diff --git a/docs/index.html b/docs/index.html index 8696d610..2d9acc70 100644 --- a/docs/index.html +++ b/docs/index.html @@ -38,7 +38,7 @@ - + @@ -179,7 +179,7 @@ 3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu or use the QR code on the right, if that's the kind of convenience you prefer =)

- +
@@ -566,7 +566,7 @@ thus 80% the distance away from the second point) then we can compute that really easily:

@@ -631,7 +631,7 @@ f(x) = cos (x)

So far, so good. Now, let's look at parametric functions, and how they cheat. Let's take the following two functions:

@@ -642,9 +642,9 @@ f(b) = sin (b) function. Parametric functions cheat by changing that. In a parametric function all the different functions share a variable, like this:

@@ -662,8 +662,8 @@ f(b) = sin (b) obvious:

@@ -691,7 +691,7 @@ f(b) = sin (b)

You may remember polynomials from high school. They're those sums that look like this:

@@ -712,12 +712,12 @@ f(x) = a · x + b · x + c · x + d values:

@@ -1083,11 +1083,11 @@ Bézier(n,t) = ❯ \underset binomial term\underbrace\binomni · \ \unders (210,30), we use this Bézier curve:

Which gives us the curve we saw at the top of the article:

@@ -1208,9 +1208,9 @@ function Bezier(3,t,w[]):

Adding these ratio values to the regular Bézier curve function is fairly easy. Where the regular function is the following:

The function for rational Bézier curves has two more terms:

@@ -1423,8 +1423,8 @@ mixture = a · value + b · value And that's easy:

@@ -1494,61 +1494,61 @@ m = a · value + (1 - a) · value refer to coordinate values "in one or more dimensions":

Disregarding our actual coordinates for a moment, we have:

We can write this as a sum of four expressions:

And we can expand these expressions:

Furthermore, we can make all the 1 and 0 factors explicit:

And that, we can view as a series of four matrix operations:

@@ -2535,7 +2535,7 @@ Bézier(n,t) = ❯ w B (t) , where B (t) = \binomni · (1-t) · t t and 1-t:

@@ -2544,9 +2544,9 @@ x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t t component:

@@ -2556,18 +2556,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t) I promise, it's about to make sense. We start with (1-t):

t part, but that's not a problem:

Let's do this:

where the matrix M is an n+1 by n matrix, and looks like:

First, let's look at the derivative rule for Bézier curves, which is:

n) as:

n-1 and k-1, we'll be on the right track.

And that's just a summation of lower order curves:

@@ -2979,7 +2979,7 @@ Bézier (t) ── = n · B (t) · (w - w ) + n · B />

We can rewrite this as a normal summation, and we're done:

@@ -3084,14 +3084,14 @@ tangent (t) = B' (t) point, and then do whatever it is we want to do based on those directions:

@@ -3153,7 +3153,7 @@ y' = x · sin (\phi) + y · cos (\phi) />

Which is the "long" version of the following matrix transformation:

@@ -3616,10 +3616,10 @@ generateRMFrames(steps) -> frames: derivatives section:

And then we turn this into our solution for t using basic arithmetics:

@@ -3656,7 +3656,7 @@ B'(t) = a(1-t) + b(t)= 0, you haven't, it looks like this:

@@ -3696,21 +3696,21 @@ B'(t) uses { v ,v ,v }, where v = 3(p -p ), v = 3(p -p ), v = 3(p -p ) />

And then, using these v values, we can find out what our a, b, and c should be:

fy(x) is zero, so that the next t is the same as the one we already have:

@@ -4312,11 +4312,11 @@ x = x - ─────── by -160, gives us:

@@ -4324,20 +4324,20 @@ x = x - ─────── become:

If we drop all the zero-terms, this gives us:

@@ -4418,7 +4418,7 @@ x = x - ───────

What we need to do is solve a simple equation:

@@ -4428,7 +4428,7 @@ C(t) = 0 circle being on the other side of the curve. So what does C(t) look like? Actually something that seems not too hard:

@@ -4457,15 +4457,15 @@ C(t) = Bézier \prime(t) · Bézier \prime\prime(t) - Bézier \prime(t) with the first and second derivatives, given our basis functions:

And of course the same functions for y:

t values, we end up with the following simple term function for C(t):

@@ -4981,16 +4981,16 @@ mapped = (x) = │ 1 2

First, let's look at the function for x(t):

