What shape does a "cycloid egg" trace as it rolls on flat surface?

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A point on a circle rolling on a flat surface traces a cycloid.

cycloid

What curve does point on a rolling "cycloid egg" trace?

cycloid egg/ellipse

I tried to draw what it would trace, but failed.

Also, I think the more interesting question here is what shape with a point on its edge traces a half circle as it roll on a flat surface?

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Identify the Euclidean plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$.

Given any geometric shape $K$ lying on the upper half plane touching the real axis at origin, nice enough so that following description make sense. Let $\gamma_K : \mathbb{R} \to \partial K \subset \mathbb{C}$ be a parametrization of $\partial K$, the boundary of $K$, by arc length $s$ subject to the initial condition: $\gamma_K(0) = 0$ and $\gamma_K'(0) = 1$.

When we roll $K$ along the real axis in the positive $x$-direction, the roulette, the locus of the point on $\partial K$ originally at origin will be given by the formula

$$z_K(s) = s - \frac{\gamma_K(s)}{\frac{d\gamma_K(s)}{ds}}\tag{*1}$$

As an example, consider the case $K$ is the unit disk $D$. We can arc-length parametrize $\partial D$ as $$\gamma_D(\theta) = i(1 - e^{i\theta})$$ This corresponding roulette will be a cycloid given by the formula:

$$z_D(\theta) = \theta - \frac{i(1-e^{i\theta})}{e^{i\theta}} = \theta -i( e^{-i\theta} - 1)\tag{*2}$$

When $B$ is a "cycloid egg" $E$ corresponds to a unit circle and the bottom of it is touching the origin, we can use $(*2)$ to work out following parametrization of $\partial E$: $$\gamma_E(\theta) = \begin{cases} \theta - i(e^{i\theta} - 1), & \theta \in [-\pi,\pi]\\ 2\pi - \theta + i(e^{i\theta} + 3),&\theta \in [ \pi, 3\pi] \end{cases} $$ For $\theta$ outside $[-\pi,3\pi]$, we can extend this parametrization by periodicity.

Notice $$\frac{d\gamma_E(\theta)}{d\theta} = \begin{cases} +(1 + e^{i\theta}), & \theta \in [-\pi,\pi]\\ -(1+e^{i\theta}), & \theta \in [ \pi,3\pi] \end{cases} = 2\left|\cos\frac{\theta}{2}\right|e^{i\frac{\theta}{2}} $$ The arc-length $s$ for $\partial E$ is

$$s(\theta) = \int_0^\theta 2\left|\cos\frac{t}{2}\right| dt = \begin{cases} 4\sin\frac{\theta}{2}, &\theta \in [-\pi,\pi]\\ 8 - 4\sin\frac{\theta}{2},&\theta \in [\pi,3\pi] \end{cases} \quad\text{ and }\quad \frac{d\gamma_E(\theta)}{ds(\theta)} = e^{i\frac{\theta}{2}} $$ Substitute this back into $(*1)$, we get

$$z_E(\theta) = \begin{cases} 2\sin\frac{\theta}{2} - \theta e^{-i\frac{\theta}{2}},&\theta \in [-\pi,\pi]\\ 8 - 4\sin\frac{\theta}{2} - (2\pi - \theta + 3i)e^{-i\frac{\theta}{2}} - i e^{i\frac{\theta}{2}},&\theta \in [\pi,3\pi] \end{cases} $$ For $\theta \not\in [-\pi,3\pi]$, pick an integer $N$ such that $\theta_0 = \theta - 4\pi N \in [-\pi,3\pi]$, then periodicity of $\gamma_{E}$ implies $$z(\theta) = z(\theta_0 + 4\pi N) = z(\theta_0) + 16N$$

In terms of real coordinates, we have

$$(x(\theta),y(\theta)) = \begin{cases} \left( 2\sin\frac{\theta}{2} - \theta\cos\frac{\theta}{2}, \theta\sin\frac{\theta}{2}\right), &\theta \in [-\pi,\pi]\\ \left( 8 - 6\sin\frac{\theta}{2} + (\theta-2\pi)\cos\frac{\theta}{2}, -4\cos\frac{\theta}{2} - (\theta-2\pi)\sin\frac{\theta}{2} \right), & \theta \in [\pi, 3\pi] \end{cases} $$ and similarly, $$x(\theta) = x(\theta_0) + 16N\quad\text{ and }\quad y(\theta) = y(\theta_0)$$ for $\theta = \theta_0 + 4\pi N \not\in [-\pi, 3\pi]$, $\theta_0 \in [-\pi,3\pi]$ and $N \in \mathbb{Z}$.

Following is a picture of the roulette of the "cycloid egg" for $\theta \in [0,4\pi]$.

$\hspace1in$ Cycloid of Cycloid egg

As one can see, qualitatively it is very similar to the ordinary cycloid except it is flattened at the top. Aside from that, there doesn't seem anything special.

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Concerning the 2nd question, Oscar's comment is on the right .."track". A circular arc will have its curvature center in a constant point, so the roulette generating it shall have a constant point of contact with the "ground line", i.e. it cannot have a smooth shape.

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Maybe a line segment is the only shape

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