Rational Quadratic Bézier

This was a homework question from the class that I found interesting. The proof method here is quite different from what was outlined in class by Dr. Wang.

§ Question 1

§ Solution

Let $P_0 = A$ and $P_2 = B$ so that $P(0) = A$ and $P(1) = B.$ $A$ and $B$ being points on the unit circle are represented by $A = \langle\cos\alpha, \sin\alpha\rangle$ and $B = \langle\cos\beta, \sin\beta\rangle.$ Assume wlog. that $\beta > \alpha$ so that the subtended angle is given by $\theta = \beta - \alpha,$ and that the origin of our local coordinate frame is at the centre of the circle.

At $t=0.5$ the curve should be the midpoint of our circular arc $M$,

$$\begin{align*} P(0.5) = M &= \frac{0.5P_0 + wP_1 + 0.5P_2}{1+w}\\ (1+w)M &= \frac{1}{2}(P_0 + P_2) + wP_1\\ (1+w)M - wP_1 &= \underbrace{\frac{1}{2}(P_0 + P_2)}_{X}. \end{align*}$$

Observe that $X$ is midpoint of the chord connecting the end-points of the circular arc. From the above $X$, $M$, and $P_1$ are colinear implying that there exists scalars $z,k \in \mathbb{R}$ such that $P_1 = zM = zkX$, yielding the constraint

$$\|X(1 - wk - wzk)\| = (1 - wk - wzk)\|X\| = 1. \tag{\dag}$$

We have that $\|M\| = 1$ from the fact that $M$ is on the circle, this implies $k = \|X\|^{-1}.$ From the illustration, we have that $\angle{XOP_0} = \theta / 2$ thus $\cos(\theta / 2) = \|X\|$ and $k~=~\sec(\theta / 2)$.

Next, notice that $\|P_1\| = \|zM\| = z\|M\| = z$ is equivalent to finding $\|X\| + \|\overline{XP_1}\|$. We found $\|X\|$ previously, enabling us to solve for $\|\overline{XP_1}\|$ — it is the tangent of the angle formed by $\angle P_1P_0X = \gamma$ times the length $\overline{XP_0}$.

To find $\gamma$, we can use the fact the inner product of two direction vectors is proportional to the cosine of the angle, or $\cos\theta = (A \cdot B) / (\|A\|\|B\|)$.

The direction of $\overline{P_1P_0}$ is given by the tangent of the circle at $P(0)$ by definition of the Bézier curve, and the direction of $\overline{P_0P_2}$ is the difference between points.

$$\begin{align*} \gamma &= \arccos\left(\frac{P'(0) \cdot \overline{P_0P_1}}{\|P'(0)\|\|\overline{P_0P_1}\|}\right)\\ &= \arccos\left(\frac{\langle -\sin\alpha, \cos\alpha\rangle \cdot \langle \cos\beta - \cos\alpha, \sin\beta - \sin\alpha \rangle}{\|\langle \cos\beta - \cos\alpha, \sin\beta - \sin\alpha \rangle\|}\right)\\ &= \arccos\left(\frac{-\sin(\alpha)(\cos\beta - \cos\alpha) + \cos(\alpha)(\sin\beta - \sin\alpha)}{\sqrt{(\cos\beta - \cos\alpha)^2 + (\sin\beta - \sin\alpha)^2}}\right)\\ &= \arccos\left(\frac{\cos\alpha\sin\beta - \sin\alpha\cos\beta}{\sqrt{2 - 2\cos(\beta - \alpha)}}\right)\\ &= \arccos\left(\frac{\sin(\beta - \alpha)}{\sqrt{2 - 2\cos(\beta - \alpha)}}\right)\\ &= \arccos\left(\left\|\frac{\sin(\beta - \alpha)}{\sqrt{2 - 2\cos(\beta - \alpha)}}\right\|\right)\tag{$\gamma \in \mathbb{R}^{+}$ by construction}\\ &= \arccos\left[\sqrt{\left(\frac{1}{\sqrt{2}}\cdot\frac{\sin(\beta - \alpha)}{\sqrt{1 - \cos(\beta - \alpha)}}\right)^2}\right]\tag{definition of abs. value}\\ &= \arccos\left[\sqrt{\frac{1}{2}\cdot\frac{\sin^2(\beta - \alpha)}{1 - \cos(\beta - \alpha)}}\right]\\ &= \arccos\left[\sqrt{\frac{1}{2}\cdot\frac{1 - \cos^2(\beta - \alpha)}{1 - \cos(\beta - \alpha)}}\right]\tag{using $\cos^2\theta + \sin^2\theta = 1$}\\ &= \arccos\left[\sqrt{\frac{1}{2}\cdot\frac{(1 - \cos(\beta - \alpha)) (1 + \cos(\beta - \alpha))}{1 - \cos(\beta - \alpha)}}\right]\tag{difference of squares}\\ &= \arccos\left[\sqrt{\frac{1 + \cos(\beta - \alpha)}{2}}\right]\tag{half angle identity}\\ &= \arccos\left[\cos\left(\frac{\beta - \alpha}{2}\right)\right]\\ &= \frac{\beta - \alpha}{2}\\ &= \theta / 2 \end{align*}$$

