info/BezierInfo-2
1
0
Fork 0
mirror of https://github.com/Pomax/BezierInfo-2.git synced 2025-08-29 02:59:58 +02:00
Code Issues Releases Wiki Activity

full regeneration

Browse Source
This commit is contained in:
Pomax
2021-02-28 12:28:41 -08:00
parent d8fab98ec7
commit 45394f6d0f
329 changed files with 3135 additions and 2361 deletions
Download Patch File Download Diff File Expand all files Collapse all files

26
docs/chapters/explanation/content.en-GB.md
View File

@@ -59,9 +59,9 @@ Bézier curves are polynomials of <i>t</i>, rather than <i>x</i>, with the value
\[
\begin{aligned}
linear &= (1-t) + t \\
square &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
cubic &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\textit{linear} &= (1-t) + t \\
\textit{square} &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
\textit{cubic} &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\end{aligned}
\]
@@ -69,10 +69,10 @@ I know what you're thinking: that doesn't look too simple! But if we remove <i>t
\[
\begin{aligned}
linear &= \hspace{2.5em} 1 + 1 \\
square &= \hspace{1.7em} 1 + 2 + 1\\
cubic &= \hspace{0.85em} 1 + 3 + 3 + 1\\
quartic &= 1 + 4 + 6 + 4 + 1
\textit{linear} &= \hspace{2.5em} 1 + 1 \\
\textit{square} &= \hspace{1.7em} 1 + 2 + 1\\
\textit{cubic} &= \hspace{0.85em} 1 + 3 + 3 + 1\\
\textit{quartic} &= 1 + 4 + 6 + 4 + 1
\end{aligned}
\]
@@ -82,19 +82,19 @@ There's an equally simple way to figure out how the polynomial terms work: if we
\[
\begin{aligned}
linear &= BLUE[a] + RED[b] \\
square &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
cubic &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\textit{linear} &= BLUE[a] + RED[b] \\
\textit{square} &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
\textit{cubic} &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\end{aligned}
\]
It's basically just a sum of "every combination of <i>a</i> and <i>b</i>", progressively replacing <i>a</i>'s with <i>b</i>'s after every + sign. So that's actually pretty simple too. So now you know binomial polynomials, and just for completeness I'm going to show you the generic function for this:
\[
Bézier(n,t) = \sum_{i=0}^{n}
\underset{binomial~term}{\underbrace{\binom{n}{i}}}
\textit{Bézier}(n,t) = \sum_{i=0}^{n}
\underset{\textit{binomial term}}{\underbrace{\binom{n}{i}}}
\cdot\
\underset{polynomial~term}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\underset{\textit{polynomial term}}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\]
And that's the full description for Bézier curves. Σ in this function indicates that this is a series of additions (using the variable listed below the Σ, starting at ...=&lt;value&gt; and ending at the value listed on top of the Σ).

26
docs/chapters/explanation/content.ja-JP.md
View File

@@ -57,9 +57,9 @@
\[
\begin{aligned}
1次 &= (1-t) + t \\
2次 &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
3次 &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\textit{1次} &= (1-t) + t \\
\textit{2次} &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
\textit{3次} &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\end{aligned}
\]
@@ -67,10 +67,10 @@
\[
\begin{aligned}
1次 &= \hspace{2.5em} 1 + 1 \\
2次 &= \hspace{1.7em} 1 + 2 + 1\\
3次 &= \hspace{0.85em} 1 + 3 + 3 + 1\\
4次 &= 1 + 4 + 6 + 4 + 1
\textit{1次} &= \hspace{2.5em} 1 + 1 \\
\textit{2次} &= \hspace{1.7em} 1 + 2 + 1\\
\textit{3次} &= \hspace{0.85em} 1 + 3 + 3 + 1\\
\textit{4次} &= 1 + 4 + 6 + 4 + 1
\end{aligned}
\]
@@ -80,19 +80,19 @@
\[
\begin{aligned}
1次 &= BLUE[a] + RED[b] \\
2次 &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
3次 &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\textit{1次} &= BLUE[a] + RED[b] \\
\textit{2次} &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
\textit{3次} &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\end{aligned}
\]
これは要するに、「<i>a</i>と<i>b</i>のすべての組み合わせ」の単なる和です。プラスが出てくるたびに、<i>a</i>を<i>b</i>へと1つずつ置き換えていけばよいのです。こちらも本当に単純です。さて、これで「二項係数多項式」がわかりました。完璧を期するため、この関数の一般の形を示しておきます。
\[
Bézier(n,t) = \sum_{i=0}^{n}
\underset{二項係数部分の項}{\underbrace{\binom{n}{i}}}
\textit{Bézier}(n,t) = \sum_{i=0}^{n}
\underset{\textit{二項係数部分の項}}{\underbrace{\binom{n}{i}}}
\cdot\
\underset{多項式部分の項}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\underset{\textit{多項式部分の項}}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\]
そして、これがベジエ曲線の完全な表現です。この関数中のΣは、加算の繰り返し（Σの下にある変数を使って、...=<値>から始めてΣの下にある値まで）を表します。

