start simple, by
implementing a line-line intersection checker. While we can solve this the traditional calculus way (determine the functions for both
lines, then compute the intersection by equating them and solving for two unknowns), linear algebra actually offers a nicer solution.
</p>
<h3>Line-line intersections</h3>
<p>
If we have two line segments with two coordinates each, segments A-B and C-D, we can find the intersection of the lines these segments are
an intervals on by linear algebra, using the procedure outlined in this
<a href="https://www.topcoder.com/community/competitive-programming/tutorials/geometry-concepts-line-intersection-and-its-applications/"
>top coder</a
>
article. Of course, we need to make sure that the intersection isn't just on the lines our line segments lie on, but actually on our line
segments themselves. So after we find the intersection, we need to verify that it lies without the bounds of our original line segments.
</p>
<p>
The following graphic implements this intersection detection, showing a red point for an intersection on the lines our segments lie on
(thus being a virtual intersection point), and a green point for an intersection that lies on both segments (being a real intersection
point).
</p>
<graphics-element
title="Line/line intersections"
width="275"
height="275"
src="./chapters/intersections/line-line.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/intersections/b3f61036d8dc9888a6a64a1171583dd1.png" loading="lazy" />
<label>Line/line intersections</label>
</fallback-image></graphics-element
>
<div class="howtocode">
<h3>Implementing line-line intersections</h3>
<p>
Let's have a look at how to implement a line-line intersection checking function. The basics are covered in the article mentioned above,
but sometimes you need more function signatures, because you might not want to call your function with eight distinct parameters. Maybe
you're using point structs for the line. Let's get coding:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="17">
<textarea disabled rows="17" role="doc-example">
lli8 = function(x1,y1,x2,y2,x3,y3,x4,y4):
var nx=(x1*y2-y1*x2)*(x3-x4)-(x1-x2)*(x3*y4-y3*x4),
ny=(x1*y2-y1*x2)*(y3-y4)-(y1-y2)*(x3*y4-y3*x4),
d=(x1-x2)*(y3-y4)-(y1-y2)*(x3-x4);
if d=0:
return false
return point(nx/d, ny/d)
lli4 = function(p1, p2, p3, p4):
var x1 = p1.x, y1 = p1.y,
x2 = p2.x, y2 = p2.y,
x3 = p3.x, y3 = p3.y,
x4 = p4.x, y4 = p4.y;
return lli8(x1,y1,x2,y2,x3,y3,x4,y4)
lli = function(line1, line2):
return lli4(line1.p1, line1.p2, line2.p1, line2.p2)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
</table>
</div>
<h3>What about curve-line intersections?</h3>
<p>
Curve/line intersection is more work, but we've already seen the techniques we need to use in order to perform it: first we
translate/rotate both the line and curve together, in such a way that the line coincides with the x-axis. This will position the curve in
a way that makes it cross the line at points where its y-function is zero. By doing this, the problem of finding intersections between a
curve and a line has now become the problem of performing root finding on our translated/rotated curve, as we already covered in the
section on finding extremities.
</p>
<div class="figure">
<graphics-element
title="Quadratic curve/line intersections"
width="275"
height="275"
src="./chapters/intersections/curve-line.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/intersections/9b70fb7b03f082882515e55c0a1eacff.png" loading="lazy" />
<label>Quadratic curve/line intersections</label>
</fallback-image></graphics-element
>
<graphics-element
title="Cubic curve/line intersections"
width="275"
height="275"
src="./chapters/intersections/curve-line.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/intersections/7196d3dec75d53f5df9d9c832ac3c493.png" loading="lazy" />
<label>Cubic curve/line intersections</label>
</fallback-image></graphics-element
>
</div>
<p>
Curve/curve intersection, however, is more complicated. Since we have no straight line to align to, we can't simply align one of the
curves and be left with a simple procedure. Instead, we'll need to apply two techniques we've met before: de Casteljau's algorithm, and
curve splitting.
</p>
</section>
<section id="curveintersection">
<h1>
<div class="nav"><a href="ko-KR/index.html#intersections">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#abc">다음</a></div>
<a href="ko-KR/index.html#curveintersection">Curve/curve intersection</a>
</h1>
<p>
Using de Casteljau's algorithm to split the curve we can now implement curve/curve intersection finding using a "divide and conquer"
technique:
</p>
<ol>
<li>
Take two curves <em>C<sub>1</sub></em> and <em>C<sub>2</sub></em
>, and treat them as a pair.
</li>
<li>If their bounding boxes overlap, split up each curve into two sub-curves</li>
<li>
With <em>C<sub>1.1</sub></em
>, <em>C<sub>1.2</sub></em
>, <em>C<sub>2.1</sub></em> and <em>C<sub>2.2</sub></em
>, form four new pairs (<em>C<sub>1.1</sub></em
>,<em>C<sub>2.1</sub></em
>), (<em>C<sub>1.1</sub></em
>, <em>C<sub>2.2</sub></em
>), (<em>C<sub>1.2</sub></em
>,<em>C<sub>2.1</sub></em
>), and (<em>C<sub>1.2</sub></em
>,<em>C<sub>2.2</sub></em
>).
</li>
<li>
For each pair, check whether their bounding boxes overlap.
<ol>
<li>If their bounding boxes do not overlap, discard the pair, as there is no intersection between this pair of curves.</li>
<li>If there <em>is</em> overlap, rerun all steps for this pair.</li>
</ol>
</li>
<li>
Once the sub-curves we form are so small that they effectively occupy sub-pixel areas, we consider an intersection found, noting that we
might have a cluster of multiple intersections at the sub-pixel level, out of which we pick one to act as "found" <code>t</code> value
(we can either throw all but one away, we can average the cluster's <code>t</code> values, or you can do something even more creative).
</li>
</ol>
<p>
This algorithm will start with a single pair, "balloon" until it runs in parallel for a large number of potential sub-pairs, and then
taper back down as it homes in on intersection coordinates, ending up with as many pairs as there are intersections.
</p>
<p>
The following graphic applies this algorithm to a pair of cubic curves, one step at a time, so you can see the algorithm in action. Click
the button to run a single step in the algorithm, after setting up your curves in some creative arrangement. You can also change the value
that is used in step 5 to determine whether the curves are small enough. Manipulating the curves or changing the threshold will reset the
algorithm, so you can try this with lots of different curves.
</p>
<p>(can you find the configuration that yields the maximum number of intersections between two cubic curves? Nine intersections!)</p>
<graphics-element
title="Curve/curve intersections"
width="825"
height="275"
src="./chapters/curveintersection/curve-curve.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/curveintersection/b155682162a5b6da6d40c7b531164a7e.png" loading="lazy" />
<label></label>
</fallback-image>
<button class="next">Advance one step</button>
<input type="range" min="0.01" max="5" step="0.01" value="1" class="slide-control" />
</graphics-element>
<p>
Finding self-intersections is effectively the same procedure, except that we're starting with a single curve, so we need to turn that into
two separate curves first. This is trivially achieved by splitting at an inflection point, or if there are none, just splitting at
<code>t=0.5</code> first, and then running the exact same algorithm as above, with all non-overlapping curve pairs getting removed at each
iteration, and each successive step homing in on the curve's self-intersection points.
</p>
</section>
<section id="abc">
<h1>
<div class="nav">
<a href="ko-KR/index.html#curveintersection">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#pointcurves">다음</a>
</div>
<a href="ko-KR/index.html#abc">The projection identity</a>
</h1>
<p>
De Casteljau's algorithm is the pivotal algorithm when it comes to Bézier curves. You can use it not just to split curves, but also to
draw them efficiently (especially for high-order Bézier curves), as well as to come up with curves based on three points and a tangent.
Particularly this last thing is really useful because it lets us "mold" a curve, by picking it up at some point, and dragging that point
around to change the curve's shape.
</p>
<p>How does that work? Succinctly: we run de Casteljau's algorithm in reverse!</p>
<p>
In order to run de Casteljau's algorithm in reverse, we need a few basic things: a start and end point, a point on the curve that we want
to be moving around, which has an associated <em>t</em> value, and a point we've not explicitly talked about before, and as far as I know
has no explicit name, but lives one iteration higher in the de Casteljau process then our on-curve point does. I like to call it "A" for
reasons that will become obvious.
</p>
<p>
So let's use graphics instead of text to see where this "A" is, because text only gets us so far: move the sliders for the following
graphics to see what, given a specific <code>t</code> value, our <code>A</code> coordinate is. As well as some other coordinates, which
taken together let us derive a value that the graphics call "ratio": if you move the curve's points around, A, B, and C will move, what
happens to that value?
</p>
<div class="figure">
<graphics-element
inline="{true}"
title="Projections in a quadratic Bézier curve"
width="275"
height="275"
src="./chapters/abc/abc.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/abc/7a69dd4350ddda5701712e1d3b46b863.png" loading="lazy" />
<label>Projections in a quadratic Bézier curve</label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0.5" class="slide-control" />
</graphics-element>
<graphics-element
inline="{true}"
title="Projections in a cubic Bézier curve"
width="275"
height="275"
src="./chapters/abc/abc.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/abc/eeec7cf16fb22c666e0143a3a030731f.png" loading="lazy" />
<label>Projections in a cubic Bézier curve</label>
</fallback-image>
<input type="range" min="0" max="1" step="0.01" value="0.5" class="slide-control" />
</graphics-element>
</div>
<p>So these graphics show us several things:</p>
<ol>
<li>a point at the tip of the curve construction's "hat": let's call that <code>A</code>, as well as</li>
<li>our on-curve point give our chosen <code>t</code> value: let's call that <code>B</code>, and finally,</li>
<li>
a point that we get by projecting A, through B, onto the line between the curve's start and end points: let's call that <code>C</code>.
</li>
<li>
for both quadratic and cubic curves, two points <code>e1</code> and <code>e2</code>, which represent the single-to-last step in de
Casteljau's algorithm: in the last step, we find <code>B</code> at <code>(1-t) * e1 + t * e2</code>.
</li>
<li>
for cubic curves, also the points <code>v1</code> and <code>v2</code>, which together with <code>A</code> represent the first step in de
Casteljau's algorithm: in the next step, we find <code>e1</code> and <code>e2</code>.
</li>
</ol>
<p>
These three values A, B, and C allow us to derive an important identity formula for quadratic and cubic Bézier curves: for any point on
the curve with some <code>t</code> value, the ratio of distances from A to B and B to C is fixed: if some <code>t</code> value sets up a C
that is 20% away from the start and 80% away from the end, then <em>it doesn't matter where the start, end, or control points are</em>;
for that <code>t</code> value, <code>C</code> will <em>always</em> lie at 20% from the start and 80% from the end point. Go ahead, pick an
on-curve point in either graphic and then move all the other points around: if you only move the control points, start and end won't move,
and so neither will C, and if you move either start or end point, C will move but its relative position will not change.
</p>
<p>
So, how can we compute <code>C</code>? We start with our observation that <code>C</code> always lies somewhere between the start and end
points, so logically <code>C</code> will have a function that interpolates between those two coordinates:
</p>
<!--
C = u(t) · P + (1-u(t)) · P
start end
-->
<img class="LaTeX SVG" src="./images/chapters/abc/8c6662f605722fb2ff6cd7f65243a126.svg" width="229px" height="16px" loading="lazy" />
<p>
If we can figure out what the function <code>u(t)</code> looks like, we'll be done. Although we do need to remember that this
<code>u(t)</code> will have a different form depending on whether we're working with quadratic or cubic curves.
<a href="https://mathoverflow.net/questions/122257/finding-the-formula-for-bezier-curve-ratios-hull-point-point-baseline"
>Running through the maths</a
>
(with thanks to Boris Zbarsky) shows us the following two formulae:
</p>
<!--
2
(1-t)
u(t) = ───────────
quadratic 2 2
t + (1-t)
-->
<img class="LaTeX SVG" src="./images/chapters/abc/8e7cfee39c98f2ddf9b635a914066cf6.svg" width="183px" height="41px" loading="lazy" />
<p>And</p>
<!--
3
(1-t)
u(t) = ───────────
cubic 3 3
t + (1-t)
-->
<img class="LaTeX SVG" src="./images/chapters/abc/a0b99054cc82ca1fb147f077e175ef10.svg" width="161px" height="41px" loading="lazy" />
<p>
So, if we know the start and end coordinates and the <em>t</em> value, we know C without having to calculate the <code>A</code> or even
<code>B</code> coordinates. In fact, we can do the same for the ratio function. As another function of <code>t</code>, we technically
don't need to know what <code>A</code> or <code>B</code> or <code>C</code> are. It, too, can be expressed as a pure function of
<code>t</code>.
</p>
<p>We start by observing that, given <code>A</code>, <code>B</code>, and <code>C</code>, the following always holds:</p>
<!--
distance(B,C)
ratio(t) = ────────────── = Constant
distance(A,B)
-->
<img class="LaTeX SVG" src="./images/chapters/abc/131454dcbac04e567f322979f4af80c6.svg" width="245px" height="39px" loading="lazy" />
<p>Working out the maths for this, we see the following two formulae for quadratic and cubic curves:</p>
<!--
2 2
t + (1-t) - 1
ratio(t) = |───────────────|
quadratic 2 2
t + (1-t)
-->
<img class="LaTeX SVG" src="./images/chapters/abc/5924e162b50272c40c842fad14b8fa48.svg" width="239px" height="41px" loading="lazy" />
<p>And</p>
<!--
3 3
t + (1-t) - 1
ratio(t) = |───────────────|
cubic 3 3
t + (1-t)
-->
<img class="LaTeX SVG" src="./images/chapters/abc/8cd992c1ceaae2e67695285beef23a24.svg" width="219px" height="41px" loading="lazy" />
<p>
Which now leaves us with some powerful tools: given three points (start, end, and "some point on the curve"), as well as a
<code>t</code> value, we can <em>construct</em> curves. We can compute <code>C</code> using the start and end points and our
<code>u(t)</code> function, and once we have <code>C</code>, we can use our on-curve point (<code>B</code>) and the
<code>ratio(t)</code> function to find <code>A</code>:
</p>
<!--
C - B B - C
A = B - ───────── = B + ─────────
ratio(t) ratio(t)
-->
<img class="LaTeX SVG" src="./images/chapters/abc/51a9d0588be822a5c80ea38f7d348641.svg" width="215px" height="37px" loading="lazy" />
<p>
With <code>A</code> found, finding <code>e1</code> and <code>e2</code> for quadratic curves is a matter of running the linear
interpolation with <code>t</code> between start and <code>A</code> to yield <code>e1</code>, and between <code>A</code> and end to yield
<code>e2</code>. For cubic curves, there is no single pair of points that can act as <code>e1</code> and <code>e2</code> (there are
infinitely many, because the tangent at B is a free parameter for cubic curves) so as long as the distance ratio between
<code>e1</code> to <code>B</code> and <code>B</code> to <code>e2</code> is the Bézier ratio <code>(1-t):t</code>, we are free to pick any
pair, after which we can reverse engineer <code>v1</code> and <code>v2</code>:
</p>
<!--
╭ e - t · A
│ 1
╭ e = (1-t) · v + t · A │ v = ───────────
╡ 1 1 ==> ╡ 1 1-t
│ e = (1-t) · A + t · v │ e - (1-t) · A
╰ 2 2 │ 2
│ v = ───────────────
╰ 2 t
-->
<img class="LaTeX SVG" src="./images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg" width="339px" height="73px" loading="lazy" />
<p>And then reverse engineer the curve's control points:</p>
<!--
╭ v - (1-t) · start
│ 1
╭ v = (1-t) · start + t · C │ C = ────────────────────
╡ 1 1 ==> ╡ 1 t
│ v = (1-t) · C + t · end │ v - t · end
╰ 2 2 │ 2
│ C = ──────────────
╰ 2 1-t
-->
<img class="LaTeX SVG" src="./images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg" width="383px" height="75px" loading="lazy" />
<p>
So: if we have a curve's start and end points, as well as some third point B that we want the curve to pass through, then for any
<code>t</code> value we implicitly know all the ABC values, which (combined with an educated guess on appropriate <code>e1</code> and
<code>e2</code> coordinates for cubic curves) gives us the necessary information to reconstruct a curve's "de Casteljau skeleton". Which
means that we can now do several things: we can "fit" curves using only three points, which means we can also "mold" curves by moving an
on-curve point but leaving its start and end points, and then reconstruct the curve based on where we moved the on-curve point to. These
are very useful things, and we'll look at both in the next few sections.
</p>
</section>
<section id="pointcurves">
<h1>
<div class="nav"><a href="ko-KR/index.html#abc">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#projections">다음</a></div>
<a href="ko-KR/index.html#pointcurves">Creating a curve from three points</a>
</h1>
<p>
Given the preceding section, you might be wondering if we can use that knowledge to just "create" curves by placing some points and having
the computer do the rest, to which the answer is: that's exactly what we can now do!
</p>
<p>
For quadratic curves, things are pretty easy. Technically, we'll need a <code>t</code> value in order to compute the ratio function used
in computing the ABC coordinates, but we can just as easily approximate one by treating the distance between the start and
<code>B</code> point, and <code>B</code> and end point as a ratio, using
</p>
<!--
╭ d = || Start - B||
│ 1
│ d = || End - B||
│ 2
╡ d
│ 1
│ t= ─────
│ d +d
╰ 1 2
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointcurves/f8182445c1cd7ae9f368b88fa7090e53.svg"
width="119px"
height="84px"
loading="lazy"
/>
<p>
With this code in place, creating a quadratic curve from three points is literally just computing the ABC values, and using
<code>A</code> as our curve's control point:
</p>
<graphics-element
title="Fitting a quadratic Bézier curve"
width="275"
height="275"
src="./chapters/pointcurves/quadratic.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/pointcurves/067a3df30e32708fc0d13f8eb78c0b05.png" loading="lazy" />
<label>Fitting a quadratic Bézier curve</label>
</fallback-image></graphics-element
>
<p>
For cubic curves we need to do a little more work, but really only just a little. We're first going to assume that a decent curve through
the three points should approximate a circular arc, which first requires knowing how to fit a circle to three points. You may remember (if
you ever learned it!) that a line between two points on a circle is called a
<a href="https://en.wikipedia.org/wiki/Chord_%28geometry%29">chord</a>, and that one property of chords is that the line from the center
of any chord, perpendicular to that chord, passes through the center of the circle.
