Reparametrization of rational Bézier curve

Question

I am trying to solve the following task

Using rational Bézier curve find the control points and weights of one sixth of a circle $c_1$, such that $$c_1(0)=\{3,0\},c_1(1)=\{\frac{3}{2},\frac{3\sqrt{3}}{2}\},c_1(\pi/4)=\{\frac{3}{\sqrt{2}},\frac{3}{\sqrt2}\}.$$

I have calculated the control points $\{3,0\},\{3,\sqrt{3}\},\{\frac{3}{\sqrt2},\frac{3}{\sqrt2}\}$,their respective weights $\{1,\frac{\sqrt3}{2},1\}$ and their knots $\{0,0,0,\pi/4,\pi/4,\pi/4\}.$

I don't know how to proceed from here with reparametrizing the curve so that the wanted equalities hold.

Answer 1

A different hint …

You have constructed a curve that has the correct shape (it’s the desired circular arc), but its parameterization is not what you want.

You can change the parameterization by changing the weights. But you need the shape of the curve to remain unchanged. The key point is this: the shape of the curve will not change long as you don’t change the quantity $$ \sigma = \frac{w_0w_2}{w_1^2} $$ You can arbitrarily set $w_0 = 1$, and then adjust $w_1$ and $w_2$ to get the parameterization you need. Specifically, you just have to find $w_1$ and $w_2$ such that $\sigma = \tfrac43$, as before, and such that $c(\pi/4)$ has the desired value.

Answer 2

I will proceed by successive logical steps.

1. As you are given 3 points connected by an arc of circle : this (unique!) circle $(C)$ is easily found to have its center in $(0,0)$ and its radius $3$.

2. Now remember that the classical parametrization of the unit circle is (so called "half-angle formulas" with $t=\tan(\alpha/2)$ ) is given by

$$x=cos(\alpha)=\frac{1-t^2}{1+t^2}, \ \ y=\sin(\alpha)=\frac{2t}{1+t^2}$$

(with uniform speed). These formulas are naturaly rational Bezier formulas

Therefore we can take as initial rational Bezier formulas for (C) are :

$$x=3 \frac{1-t^2}{1+t^2}, \ \ y=3\frac{2t}{1+t^2} \ \text{for} \ 0 \le t \le 1/\sqrt{3}$$

3. Why these values of $t$ ?

No problem for value $t=0$ for which $x=3,y=0$ as awaited.

For value $t=1/\sqrt{3}$, it is obtained by solving :

$$\frac32=3 \frac{1-t^2}{1+t^2} \ \text{or the equivalent eq.} \ \frac{3\sqrt{3}}{2}=3\frac{2t}{1+t^2}$$

The same operation for the intermediate point gives $t=\sqrt{2}-1$.

4. Therefore the re-parametrization issue boils down to find an homographic transform (linear fractional transformation) : $$y=\frac{ax}{x-b}$$

passing through points $$(0,0), (\sqrt{2}-1,\pi/4),(1/\sqrt{3},1)$$

with : $$a = (\pi/4)\frac{(\sqrt{6} - \sqrt{3} - 1)}{(\sqrt{6}-\sqrt{3} - \pi/4)}\approx 3.2655$$

$$ b =-\frac{(\pi/4)(\sqrt{2} - 1) - \sqrt{2} + 1}{\sqrt{6} - \sqrt{3} -\pi/4}\approx-1.3080$$