# Rolling an elliptical disc on the $x$ axis

You're given the elliptical disc bounded by

$$\dfrac{x^2}{a^2} + \dfrac{(y - b)^2}{b^2} = 1$$

where $$a = 5, b = 2$$. You roll this ellipse to the right along the positive $$x$$ axis, such that it is always tangent to the $$x$$ axis. You stop rolling when the elliptical boundary of the disc becomes tangent to the $$x$$ axis at $$(5, 0)$$. If a point $$P_0 = (-2, 3)$$ is on the disc before rolling, find its position after rolling.

My attempt:

The distance travelled on the $$x$$ axis, is the same traversed on the boundary of the ellipse. From the parametric equation of the ellipse, one can determine the eccentric angle of the point of tangency on the rolled ellipse using the arc length (which is equal to $$5$$). Then, the initial and final normal vectors to the circumference of the ellipse can be computed, and the angle between them determines the rotation of the ellipse. Having the tangency point and the rotation determines the center of the rolled ellipse. The affine transformation between the initial points on the disc and rolled images is given by

$$p' = C + R (p - C_0)$$

where $$R$$ is the rotation matrix resulting from rolling, $$C_0$$ is the initial center (equal to $$(0, b)$$) and $$C$$ is the final center. Using this equation with $$p = (-2, 3)$$, one can compute its image $$p'$$.

• To do your task one needs to find the central angle by knowing that the length of the subtended elliptical arc is $5$. But that requires elliptic functions, I'm afraid. Commented May 31 at 16:14

A point on the ellipse $$\frac{x^2}{25}+\frac{(y-2)^2}{4}=1$$ can be written as $$(5\cos\theta,2\sin\theta+2)$$ where $$0\le\theta\lt 2\pi$$.

Let $$A(5c,2s+2)$$ (where $$c:=\cos\alpha,s:=\sin\alpha$$ and $$\frac{3}{2}\pi\lt\alpha\lt 2\pi$$) be a point satisfying $$5=\int_{\frac{3}{2}\pi}^{\alpha}\sqrt{(-5\sin\theta)^2+(2\cos\theta)^2}\ d\theta$$

We have $$\int \sqrt{(-5\sin\theta)^2+(2\cos\theta)^2}\ d\theta=2 E\bigg(\theta\ \bigg|-\frac{21}{4}\bigg) + C$$ where $$E(\theta|m)$$ is the elliptic integral of the second kind with parameter $$m=k^2$$.

According to WolframAlpha, we get $$\alpha\approx 5.9388$$

Let $$\ell$$ be the tangent line at $$A$$. Then, the equation of $$\ell$$ is given by $$\ell : y-(2s+2)=\frac{-2c}{5s}(x-5c)$$ i.e. $$2cx+5sy-10s-10=0$$

Let $$B$$ be a point on $$\ell$$ such that $$P_0B\perp \ell$$.

Then, the equation of the line $$P_0B$$ is given by $$y-3=\frac{5s}{2c}(x+2)$$ i.e. $$5sx-2cy+10s+6c=0$$

Since we can see that $$P_0$$ is under the normal at $$A$$, we finally get \begin{align}x&=5\color{red}-AB \\\\&=5-(\text{distance between A and P_0B}) \\\\&=5-\frac{|5s\times 5c-2c(2s+2)+10s+6c|}{\sqrt{(5s)^2+(-2c)^2}} \\\\&=5-\frac{|10s+21cs+2c|}{\sqrt{4 c^2 + 25 s^2}} \\\\&\approx \color{red}{1.7701}\end{align} and \begin{align}y&=P_0B \\\\&=(\text{distance between P_0 and \ell}) \\\\&=\frac{|2c\times (-2)+5s\times 3-10s-10|}{\sqrt{(2c)^2+(5s)^2}} \\\\&=\frac{|-4c+5s-10|}{\sqrt{4 c^2 + 25 s^2}} \\\\&\approx \color{red}{6.1115}\end{align}

Let $$A(a,b)$$ be a point on the ellipse $$\frac{x^2}{5^2}+\frac{(y-2)^2}{2^2}=1$$ such that the arc length from the origin $$O(0,0)$$ to $$A(a,b)$$ is $$5$$ units as shown in the figure below.

We have,

$$\frac{\mathrm dy}{\mathrm dx}=-\frac{2^2x}{5^2(y-2)}=-\frac{4x}{25(y-2)}\tag{1}$$ Arc length from $$x=0$$ to $$x=a$$, $$\int_0^a\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\ \mathrm dx=5\int_0^a\dfrac{\sqrt{1-\frac{21}{25}\left(\frac x5\right)^2}}{\sqrt{1-\left(\frac x5\right)^2}}\ \mathrm dx=5E\left(\varphi\ \bigg|\ \frac{21}{25}\right)\tag{2}$$ where $$\sin\varphi=\frac a5$$, and $$E(\varphi\ |\ k^2)$$ is the incomplete elliptic integral of the second kind with parameter $$m=k^2$$. We have to solve $$5E\left(\varphi\ |\ \frac{21}{25}\right)=5$$ or $$E\left(\sin^{-1}\left(\frac a5\right)\ |\ \frac{21}{25}\right)=1$$ for $$a$$. Since the elliptic integral cannot be expressed in elementary functions, it is often computed numerically. According to WolframAlpha, $$\color{red}{a\approx4.7064441415}$$. And $$\color{red}{b=2-2\sqrt{1-\frac{a^2}{25}}\approx1.3247953148}$$.