</p>
<p>
That means that if we have three points on a circle, we have three (different) chords, and consequently, three (different) lines that go
from those chords through the center of the circle: if we find two of those lines, then their intersection will be our circle's center,
and the circle's radius will—by definition!—be the distance from the center to any of our three points:
</p>
<graphics-element
title="Finding a circle through three points"
width="275"
height="275"
src="./chapters/pointcurves/circle.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/pointcurves/43875f6ad588bfd04cdb65b591a62052.png" loading="lazy" />
<label>Finding a circle through three points</label>
</fallback-image></graphics-element
>
<p>
With that covered, we now also know the tangent line to our point <code>B</code>, because the tangent to any point on the circle is a line
through that point, perpendicular to the line from that point to the center. That just leaves marking appropriate points
<code>e1</code> and <code>e2</code> on that tangent, so that we can construct a new cubic curve hull. We use the approach as we did for
quadratic curves to automatically determine a reasonable <code>t</code> value, and then our <code>e1</code> and
<code>e2</code> coordinates must obey the standard de Casteljau rule for linear interpolation:
</p>
<!--
╭ e = B + t · d
╡ 1
│ e = B - (1-t) · d
╰ 2
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointcurves/8ffdd4a58cbd0fc24caef781f23a7950.svg"
width="139px"
height="40px"
loading="lazy"
/>
<p>
Where <code>d</code> is the total length of the line segment from <code>e1</code> to <code>e2</code>. So how long do we make that? There
are again all kinds of approaches we can take, and a simple-but-effective one is to set the length of that segment to "one third the
length of the baseline". This forces <code>e1</code> and <code>e2</code> to always be the "linear curve" distance apart, which means if we
place our three points on a line, it will actually <em>look</em> like a line. Nice! The last thing we'll need to do is make sure to flip
the sign of <code>d</code> depending on which side of the baseline our <code>B</code> is located, so we don't end up creating a funky
curve with a loop in it. To do this, we can use the <a href="https://en.wikipedia.org/wiki/Atan2">atan2</a> function:
</p>
<!--
\phi = (atan2(E -S , E -S ) - atan2(B -S , B -S ) + 2 \pi) mod 2 \pi
y y x x y y x x
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointcurves/a5cd63b54be6b554290c38787cfbbabd.svg"
width="488px"
height="24px"
loading="lazy"
/>
<p>
This angle φ will be between 0 and π if <code>B</code> is "above" the baseline (rotating all three points so that the start is on the left
and the end is the right), so we can use a relatively straight forward check to make sure we're using the correct sign for our value
<code>d</code>:
</p>
<!--
d = { d if 0 ≤\phi ≤\pi
-d if \phi < 0 \lor \phi > \pi
-->
<img
class="LaTeX SVG"
src="./images/chapters/pointcurves/4fe687c8a65265a2a755ba5841d0e31d.svg"
width="177px"
height="40px"
loading="lazy"
/>
<p>The result of this approach looks as follows:</p>
<graphics-element
title="Finding the cubic e₁ and e₂ given three points "
width="275"
height="275"
src="./chapters/pointcurves/circle.js"
data-show-curve="true"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/pointcurves/75f7b5b31e98444e13f17e5c3e5b7322.png" loading="lazy" />
<label>Finding the cubic e₁ and e₂ given three points </label>
</fallback-image></graphics-element
>
<p>
It is important to remember that even though we're using a circular arc to come up with decent <code>e1</code> and <code>e2</code> terms,
we're <em>not</em> trying to perfectly create a circular arc with a cubic curve (which is good, because we can't;
<a href="#arcapproximation">more on that later</a>), we're <em>only</em> trying to come up with some reasonable <code>e1</code> and
<code>e2</code> points so we can construct a new cubic curve... so now that we have those: let's see what kind of cubic curve that gives
us:
</p>
<graphics-element
title="Fitting a cubic Bézier curve"
width="275"
height="275"
src="./chapters/pointcurves/cubic.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/pointcurves/eab6ea46fa93030e03ec0ef7deb571dc.png" loading="lazy" />
<label>Fitting a cubic Bézier curve</label>
</fallback-image></graphics-element
>
<p>That looks perfectly serviceable!</p>
<p>
Of course, we can take this one step further: we can't just "create" curves, we also have (almost!) all the tools available to "mold"
curves, where we can reshape a curve by dragging a point on the curve around while leaving the start and end fixed, effectively molding
the shape as if it were clay or the like. We'll see the last tool we need to do that in the next section, and then we'll look at
implementing curve molding in the section after that, so read on!
</p>
</section>
<section id="projections">
<h1>
<div class="nav">
<a href="ko-KR/index.html#pointcurves">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#circleintersection">다음</a>
</div>
<a href="ko-KR/index.html#projections">Projecting a point onto a Bézier curve</a>
</h1>
<p>
Before we can move on to actual curve molding, it'll be good if know how to actually be able to find "some point on the curve" that we're
trying to click on. After all, if all we have is our Bézier coordinates, that is not in itself enough to figure out which point on the
curve our cursor will be closest to. So, how do we project points onto a curve?
</p>
<p>
If the Bézier curve is of low enough order, we might be able to
<a href="https://web.archive.org/web/20140713004709/http://jazzros.blogspot.com/2011/03/projecting-point-on-bezier-curve.html"
>work out the maths for how to do this</a
>, and get a perfect <code>t</code> value back, but in general this is an incredibly hard problem and the easiest solution is, really, a
numerical approach again. We'll be finding our ideal <code>t</code> value using a
<a href="https://en.wikipedia.org/wiki/Binary_search_algorithm">binary search</a>. First, we do a coarse distance-check based on
<code>t</code> values associated with the curve's "to draw" coordinates (using a lookup table, or LUT). This is pretty fast:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="8">
<textarea disabled rows="8" role="doc-example">
p = some point to project onto the curve
d = some initially huge value
i = 0
for (coordinate, index) in LUT:
q = distance(coordinate, p)
if q < d:
d = q
i = index</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
</table>
<p>
After this runs, we know that <code>LUT[i]</code> is the coordinate on the curve <em>in our LUT</em> that is closest to the point we want
to project, so that's a pretty good initial guess as to what the best projection onto our curve is. To refine it, we note that
<code>LUT[i]</code> is a better guess than both <code>LUT[i-1]</code> and <code>LUT[i+1]</code>, but there might be an even better
projection <em>somewhere else</em> between those two values, so that's what we're going to be testing for, using a variation of the binary
search.
</p>
<ol>
<li>
we start with our point <code>p</code>, and the <code>t</code> values <code>t1=LUT[i-1].t</code> and <code>t2=LUT[i+1].t</code>, which
span an interval <code>v = t2-t1</code>.
</li>
<li>
we test this interval in five spots: the start, middle, and end (which we already have), and the two points in between the middle and
start/end points
</li>
<li>
we then check which of these five points is the closest to our original point <code>p</code>, and then repeat step 1 with the points
before and after the closest point we just found.
</li>
</ol>
<p>
This makes the interval we check smaller and smaller at each iteration, and we can keep running the three steps until the interval becomes
so small as to lead to distances that are, for all intents and purposes, the same for all points.
</p>
<p>
So, let's see that in action: in this case, I'm going to arbitrarily say that if we're going to run the loop until the interval is smaller
than 0.001, and show you what that means for projecting your mouse cursor or finger tip onto a rather complex Bézier curve (which, of
course, you can reshape as you like). Also shown are the original three points that our coarse check finds.
</p>
<graphics-element
title="Projecting a point onto a Bézier curve"
width="400"
height="400"
src="./chapters/projections/project.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/projections/536a94885dc1637c066d0ef4f6e4e650.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
</section>
<section id="circleintersection">
<h1>
<div class="nav"><a href="ko-KR/index.html#projections">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#molding">다음</a></div>
<a href="ko-KR/index.html#circleintersection">Intersections with a circle</a>
</h1>
<p>
It might seem odd to cover this subject so much later than the line/line, line/curve, and curve/curve intersection topics from several
sections earlier, but the reason we can't cover circle/curve intersections is that we can't really discuss circle/curve intersection until
we've covered the kind of lookup table (LUT) walking that the section on projecting a point onto a curve uses. To see why, let's look at
what we would have to do if we wanted to find the intersections between a curve and a circle using calculus.
</p>
<p>
First, we observe that "finding intersections" in this case means that, given a circle defined by a center point
<code>c = (x,y)</code> and a radius <code>r</code>, we want to find all points on the Bezier curve for which the distance to the circle's
center point is equal to the circle radius, which by definition means those points lie on the circle, and so count as intersections. In
maths, that means we're trying to solve:
</p>
<!--
dist(B(t), c) = r
-->
<img
class="LaTeX SVG"
src="./images/chapters/circleintersection/373248ec6a579bacf6c6a317e6db597a.svg"
width="107px"
height="16px"
loading="lazy"
/>
<p>Which seems simple enough. Unfortunately, when we expand that <code>dist</code> function, things get a lot more problematic:</p>
<!--
r= dist(B(t), c)
┌─────────────────────────┐
│ 2 2
= │(B t - c ) + (B t - c )
⟍│ x x y y
┌─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
│╭ 3 2 2 3 ╮2 ╭ 3 2 2 3 ╮2
= ││ x (1-t) + 3 x (1-t) t + 2 x (1-t) t + x t - c │ + │ y (1-t) + 3 y (1-t) t + 2 y (1-t) t + y t - c │
⟍│╰ 1 2 3 4 x ╯ ╰ 1 2 3 4 y ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/circleintersection/2f42c862a0a9d0764727d42b16cf68a0.svg"
width="811px"
height="103px"
loading="lazy"
/>
<p>
And now we have a problem because that's a sixth degree polynomial inside the square root. So, thanks to the
<a href="https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem">Abel-Ruffini theorem</a> that we saw before, we can't solve this by
just going "square both sides because we don't care about signs"... we can't solve a sixth degree polynomial. So, we're going to have to
actually evaluate that expression. We can "simplify" this by translating all our coordinates so that the center of the circle is (0,0) and
all our coordinates are shifted accordingly, which makes the c<sub>x</sub> and c<sub>y</sub> terms fall away, but then we're still left
with a monstrous function to solve.
</p>
<p>
So instead, we turn to the same kind of "LUT walking" that we saw for projecting points onto a curve, with a twist: instead of finding the
on-curve point with the smallest distance to our projection point, we want to find the on-curve point that has the exact distance
<code>r</code> to our projection point (namely, our circle center). Of course, there can be more than one such point, so there's also a
bit more code to make sure we find all of them, but let's look at the steps involved:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="9">
<textarea disabled rows="9" role="doc-example">
p = our circle's center point
r = our circle's radius
d = some initially huge value
i = 0
for (coordinate, index) in LUT:
q = abs(distance(coordinate, p) - r)
if q < d:
d = q
i = index</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
</table>
<p>
This is <em>very</em> similar to the code in the previous section, with an extra input <code>r</code> for the circle radius, and a minor
change in the "distance for this coordinate": rather than just <code>distance(coordinate, p)</code> we want to know the difference between
that distance and the circle radius. After all, if that difference is zero, then the distance from the coordinate to the circle center is
exactly the radius, so the coordinate lies on both the curve and the circle.
</p>
<p>So far so good.</p>
<p>However, we also want to make sure we find <em>all</em> the points, not just a single one, so we need a little more code for that:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="10">
<textarea disabled rows="10" role="doc-example">
p = our circle's center point
r = our circle's radius
d = some initially huge value
start = 0
values = []
do:
i = findClosest(start, p, r, LUT)
if i < start, or i>0 but the same as start: stop
values.add(i);
start = i + 2;</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
</table>
<p>
After running this code, <code>values</code> will be the list of all LUT coordinates that are closest to the distance <code>r</code>: we
can use those values to run the same kind of refinement lookup we used for point projection (with the caveat that we're now
<em>not</em> checking for smallest distance, but for "distance closest to <code>r</code>"), and we'll have all our intersection points. Of
course, that does require explaining what <code>findClosest</code> does: rather than looking for a global minimum, we're now interested in
finding a <em>local</em> minimum, so instead of checking a single point and looking at its distance value, we check <em>three</em> points
("current", "previous" and "before previous") and then check whether they form a local minimum:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="19">
<textarea disabled rows="19" role="doc-example">
findClosest(start, p, r, LUT):
minimizedDistance = some very large number
pd2 = LUT[start-2], if it exists. Otherwise use minimizedDistance
pd1 = LUT[start-1], if it exists. Otherwise use minimizedDistance
slice = LUT.subset(start, LUT.length)
epsilon = the largest point-to-point distance in our LUT
i = -1;
for (coordinate, index) in slice:
q = abs(dist(coordinate, p) - r);
if pd1 less than all three values epsilon, pd2, and q:
i = index - 1
break
minimizedDistance = min(q, minimizedDistance)
pd2 = pd1
pd1 = q
return start + i</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
<tr>
<td>18</td>
</tr>
<tr>
<td>19</td>
</tr>
</table>
<p>
In words: given a <code>start</code> index, the circle center and radius, and our LUT, we check where (closest to our
<code>start</code> index) we can find a local minimum for the difference between "the distance from the curve to the circle center", and
the circle's radius. We track this by looking at three values (associated with the indices <code>index-2</code>, <code>index-1</code>, and
<code>index</code>), and we know we've found a local minimum if the three values show that the middle value (<code>pd1</code>) is less
than either value beside it. When we do, we can set our "best guess, relative to <code>start</code>" as <code>index-1</code>. Of course,
since we're now checking values relative to some <code>start</code> value, we might not find another candidate value at all, in which case
we return <code>start - 1</code>, so that a simple "is the result less than <code>start</code>?" lets us determine that there are no more
intersections to find.
</p>
<p>
Finally, while not necessary for point projection, there is one more step we need to perform when we run the binary refinement function on
our candidate LUT indices, because we've so far only been testing using distances "closest to the radius of the circle", and that's
actually not good enough... we need distances that <em>are</em> the radius of the circle. So, after running the refinement for each of
these indices, we need to discard any final value that isn't the circle radius. And because we're working with floating point numbers,
what this really means is that we need to discard any value that's a pixel or more "off". Or, if we want to get really fancy, "some small
<code>epsilon</code> value".
</p>
<p>
Based on all of that, the following graphic shows this off for the standard cubic curve (which you can move the coordinates around for, of
course) and a circle with a controllable radius centered on the graphic's center, using the code approach described above.
</p>
<graphics-element
title="circle intersection"
width="275"
height="275"
src="./chapters/circleintersection/circle.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/circleintersection/3a3ea30971de247da93034de614a63a4.png" loading="lazy" />
<label>circle intersection</label>
</fallback-image>
<input type="range" min="1" max="150" step="1" value="70" class="slide-control" />
</graphics-element>
<p>And of course, for the full details, click that "view source" link.</p>
</section>
<section id="molding">
<h1>
<div class="nav">
<a href="ko-KR/index.html#circleintersection">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#curvefitting">다음</a>
</div>
<a href="ko-KR/index.html#molding">Molding a curve</a>
</h1>
<p>
Armed with knowledge of the "ABC" relation, point-on-curve projection, and guestimating reasonable looking helper values for cubic curve
construction, we can finally cover curve molding: updating a curve's shape interactively, by dragging points on the curve around.
</p>
<p>
For quadratic curve, this is a really simple trick: we project our cursor onto the curve, which gives us a <code>t</code> value and
initial <code>B</code> coordinate. We don't even need the latter: with our <code>t</code> value and "wherever the cursor is" as target
<code>B</code>, we can compute the associated <code>C</code>:
</p>
<!--
C = u(t) · Start + (1-u(t) ) · End
q q
-->
<img class="LaTeX SVG" src="./images/chapters/molding/48887d68a861a0acdf8313e23fb19880.svg" width="247px" height="24px" loading="lazy" />
<p>And then the associated <code>A</code>:</p>
<!--
C - B B - C
A = B - ────────── = B + ──────────
ratio(t) ratio(t)
q q
-->
<img class="LaTeX SVG" src="./images/chapters/molding/2e65bc9c934380c2de6a24bcd5c1c7b7.svg" width="228px" height="41px" loading="lazy" />
<p>And we're done, because that's our new quadratic control point!</p>
<graphics-element
title="Molding a quadratic Bézier curve"
width="825"
height="275"
src="./chapters/molding/molding.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/molding/2bd215d1db191b52a89a94727b6aa5ce.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>
As before, cubic curves are a bit more work, because while it's easy to find our initial <code>t</code> value and ABC values, getting
those all-important <code>e1</code> and <code>e2</code> coordinates is going to pose a bit of a problem... in the section on curve
creation, we were free to pick an appropriate <code>t</code> value ourselves, which allowed us to find appropriate <code>e1</code> and
<code>e2</code> coordinates. That's great, but when we're curve molding we don't have that luxury: whatever point we decide to start
moving around already has its own <code>t</code> value, and its own <code>e1</code> and <code>e2</code> values, and those may not make
sense for the rest of the curve.
</p>
<p>
For example, let's see what happens if we just "go with what we get" when we pick a point and start moving it around, preserving its
<code>t</code> value and <code>e1</code>/<code>e2</code> coordinates:
</p>
<graphics-element
title="Molding a cubic Bézier curve"
width="825"
height="275"
src="./chapters/molding/molding.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/molding/a54f990aa49fb9ea8200b59259f955f3.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>
That looks reasonable, close to the original point, but the further we drag our point, the less "useful" things become. Especially if we
drag our point across the baseline, rather than turning into a nice curve.
</p>
<p>
One way to combat this might be to combine the above approach with the approach from the
<a href="#pointcurves">creating curves</a> section: generate both the "unchanged <code>t</code>/<code>e1</code>/<code>e2</code>" curve, as
well as the "idealized" curve through the start/cursor/end points, with idealized <code>t</code> value, and then interpolating between
those two curves:
</p>
<graphics-element
title="Molding a cubic Bézier curve"
width="825"
height="275"
src="./chapters/molding/molding.js"
data-type="cubic"
data-interpolated="true"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="825px" height="275px" src="./images/chapters/molding/1039d0bb0e49cfb472c2fa37f9010190.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="10" max="200" step="1" value="100" class="slide-control" />
</graphics-element>
<p>
The slide controls the "falloff distance" relative to where the original point on the curve is, so that as we drag our point around, it
interpolates with a bias towards "preserving <code>t</code>/<code>e1</code>/<code>e2</code>" closer to the original point, and bias
towards "idealized" form the further away we move our point, with anything that's further than our falloff distance simply
<em>being</em> the idealized curve. We don't even try to interpolate at that point.
</p>
<p>
A more advanced way to try to smooth things out is to implement <em>continuous</em> molding, where we constantly update the curve as we
move around, and constantly change what our <code>B</code> point is, based on constantly projecting the cursor on the curve
<em>as we're updating it</em> - this is, you won't be surprised to learn, tricky, and beyond the scope of this section: interpolation
(with a reasonable distance) will do for now!
</p>
</section>
<section id="curvefitting">
<h1>
<div class="nav"><a href="ko-KR/index.html#molding">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#catmullconv">다음</a></div>
<a href="ko-KR/index.html#curvefitting">Curve fitting</a>
</h1>
<p>
Given the previous section, one question you might have is "what if I don't want to guess <code>t</code> values?". After all, plenty of
graphics packages do automated curve fitting, so how can we implement that in a way that just finds us reasonable <code>t</code> values
all on its own?
</p>
<p>
And really this is just a variation on the question "how do I get the curve through these X points?", so let's look at that. Specifically,
let's look at the answer: "curve fitting". This is in fact a rather rich field in geometry, applying to anything from data modelling to
path abstraction to "drawing", so there's a fair number of ways to do curve fitting, but we'll look at one of the most common approaches:
something called a <a href="https://en.wikipedia.org/wiki/Least_squares">least squares</a>
<a href="https://en.wikipedia.org/wiki/Polynomial_regression">polynomial regression</a>. In this approach, we look at the number of points
we have in our data set, roughly determine what would be an appropriate order for a curve that would fit these points, and then tackle the
question "given that we want an <code>nth</code> order curve, what are the coordinates we can find such that our curve is "off" by the
least amount?".
</p>
<p>
Now, there are many ways to determine how "off" points are from the curve, which is where that "least squares" term comes in. The most
common tool in the toolbox is to minimise the <em>squared distance</em> between each point we have, and the corresponding point on the
curve we end up "inventing". A curve with a snug fit will have zero distance between those two, and a bad fit will have non-zero distances
between every such pair. It's a workable metric. You might wonder why we'd need to square, rather than just ensure that distance is a
positive value (so that the total error is easy to compute by just summing distances) and the answer really is "because it tends to be a
little better". There's lots of literature on the web if you want to deep-dive the specific merits of least squared error metrics versus
least absolute error metrics, but those are <em>well</em> beyond the scope of this material.
</p>
<p>
So let's look at what we end up with in terms of curve fitting if we start with the idea of performing least squares Bézier fitting. We're
going to follow a procedure similar to the one described by Jim Herold over on his
<a href="https://web.archive.org/web/20180403213813/http://jimherold.com/2012/04/20/least-squares-bezier-fit/"
>"Least Squares Bézier Fit"</a
>
article, and end with some nice interactive graphics for doing some curve fitting.
</p>
<p>
Before we begin, we're going to use the curve in matrix form. In the <a href="#matrix">section on matrices</a>, I mentioned that some
things are easier if we use the matrix representation of a Bézier curve rather than its calculus form, and this is one of those things.
</p>
<p>
As such, the first step in the process is expressing our Bézier curve as powers/coefficients/coordinate matrix <strong>T x M x C</strong>,
by expanding the Bézier functions.
