diff --git a/docs/images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg b/docs/images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg new file mode 100644 index 00000000..c7c29e8c --- /dev/null +++ b/docs/images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg @@ -0,0 +1 @@ + \ No newline at end of file diff --git a/docs/images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg b/docs/images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg new file mode 100644 index 00000000..a71c96b3 --- /dev/null +++ b/docs/images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg @@ -0,0 +1 @@ + \ No newline at end of file diff --git a/docs/images/snippets/abc/3166afa345aec1abda432c39b68d39a0.ascii b/docs/images/snippets/abc/3166afa345aec1abda432c39b68d39a0.ascii new file mode 100644 index 00000000..5f8660ad --- /dev/null +++ b/docs/images/snippets/abc/3166afa345aec1abda432c39b68d39a0.ascii @@ -0,0 +1,10 @@ +\setmainfont[Ligatures=TeX]TeX Gyre Pagella \setmathfontTeX Gyre Pagella Math + + ╭ e - t · A + │ 1 + ╭ e = (1-t) · v + t · A │ v = ─────────── + ╡ 1 1 \Rightarrow ╡ 1 1-t + │ e = (1-t) · A + t · v │ e - (1-t) · A + ╰ 2 2 │ 2 + │ v = ─────────────── + ╰ 2 t diff --git a/docs/images/snippets/abc/8bd3e6fed5bf8d871d30221ae400fd93.ascii b/docs/images/snippets/abc/8bd3e6fed5bf8d871d30221ae400fd93.ascii new file mode 100644 index 00000000..3966253c --- /dev/null +++ b/docs/images/snippets/abc/8bd3e6fed5bf8d871d30221ae400fd93.ascii @@ -0,0 +1,10 @@ +\setmainfont[Ligatures=TeX]TeX Gyre Pagella \setmathfontTeX Gyre Pagella Math + + ╭ v - (1-t) · start + │ 1 + ╭ v = (1-t) · start + t · C │ C = ──────────────────── + ╡ 1 1 \Rightarrow ╡ 1 t + │ v = (1-t) · C + t · end │ v - t · end + ╰ 2 2 │ 2 + │ C = ────────────── + ╰ 2 1-t diff --git a/docs/index.html b/docs/index.html index ee48ee69..d72cb74f 100644 --- a/docs/index.html +++ b/docs/index.html @@ -38,7 +38,7 @@ - + @@ -6112,37 +6112,38 @@ lli = function(line1, line2):

With A found, finding e1 and e2 for quadratic curves is a matter of running the linear interpolation with t between start and A to yield e1, and between A and end to yield - e2. For cubic curves, there is no single pair of points that can act as e1 and e2: as long as the - distance ratio between e1 to B and B to e2 is the Bézier ratio (1-t):t, - we can reverse engineer v1 and v2: + e2. For cubic curves, there is no single pair of points that can act as e1 and e2 (there are + infinitely many, because the tangent at B is a free parameter for cubic curves) so as long as the distance ratio between + e1 to B and B to e2 is the Bézier ratio (1-t):t, we are free to pick any + pair, after which we can reverse engineer v1 and v2:

- +

And then reverse engineer the curve's control points:

- +

So: if we have a curve's start and end points, then for any t value we implicitly know all the ABC values, which (combined with an educated guess on appropriate e1 and e2 coordinates for cubic curves) gives us the necessary information diff --git a/docs/ja-JP/index.html b/docs/ja-JP/index.html index 737e9e87..3f9526dd 100644 --- a/docs/ja-JP/index.html +++ b/docs/ja-JP/index.html @@ -41,7 +41,7 @@ - + @@ -6227,37 +6227,38 @@ lli = function(line1, line2):

With A found, finding e1 and e2 for quadratic curves is a matter of running the linear interpolation with t between start and A to yield e1, and between A and end to yield - e2. For cubic curves, there is no single pair of points that can act as e1 and e2: as long as the - distance ratio between e1 to B and B to e2 is the Bézier ratio (1-t):t, - we can reverse engineer v1 and v2: + e2. For cubic curves, there is no single pair of points that can act as e1 and e2 (there are + infinitely many, because the tangent at B is a free parameter for cubic curves) so as long as the distance ratio between + e1 to B and B to e2 is the Bézier ratio (1-t):t, we are free to pick any + pair, after which we can reverse engineer v1 and v2:

- +

And then reverse engineer the curve's control points:

- +

So: if we have a curve's start and end points, then for any t value we implicitly know all the ABC values, which (combined with an educated guess on appropriate e1 and e2 coordinates for cubic curves) gives us the necessary information diff --git a/docs/news/2020-09-18.html b/docs/news/2020-09-18.html index 6390ebc7..6aed392f 100644 --- a/docs/news/2020-09-18.html +++ b/docs/news/2020-09-18.html @@ -34,7 +34,7 @@ - + diff --git a/docs/news/2020-11-22.html b/docs/news/2020-11-22.html index 0069ef99..35ea6249 100644 --- a/docs/news/2020-11-22.html +++ b/docs/news/2020-11-22.html @@ -34,7 +34,7 @@ - + diff --git a/docs/news/index.html b/docs/news/index.html index 841da1cb..01c0d6e5 100644 --- a/docs/news/index.html +++ b/docs/news/index.html @@ -33,7 +33,7 @@ - + diff --git a/docs/news/rss.xml b/docs/news/rss.xml index 0547f99c..1ee07f2d 100644 --- a/docs/news/rss.xml +++ b/docs/news/rss.xml @@ -6,7 +6,7 @@ News updates for the primer on Bézier Curves by Pomax en-GB - Mon Aug 30 2021 15:13:10 +00:00 + Mon Aug 30 2021 21:59:19 +00:00 https://pomax.github.io/bezierinfo/images/og-image.png A Primer on Bézier Curves diff --git a/docs/ru-RU/index.html b/docs/ru-RU/index.html index 7decb9b5..ef1677f5 100644 --- a/docs/ru-RU/index.html +++ b/docs/ru-RU/index.html @@ -34,7 +34,7 @@ - + @@ -6384,37 +6384,38 @@ lli = function(line1, line2):

With A found, finding e1 and e2 for quadratic curves is a matter of running the linear interpolation with t between start and A to yield e1, and between A and end to yield - e2. For cubic curves, there is no single pair of points that can act as e1 and e2: as long as the - distance ratio between e1 to B and B to e2 is the Bézier ratio (1-t):t, - we can reverse engineer v1 and v2: + e2. For cubic curves, there is no single pair of points that can act as e1 and e2 (there are + infinitely many, because the tangent at B is a free parameter for cubic curves) so as long as the distance ratio between + e1 to B and B to e2 is the Bézier ratio (1-t):t, we are free to pick any + pair, after which we can reverse engineer v1 and v2:

- +

And then reverse engineer the curve's control points:

- +

So: if we have a curve's start and end points, then for any t value we implicitly know all the ABC values, which (combined with an educated guess on appropriate e1 and e2 coordinates for cubic curves) gives us the necessary information diff --git a/docs/uk-UA/index.html b/docs/uk-UA/index.html index 6f10c312..69725d46 100644 --- a/docs/uk-UA/index.html +++ b/docs/uk-UA/index.html @@ -39,7 +39,7 @@ - + @@ -6360,37 +6360,38 @@ lli = function(line1, line2):

With A found, finding e1 and e2 for quadratic curves is a matter of running the linear interpolation with t between start and A to yield e1, and between A and end to yield - e2. For cubic curves, there is no single pair of points that can act as e1 and e2: as long as the - distance ratio between e1 to B and B to e2 is the Bézier ratio (1-t):t, - we can reverse engineer v1 and v2: + e2. For cubic curves, there is no single pair of points that can act as e1 and e2 (there are + infinitely many, because the tangent at B is a free parameter for cubic curves) so as long as the distance ratio between + e1 to B and B to e2 is the Bézier ratio (1-t):t, we are free to pick any + pair, after which we can reverse engineer v1 and v2:

- +

And then reverse engineer the curve's control points:

- +

So: if we have a curve's start and end points, then for any t value we implicitly know all the ABC values, which (combined with an educated guess on appropriate e1 and e2 coordinates for cubic curves) gives us the necessary information diff --git a/docs/zh-CN/index.html b/docs/zh-CN/index.html index 1fc45084..5edb9ee6 100644 --- a/docs/zh-CN/index.html +++ b/docs/zh-CN/index.html @@ -41,7 +41,7 @@ - + @@ -6203,37 +6203,38 @@ lli = function(line1, line2):

With A found, finding e1 and e2 for quadratic curves is a matter of running the linear interpolation with t between start and A to yield e1, and between A and end to yield - e2. For cubic curves, there is no single pair of points that can act as e1 and e2: as long as the - distance ratio between e1 to B and B to e2 is the Bézier ratio (1-t):t, - we can reverse engineer v1 and v2: + e2. For cubic curves, there is no single pair of points that can act as e1 and e2 (there are + infinitely many, because the tangent at B is a free parameter for cubic curves) so as long as the distance ratio between + e1 to B and B to e2 is the Bézier ratio (1-t):t, we are free to pick any + pair, after which we can reverse engineer v1 and v2:

- +

And then reverse engineer the curve's control points:

- +

So: if we have a curve's start and end points, then for any t value we implicitly know all the ABC values, which (combined with an educated guess on appropriate e1 and e2 coordinates for cubic curves) gives us the necessary information