• First calculate the angle $$(\alpha)$$ made by $$AP_0$$ with the positive $$x-$$axis. $$\tan\alpha=\frac{3-b}{-2-a}\approx-0.2497902987\implies\color{red}{\alpha\approx2.8968113661} \text{ rad}$$. And $$\color{red}{|AP_0|}=\sqrt{(-2-a)^2+(3-b)^2}=\color{red}{|A'P_0'|\approx6.9125034365}$$.

• At $$A(a,b)$$, $$\tan\beta=\frac{4a}{25(2-b)}\approx1.1152633848\implies\color{red}{\beta\approx0.8398355941} \text{ rad}$$ (angle made by the tangent with $$x-$$axis at $$A$$).

Observe that $$A'P_0'$$ makes an angle of $$\color{red}{(\alpha-\beta)\approx2.056975772}\text{ rad}$$ with the positive $$x-$$axis.

So, by projection $$\color{blue}{\boxed{p=5+|A'P_0'|\cos(\alpha-\beta)\approx1.77012267}}$$ $$\color{blue}{\boxed{q=|A'P_0'|\sin(\alpha-\beta)\approx6.11151341}}$$

Hence, the position of $$P_0$$ after rolling will be $$\color{blue}{P_0'(1.77012267,6.11151341)}$$ approximately.

Hope this helps!

The transformation can be executed via a rotation and translation. That means 3 degrees of freedom must be determined, the angle of rotation and the magnitude and direction of the translation.

We are given $$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$$

This can be parameterized as:

$$x=x_0+a\sin \theta$$

$$y=y_0-b \cos \theta$$

Rolling means: $$x_c= \int_0^\lambda \sqrt{a^2\cos^2\theta + b^2 \sin^2\theta}\ \mathrm d\theta=aE\left(\lambda\ \bigg|\ \left(1-\frac{b^2}{a^2}\right)\right)$$

That means that the transformation is mapping $$(x_0+a\sin \lambda, y_0-b \cos \lambda) \to (x_c,0)$$

$$\frac{-(x_\lambda-x_0) b^2}{a^2(y_\lambda-y_0)}=\frac{-ab^2\sin \lambda}{-a^2b\cos \lambda}=\frac{b\sin \lambda}{a \cos \lambda}=(b/a)\tan \lambda$$

The slope of the tangent line at the point on the original curve becomes 0 under a rotation of $$\theta_c$$

$$\tan \theta_c = (b/a)\tan \lambda$$

$$a \sin \theta c \cos \lambda = b \cos \theta_c \sin \lambda$$

$$\begin{bmatrix} \cos \theta_c& \sin \theta_c \\ -\sin \theta_c & \cos \theta_c\end{bmatrix} \begin{bmatrix} (x_0+a \sin \lambda) \\ (y_0-b\cos \lambda) \end{bmatrix} + \begin{bmatrix} p \\ q \end{bmatrix}= \begin{bmatrix} x_c \\ 0 \end{bmatrix}$$

$$p= x_c-\cos \theta_c (x_0+a\sin \lambda)-\sin\theta_c (y_0-b\cos \lambda)$$

$$p=x_c- x_0 \cos \theta_c - a \cos \theta_c \sin \lambda -y_0 \sin \theta_c + b \sin\theta_c \cos \lambda$$

$$q= \sin \theta_c(x_0+a\sin \lambda)-\cos \theta_c(y_0-b\cos \lambda)$$

$$q= x_0\sin \theta_c + a \sin \theta_c \sin \lambda - y_0\cos \theta_c + b \cos \theta_c \cos \lambda$$

$$x'=x \cos \theta_c + y \sin \theta _c +x_c-\cos \theta_c (x_0+a\sin \lambda)-\sin\theta_c (y_0-b\cos \lambda)$$

$$x'=(x-x_0)\cos \theta_c + (y-y_0)\sin \theta_c +x_c - a \cos \theta_c \sin \lambda + b \sin \theta_c \cos \lambda$$

$$y'=-\sin \theta_c x + \cos \theta_c y + \sin \theta_c(x_0+a\sin \lambda)-\cos \theta_c(y_0-b\cos \lambda)$$

$$y' = - \sin \theta_c (x-x_0)+ \cos \theta_c (y-y_0)+ a \sin \theta_c \sin \lambda + b \cos \theta_c \cos \lambda$$

Summary:

$$x_c= \int_0^\lambda \sqrt{a^2\cos^2\theta + b^2 \sin^2\theta} \ \mathrm d\theta=aE\left(\lambda\ \bigg|\ \left(1-\frac{b^2}{a^2}\right)\right)$$

$$\tan \theta_c = (b/a)\tan \lambda$$

$$x'=(x-x_0)\cos \theta_c + (y-y_0)\sin \theta_c +x_c - a \cos \theta_c \sin \lambda + b \sin \theta_c \cos \lambda$$

$$y' = - \sin \theta_c (x-x_0)+ \cos \theta_c (y-y_0)+ a \sin \theta_c \sin \lambda + b \cos \theta_c \cos \lambda$$