</p>
<div class="note">
<h2>Revisiting the matrix representation</h2>
<p>
Rewriting Bézier functions to matrix form is fairly easy, if you first expand the function, and then arrange them into a multiple line
form, where each line corresponds to a power of t, and each column is for a specific coefficient. First, we expand the function:
</p>
<!--
2 2
B = a (1-t) + 2 b (1-t) t + c t
quadratic
2 2 2
= a - 2at + at + 2bt - 2bt + ct
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/dd303afb51d580fb2bf1b914c010f83d.svg"
width="287px"
height="43px"
loading="lazy"
/>
<p>And then we (trivially) rearrange the terms across multiple lines:</p>
<!--
B =a
quadratic
- 2at+ 2bt
2 2 2
+ at - 2bt + ct
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/ff701138fd7a6e35700a2e1ee3e9c020.svg"
width="209px"
height="64px"
loading="lazy"
/>
<p>
This rearrangement has "factors of t" at each row (the first row is t⁰, i.e. "1", the second row is t¹, i.e. "t", the third row is t²)
and "coefficient" at each column (the first column is all terms involving "a", the second all terms involving "b", the third all terms
involving "c").
</p>
<p>With that arrangement, we can easily decompose this as a matrix multiplication:</p>
<!--
┌ a + 0 + 0 ┐ ┌ 1 0 0 ┐ ┌ a ┐
B = T · M · C = ┌ 2 ┐ · │ -2a + 2b + 0 │ = ┌ 2 ┐ · │ -2 2 0 │ · │ b │
quadratic └ 1 t t ┘ └ a + -2b + c ┘ └ 1 t t ┘ └ 1 -2 1 ┘ └ c ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/db85a7a46b869c892662c26b6aea15a1.svg"
width="616px"
height="53px"
loading="lazy"
/>
<p>We can do the same for the cubic curve, of course. We know the base function for cubics:</p>
<!--
3 2 2 3
B =a(1-t) + 3b(1-t) t + 3c(1-t)t + dt
cubic
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/17d5fbeffcdcceca98cdba537295d258.svg"
width="351px"
height="19px"
loading="lazy"
/>
<p>So we write out the expansion and rearrange:</p>
<!--
B =a
cubic
- 3at + 3bt
2 2 2
+ 3at - 6bt +3ct
3 3 3 3
- at + 3bt -3ct + dt
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/8928f757abd1376abdc4069e1aa774f2.svg"
width="244px"
height="87px"
loading="lazy"
/>
<p>Which we can then decompose:</p>
<!--
┌ 1 0 0 0 ┐ ┌ a ┐
B = T · M · C = ┌ 2 3 ┐ · │ -3 3 0 0 │ · │ b │
cubic └ 1 t t t ┘ │ 3 -6 3 0 │ │ c │
└ -1 3 -3 1 ┘ └ d ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/4e6e20c823c8cc72e0cc00e4ab5b7556.svg"
width="401px"
height="72px"
loading="lazy"
/>
<p>And, of course, we can do this for quartic curves too (skipping the expansion step):</p>
<!--
┌ 1 0 0 0 0 ┐ ┌ a ┐
│ -4 4 0 0 0 │ │ b │
B = T · M · C = ┌ 2 3 4 ┐ · │ 6 -12 6 0 0 │ · │ c │
quartic └ 1 t t t t ┘ │ -4 12 -12 4 0 │ │ d │
└ 1 -4 6 -4 1 ┘ └ e ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/8a66af7570bac674966f6316820ea31b.svg"
width="485px"
height="92px"
loading="lazy"
/>
<p>
And so and on so on. Now, let's see how to use these <strong>T</strong>, <strong>M</strong>, and <strong>C</strong>, to do some curve
fitting.
</p>
</div>
<p>
Let's get started: we're going to assume we picked the right order curve: for <code>n</code> points we're fitting an <code>n-1</code
><sup>th</sup> order curve, so we "start" with a vector <strong>P</strong> that represents the coordinates we already know, and for which
we want to do curve fitting:
</p>
<!--
┌ p ┐
│ 1 │
│ p │
P = │ 2 │
│ ... │
│ p │
└ n ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/b017da988c9a778a4ce6a6f4ea4790d4.svg"
width="63px"
height="73px"
loading="lazy"
/>
<p>
Next, we need to figure out appropriate <code>t</code> values for each point in the curve, because we need something that lets us tie "the
actual coordinate" to "some point on the curve". There's a fair number of different ways to do this (and a large part of optimizing "the
perfect fit" is about picking appropriate <code>t</code> values), but in this case let's look at two "obvious" choices:
</p>
<ol>
<li>equally spaced <code>t</code> values, and</li>
<li><code>t</code> values that align with distance along the polygon.</li>
</ol>
<p>
The first one is really simple: if we have <code>n</code> points, then we'll just assign each point <code>i</code> a <code>t</code> value
of <code>(i-1)/(n-1)</code>. So if we have four points, the first point will have <code>t=(1-1)/(4-1)=0/3</code>, the second point will
have <code>t=(2-1)/(4-1)=1/3</code>, the third point will have <code>t=2/3</code>, and the last point will be <code>t=1</code>. We're just
straight up spacing the <code>t</code> values to match the number of points we have.
</p>
<p>
The second one is a little more interesting: since we're doing polynomial regression, we might as well exploit the fact that our base
coordinates just constitute a collection of line segments. At the first point, we're fixing t=0, and the last point, we want t=1, and
anywhere in between we're simply going to say that <code>t</code> is equal to the distance along the polygon, scaled to the [0,1] domain.
</p>
<p>To get these values, we first compute the general "distance along the polygon" matrix:</p>
<!--
╭ d = 0
D = ┌ d d ... d ┐, where ╡ 1
└ 1 2 n ┘ │ d = d + length(p , p )
╰ i i-1 i-1 i
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/2f82371abb7835f9b9d440dc5dd151a8.svg"
width="395px"
height="40px"
loading="lazy"
/>
<p>
Where <code>length()</code> is literally just that: the length of the line segment between the point we're looking at, and the previous
point. This isn't quite enough, of course: we still need to make sure that all the values between <code>i=1</code> and
<code>i=n</code> fall in the [0,1] interval, so we need to scale all values down by whatever the total length of the polygon is:
</p>
<!--
╭ s = 0
│ 1
S = ┌ s s ... s ┐, where ╡ s = d / d
└ 1 2 n ┘ │ i i n
│ s = 1
╰ n
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/6f734d319a1cfe0de76574a65abb07e1.svg"
width="272px"
height="55px"
loading="lazy"
/>
<p>
And now we can move on to the actual "curve fitting" part: what we want is a function that lets us compute "ideal" control point values
such that if we build a Bézier curve with them, that curve passes through all our original points. Or, failing that, have an overall error
distance that is as close to zero as we can get it. So, let's write out what the error distance looks like.
</p>
<p>
As mentioned before, this function is really just "the distance between the actual coordinate, and the coordinate that the curve evaluates
to for the associated <code>t</code> value", which we'll square to get rid of any pesky negative signs:
</p>
<!--
2
E(C) = (p - Bézier(s ))
i i i
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/d39ca235454ced9681b523be056864d2.svg"
width="177px"
height="23px"
loading="lazy"
/>
<p>
Since this function only deals with individual coordinates, we'll need to sum over all coordinates in order to get the full error
function. So, we literally just do that; the total error function is simply the sum of all these individual errors:
</p>
<!--
__ n 2
E(C) = ❯ (p - Bézier(s ))
‾‾ i=1 i i
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/097aa1948b6cdbf9dc7579643a7af246.svg"
width="195px"
height="41px"
loading="lazy"
/>
<p>
And here's the trick that justifies using matrices: while we can work with individual values using calculus, with matrices we can compute
as many values as we make our matrices big, all at the "same time", We can replace the individual terms p<sub>i</sub> with the full
<strong>P</strong> coordinate matrix, and we can replace Bézier(s<sub>i</sub>) with the matrix representation
<strong>T x M x C</strong> we talked about before, which gives us:
</p>
<!--
2
E(C) = (P - TMC)
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/7b0199bb515d2754c03d8f796b29febf.svg"
width="141px"
height="21px"
loading="lazy"
/>
<p>In which we can replace the rather cumbersome "squaring" operation with a more conventional matrix equivalent:</p>
<!--
T
E(C) = (P - TMC) (P - TMC)
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/8068231b915832938136d5833f74751d.svg"
width="225px"
height="21px"
loading="lazy"
/>
<p>
Here, the letter <code>T</code> is used instead of the number 2, to represent the
<a href="https://en.wikipedia.org/wiki/Transpose">matrix transpose</a>; each row in the original matrix becomes a column in the transposed
matrix instead (row one becomes column one, row two becomes column two, and so on).
</p>
<p>
This leaves one problem: <strong>T</strong> isn't actually the matrix we want: we don't want symbolic <code>t</code> values, we want the
actual numerical values that we computed for <strong>S</strong>, so we need to form a new matrix, which we'll call 𝕋, that makes use of
those, and then use that 𝕋 instead of <strong>T</strong> in our error function:
</p>
<!--
┌ 0 1 n-2 n-1 ┐
│ s s ... s s │
│ 1 1 1 1 │
│ │
𝕋 = │ \vdots ... \vdots │
│ │
│ 0 1 n-2 n-1 │
│ s s ... s s │
└ n n n n ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/4dd55c228a26bb50da912a45e8721024.svg"
width="201px"
height="96px"
loading="lazy"
/>
<p>Which, because of the first and last values in <strong>S</strong>, means:</p>
<!--
┌ 1 0 ... 0 0 ┐
│ n-2 n-1 │
│ 1 s s s │
│ 2 2 2 │
𝕋 = │ \vdots ... \vdots │
│ n-2 n-1 │
│ 1 s s s │
│ n-1 n-1 n-1 │
└ 1 1 ... 1 1 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/38bb81bdd3eaa72c2336514187aa374b.svg"
width="212px"
height="92px"
loading="lazy"
/>
<p>Now we can properly write out the error function as matrix operations:</p>
<!--
T
E(C) = (P - 𝕋MC) (P - 𝕋MC)
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/989f2ad06ae308f71cef527a5594129a.svg"
width="231px"
height="21px"
loading="lazy"
/>
<p>
So, we have our error function: we now need to figure out the expression for where that function has minimal value, e.g. where the error
between the true coordinates and the coordinates generated by the curve fitting is smallest. Like in standard calculus, this requires
taking the derivative, and determining where that derivative is zero:
</p>
<!--
∂E T
── = 0 = -2𝕋 (P - 𝕋MC)
∂C
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/ea24b0e42f0a89464bda275ac8f9bacf.svg"
width="197px"
height="36px"
loading="lazy"
/>
<div class="note">
<h2>Where did this derivative come from?</h2>
<p>
That... is a good question. In fact, when trying to run through this approach, I ran into the same question! And you know what? I
straight up had no idea. I'm decent enough at calculus, I'm decent enough at linear algebra, and I just don't know.
</p>
<p>
So I did what I always do when I don't understand something: I asked someone to help me understand how things work. In this specific
case, I
<a href="https://math.stackexchange.com/questions/2825438/how-do-you-compute-the-derivative-of-a-matrix-algebra-expression"
>posted a question</a
>
to <a href="https://math.stackexchange.com">Math.stackexchange</a>, and received a answer that goes into way more detail than I had
hoped to receive.
</p>
<p>
Is that answer useful to you? Probably: no. At least, not unless you like understanding maths on a recreational level. And I do mean
maths in general, not just basic algebra. But it does help in giving us a reference in case you ever wonder "Hang on. Why was that
true?". There are answers. They might just require some time to come to understand.
</p>
</div>
<p>
Now, given the above derivative, we can rearrange the terms (following the rules of matrix algebra) so that we end up with an expression
for <strong>C</strong>:
</p>
<!--
-1 T -1 T
C = M (𝕋 𝕋) 𝕋 P
-->
<img
class="LaTeX SVG"
src="./images/chapters/curvefitting/9dec10b81a61b456ca1550cd9b7ba513.svg"
width="168px"
height="27px"
loading="lazy"
/>
<p>
Here, the "to the power negative one" is the notation for the
<a href="https://en.wikipedia.org/wiki/Invertible_matrix">matrix inverse</a>. But that's all we have to do: we're done. Starting with
<strong>P</strong> and inventing some <code>t</code> values based on the polygon the coordinates in <strong>P</strong> define, we can
compute the corresponding Bézier coordinates <strong>C</strong> that specify a curve that goes through our points. Or, if it can't go
through them exactly, as near as possible.
</p>
<p>
So before we try that out, how much code is involved in implementing this? Honestly, that answer depends on how much you're going to be
writing yourself. If you already have a matrix maths library available, then really not that much code at all. On the other hand, if you
are writing this from scratch, you're going to have to write some utility functions for doing your matrix work for you, so it's really
anywhere from 50 lines of code to maybe 200 lines of code. Not a bad price to pay for being able to fit curves to pre-specified
coordinates.
</p>
<p>
So let's try it out! The following graphic lets you place points, and will start computing exact-fit curves once you've placed at least
three. You can click for more points, and the code will simply try to compute an exact fit using a Bézier curve of the appropriate order.
Four points? Cubic Bézier. Five points? Quartic. And so on. Of course, this does break down at some point: depending on where you place
your points, it might become mighty hard for the fitter to find an exact fit, and things might actually start looking horribly off once
there's enough points for compound
<a href="https://en.wikipedia.org/wiki/Round-off_error#Floating-point_number_system">floating point rounding errors</a> to start making a
difference (which is around 10~11 points).
</p>
<graphics-element
title="Fitting a Bézier curve"
width="550"
height="275"
src="./chapters/curvefitting/curve-fitting.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="550px" height="275px" src="./images/chapters/curvefitting/798f3d7151dfb2887c7881a08e65cdd3.png" loading="lazy" />
<label></label>
</fallback-image>
<button class="toggle">toggle</button>
<!-- additional sliders will get created on the fly -->
</graphics-element>
<p>
You'll note there is a convenient "toggle" buttons that lets you toggle between equidistant <code>t</code> values, and distance ratio
along the polygon formed by the points. Arguably more interesting is that once you have points to abstract a curve, you also get
<em>direct control</em> over the time values through sliders for each, because if the time values are our degree of freedom, you should be
able to freely manipulate them and see what the effect on your curve is.
</p>
</section>
<section id="catmullconv">
<h1>
<div class="nav">
<a href="ko-KR/index.html#curvefitting">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#catmullfitting">다음</a>
</div>
<a href="ko-KR/index.html#catmullconv">Bézier curves and Catmull-Rom curves</a>
</h1>
<p>
Taking an excursion to different splines, the other common design curve is the
<a href="https://en.wikipedia.org/wiki/Cubic_Hermite_spline#Catmull.E2.80.93Rom_spline">Catmull-Rom spline</a>, which unlike Bézier curves
pass <em>through</em> each control point, so they offer a kind of "built-in" curve fitting.
</p>
<p>
In fact, let's start with just playing with one: the following graphic has a predefined curve that you manipulate the points for, and lets
you add points by clicking/tapping the background, as well as let you control "how fast" the curve passes through its point using the
tension slider. The tenser the curve, the more the curve tends towards straight lines from one point to the next.
</p>
<graphics-element
title="A Catmull-Rom curve"
width="275"
height="275"
src="./chapters/catmullconv/catmull-rom.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/catmullconv/aa46749b9469341d9249ca452390d875.png" loading="lazy" />
<label>A Catmull-Rom curve</label>
</fallback-image>
<input type="range" min="0.1" max="1" step="0.01" value="0.5" class="slide-control tension" />
</graphics-element>
<p>
Now, it may look like Catmull-Rom curves are very different from Bézier curves, because these curves can get very long indeed, but what
looks like a single Catmull-Rom curve is actually a <a href="https://en.wikipedia.org/wiki/Spline_(mathematics)">spline</a>: a single
curve built up of lots of identically-computed pieces, similar to if you just took a whole bunch of Bézier curves, placed them end to end,
and lined up their control points so that things look like a single curve. For a Catmull-Rom curve, each "piece" between two points is
defined by the point's coordinates, and the tangent for those points, the latter of which
<a href="https://en.wikipedia.org/wiki/Cubic_Hermite_spline#Catmull%E2%80%93Rom_spline">can trivially be derived</a> from knowing the
previous and next point:
</p>
<!--
┌ V = P ┐
│ 1 2 │
┌ P ┐ │ V = P │
│ 1 │ │ 2 3 │
│ P │ │ P - P │
│ 2 │ = │ 3 1 │
│ P │points │ V' = ─────── │ point-tangent
│ 3 │ │ 1 2 │
│ P │ │ P - P │
└ 4 ┘ │ 4 2 │
│ V' = ─────── │
└ 2 2 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/639ca0b74a805c3aebac79b181eac908.svg"
width="272px"
height="88px"
loading="lazy"
/>
<p>
One downside of this is that—as you may have noticed from the graphic—the first and last point of the overall curve don't actually join up
with the rest of the curve: they don't have a previous/next point respectively, and so there is no way to calculate what their tangent
should be. Which also makes it rather tricky to fit a Catmull-Rom curve to three points like we were able to do for Bézier curves. More on
that in <a href="#catmullfitting">the next section</a>.
</p>
<p>
In fact, before we move on, let's look at how to actually draw the basic form of these curves (I say basic, because there are a number of
variations that make things
<a href="https://en.wikipedia.org/wiki/Centripetal_Catmull%E2%80%93Rom_spline#Definition">considerable</a> more
<a href="https://en.wikipedia.org/wiki/Kochanek%E2%80%93Bartels_spline">complex</a>):
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="19">
<textarea disabled rows="19" role="doc-example">
tension = some value greater than 0, defaulting to 1
points = a list of at least 4 coordinates
for p = 1 to points.length-3 (inclusive):
p0 = points[p-1]
v1 = p1 = points[p]
v2 = p2 = points[p+1]
p3 = points[p+2]
s = 2 * tension
dv1 = (p2-p0) / s
dv2 = (p3-p1) / s
for t = 0 to 1 (inclusive):
c0 = 2*t^3 - 3*t^2 + 1,
c1 = t^3 - 2*t^2 + t,
c2 = -2*t^3 + 3*t^2,
c3 = t^3 - t^2
point(c0 * v1 + c1 * dv1 + c2 * v2 + c3 * dv2)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
<tr>
<td>18</td>
</tr>
<tr>
<td>19</td>
</tr>
</table>
<p>
Now, since a Catmull-Rom curve is a form of <a href="https://en.wikipedia.org/wiki/Cubic_Hermite_spline">cubic Hermite spline</a>, and as
cubic Bézier curves are <em>also</em> a form of cubic Hermite spline, we run into an interesting bit of maths programming: we can convert
one to the other and back, and the maths for doing so is surprisingly simple!
</p>
<p>The main difference between Catmull-Rom curves and Bézier curves is "what the points mean":</p>
<ul>
<li>
A cubic Bézier curve is defined by a start point, a control point that implies the tangent at the start, a control point that implies
the tangent at the end, and an end point, plus a characterizing matrix that we can multiply by that point vector to get on-curve
coordinates.
</li>
<li>
A Catmull-Rom curve is defined by a start point, a tangent that for that starting point, an end point, and a tangent for that end point,
plus a characteristic matrix that we can multiple by the point vector to get on-curve coordinates.
</li>
</ul>
<p>
Those are <em>very</em> similar, so let's see exactly <em>how</em> similar they are. We've already see the matrix form for Bézier curves,
so how different is the matrix form for Catmull-Rom curves?:
</p>
<!--
┌ V ┐
│ 1 │
┌ 1 0 0 0 ┐ │ V │
CatmullRom(t) = ┌ 2 3 ┐ · │ 0 0 1 0 │ · │ 2 │
└ 1 t t t ┘ │ -3 3 -2 -1 │ │ V' │
└ 2 -2 1 1 ┘ │ 1 │
│ V' │
└ 2 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/98ddf6415bd9827a6d899b21d0a5f736.svg"
width="409px"
height="75px"
loading="lazy"
/>
<p>
That's pretty dang similar. So the question is: how can we convert that expression with Catmull-Rom matrix and vector into an expression
of the Bézier matrix and vector? The short answer is of course "by using linear algebra", but the longer answer is the rest of this
section, and involves some maths that you may not even care for: if you just want to know the (incredibly simple) conversions between the
two curve forms, feel free to skip to the end of the following explanation, but if you want to <em>how</em> we can get one from the
other... let's get mathing!