We can rewrite this to a plain polynomial form, by just fully writing out the expansion and then collecting the polynomial factors, as:

@@ -5001,7 +5001,7 @@ x(t) = (-a + 3b- 3c + d)t + (3a - 6b + 3c)t + (-3a + 3b)t + a known constants, we can trivially rewrite this (by moving the x(t) across the equal sign) as:

@@ -5085,7 +5085,7 @@ y = curve.get(t).y @@ -5197,16 +5197,16 @@ length = | ⟍│(dx/dt) +(dy/dt) dt shifting and scaling the inputs. Doing so, we get the following:

these tables would tell us that for n=2 we must use the following values:

@@ -5410,14 +5410,14 @@ t = + ──── tiny bit:

@@ -5494,7 +5494,7 @@ function kappa(t, B): circle that "fits" the curve's curvature at any point, using what is possibly the simplest bit of maths found in this entire primer:

@@ -5912,22 +5912,22 @@ C = u(t) · P + (1-u(t)) · P (with thanks to Boris Zbarsky) shows us the following two formulae:

And

@@ -5938,7 +5938,7 @@ u(t) = ───────────

We start by observing that, given A, B, and C, the following always holds:

And

@@ -5971,7 +5971,7 @@ ratio(t) = |───────────────| ratio(t) function to find A:

@@ -6034,15 +6034,15 @@ A = B - ───────── = B + ───────── B point, and B and end point as a ratio, using

e2 coordinates must obey the standard de Casteljau rule for linear interpolation:

@@ -6112,7 +6112,7 @@ A = B - ───────── = B + ───────── curve with a loop in it. To do this, we can use the atan2 function:

@@ -6129,8 +6129,8 @@ A = B - ───────── = B + ───────── d:

Which seems simple enough. Unfortunately, when we expand that dist function, things get a lot more problematic:

B, we can compute the associated C:

And then the associated A:

@@ -6708,12 +6708,12 @@ A = B - ────────── = B + ────────── form, where each line corresponds to a power of t, and each column is for a specific coefficient. First, we expand the function:

And then we (trivially) rearrange the terms across multiple lines:

With that arrangement, we can easily decompose this as a matrix multiplication:

So we write out the expansion and rearrange:

Which we can then decompose:

@@ -6889,10 +6889,10 @@ D = ┌ d d ... d ┐, where ╡ 1 i=n fall in the [0,1] interval, so we need to scale all values down by whatever the total length of the polygon is:

T x M x C we talked about before, which gives us:

In which we can replace the rather cumbersome "squaring" operation with a more conventional matrix equivalent:

@@ -6985,7 +6985,7 @@ E(C) = (P - TMC) (P - TMC) those, and then use that 𝕋 instead of T in our error function:

@@ -7042,7 +7042,7 @@ E(C) = (P - 𝕋MC) (P - 𝕋MC) taking the derivative, and determining where that derivative is zero:

@@ -7164,13 +7164,13 @@ C = M (𝕋 𝕋) 𝕋 P previous and next point:

@@ -7951,7 +7951,7 @@ B'(1) = B'(0) last control point through the last on-curve point. And mirroring any point A through any point B is really simple:

B(t). Easy enough:

N(t) always has length 1, which we can achieve by dividing B'(t) by its magnitude:

B'(t), with t = 0 as start and t = 1 as end:

φ:

First we solve for b:

which yields:

@@ -8450,7 +8450,7 @@ b = ──────── polynomial, and we know where the circle arc's actual point P is for angle φ/2:

2" (which is "half height" P2), and so forth:

Wolfram Alpha is definitely the way to go. That said, let's get going:

k:

Which also gives us the commonly found value of 0.55228 for quarter circles, based on them having an angle of half π:

First, let's express the total error (given circular arc angle θ, and some k) using standard calculus notation:

@@ -9541,7 +9541,7 @@ N (t) = ╡ i i+1 P(t), we can compute this point by evaluating d(t) on a curve section between knots i and i+1:

@@ -9582,34 +9582,34 @@ d (t) = 0, d (t) = N (t) = ╡ i i+1 recursion diagram:

diff --git a/docs/ja-JP/index.html b/docs/ja-JP/index.html index 2f9a514e..02148129 100644 --- a/docs/ja-JP/index.html +++ b/docs/ja-JP/index.html @@ -41,7 +41,7 @@ - + @@ -194,7 +194,7 @@
-

— Pomax

+

— Pomax