Numerical computation:

$$x_c=5,\ x_0=0,\ y_0=2,\ a=5,\ b=2,\ x=-2,\ y=3$$.

$$5=5E\left(\lambda\ \bigg|\ \frac{21}{25}\right)\implies\color{red}{\lambda\approx1.226428}$$

$$\theta_c=\tan^{-1}\left(\frac25\tan(\lambda)\right)\implies\color{red}{\theta_c\approx0.839836}$$

$$\color{red}{\boxed{x'\approx1.7701,\ y'\approx6.1115}}$$

The distance traversed is $$5$$ in length. This distance is the arc length of the ellipse circumference from an eccentric angle of $$t = - \dfrac{\pi}{2}$$ to $$t = \alpha$$. Now the arc length between eccentric angles $$t_1$$ and $$t_2$$ is given by

$$s = \displaystyle \int_{t_1}^{t_2} \| p'(t) \| \ dt$$

where $$p(t)$$ is the position on the ellipse and is given by

$$p(t) = ( 5 \cos t , 2 + 2 \sin t )$$

So that

$$p'(t) = (- 5 \sin t, 2 \cos t )$$

Therefore,

$$s = 5 = \displaystyle \int_{-\dfrac{\pi}{2}}^\alpha \sqrt{ 25 \sin^2 t + 4 \cos^2 t } \ d t$$

To find $$\alpha$$, I used the single variable Newton's method coupled with a simple Simpson's-rule numerical integrator, and this gave me

$$\alpha = -0.344368568$$

which is consistent with the angle found by @mathlove (the difference being $$2 \pi$$; because he took $$t_1$$ to $$\dfrac{3 \pi}{2}$$ ).

Having found the final angle $$\alpha$$, now we have to calculate the rotation of the ellipse, and for that, we use the normal vectors at $$(0, 0)$$ and at $$P_\alpha = ( 5 \cos \alpha, 2 + 2 \sin \alpha )$$.

The normal vector is given by

$$N = Q (P - C_0)$$

where $$Q = \begin{bmatrix} \dfrac{1}{5^2} && 0 \\ 0 && \dfrac{1}{2^2} \end{bmatrix}$$ And $$C_0 = (0, 2)$$

and then we normalize both vectors, this gives

$$N_0 = (0, -1)$$

$$N_1 = ( 0.744533328, -0.667585293 )$$

Now, in the rolled ellipse, we want $$N_1$$ to point in the $$(0, -1)$$ direction, and this means (since both vectors are unit vectors) that

$$\sin \theta = N_1 \times N_0 = -0.744533328$$

Therefore, the angle of rotation is $$\theta = - 0.839835524 = -48.119031^\circ$$

With this we can compute the rotation matrix for the ellipse as follows

$$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix}$$

And consequently, we can compute the new ellipse matrix as follows

$$Q = R D R^T$$

where $$D = \begin{bmatrix} \dfrac{1}{5^2} && 0 \\ 0 && \dfrac{1}{2^2} \end{bmatrix}$$

Finally, we have to compute the center of the new ellipse.

The equation of the ellipse is

$$(r - C)^T Q (r - C) = 1$$

Where the unknown is vector $$C$$. But we do know that $$r_1 = (5, 0)$$ is on this ellipse, and that the normal vector points in the $$- \mathbf{j}$$ direction ( $$\mathbf{j} = [0, 1]^T$$ ).

This normal vector direction is nothing but the direction of the gradient of the ellipse which is given by

$$\nabla = 2Q (r_1 - C)$$

Therefore,

$$Q (r_1 - C) = - k \mathbf{j}$$

So that

$$(r_1 - C) = - k Q^{-1} \mathbf{j}$$

Substituting this expression into the equation of the ellipse, gives us

$$k = \dfrac{1}{\sqrt{ \mathbf{j}^T Q^{-1} \mathbf{j} } } = \dfrac{1}{ \sqrt{ Q^{-1}_{22} }}$$

Since we now have $$k$$ (remember that $$Q$$ is known), then from the above equation, it follows that

$$C = r_1 - k Q^{-1} \mathbf{j} = r_1 - \dfrac{Q^{-1} \mathbf{j}}{ \sqrt{ Q^{-1}_{22} } }$$

Hence, the rolling of the ellipse is equivalent to the following three steps: 1. Shift the original ellipse so that it center becomes at $$(0,0)$$. 2. Rotate the ellipse by the angle $$\theta$$ found above. 3. Shift the rotated ellipse so that its center moves from the origin to $$C$$ found above.

It follows that we have a point $$P_0$$ attached to the ellipse frame, then its final location is given by

$$P_1 = C + R (P_0 - C_0)$$

Substituting $$C_0, P_0, R, C$$ gives

$$P_1 = (1.770122358, 6.111513188)$$