</p>
<div class="note">
<h2>Deriving the conversion formulae</h2>
<p>
In order to convert between Catmull-Rom curves and Bézier curves, we need to know two things. Firstly, how to express the Catmull-Rom
curve using a "set of four coordinates", rather than a mix of coordinates and tangents, and secondly, how to convert those Catmull-Rom
coordinates to and from Bézier form.
</p>
<p>
We start with the first part, to figure out how we can go from Catmull-Rom <strong>V</strong> coordinates to Bézier
<strong>P</strong> coordinates, by applying "some matrix <strong>T</strong>". We don't know what that <strong>T</strong> is yet, but
we'll get to that:
</p>
<!--
┌ P ┐
│ 2 │
┌ V ┐ ┌ P ┐ │ P │
│ 1 │ │ 1 │ │ 3 │
│ V │ │ P │ │ P - P │
│ 2 │ = T · │ 2 │ = │ 3 1 │
│ V' │ │ P │ │ ─────── │
│ 1 │ │ 3 │ │ 2 │
│ V' │ │ P │ │ P - P │
└ 2 ┘ └ 4 ┘ │ 4 2 │
│ ─────── │
└ 2 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/b94a4dafc12ba7e4fbf3aff924f55464.svg"
width="187px"
height="83px"
loading="lazy"
/>
<p>
So, this mapping says that in order to map a Catmull-Rom "point + tangent" vector to something based on an "all coordinates" vector, we
need to determine the mapping matrix such that applying <em>T</em> yields P2 as start point, P3 as end point, and two tangents based on
the lines between P1 and P3, and P2 nd P4, respectively.
</p>
<p>Computing <em>T</em> is really more "arranging the numbers":</p>
<!--
┌ P ┐
│ 2 │
┌ P ┐ │ P │ ┌ 0 · P1 + 1 · P2 + 0 · P3 + 0 · P4 ┐ ┌ 0 1 0 0 ┐ ┌ P ┐
│ 1 │ │ 3 │ │ 0 · P1 + 0 · P2 + 1 · P3 + 0 · P4 │ │ 0 0 1 0 │ │ 1 │
│ P │ │ P - P │ │ -1 1 │ │ -1 1 │ │ P │
T · │ 2 │ = │ 3 1 │ = │ ── · P1 + 0 · P2 + ─ · P3 + 0 · P4 │ = │ ── 0 ─ 0 │ · │ 2 │
│ P │ │ ─────── │ │ 2 2 │ │ 2 2 │ │ P │
│ 3 │ │ 2 │ │ -1 1 │ │ -1 1 │ │ 3 │
│ P │ │ P - P │ │ 0 · P1 ── · P2 + 0 · P3 + ─ · P4 │ │ 0 ── 0 ─ │ │ P │
└ 4 ┘ │ 4 2 │ └ 2 2 ┘ └ 2 2 ┘ └ 4 ┘
│ ─────── │
└ 2 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/a55773fdcdfd99947acc4f86ad2d4a3d.svg"
width="591px"
height="83px"
loading="lazy"
/>
<p>Thus:</p>
<!--
┌ 0 1 0 0 ┐
│ 0 0 1 0 │
│ -1 1 │
T = │ ── 0 ─ 0 │
│ 2 2 │
│ -1 1 │
│ 0 ── 0 ─ │
└ 2 2 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/7bab9dd3da654b05fa065076894e2d82.svg"
width="143px"
height="81px"
loading="lazy"
/>
<p>
However, we're not <em>quite</em> done, because Catmull-Rom curves have that "tension" parameter, written as τ (a lowercase"tau"), which
is a scaling factor for the tangent vectors: the bigger the tension, the smaller the tangents, and the smaller the tension, the bigger
the tangents. As such, the tension factor goes in the denominator for the tangents, and before we continue, let's add that tension
factor into both our coordinate vector representation, and mapping matrix <em>T</em>:
</p>
<!--
┌ P ┐
│ 2 │
┌ V ┐ │ P │ ┌ 0 1 0 0 ┐
│ 1 │ │ 3 │ │ 0 0 1 0 │
│ V │ │ P - P │ │ -1 1 │
│ 2 │ = │ 3 1 │ ,\ T = │ ── 0 ── 0 │
│ V' │ │ ─────── │ │ 2τ 2τ │
│ 1 │ │ 2τ │ │ -1 1 │
│ V' │ │ P - P │ │ 0 ── 0 ── │
└ 2 ┘ │ 4 2 │ └ 2τ 2τ ┘
│ ─────── │
└ 2τ ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/8de53f207d68b25854a5f0b924ac6010.svg"
width="285px"
height="84px"
loading="lazy"
/>
<p>
With the mapping matrix properly done, let's rewrite the "point + tangent" Catmull-Rom matrix form to a matrix form in terms of four
coordinates, and see what we end up with:
</p>
<!--
┌ V ┐
│ 1 │
┌ 1 0 0 0 ┐ │ V │
CatmullRom(t) = ┌ 2 3 ┐ · │ 0 0 1 0 │ · │ 2 │
└ 1 t t t ┘ │ -3 3 -2 -1 │ │ V' │
└ 2 -2 1 1 ┘ │ 1 │
│ V' │
└ 2 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/902c290a790b4d44d10236f4a1456cdc.svg"
width="409px"
height="75px"
loading="lazy"
/>
<p>Replace point/tangent vector with the expression for all-coordinates:</p>
<!--
┌ 0 1 0 0 ┐ ┌ P ┐
│ 0 0 1 0 │ │ 1 │
┌ 1 0 0 0 ┐ │ -1 1 │ │ P │
CatmullRom(t) = ┌ 2 3 ┐ · │ 0 0 1 0 │ · │ ── 0 ── 0 │ · │ 2 │
└ 1 t t t ┘ │ -3 3 -2 -1 │ │ 2τ 2τ │ │ P │
└ 2 -2 1 1 ┘ │ -1 1 │ │ 3 │
│ 0 ── 0 ── │ │ P │
└ 2τ 2τ ┘ └ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/032409c03915a6ba75864e1dceae416d.svg"
width="549px"
height="81px"
loading="lazy"
/>
<p>and merge the matrices:</p>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │ ┌ P ┐
│ ── 0 ── 0 │ │ 1 │
│ 2τ 2τ │ │ P │
CatmullRom(t) = ┌ 2 3 ┐ · │ 1 1 1 -1 │ · │ 2 │
└ 1 t t t ┘ │ ─ ── - 3 3 - ─ ── │ │ P │
│ τ 2t t 2t │ │ 3 │
│ -1 1 1 1 │ │ P │
│ ── 2 - ── ── - 2 ── │ └ 4 ┘
└ 2t 2τ 2τ 2t ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/e653724c11600cbf682f1c809c8c6508.svg"
width="455px"
height="84px"
loading="lazy"
/>
<p>This looks a lot like the Bézier matrix form, which as we saw in the chapter on Bézier curves, should look like this:</p>
<!--
┌ P ┐
│ 1 │
┌ 1 0 0 0 ┐ │ P │
Bézier(t) = ┌ 2 3 ┐ · │ -3 3 0 0 │ · │ 2 │
└ 1 t t t ┘ │ 3 -6 3 0 │ │ P │
└ -1 3 -3 1 ┘ │ 3 │
│ P │
└ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/389a1ea8c9e92df9a2b38718e34bae7b.svg"
width="353px"
height="73px"
loading="lazy"
/>
<p>So, if we want to express a Catmull-Rom curve using a Bézier curve, we'll need to turn this Catmull-Rom bit:</p>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │ ┌ P ┐
│ ── 0 ── 0 │ │ 1 │
│ 2τ 2τ │ │ P │
│ 1 1 1 -1 │ · │ 2 │
│ ─ ── - 3 3 - ─ ── │ │ P │
│ τ 2t t 2t │ │ 3 │
│ -1 1 1 1 │ │ P │
│ ── 2 - ── ── - 2 ── │ └ 4 ┘
└ 2t 2τ 2τ 2t ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/574bed6665be06b309b8da722c616a41.svg"
width="227px"
height="84px"
loading="lazy"
/>
<p>Into something that looks like this:</p>
<!--
┌ P ┐
│ 1 │
┌ 1 0 0 0 ┐ │ P │
│ -3 3 0 0 │ · │ 2 │
│ 3 -6 3 0 │ │ P │
└ -1 3 -3 1 ┘ │ 3 │
│ P │
└ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/fe2e6fd487df224b2f55a601898ce333.svg"
width="167px"
height="73px"
loading="lazy"
/>
<p>And the way we do that is with a fairly straight forward bit of matrix rewriting. We start with the equality we need to ensure:</p>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │ ┌ P ┐ ┌ P ┐
│ ── 0 ── 0 │ │ 1 │ │ 1 │
│ 2τ 2τ │ │ P │ ┌ 1 0 0 0 ┐ │ P │
│ 1 1 1 -1 │ · │ 2 │ = │ -3 3 0 0 │ · V · │ 2 │
│ ─ ── - 3 3 - ─ ── │ │ P │ │ 3 -6 3 0 │ │ P │
│ τ 2t t 2t │ │ 3 │ └ -1 3 -3 1 ┘ │ 3 │
│ -1 1 1 1 │ │ P │ │ P │
│ ── 2 - ── ── - 2 ── │ └ 4 ┘ └ 4 ┘
└ 2t 2τ 2τ 2t ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/b59ff8d654e65df4c874901983208893.svg"
width="440px"
height="84px"
loading="lazy"
/>
<p>Then we remove the coordinate vector from both sides without affecting the equality:</p>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │
│ ── 0 ── 0 │
│ 2τ 2τ │ ┌ 1 0 0 0 ┐
│ 1 1 1 -1 │ = │ -3 3 0 0 │ · V
│ ─ ── - 3 3 - ─ ── │ │ 3 -6 3 0 │
│ τ 2t t 2t │ └ -1 3 -3 1 ┘
│ -1 1 1 1 │
│ ── 2 - ── ── - 2 ── │
└ 2t 2τ 2τ 2t ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/65e589eafae8ff2f39392d8143d2845c.svg"
width="353px"
height="84px"
loading="lazy"
/>
<p>Then we can "get rid of" the Bézier matrix on the right by left-multiply both with the inverse of the Bézier matrix:</p>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │
│ ── 0 ── 0 │
┌ 1 0 0 0 ┐ -1 │ 2τ 2τ │ ┌ 1 0 0 0 ┐ -1 ┌ 1 0 0 0 ┐
│ -3 3 0 0 │ · │ 1 1 1 -1 │ = │ -3 3 0 0 │ · │ -3 3 0 0 │ · V
│ 3 -6 3 0 │ │ ─ ── - 3 3 - ─ ── │ │ 3 -6 3 0 │ │ 3 -6 3 0 │
└ -1 3 -3 1 ┘ │ τ 2t t 2t │ └ -1 3 -3 1 ┘ └ -1 3 -3 1 ┘
│ -1 1 1 1 │
│ ── 2 - ── ── - 2 ── │
└ 2t 2τ 2τ 2t ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/917b176a45959b026c56f81999505dc7.svg"
width="657px"
height="88px"
loading="lazy"
/>
<p>
A matrix times its inverse is the matrix equivalent of 1, and because "something times 1" is the same as "something", so we can just
outright remove any matrix/inverse pair:
</p>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │
│ ── 0 ── 0 │
┌ 1 0 0 0 ┐ -1 │ 2τ 2τ │
│ -3 3 0 0 │ · │ 1 1 1 -1 │ = V
│ 3 -6 3 0 │ │ ─ ── - 3 3 - ─ ── │
└ -1 3 -3 1 ┘ │ τ 2t t 2t │
│ -1 1 1 1 │
│ ── 2 - ── ── - 2 ── │
└ 2t 2τ 2τ 2t ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/4e0da16710a7339f04dd844c7705423e.svg"
width="369px"
height="88px"
loading="lazy"
/>
<p>And now we're <em>basically</em> done. We just multiply those two matrices and we know what <em>V</em> is:</p>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │
│ ── 1 ── 0 │
│ 6τ 6τ │ = V
│ 1 -1 │
│ 0 ── 1 ── │
│ 6τ 6τ │
└ 0 0 1 0 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/1cef6bbf7b3d10e8c0aaecfac816cc86.svg"
width="161px"
height="77px"
loading="lazy"
/>
<p>We now have the final piece of our function puzzle. Let's run through each step.</p>
<ol>
<li>Start with the Catmull-Rom function:</li>
</ol>
<!--
┌ V ┐
│ 1 │
┌ 1 0 0 0 ┐ │ V │
CatmullRom(t) = ┌ 2 3 ┐ · │ 0 0 1 0 │ · │ 2 │
└ 1 t t t ┘ │ -3 3 -2 -1 │ │ V' │
└ 2 -2 1 1 ┘ │ 1 │
│ V' │
└ 2 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/902c290a790b4d44d10236f4a1456cdc.svg"
width="409px"
height="75px"
loading="lazy"
/>
<ol start="2">
<li>rewrite to pure coordinate form:</li>
</ol>
<!--
┌ P ┐
│ 2 │
│ P │
│ 3 │
┌ 1 0 0 0 ┐ │ P - P │
= ┌ 2 3 ┐ · │ 0 0 1 0 │ · │ 3 1 │
└ 1 t t t ┘ │ -3 3 -2 -1 │ │ ─────── │
└ 2 -2 1 1 ┘ │ 2τ │
│ P - P │
│ 4 2 │
│ ─────── │
└ 2τ ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/211dadbb9d0f6b2e381f18ea3c4d12fb.svg"
width="324px"
height="84px"
loading="lazy"
/>
<ol start="3">
<li>rewrite for "normal" coordinate vector:</li>
</ol>
<!--
┌ 0 1 0 0 ┐ ┌ P ┐
│ 0 0 1 0 │ │ 1 │
┌ 1 0 0 0 ┐ │ -1 1 │ │ P │
= ┌ 2 3 ┐ · │ 0 0 1 0 │ · │ ── 0 ── 0 │ · │ 2 │
└ 1 t t t ┘ │ -3 3 -2 -1 │ │ 2τ 2τ │ │ P │
└ 2 -2 1 1 ┘ │ -1 1 │ │ 3 │
│ 0 ── 0 ── │ │ P │
└ 2τ 2τ ┘ └ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/a8158b35ec221cccff51a53cdc7f440b.svg"
width="441px"
height="81px"
loading="lazy"
/>
<ol start="4">
<li>merge the inner matrices:</li>
</ol>
<!--
┌ 0 1 0 0 ┐
│ -1 1 │ ┌ P ┐
│ ── 0 ── 0 │ │ 1 │
│ 2τ 2τ │ │ P │
= ┌ 2 3 ┐ · │ 1 1 1 -1 │ · │ 2 │
└ 1 t t t ┘ │ ─ ── - 3 3 - ─ ── │ │ P │
│ τ 2t t 2t │ │ 3 │
│ -1 1 1 1 │ │ P │
│ ── 2 - ── ── - 2 ── │ └ 4 ┘
└ 2t 2τ 2τ 2t ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/53f216327c0bbcf02b2a331fbf44d389.svg"
width="348px"
height="84px"
loading="lazy"
/>
<ol start="5">
<li>rewrite for Bézier matrix form:</li>
</ol>
<!--
┌ 0 1 0 0 ┐ ┌ P ┐
│ -1 1 │ │ 1 │
┌ 1 0 0 0 ┐ │ ── 1 ── 0 │ │ P │
= ┌ 2 3 ┐ · │ -3 3 0 0 │ · │ 6τ 6τ │ · │ 2 │
└ 1 t t t ┘ │ 3 -6 3 0 │ │ 1 -1 │ │ P │
└ -1 3 -3 1 ┘ │ 0 ── 1 ── │ │ 3 │
│ 6τ 6τ │ │ P │
└ 0 0 1 0 ┘ └ 4 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/4c8684109149b0dc79f5583a5912fcd9.svg"
width="431px"
height="77px"
loading="lazy"
/>
<ol start="6">
<li>and transform the coordinates so we have a "pure" Bézier expression:</li>
</ol>
<!--
┌ P ┐
│ 2 │
│ P -P │
│ 3 1 │
┌ 1 0 0 0 ┐ │ P + ────── │
= ┌ 2 3 ┐ · │ -3 3 0 0 │ · │ 2 6 · τ │
└ 1 t t t ┘ │ 3 -6 3 0 │ │ P -P │
└ -1 3 -3 1 ┘ │ 4 2 │
│ P - ────── │
│ 3 6 · τ │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/157b287d6b74109d8c8b634990ea6549.svg"
width="348px"
height="81px"
loading="lazy"
/>
<p>And we're done: we finally know how to convert these two curves!</p>
</div>
<p>
If we have a Catmull-Rom curve defined by four coordinates P<sub>1</sub> through P<sub>4</sub>, then we can draw that curve using a Bézier
curve that has the vector:
</p>
<!--
┌ P ┐
│ 2 │
┌ P ┐ │ P -P │
│ 1 │ │ 3 1 │
│ P │ │ P + ────── │
│ 2 │ ==> │ 2 6 · τ │
│ P │ CatmullRom │ P -P │ Bézier
│ 3 │ │ 4 2 │
│ P │ │ P - ────── │
└ 4 ┘ │ 3 6 · τ │
│ P │
└ 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/a323848e706c473833cda0b02bc220ef.svg"
width="241px"
height="85px"
loading="lazy"
/>
<p>
Similarly, if we have a Bézier curve defined by four coordinates P<sub>1</sub> through P<sub>4</sub>, we can draw that using a standard
tension Catmull-Rom curve with the following coordinate values:
</p>
<!--
┌ P ┐ ┌ P ┐
│ 1 │ │ 1 │
│ P │ │ P │
│ 2 │ ==> │ 4 │
│ P │ Bézier │ P + 3(P - P ) │ CatmullRom
│ 3 │ │ 4 1 2 │
│ P │ │ P + 3(P - P ) │
└ 4 ┘ └ 1 4 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/9ae99b090883023a485be7be098858e9.svg"
width="276px"
height="77px"
loading="lazy"
/>
<p>Or, if your API allows you to specify Catmull-Rom curves using plain coordinates:</p>
<!--
┌ P ┐ ┌ P + 6(P - P ) ┐
│ 1 │ │ 4 1 2 │
│ P │ │ P │
│ 2 │ ==> │ 1 │
│ P │ Bézier │ P │ CatmullRom
│ 3 │ │ 4 │
│ P │ │ P + 6(P - P ) │
└ 4 ┘ └ 1 4 3 ┘
-->
<img
class="LaTeX SVG"
src="./images/chapters/catmullconv/012a8ab7a4de935c1c8d61dcd14fc62c.svg"
width="276px"
height="77px"
loading="lazy"
/>
</section>
<section id="catmullfitting">
<h1>
<div class="nav">
<a href="ko-KR/index.html#catmullconv">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#polybezier">다음</a>
</div>
<a href="ko-KR/index.html#catmullfitting">Creating a Catmull-Rom curve from three points</a>
</h1>
<p>
Much shorter than the previous section: we saw that Catmull-Rom curves need at least 4 points to draw anything sensible, so how do we
create a Catmull-Rom curve from three points?
</p>
<p>Short and sweet: we don't.</p>
<p>
We run through the maths that lets us <a href="#pointcurves">create a cubic Bézier curve</a>, and then convert its coordinates to
Catmull-Rom form using the conversion formulae we saw above.