26
docs/chapters/explanation/content.ru-RU.md
View File

@@ -59,9 +59,9 @@
\[
\begin{aligned}
линейный &= (1-t) + t \\
квадратный &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
кубический &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\textit{линийный} &= (1-t) + t \\
\textit{квадратый} &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
\textit{кубический} &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\end{aligned}
\]
@@ -69,10 +69,10 @@
\[
\begin{aligned}
линийный &= \hspace{2.5em} 1 + 1 \\
квадратый &= \hspace{1.7em} 1 + 2 + 1\\
кубический &= \hspace{0.85em} 1 + 3 + 3 + 1\\
квартический &= 1 + 4 + 6 + 4 + 1
\textit{линийный} &= \hspace{2.5em} 1 + 1 \\
\textit{квадратый} &= \hspace{1.7em} 1 + 2 + 1\\
\textit{кубический} &= \hspace{0.85em} 1 + 3 + 3 + 1\\
\textit{квартический} &= 1 + 4 + 6 + 4 + 1
\end{aligned}
\]
@@ -82,19 +82,19 @@
\[
\begin{aligned}
линийный &= BLUE[a] + RED[b] \\
квадратый &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
кубический &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\textit{линийный} &= BLUE[a] + RED[b] \\
\textit{квадратый} &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
\textit{кубический} &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\end{aligned}
\]
В целом это просто сума "каждого сочетания <i>a</i> и <i>b</i>", получаемая прогрессивной заменой <i>a</i> на <i>b</i> по ходу уравнения. Потому, это так-же довольно просто. Итак теперь вы знаете что такое биноминальные полиномы. Для полноты картины, ниже привожу их общую функцию:
\[
Bézier(n,t) = \sum_{i=0}^{n}
\underset{биноминальный~термин}{\underbrace{\binom{n}{i}}}
\textit{Bézier}(n,t) = \sum_{i=0}^{n}
\underset{\textit{биноминальный термин}}{\underbrace{\binom{n}{i}}}
\cdot\
\underset{полиноминальный~термин}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\underset{\textit{полиноминальный термин}}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\]
И теперь, это полное объяснение. Σ в этой функции означает, что это серия сум (с использованием переменной приведенной под Σ, со стартовым значением в ...=&lt;value&gt; и максимальным значением представленным над Σ)

26
docs/chapters/explanation/content.zh-CN.md
View File

@@ -57,9 +57,9 @@
\[
\begin{aligned}
linear &= (1-t) + t \\
square &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
cubic &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\textit{linear} &= (1-t) + t \\
\textit{square} &= (1-t)^2 + 2 \cdot (1-t) \cdot t + t^2 \\
\textit{cubic} &= (1-t)^3 + 3 \cdot (1-t)^2 \cdot t + 3 \cdot (1-t) \cdot t^2 + t^3
\end{aligned}
\]
@@ -67,10 +67,10 @@
\[
\begin{aligned}
linear &= \hspace{2.5em} 1 + 1 \\
square &= \hspace{1.7em} 1 + 2 + 1\\
cubic &= \hspace{0.85em} 1 + 3 + 3 + 1\\
quartic &= 1 + 4 + 6 + 4 + 1
\textit{linear} &= \hspace{2.5em} 1 + 1 \\
\textit{square} &= \hspace{1.7em} 1 + 2 + 1\\
\textit{cubic} &= \hspace{0.85em} 1 + 3 + 3 + 1\\
\textit{quartic} &= 1 + 4 + 6 + 4 + 1
\end{aligned}
\]
@@ -80,19 +80,19 @@
\[
\begin{aligned}
linear &= BLUE[a] + RED[b] \\
square &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
cubic &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\textit{linear} &= BLUE[a] + RED[b] \\
\textit{square} &= BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot RED[b] + RED[b] \cdot RED[b] \\
\textit{cubic} &= BLUE[a] \cdot BLUE[a] \cdot BLUE[a] + BLUE[a] \cdot BLUE[a] \cdot RED[b] + BLUE[a] \cdot RED[b] \cdot RED[b] + RED[b] \cdot RED[b] \cdot RED[b]\\
\end{aligned}
\]
基本上它就是“每个<i>a</i>和<i>b</i>结合项”的和，在每个加号后面逐步的将<i>a</i>换成<i>b</i>。因此这也很简单。现在你已经知道了二次多项式，为了叙述的完整性，我将给出一般方程：
\[
Bézier(n,t) = \sum_{i=0}^{n}
\underset{binomial~term}{\underbrace{\binom{n}{i}}}
\textit{Bézier}(n,t) = \sum_{i=0}^{n}
\underset{\textit{binomial term}}{\underbrace{\binom{n}{i}}}
\cdot\
\underset{polynomial~term}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\underset{\textit{polynomial term}}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\]
这就是贝塞尔曲线完整的描述。在这个函数中的Σ表示了这是一系列的加法（用Σ下面的变量，从...=<值>开始，直到Σ上面的数字结束）。