Solving for $z$,

$$\begin{align*} \|P_1\| = z &= \|X\| + \|\overline{XP_0}\|\tan\gamma\\ &= \cos(\theta / 2) + \sin(\theta / 2)\tan(\theta / 2)\\ &= \sec(\theta / 2). \end{align*}$$

Plugging $z = k = \sec(\theta / 2)$ into ($\dag$) gives

$$\begin{align*} 1 &= (1 - wk - wzk)\|X\|\\ &= \left[1 - w\sec(\theta / 2) - w\sec^2(\theta / 2)\right]\cos(\theta/2)\\ \sec(\theta/2) &= 1 - w\sec(\theta / 2) - w\sec^2(\theta / 2)\\ 1 &= \cos(\theta / 2) - w - w\sec(\theta / 2)\\ \cos(\theta / 2) + 1 &= w + w\sec(\theta / 2)\\ \cos(\theta / 2) + 1 &= w(1 + \sec(\theta / 2))\\ w &= \frac{\cos(\theta / 2) + 1}{\sec(\theta / 2) + 1}\\ w &= \cos(\theta / 2) \end{align*}$$

Therefore, $P_0 = A, P_1 = \frac{1}{2}\sec^2(\theta / 2) (A + B), P_2 = B, w = \cos(\theta / 2)$.

§ Question 2

§ Solution

Plugging in the results from part A, we get the following expression

$$P(t) = \frac{(1-t)^2 P_0 + \sec(\theta / 2) t (1-t) (P_0 + P_2) + t^2 P_2}{(1-t)^2 + 2 \cos(\theta / 2) t (1-t) + t^2},\ t \in [0, 1].$$

Taking the limit of the subtended angle $\theta$ as it tends towards $\pi$ yields an expression for representing a semi-circle with a rational Bézier,

$$\begin{align*} \lim_{\theta \to \pi} P(t) &= \lim_{\theta \to \pi}\frac{(1-t)^2 P_0 + \sec(\theta / 2) t (1-t) (P_0 + P_2) + t^2 P_2}{(1-t)^2 + 2 \cos(\theta / 2) t (1-t) + t^2}. \end{align*}$$

The denominator of the above expression can be solved directly yielding $(1-t)^2 + t^2$.

Next, we focus on evaluating the numerator. Following the assumptions we made about $A, B$, and $\theta$ at the beginning, we have that the limit as $\theta \to \pi$ is the same as the limit as both $\alpha \to 0$ and $\beta \to \pi.$ This enables the numerator to be expressed as the following expression

$$\begin{align*} &(1-t)^2\lim_{\alpha \to 0} \langle \cos\alpha, \sin\alpha\rangle\\ &\quad+ t (1-t) \lim_{(\alpha, \beta) \to (0, \pi)} \sec\left(\frac{\beta - \alpha}{2}\right) \left\langle \cos\alpha + \cos\beta, \sin\alpha + \sin\beta\right\rangle\\ &\quad+ t^2 \lim_{\beta \to \pi} \langle\cos\beta, \sin\beta\rangle. \end{align*}$$

The limits in the first and third line are trivial. We shift our focus to the second line as $\lim_{\theta \to \pi / 2} \sec\theta$ is indeterminant.

Using the identity $\cos \alpha - \cos \beta = \mp2 \sin(A+B) \sin(\mp A\pm B)$, we have that

$$\begin{align*} &\lim_{(\alpha, \beta) \to (0, \pi)}\sec\left(\frac{\beta - \alpha}{2}\right) \left\langle \cos\alpha + \cos\beta, \sin\alpha + \sin\beta\right\rangle\\ =&2\lim_{(\alpha, \beta) \to (0, \pi)}\sec\left(\frac{\beta - \alpha}{2}\right) \left\langle \cos\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right), \sin\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right)\right\rangle\\ =&2\lim_{(\alpha, \beta) \to (0, \pi)}\sec\left(\frac{\beta - \alpha}{2}\right) \left\langle \cos\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\beta - \alpha}{2}\right), \sin\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\beta - \alpha}{2}\right)\right\rangle\\ =&2\lim_{(\alpha, \beta) \to (0, \pi)}\left\langle \cos\left(\frac{\alpha + \beta}{2}\right), \sin\left(\frac{\alpha + \beta}{2}\right)\right\rangle\\ =&\ 2\Big\langle\cos\left(\frac{\pi}{2}\right), \sin\left(\frac{\pi}{2}\right)\Big\rangle\\ =&\ 2\langle 0, 1\rangle \end{align*}$$

Finally, the semi-circle is represented by the following rational quadratic Bézier curve

$$P(t) = \frac{(1-t)^2\langle 1, 0\rangle + 2t(1-t) \langle 0, 1\rangle + t^2 \langle -1, 0\rangle}{(1-t)^2 + t^2}.$$