</p>
</section>
<section id="polybezier">
<h1>
<div class="nav">
<a href="ko-KR/index.html#catmullfitting">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#offsetting">다음</a>
</div>
<a href="ko-KR/index.html#polybezier">Forming poly-Bézier curves</a>
</h1>
<p>
Much like lines can be chained together to form polygons, Bézier curves can be chained together to form poly-Béziers, and the only trick
required is to make sure that:
</p>
<ol>
<li>the end point of each section is the starting point of the following section, and</li>
<li>the derivatives across that dual point line up.</li>
</ol>
<p>Unless you want sharp corners, of course. Then you don't even need 2.</p>
<p>
We'll cover three forms of poly-Bézier curves in this section. First, we'll look at the kind that just follows point 1. where the end
point of a segment is the same point as the start point of the next segment. This leads to poly-Béziers that are pretty hard to work with,
but they're the easiest to implement:
</p>
<graphics-element
title="Unlinked quadratic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="quadratic"
data-link="coordinate"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/41522a397171423e8a465dc8c74f6e87.png" loading="lazy" />
<label>Unlinked quadratic poly-Bézier</label>
</fallback-image></graphics-element
>
<graphics-element
title="Unlinked cubic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="cubic"
data-link="coordinate"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/6fb33629373d7a731b6ac3f5365cb9f0.png" loading="lazy" />
<label>Unlinked cubic poly-Bézier</label>
</fallback-image></graphics-element
>
<p>
Dragging the control points around only affects the curve segments that the control point belongs to, and moving an on-curve point leaves
the control points where they are, which is not the most useful for practical modelling purposes. So, let's add in the logic we need to
make things a little better. We'll start by linking up control points by ensuring that the "incoming" derivative at an on-curve point is
the same as it's "outgoing" derivative:
</p>
<!--
B'(1) = B'(0)
n n+1
-->
<img class="LaTeX SVG" src="./images/chapters/polybezier/a37252ff55837b918d9d64078ae92ae7.svg" width="124px" height="17px" loading="lazy" />
<p>
We can effect this quite easily, because we know that the vector from a curve's last control point to its last on-curve point is equal to
the derivative vector. If we want to ensure that the first control point of the next curve matches that, all we have to do is mirror that
last control point through the last on-curve point. And mirroring any point A through any point B is really simple:
</p>
<!--
┌ B + (B - A ) ┐ ┌ 2B - A ┐
Mirrored = │ x x x │ = │ x x │
│ B + (B - A ) │ │ 2B - A │
└ y y y ┘ └ y y ┘
-->
<img class="LaTeX SVG" src="./images/chapters/polybezier/2249056953a47ab1944bb5a41dcbed8c.svg" width="301px" height="40px" loading="lazy" />
<p>
So let's implement that and see what it gets us. The following two graphics show a quadratic and a cubic poly-Bézier curve again, but this
time moving the control points around moves others, too. However, you might see something unexpected going on for quadratic curves...
</p>
<graphics-element
title="Connected quadratic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="quadratic"
data-link="derivative"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/db04f805f42bdc9a1b7ec4d6b401d853.png" loading="lazy" />
<label>Connected quadratic poly-Bézier</label>
</fallback-image></graphics-element
>
<graphics-element
title="Connected cubic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="cubic"
data-link="derivative"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/fe41b628f46f7035d151a8210d30111f.png" loading="lazy" />
<label>Connected cubic poly-Bézier</label>
</fallback-image></graphics-element
>
<p>
As you can see, quadratic curves are particularly ill-suited for poly-Bézier curves, as all the control points are effectively linked.
Move one of them, and you move all of them. Not only that, but if we move the on-curve points, it's possible to get a situation where a
control point cannot satisfy the constraint that it's the reflection of its two neighbouring control points... This means that we cannot
use quadratic poly-Béziers for anything other than really, really simple shapes. And even then, they're probably the wrong choice. Cubic
curves are pretty decent, but the fact that the derivatives are linked means we can't manipulate curves as well as we might if we relaxed
the constraints a little.
</p>
<p>So: let's relax the requirement a little.</p>
<p>
We can change the constraint so that we still preserve the <em>angle</em> of the derivatives across sections (so transitions from one
section to the next will still look natural), but give up the requirement that they should also have the same <em>vector length</em>.
Doing so will give us a much more useful kind of poly-Bézier curve:
</p>
<graphics-element
title="Angularly connected quadratic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="quadratic"
data-link="direction"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/777f3814965c39ec3cdbb13eab0c4eeb.png" loading="lazy" />
<label>Angularly connected quadratic poly-Bézier</label>
</fallback-image></graphics-element
>
<graphics-element
title="Angularly connected cubic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="cubic"
data-link="direction"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/f6c55cbc66333b6630939f67fc20e086.png" loading="lazy" />
<label>Angularly connected cubic poly-Bézier</label>
</fallback-image></graphics-element
>
<p>
Cubic curves are now better behaved when it comes to dragging control points around, but the quadratic poly-Bézier still has the problem
that moving one control points will move the control points and may ending up defining "the next" control point in a way that doesn't
work. Quadratic curves really aren't very useful to work with...
</p>
<p>
Finally, we also want to make sure that moving the on-curve coordinates preserves the relative positions of the associated control points.
With that, we get to the kind of curve control that you might be familiar with from applications like Photoshop, Inkscape, Blender, etc.
</p>
<graphics-element
title="Standard connected quadratic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="quadratic"
data-link="conventional"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/b3aebf7803f4430187c249a891095062.png" loading="lazy" />
<label>Standard connected quadratic poly-Bézier</label>
</fallback-image></graphics-element
>
<graphics-element
title="Standard connected cubic poly-Bézier"
width="275"
height="275"
src="./chapters/polybezier/poly.js"
data-type="cubic"
data-link="conventional"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/polybezier/1b94c6ada011bd8e50330e31a851a62e.png" loading="lazy" />
<label>Standard connected cubic poly-Bézier</label>
</fallback-image></graphics-element
>
<p>
Again, we see that cubic curves are now rather nice to work with, but quadratic curves have a new, very serious problem: we can move an
on-curve point in such a way that we can't compute what needs to "happen next". Move the top point down, below the left and right points,
for instance. There is no way to preserve correct control points without a kink at the bottom point. Quadratic curves: just not that
good...
</p>
<p>
A final improvement is to offer fine-level control over which points behave which, so that you can have "kinks" or individually controlled
segments when you need them, with nicely well-behaved curves for the rest of the path. Implementing that, is left as an exercise for the
reader.
</p>
</section>
<section id="offsetting">
<h1>
<div class="nav">
<a href="ko-KR/index.html#polybezier">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#graduatedoffset">다음</a>
</div>
<a href="ko-KR/index.html#offsetting">Curve offsetting</a>
</h1>
<p>
Perhaps you're like me, and you've been writing various small programs that use Bézier curves in some way or another, and at some point
you make the step to implementing path extrusion. But you don't want to do it pixel based; you want to stay in the vector world. You find
that extruding lines is relatively easy, and tracing outlines is coming along nicely (although junction caps and fillets are a bit of a
hassle), and then you decide to do things properly and add Bézier curves to the mix. Now you have a problem.
</p>
<p>
Unlike lines, you can't simply extrude a Bézier curve by taking a copy and moving it around, because of the curvatures; rather than a
uniform thickness, you get an extrusion that looks too thin in places, if you're lucky, but more likely will self-intersect. The trick,
then, is to scale the curve, rather than simply copying it. But how do you scale a Bézier curve?
</p>
<p>
Bottom line: <strong>you can't</strong>. So you cheat. We're not going to do true curve scaling, or rather curve offsetting, because
that's impossible. Instead we're going to try to generate 'looks good enough' offset curves.
</p>
<div class="note">
<h3>"What do you mean, you can't? Prove it."</h3>
<p>
First off, when I say "you can't," what I really mean is "you can't offset a Bézier curve with another Bézier curve", not even by using
a really high order curve. You can find the function that describes the offset curve, but it won't be a polynomial, and as such it
cannot be represented as a Bézier curve, which <strong>has</strong> to be a polynomial. Let's look at why this is:
</p>
<p>
From a mathematical point of view, an offset curve <code>O(t)</code> is a curve such that, given our original curve <code>B(t)</code>,
any point on <code>O(t)</code> is a fixed distance <code>d</code> away from coordinate <code>B(t)</code>. So let's math that:
</p>
<!--
O(t) = B(t) + d
-->
<img
class="LaTeX SVG"
src="./images/chapters/offsetting/3c80407cfd0bd8c8ebea239272aeabe5.svg"
width="108px"
height="16px"
loading="lazy"
/>
<p>
However, we're working in 2D, and <code>d</code> is a single value, so we want to turn it into a vector. If we want a point distance
<code>d</code> "away" from the curve <code>B(t)</code> then what we really mean is that we want a point at <code>d</code> times the
"normal vector" from point <code>B(t)</code>, where the "normal" is a vector that runs perpendicular ("at a right angle") to the tangent
at <code>B(t)</code>. Easy enough:
</p>
<!--
O(t) = B(t) + d · N(t)
-->
<img
class="LaTeX SVG"
src="./images/chapters/offsetting/de8cdb128273beff2d98534b9f090b85.svg"
width="151px"
height="16px"
loading="lazy"
/>
<p>
Now this still isn't very useful unless we know what the formula for <code>N(t)</code> is, so let's find out. <code>N(t)</code> runs
perpendicular to the original curve tangent, and we know that the tangent is simply <code>B'(t)</code>, so we could just rotate that 90
degrees and be done with it. However, we need to ensure that <code>N(t)</code> has the same magnitude for every <code>t</code>, or the
offset curve won't be at a uniform distance, thus not being an offset curve at all. The easiest way to guarantee this is to make sure
<code>N(t)</code> always has length 1, which we can achieve by dividing <code>B'(t)</code> by its magnitude:
</p>
<!--
╭ B'(t) ╮
N(t) = \bot │ ────────── │
╰ || B'(t)|| ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/offsetting/e7e8e1f5727387079dbc5770181187c2.svg"
width="141px"
height="40px"
loading="lazy"
/>
<p>The magnitude of <code>B'(t)</code>, usually denoted with double vertical bars, is given by the following formula:</p>
<!--
┌─────────────────┐
│ 2 2
|| B'(t)|| = │B '(t) + B '(t)
⟍│ x y
-->
<img
class="LaTeX SVG"
src="./images/chapters/offsetting/7fd3e895c0eea0965470dd619450b679.svg"
width="188px"
height="25px"
loading="lazy"
/>
<p>
And that's where things go wrong: that square root is screwing everything up, because it turns our nice polynomials into things that are
no longer polynomials.
</p>
<p>
There is a small class of polynomials where the square root is also a polynomial, but they're utterly useless to us: any polynomial with
unweighted binomial coefficients has a square root that is also a polynomial. Now, you might think that Bézier curves are just fine
because they do, but they don't; remember that only the <strong>base</strong> function has binomial coefficients. That's before we
factor in our coordinates, which turn it into a non-binomial polygon. The only way to make sure the functions stay binomial is to make
all our coordinates have the same value. And that's not a curve, that's a point. We can already create offset curves for points, we call
them circles, and they have much simpler functions than Bézier curves.
</p>
<p>
So, since the tangent length isn't a polynomial, the normalised tangent won't be a polynomial either, which means
<code>N(t)</code> won't be a polynomial, which means that <code>d</code> times <code>N(t)</code> won't be a polynomial, which means
that, ultimately, <code>O(t)</code> won't be a polynomial, which means that even if we can determine the function for
<code>O(t)</code> just fine (and that's far from trivial!), it simply cannot be represented as a Bézier curve.
</p>
<p>
And that's one reason why Bézier curves are tricky: there are actually a <em>lot</em> of curves that cannot be represented as a Bézier
curve at all. They can't even model their own offset curves. They're weird that way. So how do all those other programs do it? Well,
much like we're about to do, they cheat. We're going to approximate an offset curve in a way that will look relatively close to what the
real offset curve would look like, if we could compute it.
</p>
</div>
<p>
So, you cannot offset a Bézier curve perfectly with another Bézier curve, no matter how high-order you make that other Bézier curve.
However, we can chop up a curve into "safe" sub-curves (where "safe" means that all the control points are always on a single side of the
baseline, and the midpoint of the curve at <code>t=0.5</code> is roughly in the center of the polygon defined by the curve coordinates)
and then point-scale each sub-curve with respect to its scaling origin (which is the intersection of the point normals at the start and
end points).
</p>
<p>
A good way to do this reduction is to first find the curve's extreme points, as explained in the earlier section on curve extremities, and
use these as initial splitting points. After this initial split, we can check each individual segment to see if it's "safe enough" based
on where the center of the curve is. If the on-curve point for <code>t=0.5</code> is too far off from the center, we simply split the
segment down the middle. Generally this is more than enough to end up with safe segments.
</p>
<p>
The following graphics show off curve offsetting, and you can use the slider to control the distance at which the curve gets offset. The
curve first gets reduced to safe segments, each of which is then offset at the desired distance. Especially for simple curves,
particularly easily set up for quadratic curves, no reduction is necessary, but the more twisty the curve gets, the more the curve needs
to be reduced in order to get segments that can safely be scaled.
</p>
<graphics-element
title="Offsetting a quadratic Bézier curve"
width="275"
height="275"
src="./chapters/offsetting/offsetting.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/offsetting/03b8e0849e7c8ba64d8c076f47fe2ec7.png" loading="lazy" />
<label>Offsetting a quadratic Bézier curve</label>
</fallback-image>
<input type="range" min="5" max="50" step="1" value="20" class="slide-control" />
</graphics-element>
<graphics-element
title="Offsetting a cubic Bézier curve"
width="275"
height="275"
src="./chapters/offsetting/offsetting.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/offsetting/4c4738b6bf9f83eded12d680a29e337b.png" loading="lazy" />
<label>Offsetting a cubic Bézier curve</label>
</fallback-image>
<input type="range" min="5" max="50" step="1" value="20" class="slide-control" />
</graphics-element>
<p>
You may notice that this may still lead to small 'jumps' in the sub-curves when moving the curve around. This is caused by the fact that
we're still performing a naive form of offsetting, moving the control points the same distance as the start and end points. If the curve
is large enough, this may still lead to incorrect offsets.
</p>
</section>
<section id="graduatedoffset">
<h1>
<div class="nav"><a href="ko-KR/index.html#offsetting">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#circles">다음</a></div>
<a href="ko-KR/index.html#graduatedoffset">Graduated curve offsetting</a>
</h1>
<p>
What if we want to do graduated offsetting, starting at some distance <code>s</code> but ending at some other distance <code>e</code>?
Well, if we can compute the length of a curve (which we can if we use the Legendre-Gauss quadrature approach) then we can also determine
how far "along the line" any point on the curve is. With that knowledge, we can offset a curve so that its offset curve is not uniformly
wide, but graduated between with two different offset widths at the start and end.
</p>
<p>
Like normal offsetting we cut up our curve in sub-curves, and then check at which distance along the original curve each sub-curve starts
and ends, as well as to which point on the curve each of the control points map. This gives us the distance-along-the-curve for each
interesting point in the sub-curve. If we call the total length of all sub-curves seen prior to seeing "the current" sub-curve
<code>S</code> (and if the current sub-curve is the first one, <code>S</code> is zero), and we call the full length of our original curve
<code>L</code>, then we get the following graduation values:
</p>
<ul>
<li>start: map <code>S</code> from interval (<code>0,L</code>) to interval <code>(s,e)</code></li>
<li>c1: <code>map(&lt;strong&gt;S+d1&lt;/strong&gt;, 0,L, s,e)</code>, d1 = distance along curve to projection of c1</li>
<li>c2: <code>map(&lt;strong&gt;S+d2&lt;/strong&gt;, 0,L, s,e)</code>, d2 = distance along curve to projection of c2</li>
<li>...</li>
<li>end: <code>map(&lt;strong&gt;S+length(subcurve)&lt;/strong&gt;, 0,L, s,e)</code></li>
</ul>
<p>
At each of the relevant points (start, end, and the projections of the control points onto the curve) we know the curve's normal, so
offsetting is simply a matter of taking our original point, and moving it along the normal vector by the offset distance for each point.
Doing so will give us the following result (these have with a starting width of 0, and an end width of 40 pixels, but can be controlled
with your up and down arrow keys):
</p>
<graphics-element
title="Offsetting a quadratic Bézier curve"
width="275"
height="275"
src="./chapters/graduatedoffset/offsetting.js"
data-type="quadratic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/graduatedoffset/8cc724d5343c65685d88c92b2d069a2a.png" loading="lazy" />
<label>Offsetting a quadratic Bézier curve</label>
</fallback-image>
<input type="range" min="5" max="50" step="1" value="20" class="slide-control" />
</graphics-element>
<graphics-element
title="Offsetting a cubic Bézier curve"
width="275"
height="275"
src="./chapters/graduatedoffset/offsetting.js"
data-type="cubic"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/graduatedoffset/17bf62e05a1fc3387b0c210f2decff45.png" loading="lazy" />
<label>Offsetting a cubic Bézier curve</label>
</fallback-image>
<input type="range" min="5" max="50" step="1" value="20" class="slide-control" />
</graphics-element>
</section>
<section id="circles">
<h1>
<div class="nav">
<a href="ko-KR/index.html#graduatedoffset">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#circles_cubic">다음</a>
</div>
<a href="ko-KR/index.html#circles">Circles and quadratic Bézier curves</a>
</h1>
<p>
Circles and Bézier curves are very different beasts, and circles are infinitely easier to work with than Bézier curves. Their formula is
much simpler, and they can be drawn more efficiently. But, sometimes you don't have the luxury of using circles, or ellipses, or arcs.
Sometimes, all you have are Bézier curves. For instance, if you're doing font design, fonts have no concept of geometric shapes, they only
know straight lines, and Bézier curves. OpenType fonts with TrueType outlines only know quadratic Bézier curves, and OpenType fonts with
Type 2 outlines only know cubic Bézier curves. So how do you draw a circle, or an ellipse, or an arc?
</p>
<p>You approximate.</p>
<p>
We already know that Bézier curves cannot model all curves that we can think of, and this includes perfect circles, as well as ellipses,
and their arc counterparts. However, we can certainly approximate them to a degree that is visually acceptable. Quadratic and cubic curves
offer us different curvature control, so in order to approximate a circle we will first need to figure out what the error is if we try to
approximate arcs of increasing degree with quadratic and cubic curves, and where the coordinates even lie.
</p>
<p>
Since arcs are mid-point-symmetrical, we need the control points to set up a symmetrical curve. For quadratic curves this means that the
control point will be somewhere on a line that intersects the baseline at a right angle. And we don't get any choice on where that will
be, since the derivatives at the start and end point have to line up, so our control point will lie at the intersection of the tangents at
the start and end point.
</p>
<p>
First, let's try to fit the quadratic curve onto a circular arc. In the following sketch you can move the mouse around over a unit circle,
to see how well, or poorly, a quadratic curve can approximate the arc from (1,0) to where your mouse cursor is:
</p>
<graphics-element
title="Quadratic Bézier arc approximation"
width="400"
height="400"
src="./chapters/circles/arc-approximation.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/circles/08ca09aacb271735e063e7e8d941a195.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="-3.1415" max="3.1415" step="0.01" value="-0.7854" class="slide-control" />
</graphics-element>
<p>
As you can see, things go horribly wrong quite quickly; even trying to approximate a quarter circle using a quadratic curve is a bad idea.
An eighth of a turns might look okay, but how okay is okay? Let's apply some maths and find out. What we're interested in is how far off
our on-curve coordinates are with respect to a circular arc, given a specific start and end angle. We'll be looking at how much space
there is between the circular arc, and the quadratic curve's midpoint.
</p>
<p>
We start out with our start and end point, and for convenience we will place them on a unit circle (a circle around 0,0 with radius 1), at
some angle <em>φ</em>:
</p>
<!--
S = \begin{pmatrix} 1
0 \end{pmatrix} , \ E = \begin{pmatrix} cos(φ)
sin(φ) \end{pmatrix}
-->
<img class="LaTeX SVG" src="./images/chapters/circles/7ab3da0922477af4cc09f5852100976b.svg" width="175px" height="40px" loading="lazy" />
<p>What we want to find is the intersection of the tangents, so we want a point C such that:</p>
<!--
C = S + a · \begin{pmatrix} 0
1 \end{pmatrix} , \ C = E + b · \begin{pmatrix} -sin(φ)
cos(φ) \end{pmatrix}
-->
<img class="LaTeX SVG" src="./images/chapters/circles/d4bfb47b623c968e3231566c9705c6c4.svg" width="284px" height="40px" loading="lazy" />
<p>
i.e. we want a point that lies on the vertical line through S (at some distance <em>a</em> from S) and also lies on the tangent line
through E (at some distance <em>b</em> from E). Solving this gives us:
</p>
<!--
╭ C = 1 = cos(φ) + b · -sin(φ)
╡ x
│ C = a = sin(φ) + b · cos(φ)
╰ y
-->
<img class="LaTeX SVG" src="./images/chapters/circles/8237af1396bb567d70c8b5e4dd7f8115.svg" width="219px" height="40px" loading="lazy" />
<p>First we solve for <em>b</em>:</p>
<!--
1 = cos(φ) + b · -sin(φ) → \ 1 - cos(φ) = -b · sin(φ) → \ -1 + cos(φ) = b · sin(φ)
-->
<img class="LaTeX SVG" src="./images/chapters/circles/1829e42ea956ee4df0e45d9ac5334ef7.svg" width="560px" height="17px" loading="lazy" />
<p>which yields:</p>
<!--
cos(φ)-1
b = ────────
sin(φ)
-->
<img class="LaTeX SVG" src="./images/chapters/circles/06369b00338310df0a810c592485aa0a.svg" width="101px" height="39px" loading="lazy" />
<p>which we can then substitute in the expression for <em>a</em>:</p>
<!--
a= sin(φ) + b · cos(φ)
-1 + cos(φ)
..= sin(φ) + ─────────── · cos(φ)
sin(φ)
2
-cos(φ) + cos (φ)
..= sin(φ) + ─────────────────
sin(φ)
2 2
sin (φ) + cos (φ) - cos(φ)
..= ──────────────────────────
sin(φ)
1 - cos(φ)
a= ──────────
sin(φ)
-->
<img class="LaTeX SVG" src="./images/chapters/circles/5a23f3bc298c85540c6dd18e304d9224.svg" width="231px" height="195px" loading="lazy" />
<p>
A quick check shows that plugging these values for <em>a</em> and <em>b</em> into the expressions for C<sub>x</sub> and C<sub>y</sub> give
the same x/y coordinates for both "<em>a</em> away from A" and "<em>b</em> away from B", so let's continue: now that we know the
coordinate values for C, we know where our on-curve point T for <em>t=0.5</em> (or angle φ/2) is, because we can just evaluate the Bézier
polynomial, and we know where the circle arc's actual point P is for angle φ/2:
</p>
<!--
φ φ
P = cos(─) , \ P = sin(─)
x 2 y 2
-->
<img class="LaTeX SVG" src="./images/chapters/circles/8b4e1d0a62380ed011f27c645ed13b28.svg" width="188px" height="32px" loading="lazy" />
<p>We compute T, observing that if <em>t=0.5</em>, the polynomial values (1-t)², 2(1-t)t, and t² are 0.25, 0.5, and 0.25 respectively:</p>
<!--
1 2 1 1
T = ─S + ─C + ─E = ─(S + 2C + E)
4 4 4 4
-->
<img class="LaTeX SVG" src="./images/chapters/circles/a04bd1558a76e60b8ca6e1fe4fa38c00.svg" width="252px" height="35px" loading="lazy" />
<p>Which, worked out for the x and y components, gives:</p>
<!--
╭ 1
│ T = ─(3 + cos(φ))
╡ x 4
│ 1╭ 2-2cos(φ) ╮ 1╭ ╭ φ ╮ ╮
│ T = ─│ ───────── + sin(φ) │ = ─│ 2tan│ ─ │ + sin(φ) │
╰ y 4╰ sin(φ) ╯ 4╰ ╰ 2 ╯ ╯
-->
<img class="LaTeX SVG" src="./images/chapters/circles/986ae9104e0bc52e95689ae7ae4504db.svg" width="408px" height="77px" loading="lazy" />
<p>And the distance between these two is the standard Euclidean distance:</p>
<!--
1 φ 4╭ φ ╮
d (φ)= T - P = ─(3 + cos(φ)) - cos(─) = 2sin │ ─ │ ,
x x x 4 2 ╰ 4 ╯
1╭ ╭ φ ╮ ╮ φ
d (φ)= T - P = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─) ,
y y y 4╰ ╰ 2 ╯ ╯ 2
⇓
┌───────┐ ┌───────┐
│ 2 2 4 φ │ 1
d(φ)= │d + d = ... = 2sin (─) │───────
⟍│ x y 4 │ 2 φ
│cos (─)
⟍│ 2
-->
<img class="LaTeX SVG" src="./images/chapters/circles/942c90bc8311e49d94059b3127fc78d5.svg" width="399px" height="153px" loading="lazy" />
<p>
So, what does this distance function look like when we plot it for a number of ranges for the angle φ, such as a half circle, quarter
circle and eighth circle?
</p>
<table>
<tbody>
<tr>
<td>
<img src="images/arc-q-pi.gif" height="190" />
plotted for 0 ≤ φ ≤ π:
</td>
<td>
<img src="images/arc-q-pi2.gif" height="187" />
plotted for 0 ≤ φ ≤ ½π:
</td>
<td>
<a
href="https://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi%2F4"
>
<img src="images/arc-q-pi4.gif" height="174" />
</a>
plotted for 0 ≤ φ ≤ ¼π:
</td>
</tr>
</tbody>
</table>
<p>
We now see why the eighth circle arc looks decent, but the quarter circle arc doesn't: an error of roughly 0.06 at <em>t=0.5</em> means
we're 6% off the mark... we will already be off by one pixel on a circle with pixel radius 17. Any decent sized quarter circle arc, say
with radius 100px, will be way off if approximated by a quadratic curve! For the eighth circle arc, however, the error is only roughly
0.003, or 0.3%, which explains why it looks so close to the actual eighth circle arc. In fact, if we want a truly tiny error, like 0.001,
we'll have to contend with an angle of (rounded) 0.593667, which equates to roughly 34 degrees. We'd need 11 quadratic curves to form a
full circle with that precision! (technically, 10 and ten seventeenth, but we can't do partial curves, so we have to round up). That's a
whole lot of curves just to get a shape that can be drawn using a simple function!
</p>
<p>
In fact, let's flip the function around, so that if we plug in the precision error, labelled ε, we get back the maximum angle for that
precision:
</p>
<!--
╭ ┌─────────────┐ ╮
│ │ ┌──────┐ │
│ ⟍│2+ε-⟍│ε(2+ε) │
φ = 4 · arccos │ ──────────────── │
│ ┌─┐ │
╰ ⟍│2 ╯
-->
<img class="LaTeX SVG" src="./images/chapters/circles/f9d15462df31186feef8c3d53c0f6163.svg" width="247px" height="53px" loading="lazy" />
<p>
And frankly, things are starting to look a bit ridiculous at this point, we're doing way more maths than we've ever done, but thankfully
this is as far as we need the maths to take us: If we plug in the precisions 0.1, 0.01, 0.001 and 0.0001 we get the radians values 1.748,
1.038, 0.594 and 0.3356; in degrees, that means we can cover roughly 100 degrees (requiring four curves), 59.5 degrees (requiring six
curves), 34 degrees (requiring 11 curves), and 19.2 degrees (requiring a whopping nineteen curves).
</p>
<p>
The bottom line? <strong>Quadratic curves are kind of lousy</strong> if you want circular (or elliptical, which are circles that have been
squashed in one dimension) curves. We can do better, even if it's just by raising the order of our curve once. So let's try the same thing
for cubic curves.
</p>
</section>
<section id="circles_cubic">
<h1>
<div class="nav">
<a href="ko-KR/index.html#circles">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#arcapproximation">다음</a>
</div>
<a href="ko-KR/index.html#circles_cubic">Circular arcs and cubic Béziers</a>
</h1>
<p>Let's look at approximating circles and circular arcs using cubic Béziers. How much better is that?</p>
<graphics-element
title="Cubic Bézier arc approximation"
width="400"
height="400"
src="./chapters/circles_cubic/arc-approximation.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/circles_cubic/891d50a946936c9701adc855de12623d.png" loading="lazy" />
<label></label>
</fallback-image>
<input type="range" min="-3.1415" max="3.1415" step="0.01" value="1.4" class="slide-control" />
</graphics-element>
<p>At cursory glance, a fair bit better, but let's find out <em>how much</em> better by looking at how to construct the Bézier curve.</p>
<p>
<img
src="images/chapter-assets/circles/image-20210417165543902.png"
alt="A construction diagram for a cubic approximation of a circular arc"
/>
</p>
<p>
The start and end points are trivial, but the mid point requires a bit of work, but it's mostly basic trigonometry once we know the angle
θ for our circular arc: if we scale our circular arc to a unit circle, we can always start our arc, with radius 1, at (1,0) and then given
our arc angle θ, we also know that the circular arc has length θ (because unit circles are nice that way). We also know our end point,
because that's just (cos(θ), sin(θ)), and so the challenge is to figure out what control points we need in order for the curve at
<em>t</em>=0.5 to exactly touch the circular arc at the angle θ/2:
</p>
<p>So let's again formally describe this:</p>
<!--
P = (1, 0)
1
P = (1, k)
2
P = P + k · (sin(θ), -cos(θ))
3 4
P = (cos(θ), sin(θ))
4
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/9054528132317434ae2c0be27572d86b.svg"
width="208px"
height="85px"
loading="lazy"
/>
<p>
Only P<sub>3</sub> isn't quite straight-forward here, and its description is based on the fact that the triangle (origin, P<sub>4</sub>,
P<sub>3</sub>) is a right angled triangle, with the distance between the origin and P<sub>4</sub> being 1 (because we're working with a
unit circle), and the distance between P<sub>4</sub> and P<sub>3</sub> being <em>k</em>, so that we can represent P<sub>3</sub> as "The
point P<sub>4</sub> plus the vector from the origin to P<sub>4</sub> but then rotated a quarter circle, counter-clockwise, and scaled by
<em>k</em>".
</p>
<p>
With that, we can determine the <em>y</em>-coordinates for A, B, e<sub>1</sub>, and e<sub>2</sub>, after which we have all the information
we need to determine what the value of <em>k</em> is. We can find these values by using (no surprise here) linear interpolation between
known points, as A is midway between P<sub>2</sub> and P<sub>3</sub>, e<sub>1</sub> is between A and "midway between P<sub>1</sub> and
P<sub>2</sub>" (which is "half height" P<sub>2</sub>), and so forth:
</p>
<!--
P + P
2 3
y y k + sin(θ) - k · cos(θ)
A = ───────── = ────────────────────────
y 2 2
1
A + ─P k + sin(θ) - k · cos(θ)
y 2 2 ──────────────────────── + ─
y 2 2 2k + sin(θ) + k · cos(θ)
e = ───────── = ──────────────────────────── = ─────────────────────────
1 2 2 4
y
k
A + mid(P , P ) A + sin(θ) - ─ cos(θ)
y 4 3 y 2 k + 3sin(θ) 2k · cos(θ)
e = ───────────────── = ────────────────────── = ────────────────────────
2 2 2 4
y
e + e
1 2
y y 3k + 4sin(θ) - 3k · cos(θ)
B = ───────── = ───────────────────────────
y 2 8
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/e0d46f5fd4bf01e72f23495757f64448.svg"
width="505px"
height="173px"
loading="lazy"
/>
<p>
Which now gives us two identities for B, because in addition to determining B through linear interpolation, we also know that B's
<em>y</em> coordinate is just <em>sin(θ/2)</em>: we started this exercise by saying we were going to approximate the circular arc using a
Bézier curve that had its midpoint, which is point B, touching the unit circle at the arc's half-angle, by definition making B the point
at (cos(θ/2), sin(θ/2)).
</p>
<p>This means we can equate the two identities we now have for B<sub>y</sub> and solve for <em>k</em>.</p>
<div class="note">
<h2>Deriving <em>k</em></h2>
<p>
Solving for <em>k</em> is fairly straight forward, but it's a fair few steps, and if you just the immediate result: using a tool like
<a href="https://www.wolframalpha.com/">Wolfram Alpha</a> is definitely the way to go. That said, let's get going:
</p>
<!--
3k + 4sin(θ)) - 3k · cos(θ) θ
────────────────────────────= sin(─)
8 2
╭ θ ╮
3k + 4sin(θ)) - 3k · cos(θ)= 8sin│ ─ │
╰ 2 ╯
╭ θ ╮
3k - 3k · cos(θ)= 8sin│ ─ │ - 4sin(θ)
╰ 2 ╯
╭ ╭ θ ╮ ╮
3k (1 - cos(θ))= 4 │ 2sin│ ─ │ - sin(θ) │
╰ ╰ 2 ╯ ╯
θ
2sin(─) - sin(θ)
2
3k= 4 · ────────────────
1 - cos(θ)
╭ θ ╮
2sin│ ─ │ - sin(θ)
4 ╰ 2 ╯
k= ─ · ──────────────────
3 1 - cos(θ)
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/cb6686f1aff26d9f47ed4c695109fd5f.svg"
width="357px"
height="263px"
loading="lazy"
/>
<p>
And finally, we can take further advantage of several trigonometric identities to <em>drastically</em> simplify our formula for
<em>k</em>:
</p>
<!--
╭ θ ╮
2sin│ ─ │ - sin(θ)
4 ╰ 2 ╯
k= ─ · ──────────────────
3 1 - cos(θ)
╭ ╭ θ ╮ ╮
│ 2sin│ ─ │ │
4 │ ╰ 2 ╯ sin(θ) │
k= ─ · │ ────────── - ────────── │
3 ╰ 1 - cos(θ) 1 - cos(θ) ╯
4 ╭ ╭ θ ╮ ╭ θ ╮ ╮
k= ─ · │ csc│ ─ │ - cot│ ─ │ │
3 ╰ ╰ 2 ╯ ╰ 2 ╯ ╯
4 ╭ θ ╮
k= ─ · tan│ ─ │
3 ╰ 4 ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/f65f4e30a9f7a08c5c0092a1a3853922.svg"
width="235px"
height="188px"
loading="lazy"
/>
<p>And we're done.</p>
</div>
<p>
So, the distance of our control points to the start/end points can be expressed as a number that we get from an almost trivial expression
involving the circular arc's angle:
</p>
<!--
4 ╭ θ ╮
k = f(θ) = ─ tan│ ─ │
3 ╰ 4 ╯
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/f47561c3870425499e31c7527c38dc94.svg"
width="145px"
height="40px"
loading="lazy"
/>
<p>
Which means that for any circular arc with angle θ and radius <em>r</em>, our Bézier approximation based on three points of incidence is:
</p>
<!--
start= (r, 0)
control = (r, k)
1
control = r · (cos(θ) + k · sin(θ), sin(θ) - k · cos(θ))
2
end= r · (cos(θ), sin(θ))
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/d880c4a1b3d7b651b054b008e952b493.svg"
width="359px"
height="85px"
loading="lazy"
/>
<p>Which also gives us the commonly found value of 0.55228 for quarter circles, based on them having an angle of half π:</p>
<!--
╭ \pi ╮ 4 ╭ \pi ╮ 4 ┌─┐
f│ ─── │ = ─ · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.55228474983[...]
╰ 2 ╯ 3 ╰ 8 ╯ 3
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/acbde4be3cde3838b99b0ffc933f1f89.svg"
width="387px"
height="36px"
loading="lazy"
/>
<p>And thus giving us the following Bézier coordinates for a quarter circle of radius <em>r</em>:</p>
<!--
start= (r, 0)
control = (r, 0.55228 · r)
1
control = (0.55228 · r, r)
2
end= (0, r)
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/ad4d512cb92280ac88af531309cdbb8c.svg"
width="177px"
height="85px"
loading="lazy"
/>
<div class="note">
<h2>So, how accurate is this?</h2>
<p>
Unlike for the quadratic curve, we can't use <i>t=0.5</i> as our reference point because by its very nature it's one of the three points
that are actually guaranteed to be on the circular arc itself. Instead, we need a different <i>t</i> value that will give us the maximum
deflection - there are two possible choices (as our curve is still strictly "overshoots" the circular arc, and it's symmetrical) but
rather than trying to use calculus to find the perfect <em>t</em> value—which we could! the maths is perfectly reasonable as long as we
get to use computers—we can also just perform a binary search for the biggest deflection and not bother with all this maths stuff.
</p>
<p>
So let's do that instead: we can run a maximum deflection check that just runs through <em>t</em> from 0 to 1 at some coarse interval,
finds a <em>t</em> value that has "the highest deflection of the bunch", then reruns the same check with a much smaller interval around
that <em>t</em> value, repeating as many times as necessary to get us an arbitrarily precise value of <em>t</em>:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="13">
<textarea disabled rows="13" role="doc-example">
getMostWrongT(radius, bezier, start, end, epsilon=1e-15):
if end-start < epsilon:
return (start+end)/2
worst_t = 0
max = 0
stepsize = (end-start)/10
for t=start to end, using stepsize:
p = bezier.get(t)
diff = p.magnitude() - radius
if diff > max:
worst_t = t
max = diff
return getMostWrongT(radius, bezier, worst_t - stepsize, worst_t + stepsize)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
</table>
<p>Plus, how often do you get to write a function with that name?</p>
<p>
Using this code, we find that our <em>t</em> values are approximately 0.211325 and 0.788675, so let's pick the lower of the two and see
what the maximum deflection is across our domain of angles, with the original quadratic error show in green (rocketing off to infinity
first, and then coming back down as we approach 2π)
</p>
<table>
<tbody>
<tr>
<td>
<img src="images/chapter-assets/circles/image-20210417173811587.png" width="95%" />
</td>
<td>
<img src="images/chapter-assets/circles/image-20210417174019035.png" width="95%" />
</td>
<td>
<img src="images/chapter-assets/circles/image-20210417174100036.png" width="95%" />
</td>
</tr>
<tr>
<td>
error plotted for 0 ≤ φ ≤ 2π
</td>
<td>
error plotted for 0 ≤ φ ≤ π
</td>
<td>
error plotted for 0 ≤ φ ≤ ½π
</td>
</tr>
</tbody>
</table>
<p>
That last image is probably not quite clear enough: the cubic approximation of a quarter circle is so incredibly much better that we
can't even really see it at the same scale of our quadratic curve. Let's scale the y-axis a little, and try that again:
</p>
<p style="text-align: center;"><img src="images/chapter-assets/circles/image-20210417174215876.png" height="350px" /></p>
<p>
Yeah... the error of a cubic approximation for a quarter circle turns out to be <em>two orders of magnitude</em> better. At
approximately 0.00027 (or: just shy of being 2.7 pixels off for a circle with a radius of 10,000 pixels) the increase in precision over
quadratic curves is quite spectacular - certainly good enough that no one in their right mind should ever use quadratic curves.
</p>
</div>
<p>So that's it, kappa is <em>4/3 · tan(θ/4)</em> , we're done! ...or are we?</p>
<h2>Can we do better?</h2>
<p>
Technically: yes, we can. But I'm going to prefix this section with "we can, and we should investigate that possibility, but let me warn
you up front that the result is <em>only</em> better if we're going to hard-code the values". We're about to get into the weeds and the
standard three-points-of-incidence value is so good already that for most applications, trying to do better won't make any sense at all.
</p>
<p>
So with that said: what we calculated above is an <em>upper bound</em> for a best fit Bézier curve for a circular arc: anywhere we don't
touch the circular arc in our approximation, we've "overshot" the arc. What if we dropped our value for <em>k</em> just a little, so that
the curve starts out as an over-estimation, but then crosses the circular arc, yielding an region of underestimation, and then crosses the
circular arc again, with another region of overestimation. This might give us a lower overall error, so let's see what we can do.
</p>
<p>First, let's express the total error (given circular arc angle θ, and some <em>k</em>) using standard calculus notation:</p>
<!--
┌───────────────────────┐
╭ 1 │ 2 2
erf (θ, k) = | \| │B (t,θ,k) + B (t,θ,k) - r\|dt
╯ 0 ⟍│ x y
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/85c3dc0187f786b37f5fa5a7cc74d642.svg"
width="331px"
height="36px"
loading="lazy"
/>
<p>
This says that the error function for a given angle and value of <em>k</em> is equal to the "infinite" sum of differences between our
curve and the circular arc, as we run <em>t</em> from 0 to 1, using an infinitely small step size. between subsequent <em>t</em> values.
</p>
<p>
Now, since we want to find the minimal error, that means we want to know where along this function things go from "error is getting
progressively less" to "error is increasing again", which means we want to know where its derivative is zero, which as mathematical
expression looks like:
</p>
<!--
╭ ┌───────┐ ╮
│ ╭ 1 │ 2 2 │ d
│ | \| │B + B - r\|dt │ ── = 0
╰ ╯ 0 ⟍│ x y ╯ dt
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/a7dc2e51b90e89ec62e4a328d2b24635.svg"
width="209px"
height="40px"
loading="lazy"
/>
<p>
And here we have the most direct application of the
<a href="https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus">Fundamental Theorem of Calculus</a>: the derivative and integral
are each other's inverse operations, so they cancel out, leaving us with our original function:
</p>
<!--
┌───────┐
│ 2 2
\| │B + B - r\| = 0, t ∈[0,1]
⟍│ x y
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/b996b7c1af4c9187004af7d04b3740a5.svg"
width="207px"
height="27px"
loading="lazy"
/>
<p>And now we just solve for that... oh wait. We've seen this before. In order to solve this, we'd end up needing to solve this:</p>
<!--
2 2
B + B = r
x y
-->
<img
class="LaTeX SVG"
src="./images/chapters/circles_cubic/610a69d8be6f86744ffb88d12eda701b.svg"
width="81px"
height="23px"
loading="lazy"
/>
<p>
And both of those terms on the left of the equal sign are 6<sup>th</sup> degree polynomials, which means—as we've covered in the section
on arc lengths—<a href="https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem">there is no symbolic solution for this equasion</a>.
Instead, we'll have to use a numerical approach to find the solutions here, so... to the computer!
</p>
<div class="note">
<h2>Iterating on a solution</h2>
<p>
By which I really mean "to the binary search algorithm", because we're dealing with a reasonably well behaved function: depending on the
value for <em>k</em> , we're either going to end up with a Bézier curve that's on average "not at distance <em>r</em> from the arc's
center", "exactly distance <em>r</em> from the arc's center", or "more than distance <em>r</em> from the arc's center", so we can just
binary search our way to the most accurate value for <em>c</em> that gets us that middle case.
</p>
<p>First our setup, where we determine our upper and lower bounds, before entering our binary search:</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="4">
<textarea disabled rows="4" role="doc-example">
findBest(radius, angle, points[]):
lowerBound = 0
upperBound = 4.0/3.0 * Math.tan(abs(angle) / 4)
return binarySearch(radius, angle, points, lowerBound, upperBound)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
</table>
<p>
And then the binary search algorithm, which can be found in pretty much any CS textbook, as well as more online articles, tutorials, and
blog posts than you can ever read in a life time:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="19">
<textarea disabled rows="19" role="doc-example">
binarySearch(radius, angle, points[], lowerBound, upperBound, epsilon=1e-15):
value = (upperBound + lowerBound)/2
if (upperBound - lowerBound < epsilon) return value
// recompute the control points, based on our current "value"
d = (points[3].y < 0 ? -1 : 1) * value * radius
points[1] = new Point(radius, d)
points[2] = new Point(
points[3].x + d * sin(angle)
points[3].y - d * cos(angle)
)
if radialError(radius, points) > 0:
// our bezier curve is longer than we want it to be: reduce the upper bound
return binarySearch(radius, angle, points, lowerBound, value)
else:
// our bezier curve is shorter than we want it to be: increase the lower bound
return binarySearch(radius, angle, points, value, upperBound)</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
<tr>
<td>11</td>
</tr>
<tr>
<td>12</td>
</tr>
<tr>
<td>13</td>
</tr>
<tr>
<td>14</td>
</tr>
<tr>
<td>15</td>
</tr>
<tr>
<td>16</td>
</tr>
<tr>
<td>17</td>
</tr>
<tr>
<td>18</td>
</tr>
<tr>
<td>19</td>
</tr>
</table>
<p>
Using the following <code>radialError</code> function, which samples the curve's approximation of the circular arc over several points
(although the first and last point will never contribute anything, so we skip them):
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="7">
<textarea disabled rows="7" role="doc-example">
radialError(radius, points[]):
err = 0
steps = 5.0
for (int i=1; i<steps; i++):
Point p = getOnCurvePoint(points, i/steps)
err += p.magnitude()/radius - 1
return err</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
</table>
<p>
In this, <code>getOnCurvePoint</code> is just the standard Bézier evaluation function, yielding a point. Treating that point as a
vector, we can get its length to the origin using a <code>magnitude</code> call.
</p>
<h2>Examining the result</h2>
<p>
Running the above code we can get a list of <em>k</em> values associated with a list of angles θ from 0 to π, and we can use that to,
for each angle, plot what the difference between the circular arc and the Bézier approximation looks like:
</p>
<p><img src="images/chapter-assets/circles/image-20210419085430711.png" alt="image-20210419085430711" /></p>
<p>
Here we see the difference between an arc and its Bézier approximation plotted as we run <em>t</em> from 0 to 1. Just by looking at the
plot we can tell that there is maximum deflection at <em>t</em> = 0.5, so let's plot the maximum deflection "function", for angles from
0 to θ:
</p>
<p>In fact, let's plot the maximum deflections for both approaches as a functions over θ:</p>
<table>
<tbody>
<tr>
<td>
<img src="images/chapter-assets/circles/image-20210418111929371.png" width="95%" />
</td>
<td>
<img src="images/chapter-assets/circles/image-20210418112008676.png" width="95%" />
</td>
<td>
<img src="images/chapter-assets/circles/image-20210418112038613.png" width="95%" />
</td>
</tr>
<tr>
<td>
max deflection using unit scale
</td>
<td>
max deflection at 10x scale
</td>
<td>
max deflection at 100x scale
</td>
</tr>
</tbody>
</table>
<p>That doesn't actually appear to be all that much better, so let's look at some numbers, to see what the improvement actually is:</p>
<table>
<thead>
<tr>
<th>angle</th>
<th>"improved" deflection</th>
<th>"upper bound" deflection</th>
<th>difference</th>
</tr>
</thead>
<tbody>
<tr>
<td>1/8 π</td>
<td>6.202833502388927E-8</td>
<td>6.657161222278773E-8</td>
<td>4.5432771988984655E-9</td>
</tr>
<tr>
<td>1/4 π</td>
<td>3.978021202111215E-6</td>
<td>4.246252911066506E-6</td>
<td>2.68231708955291E-7</td>
</tr>
<tr>
<td>3/8 π</td>
<td>4.547652269037972E-5</td>
<td>4.8397483513262785E-5</td>
<td>2.9209608228830675E-6</td>
</tr>
<tr>
<td>1/2 π</td>
<td>2.569196199214696E-4</td>
<td>2.7251652752280364E-4</td>
<td>1.559690760133403E-5</td>
</tr>
<tr>
<td>5/8 π</td>
<td>9.877526288810667E-4</td>
<td>0.0010444175859711802</td>
<td>5.666495709011343E-5</td>
</tr>
<tr>
<td>3/4 π</td>
<td>0.00298164978679627</td>
<td>0.0031455628414580605</td>
<td>1.6391305466179062E-4</td>
</tr>
<tr>
<td>7/8 π</td>
<td>0.0076323182807019885</td>
<td>0.008047777909948373</td>
<td>4.1545962924638413E-4</td>
</tr>
<tr>
<td>π</td>
<td>0.017362185964043708</td>
<td>0.018349016519545902</td>
<td>9.86830555502194E-4</td>
</tr>
</tbody>
</table>
<p>
As we can see, the increase in precision is not particularly big: for a quarter circle (π/2) the traditional <em>k</em> will be off by
2.75 pixels on a circle with radius 10,000 pixels, whereas this "better" fit will be off by 2.56 pixels. And while that's certainly an
almost 10% improvement, it's also nowhere near enough of an improvement to make a discernible difference.
</p>
</div>
<p>
At this point it should be clear that while, yes, there are improvement to be had, they're essentially insignificant while also being
<em>much</em> more computationally expensive.
</p>
<h2>TL;DR: just tell me which value I should be using</h2>
<p>
It depends on what we need to do. If we just want the best value for quarter circles, and we're going to hard code the value for
<em>k</em>, then there is no reason to hard-code the constant <code>k=4/3*tan(pi/8)</code> when you can just as easily hard-code the
constant as <code>k=0.551784777779014</code> instead.
</p>
<p><strong>If you need "the" value for quarter circles, use 0.551785 instead of 0.55228</strong></p>
<p>
However, for dynamic arc approximation, in code that tries to fit circular paths using Bézier paths instead, it should be fairly obvious
that the simple function involving a tangent computation, two divisions, and one multiplication, is vastly more performant than running
all the code we ended writing just to get a 25% lower error value, and most certainly worth preferring over getting the "more accurate"
value.
</p>
<p>
<strong>If you need to fit Béziers to circular arcs on the fly, use <code>4/3 * tan(θ/4)</code></strong>
</p>
<p>
However, always remember that if you're writing for humans, you can typically use the best of both worlds: as the user interacts with
their curves, you should draw <em>their curves</em> instead of drawing approximations of them. If they need to draw circles or circular
arcs, draw those, and only approximate them with a Bézier curve when the data needs to be exported to a format that doesn't support those.
Ideally with a preview mechanism that highlights where the errors will be, and how large they will be.
</p>
<p><strong>If you're writing code for graphics design by humans, use circular arcs for circular arcs</strong></p>
<p>
And that's it. We have pretty well exhausted this subject. There are different metrics we could use to find "different best
<em>k</em> values", like trying to match arc length (e.g. when we're optimizing for material cost), or minimizing the area between the
circular arc and the Bézier curve (e.g. when we're optimizing for inking), or minimizing the rate of change of the Bézier's curvature
(e.g. when we're optimizing for curve traversal) and they all yield values that are so similar that it's almost certainly not worth it.
(For instance, for quarter circle approximations those values are 0.551777, 0.5533344, and 0.552184 respectively. Much like the 0.551785
we get from minimizing the maximum deflection, none of these values are significantly better enough to prefer them over the upper bound
value).
</p>
</section>
<section id="arcapproximation">
<h1>
<div class="nav">
<a href="ko-KR/index.html#circles_cubic">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#bsplines">다음</a>
</div>
<a href="ko-KR/index.html#arcapproximation">Approximating Bézier curves with circular arcs</a>
</h1>
<p>
Let's look at doing the exact opposite of the previous section: rather than approximating circular arc using Bézier curves, let's
approximate Bézier curves using circular arcs.
</p>
<p>
We already saw in the section on circle approximation that this will never yield a perfect equivalent, but sometimes you need circular
arcs, such as when you're working with fabrication machinery, or simple vector languages that understand lines and circles, but not much
else.
</p>
<p>
The approach is fairly simple: pick a starting point on the curve, and pick two points that are further along the curve. Determine the
circle that goes through those three points, and see if it fits the part of the curve we're trying to approximate. Decent fit? Try spacing
the points further apart. Bad fit? Try spacing the points closer together. Keep doing this until you've found the "good approximation/bad
approximation" boundary, record the "good" arc, and then move the starting point up to overlap the end point we previously found. Rinse
and repeat until we've covered the entire curve.
</p>
<p>
We already saw how to fit a circle through three points in the section on <a href="#pointcurves">creating a curve from three points</a>,
and finding the arc through those points is straight-forward: pick one of the three points as start point, pick another as an end point,
and the arc has to necessarily go from the start point, to the end point, over the remaining point.
</p>
<p>So, how can we convert a Bézier curve into a (sequence of) circular arc(s)?</p>
<ul>
<li>Start at <code>t=0</code></li>
<li>Pick two points further down the curve at some value <code>m = t + n</code> and <code>e = t + 2n</code></li>
<li>Find the arc that these points define</li>
<li>
Determine how close the found arc is to the curve:
<ul>
<li>Pick two additional points <code>e1 = t + n/2</code> and <code>e2 = t + n + n/2</code>.</li>
<li>
These points, if the arc is a good approximation of the curve interval chosen, should lie <code>on</code> the circle, so their
distance to the center of the circle should be the same as the distance from any of the three other points to the center.
</li>
<li>
For point points, determine the (absolute) error between the radius of the circle, and the <code>actual</code> distance from the
center of the circle to the point on the curve.
</li>
<li>If this error is too high, we consider the arc bad, and try a smaller interval.</li>
</ul>
</li>
</ul>
<p>
The result of this is shown in the next graphic: we start at a guaranteed failure: s=0, e=1. That's the entire curve. The midpoint is
simply at <code>t=0.5</code>, and then we start performing a
<a href="https://en.wikipedia.org/wiki/Binary_search_algorithm">binary search</a>.
</p>
<ol>
<li>We start with <code>low=0</code>, <code>mid=0.5</code> and <code>high=1</code></li>
<li>
That'll fail, so we retry with the interval halved: <code>{0, 0.25, 0.5}</code>
<ul>
<li>If that arc's good, we move back up by half distance: <code>{0, 0.375, 0.75}</code>.</li>
<li>However, if the arc was still bad, we move <em>down</em> by half the distance: <code>{0, 0.125, 0.25}</code>.</li>
</ul>
</li>
<li>
We keep doing this over and over until we have two arcs, in sequence, of which the first arc is good, and the second arc is bad. When we
find that pair, we've found the boundary between a good approximation and a bad approximation, and we pick the good arc.
</li>
</ol>
<p>
The following graphic shows the result of this approach, with a default error threshold of 0.5, meaning that if an arc is off by a
<em>combined</em> half pixel over both verification points, then we treat the arc as bad. This is an extremely simple error policy, but
already works really well. Note that the graphic is still interactive, and you can use your up and down arrow keys keys to increase or
decrease the error threshold, to see what the effect of a smaller or larger error threshold is.
</p>
<graphics-element
title="First arc approximation of a Bézier curve"
width="275"
height="275"
src="./chapters/arcapproximation/arc.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arcapproximation/7c9cce8142fa3e85bb124520f40645ff.png" loading="lazy" />
<label>First arc approximation of a Bézier curve</label>
</fallback-image>
<input type="range" min="0.1" max="5" step="0.1" value="0.5" class="slide-control" />
</graphics-element>
<p>
With that in place, all that's left now is to "restart" the procedure by treating the found arc's end point as the new to-be-determined
arc's starting point, and using points further down the curve. We keep trying this until the found end point is for <em>t=1</em>, at which
point we are done. Again, the following graphic allows for up and down arrow key input to increase or decrease the error threshold, so you
can see how picking a different threshold changes the number of arcs that are necessary to reasonably approximate a curve:
</p>
<graphics-element
title="Arc approximation of a Bézier curve"
width="275"
height="275"
src="./chapters/arcapproximation/arcs.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="275px" height="275px" src="./images/chapters/arcapproximation/da76341b841df1af8a39f797e85dfe3c.png" loading="lazy" />
<label>Arc approximation of a Bézier curve</label>
</fallback-image>
<input type="range" min="0.1" max="5" step="0.1" value="0.5" class="slide-control" />
</graphics-element>
<p>
So... what is this good for? Obviously, if you're working with technologies that can't do curves, but can do lines and circles, then the
answer is pretty straightforward, but what else? There are some reasons why you might need this technique: using circular arcs means you
can determine whether a coordinate lies "on" your curve really easily (simply compute the distance to each circular arc center, and if any
of those are close to the arc radii, at an angle between the arc start and end, bingo, this point can be treated as lying "on the curve").
Another benefit is that this approximation is "linear": you can almost trivially travel along the arcs at fixed speed. You can also
trivially compute the arc length of the approximated curve (it's a bit like curve flattening). The only thing to bear in mind is that this
is a lossy equivalence: things that you compute based on the approximation are guaranteed "off" by some small value, and depending on how
much precision you need, arc approximation is either going to be super useful, or completely useless. It's up to you to decide which,
based on your application!
</p>
</section>
<section id="bsplines">
<h1>
<div class="nav">
<a href="ko-KR/index.html#arcapproximation">이전</a><a href="#toc">목차</a><a href="ko-KR/index.html#comments">다음</a>
</div>
<a href="ko-KR/index.html#bsplines">B-Splines</a>
</h1>
<p>
No discussion on Bézier curves is complete without also giving mention of that other beast in the curve design space: B-Splines. Easily
confused to mean Bézier splines, that's not actually what they are; they are "basis function" splines, which makes a lot of difference,
and we'll be looking at those differences in this section. We're not going to dive as deep into B-Splines as we have for Bézier curves
(that would be an entire primer on its own) but we'll be looking at how B-Splines work, what kind of maths is involved in computing them,
and how to draw them based on a number of parameters that you can pick for individual B-Splines.
</p>
<p>
First off: B-Splines are <a href="https://en.wikipedia.org/wiki/Piecewise">piecewise</a>,
<a href="https://en.wikipedia.org/wiki/Spline_(mathematics)">polynomial interpolation curves</a>, where the "single curve" is built by
performing polynomial interpolation over a set of points, using a sliding window of a fixed number of points. For instance, a "cubic"
B-Spline defined by twelve points will have its curve built by evaluating the polynomial interpolation of four points, and the curve can
be treated as a lot of different sections, each controlled by four points at a time, such that the full curve consists of smoothly
connected sections defined by points {1,2,3,4}, {2,3,4,5}, ..., {8,9,10,11}, and finally {9,10,11,12}, for eight sections.
</p>
<p>
What do they look like? They look like this! Tap on the graphic to add more points, and move points around to see how they map to the
spline curve drawn.
</p>
<graphics-element
title="A B-Spline example"
width="600"
height="300"
src="./chapters/bsplines/basic.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="600px" height="300px" src="./images/chapters/bsplines/fe3a8ca5706f286d916e36699e237e51.png" loading="lazy" />
<label></label> </fallback-image
></graphics-element>
<p>
The important part to notice here is that we are <strong>not</strong> doing the same thing with B-Splines that we do for poly-Béziers or
Catmull-Rom curves: both of the latter simply define new sections as literally "new sections based on new points", so a 12 point cubic
poly-Bézier curve is actually impossible, because we start with a four point curve, and then add three more points for each section that
follows, so we can only have 4, 7, 10, 13, 16, etc. point Poly-Béziers. Similarly, while Catmull-Rom curves can grow by adding single
points, this addition of a single point introduces three implicit Bézier points. Cubic B-Splines, on the other hand, are smooth
interpolations of <em>each possible curve involving four consecutive points</em>, such that at any point along the curve except for our
start and end points, our on-curve coordinate is defined by four control points.
</p>
<p>Consider the difference to be this:</p>
<ul>
<li>for Bézier curves, the curve is defined as an interpolation of points, but:</li>
<li>for B-Splines, the curve is defined as an interpolation of <em>curves</em>.</li>
</ul>
<p>In fact, let's look at that again, but this time with the base curves shown, too. Each consecutive four points define one curve:</p>
<graphics-element
title="The components of a B-Spline "
width="600"
height="300"
src="./chapters/bsplines/basic.js"
data-show-curves="true"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="600px" height="300px" src="./images/chapters/bsplines/41167c64c51386414c6e62f0b45e6295.png" loading="lazy" />
<label></label>
</fallback-image>
<!-- basis curve highlighter goes here -->
</graphics-element>
<p>
In order to make this interpolation of curves work, the maths is necessarily more complex than the maths for Bézier curves, so let's have
a look at how things work.
</p>
<h2>How to compute a B-Spline curve: some maths</h2>
<p>
Given a B-Spline of degree <code>d</code> and thus order <code>k=d+1</code> (so a quadratic B-Spline is degree 2 and order 3, a cubic
B-Spline is degree 3 and order 4, etc) and <code>n</code> control points <code>P<sub>0</sub></code> through <code>P<sub>n-1</sub></code
>, we can compute a point on the curve for some value <code>t</code> in the interval [0,1] (where 0 is the start of the curve, and 1 the
end, just like for Bézier curves), by evaluating the following function:
</p>
<!--
__ n
Point(t) = ❯ P · N (t)
‾‾ i=0 i i,k
-->
<img class="LaTeX SVG" src="./images/chapters/bsplines/89f8e37237d066fa70ccf6d37b3a4922.svg" width="169px" height="41px" loading="lazy" />
<p>
Which, honestly, doesn't tell us all that much. All we can see is that a point on a B-Spline curve is defined as "a mix of all the control
points, weighted somehow", where the weighting is achieved through the <em>N(...)</em> function, subscripted with an obvious parameter
<code>i</code>, which comes from our summation, and some magical parameter <code>k</code>. So we need to know two things: 1. what does
N(t) do, and 2. what is that <code>k</code>? Let's cover both, in reverse order.
</p>
<p>
The parameter <code>k</code> represents the "knot interval" over which a section of curve is defined. As we learned earlier, a B-Spline
curve is itself an interpolation of curves, and we can treat each transition where a control point starts or stops influencing the total
curvature as a "knot on the curve". Doing so for a degree <code>d</code> B-Spline with <code>n</code> control point gives us
<code>d + n + 1</code> knots, defining <code>d + n</code> intervals along the curve, and it is these intervals that the above
<code>k</code> subscript to the N() function applies to.
</p>
<p>Then the N() function itself. What does it look like?</p>
<!--
╭ t- knot ╮ ╭ knot -t ╮
│ i │ │ (i+k) │
N (t) = │ ───────────────────── │ · N (t) + │ ─────────────────────── │ · N (t)
i,k │ knot - knot │ i,k-1 │ knot - knot │ i+1,k-1
╰ (i+k-1) i ╯ ╰ (i+k) (i+1) ╯
-->
<img class="LaTeX SVG" src="./images/chapters/bsplines/2421f47aa4fe1c0d830d53b2e6563c04.svg" width="556px" height="43px" loading="lazy" />
<p>
So this is where we see the interpolation: N(t) for an <code>(i,k)</code> pair (that is, for a step in the above summation, on a specific
knot interval) is a mix between N(t) for <code>(i,k-1)</code> and N(t) for <code>(i+1,k-1)</code>, so we see that this is a recursive
iteration where <code>i</code> goes up, and <code>k</code> goes down, so it seem reasonable to expect that this recursion has to stop at
some point; obviously, it does, and specifically it does so for the following <code>i</code>/<code>k</code> values:
</p>
<!--
╭ 1 if t ∈[ knot , knot )
N (t) = ╡ i i+1
i,1 ╰ 0 otherwise
-->
<img class="LaTeX SVG" src="./images/chapters/bsplines/2514e1aa0565840e33fde0b146e3efe2.svg" width="237px" height="40px" loading="lazy" />
<p>
And this function finally has a straight up evaluation: if a <code>t</code> value lies within a knot-specific interval once we reach a
<code>k=1</code> value, it "counts", otherwise it doesn't. We did cheat a little, though, because for all these values we need to scale
our <code>t</code> value first, so that it lies in the interval bounded by <code>knots[d]</code> and <code>knots[n]</code>, which are the
start point and end point where curvature is controlled by exactly <code>order</code> control points. For instance, for degree 3 (=order
4) and 7 control points, with knot vector [1,2,3,4,5,6,7,8,9,10,11], we map <code>t</code> from [the interval 0,1] to the interval [4,8],
and then use that value in the functions above, instead.
</p>
<h2>Can we simplify that?</h2>
<p>We can, yes.</p>
<p>
People far smarter than us have looked at this work, and two in particular —
<a href="https://www.npl.co.uk/people/maurice-cox">Maurice Cox</a> and
<a href="https://en.wikipedia.org/wiki/Carl_R._de_Boor">Carl de Boor</a> — came to a mathematically pleasing solution: to compute a point
P(t), we can compute this point by evaluating <em>d(t)</em> on a curve section between knots <code>i</code> and <code>i+1</code>:
</p>
<!--
k k-1 k-1
d (t) = α · d (t) + (1-α ) · d (t)
i i,k i i,k i-1
-->
<img class="LaTeX SVG" src="./images/chapters/bsplines/f0bf7d0f1931060cd801ff707f482c16.svg" width="281px" height="21px" loading="lazy" />
<p>
This is another recursive function, with <em>k</em> values decreasing from the curve order to 1, and the value <em>α</em> (alpha) defined
by:
</p>
<!--
t - knots[i]
α = ───────────────────────────
i,k knots[i+1+n-k] - knots[i]
-->
<img class="LaTeX SVG" src="./images/chapters/bsplines/e62558cdfd8abaf22511e8e68c7afb4a.svg" width="252px" height="39px" loading="lazy" />
<p>
That looks complicated, but it's not. Computing alpha is just a fraction involving known, plain numbers. And, once we have our alpha
value, we also have <code>(1-alpha)</code> because it's a trivial subtraction. Computing the <code>d()</code> function is thus mostly a
matter of computing pretty simple arithmetical statements, with some caching of results so we can refer to them as we recurve. While the
recursion might see computationally expensive, the total algorithm is cheap, as each step only involves very simple maths.
</p>
<p>Of course, the recursion does need a stop condition:</p>
<!--
k 0 ╭ 1 if t ∈[ knot , knot )
d (t) = 0, d (t) = N (t) = ╡ i i+1
0 i i,1 ╰ 0 otherwise
-->
<img class="LaTeX SVG" src="./images/chapters/bsplines/49af474c33ce0ee0733626ea3d988570.svg" width="367px" height="40px" loading="lazy" />
<p>
So, we actually see two stopping conditions: either <code>i</code> becomes 0, in which case <code>d()</code> is zero, or
<code>k</code> becomes zero, in which case we get the same "either 1 or 0" that we saw in the N() function above.
</p>
<p>
Thanks to Cox and de Boor, we can compute points on a B-Spline pretty easily using the same kind of linear interpolation we saw in de
Casteljau's algorithm. For instance, if we write out <code>d()</code> for <code>i=3</code> and <code>k=3</code>, we get the following
recursion diagram:
</p>
<!--
╭ ╭ ╭ 1 0 0
│ │ │ α × d , with d either 0 or 1
│ │ 2 1 1 │ 3 3 3
│ │ α × d , with d = ╡ +
│ │ 3 3 3 │ ╭ 1 ╮ 0 0
│ │ │ │ 1 - α │ × d , with d either 0 or 1
│ 3 2 2 │ ╰ ╰ 3 ╯ 2 2
│ α × d , with d = ╡ +
│ 3 3 3 │ ╭ 1 0
│ │ │ α × d
│ │ ╭ 2 ╮ 1 1 │ 2 2
│ │ │ 1 - α │ × d , with d = ╡ +
3 │ │ ╰ 3 ╯ 2 2 │ ╭ 1 ╮ 0 0
d = ╡ │ │ │ 1 - α │ × d , with d either 0 or 1
3 │ ╰ ╰ ╰ 2 ╯ 1 1
│ +
│ ╭ 2 1
│ │ α × d
│ │ 2 2
│ │
│ ╭ 3 ╮ 2 2 │ +
│ │ 1 - α │ × d , with d = ╡ ╭ 1 0
│ ╰ 3 ╯ 2 2 │ │ α × d
│ │ ╭ 2 ╮ 1 1 │ 1 1
│ │ │ 1 - α │ × d , with d = ╡ +
│ │ ╰ 2 ╯ 1 1 │ ╭ 1 ╮ 0 0
│ │ │ │ 1 - α │ × d , with d either 0 or 1
╰ ╰ ╰ ╰ 1 ╯ 0 0
-->
<img class="LaTeX SVG" src="./images/chapters/bsplines/20e910bbea2e6eff511cb13cef18ef3b.svg" width="641px" height="336px" loading="lazy" />
<p>
That is, we compute <code>d(3,3)</code> as a mixture of <code>d(2,3)</code> and <code>d(2,2)</code>, where those two are themselves a
mixture of <code>d(1,3)</code> and <code>d(1,2)</code>, and <code>d(1,2)</code> and <code>d(1,1)</code>, respectively, which are
themselves a mixture of etc. etc. We simply keep expanding our terms until we reach the stop conditions, and then sum everything back up.
It's really quite elegant.
</p>
<p>
One thing we need to keep in mind is that we're working with a spline that is constrained by its control points, so even though the
<code>d(..., k)</code> values are zero or one at the lowest level, they are really "zero or one, times their respective control point", so
in the next section you'll see the algorithm for running through the computation in a way that starts with a copy of the control point
vector and then works its way up to that single point, rather than first starting "on the left", working our way "to the right" and then
summing back up "to the left". We can just start on the right and work our way left immediately.
</p>
<h2>Running the computation</h2>
<p>
Unlike the de Casteljau algorithm, where the <code>t</code> value stays the same at every iteration, for B-Splines that is not the case,
and so we end having to (for each point we evaluate) run a fairly involving bit of recursive computation. The algorithm is discussed on
<a href="https://pages.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/de-Boor.html">this Michigan Tech</a> page, but an easier to read version
is implemented by <a href="https://github.com/thibauts/b-spline/blob/master/index.js#L59-L71">b-spline.js</a>, so we'll look at its code.
</p>
<p>
Given an input value <code>t</code>, we first map the input to a value from the domain <code>[0,1]</code> to the domain
<code>[knots[degree], knots[knots.length - 1 - degree]</code>. Then, we find the section number <code>s</code> that this mapped
<code>t</code> value lies on:
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="3">
<textarea disabled rows="3" role="doc-example">
for(s=domain[0]; s < domain[1]; s++) {
if(knots[s] <= t && t <= knots[s+1]) break;
}</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
</table>
<p>
after running this code, <code>s</code> is the index for the section the point will lie on. We then run the algorithm mentioned on the MU
page (updated to use this description's variable names):
</p>
<table class="code">
<tr>
<td>1</td>
<td rowspan="10">
<textarea disabled rows="10" role="doc-example">
let v = copy of control points
for(let L = 1; L <= order; L++) {
for(let i=s; i > s + L - order; i--) {
let numerator = t - knots[i]
let denominator = knots[i - L + order] - knots[i]
let alpha = numerator / denominator
let v[i] = alpha * v[i] + (1-alpha) * v[i-1]
}
}</textarea
>
</td>
</tr>
<tr>
<td>2</td>
</tr>
<tr>
<td>3</td>
</tr>
<tr>
<td>4</td>
</tr>
<tr>
<td>5</td>
</tr>
<tr>
<td>6</td>
</tr>
<tr>
<td>7</td>
</tr>
<tr>
<td>8</td>
</tr>
<tr>
<td>9</td>
</tr>
<tr>
<td>10</td>
</tr>
</table>
<p>
(A nice bit of behaviour in this code is that we work the interpolation "backwards", starting at <code>i=s</code> at each level of the
interpolation, and we stop when <code>i = s - order + level</code>, so we always end up with a value for <code>i</code> such that those
<code>v[i-1]</code> don't try to use an array index that doesn't exist)
</p>
<h2>Open vs. closed paths</h2>
<p>
Much like poly-Béziers, B-Splines can be either open, running from the first point to the last point, or closed, where the first and last
point are the same coordinate. However, because B-Splines are an interpolation of curves, not just points, we can't simply make the first
and last point the same, we need to link as many points as are necessary to form "a curve" that the spline performs interpolation with. As
such, for an order <code>d</code> B-Spline, we need to make the first and last <code>d</code> points the same. This is of course hardly
more work than before (simply append <code>points.splice(0,d)</code> to <code>points</code>) but it's important to remember that you need
more than just a single point.
</p>
<p>
Of course if we want to manipulate these kind of curves we need to make sure to mark them as "closed" so that we know the coordinate for
<code>points[0]</code> and <code>points[n-k]</code> etc. don't just happen to have the same x/y values, but really are the same
coordinate, so that manipulating one will equally manipulate the other, but programming generally makes this really easy by storing
references to points, rather than copies (or other linked values such as coordinate weights, discussed in the NURBS section) rather than
separate coordinate objects.
</p>
<h2>Manipulating the curve through the knot vector</h2>
<p>
The most important thing to understand when it comes to B-Splines is that they work <em>because</em> of the concept of a knot vector. As
mentioned above, knots represent "where individual control points start/stop influencing the curve", but we never looked at the
<em>values</em> that go in the knot vector. If you look back at the N() and a() functions, you see that interpolations are based on
intervals in the knot vector, rather than the actual values in the knot vector, and we can exploit this to do some pretty interesting
things with clever manipulation of the knot vector. Specifically there are four things we can do that are worth looking at:
</p>
<ol>
<li>we can use a uniform knot vector, with equally spaced intervals,</li>
<li>we can use a non-uniform knot vector, without enforcing equally spaced intervals,</li>
<li>we can collapse sequential knots to the same value, locally lowering curve complexity using "null" intervals, and</li>
<li>
we can form a special case non-uniform vector, by combining (1) and (3) to for a vector with collapsed start and end knots, with a
uniform vector in between.
</li>
</ol>
<h3>Uniform B-Splines</h3>
<p>
The most straightforward type of B-Spline is the uniform spline. In a uniform spline, the knots are distributed uniformly over the entire
curve interval. For instance, if we have a knot vector of length twelve, then a uniform knot vector would be [0,1,2,3,...,9,10,11]. Or
[4,5,6,...,13,14,15], which defines <em>the same intervals</em>, or even [0,2,3,...,18,20,22], which also defines
<em>the same intervals</em>, just scaled by a constant factor, which becomes normalised during interpolation and so does not contribute to
the curvature.
</p>
<graphics-element
title="A uniform B-Spline"
width="400"
height="400"
src="./chapters/bsplines/uniform.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/bsplines/48a30189e74658737b3a8b28bb816f8a.png" loading="lazy" />
<label></label>
</fallback-image>
<!-- knot sliders go here -->
</graphics-element>
<p>
This is an important point: the intervals that the knot vector defines are <em>relative</em> intervals, so it doesn't matter if every
interval is size 1, or size 100 - the relative differences between the intervals is what shapes any particular curve.
</p>
<p>
The problem with uniform knot vectors is that, as we need <code>order</code> control points before we have any curve with which we can
perform interpolation, the curve does not "start" at the first point, nor "ends" at the last point. Instead there are "gaps". We can get
rid of these, by being clever about how we apply the following uniformity-breaking approach instead...
</p>
<h3>Reducing local curve complexity by collapsing intervals</h3>
<p>
Collapsing knot intervals, by making two or more consecutive knots have the same value, allows us to reduce the curve complexity in the
sections that are affected by the knots involved. This can have drastic effects: for every interval collapse, the curve order goes down,
and curve continuity goes down, to the point where collapsing <code>order</code> knots creates a situation where all continuity is lost
and the curve "kinks".
</p>
<graphics-element
title="A reduced uniform B-Spline"
width="400"
height="400"
src="./chapters/bsplines/reduced.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/bsplines/ceaef2fbee05a58aa11835925403b4cd.png" loading="lazy" />
<label></label>
</fallback-image>
<!-- knot sliders go here -->
</graphics-element>
<h3>Open-Uniform B-Splines</h3>
<p>
By combining knot interval collapsing at the start and end of the curve, with uniform knots in between, we can overcome the problem of the
curve not starting and ending where we'd kind of like it to:
</p>
<p>
For any curve of degree <code>D</code> with control points <code>N</code>, we can define a knot vector of length <code>N+D+1</code> in
which the values <code>0 ... D+1</code> are the same, the values <code>D+1 ... N+1</code> follow the "uniform" pattern, and the values
<code>N+1 ... N+D+1</code> are the same again. For example, a cubic B-Spline with 7 control points can have a knot vector
[0,0,0,0,1,2,3,4,4,4,4], or it might have the "identical" knot vector [0,0,0,0,2,4,6,8,8,8,8], etc. Again, it is the relative differences
that determine the curve shape.
</p>
<graphics-element
title="An open, uniform B-Spline"
width="400"
height="400"
src="./chapters/bsplines/uniform.js"
data-open="true"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/bsplines/0215dc106e4ad51afe043c0176a595f6.png" loading="lazy" />
<label></label>
</fallback-image>
<!-- knot sliders go here -->
</graphics-element>
<h3>Non-uniform B-Splines</h3>
<p>
This is essentially the "free form" version of a B-Spline, and also the least interesting to look at, as without any specific reason to
pick specific knot intervals, there is nothing particularly interesting going on. There is one constraint to the knot vector, other than
that any value <code>knots[k+1]</code> should be greater than or equal to <code>knots[k]</code>.
</p>
<h2>One last thing: Rational B-Splines</h2>
<p>
While it is true that this section on B-Splines is running quite long already, there is one more thing we need to talk about, and that's
"Rational" splines, where the rationality applies to the "ratio", or relative weights, of the control points themselves. By introducing a
ratio vector with weights to apply to each control point, we greatly increase our influence over the final curve shape: the more weight a
control point carries, the closer to that point the spline curve will lie, a bit like turning up the gravity of a control point, just like
for rational Bézier curves.
</p>
<graphics-element
title="A (closed) rational, uniform B-Spline"
width="400"
height="400"
src="./chapters/bsplines/rational-uniform.js"
reset="초기화"
viewSource="소스 보기"
>
<fallback-image>
<span class="view-source">스크립트가 꺼져 있어 대체 이미지를 대신 표시합니다.</span>
<img width="400px" height="400px" src="./images/chapters/bsplines/0d9c2186423466a32bb8fbd187409f82.png" loading="lazy" />
<label></label>
</fallback-image>
<!-- knot sliders go here -->
</graphics-element>
<p>
Of course this brings us to the final topic that any text on B-Splines must touch on before calling it a day: the
<a href="https://en.wikipedia.org/wiki/Non-uniform_rational_B-spline">NURBS</a>, or Non-Uniform Rational B-Spline (NURBS is not a plural,
the capital S actually just stands for "spline", but a lot of people mistakenly treat it as if it is, so now you know better). NURBS is an
important type of curve in computer-facilitated design, used a lot in 3D modelling (typically as NURBS surfaces) as well as in
arbitrary-precision 2D design due to the level of control a NURBS curve offers designers.
</p>
<p>
While a true non-uniform rational B-Spline would be hard to work with, when we talk about NURBS we typically mean the Open-Uniform
Rational B-Spline, or OURBS, but that doesn't roll off the tongue nearly as nicely, and so remember that when people talk about NURBS,
they typically mean open-uniform, which has the useful property of starting the curve at the first control point, and ending it at the
last.
</p>
<h2>Extending our implementation to cover rational splines</h2>
<p>
The algorithm for working with Rational B-Splines is virtually identical to the regular algorithm, and the extension to work in the
control point weights is fairly simple: we extend each control point from a point in its original number of dimensions (2D, 3D, etc.) to
one dimension higher, scaling the original dimensions by the control point's weight, and then assigning that weight as its value for the
extended dimension.
</p>
<p>For example, a 2D point <code>(x,y)</code> with weight <code>w</code> becomes a 3D point <code>(w * x, w * y, w)</code>.</p>
<p>
We then run the same algorithm as before, which will automatically perform weight interpolation in addition to regular coordinate
interpolation, because all we've done is pretended we have coordinates in a higher dimension. The algorithm doesn't really care about how
many dimensions it needs to interpolate.
</p>
<p>
In order to recover our "real" curve point, we take the final result of the point generation algorithm, and "unweigh" it: we take the
final point's derived weight <code>w'</code> and divide all the regular coordinate dimensions by it, then throw away the weight
information.
</p>
<p>
Based on our previous example, we take the final 3D point <code>(x', y', w')</code>, which we then turn back into a 2D point by computing
<code>(x'/w', y'/w')</code>. And that's it, we're done!
</p>
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