what is the lowest point of a tilted elliptical plate?

Question

I'd like to know the lowest point $z_\min$ of an ellipse with radius $r_x, r_y$ in (Euclidian) XY that's tilted in XYZ - first rotated around X axis by $\gamma$, then rotated around Y axis by $\alpha$.

If it's only tilted around one axis, it's easy.

$$-z_\min = r_x|\sin(\alpha)|$$ (tx for pointing out Achille)

For the case $\alpha \neq 0$ and $\gamma \neq 0$, I thought about identifying the new direction and corresponding radius that lead to lowest point, but not sure exactly how to go about it.

What's the way to find $z_\min$ with both $\alpha \neq 0$ and $\gamma \neq 0$?

Answer 1

Instead of planar ellipse, let's us recall some results we know about ellipsoid.

Given any ellipsoid with semi-major axis $a,b,c$ and the corresponding symmetry axis pointing along direction $\hat{p},\hat{q},\hat{r}$ (represented as unit vectors). The equation of ellipsoid is given by

$$\frac{(\hat{p}\cdot\vec{x})^2}{a^2} + \frac{(\hat{q}\cdot\vec{x})^2}{b^2} + \frac{(\hat{r}\cdot\vec{x})^2}{c^2} = 1\tag{*1}$$

If $\hat{n}$ is any unit vector, the maximum extent of ellipsoid $(*1)$ along direction $\hat{n}$ is equal to

$$\max\big\{\; \hat{n}\cdot\vec{x} : \vec{x} \text{ satisfies } (*1) \;\big\} = \sqrt{ a^2 (\hat{p}\cdot\hat{n})^2 + b^2 (\hat{q}\cdot\hat{n})^2 + c^2 (\hat{r}\cdot\hat{n})^2 }\tag{*2}$$

If we treat the planar ellipse at hand as the limiting case of a flat ellipsoid and let $\hat{n}$ be the unit vector pointing towards the $-ve$ $z$-direction. We can read off the answer using $(*2)$ once we know the unit vectors $\hat{p}, \hat{q}, \hat{r}$ corresponds to the semi-major axis $a = r_x, b = r_y, c = 0$.

Start from same information as other answer, the transformation matrices for rotation about the $x$-axis for angle $\gamma$ and rotation about the $y$-axis for angle $\alpha$ are:

$$\mathcal{R}_{x,\gamma} = \begin{bmatrix} 1&0&0\\ 0&\cos(\gamma)&-\sin(\gamma)\\ 0&\sin(\gamma)&\cos(\gamma) \end{bmatrix} \quad\text{ and }\quad \mathcal{R}_{y,\alpha} = \begin{bmatrix} \cos(\alpha)&0&\sin(\alpha)\\ 0&1&0\\ -\sin(\alpha)&0&\cos(\alpha) \end{bmatrix}. $$ Multiply $\mathcal{R}_{x,\gamma}$ by $\mathcal{R}_{y,\alpha}$ from the left, the $3$ column vectors of the resulting matrix $$\mathcal{R}_{y,\alpha} \mathcal{R}_{x,\gamma} = \begin{bmatrix} \cos\alpha & \sin\alpha\sin\gamma & \sin\alpha\cos\gamma\\ 0 & \cos\gamma & -\sin\gamma\\ -\sin\alpha & \cos\alpha\sin\gamma & \cos\alpha\cos\gamma \end{bmatrix} $$ will be the $3$ unit vectors $\hat{p}$, $\hat{q}$, $\hat{r}$ we seek. i.e

$$ \hat{p} = \mathcal{R}_{y,\alpha} \mathcal{R}_{x,\gamma} \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} = \begin{bmatrix} \cos\alpha \\ 0 \\ -\sin\alpha \end{bmatrix} \quad\text{ and }\quad \hat{q} = \mathcal{R}_{y,\alpha} \mathcal{R}_{x,\gamma} \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix} \sin\alpha\sin\gamma\\ \cos\gamma\\ \cos\alpha\sin\gamma \end{bmatrix} $$

Throw this to $(*2)$ and recall $\hat{n} = (0,0,-1)$, we will obtain

$$ \begin{cases} \hat{p} \cdot \hat{n} &= \sin\alpha,\\ \hat{q} \cdot \hat{n} &= -\cos\alpha\sin\gamma \end{cases} \quad\implies\quad z_{min} = -\sqrt{ (r_x \sin\alpha)^2 + (r_y\cos\alpha\sin\gamma)^2} $$ Please note that if you tilt along one axis (say $\gamma = 0$) without any "shearing", then $z_{min} = -|r_x \sin\alpha|$ instead of $ -r_x\tan\alpha$.

Answer 2

Consider a parametric version of an ellipse laying in the $xy$ plane is centered at the origin. The following vector will describe the ellipse: $$\begin{bmatrix} r_x\cos(\theta)\\ r_y\sin(\theta)\\0 \end{bmatrix}$$ and $0\le \theta <2\pi.$

A rotation by angle $\gamma$ around the $x$ axis can be performed by the following matrix: $$\begin{bmatrix} 1&0&0\\ 0&\cos(\gamma)&-\sin(\gamma)\\ 0&\cos(\gamma)&\sin(\gamma) \end{bmatrix}$$

With this the vector describing the ellipse canges

$$\begin{bmatrix} 1&0&0\\ 0&\cos(\gamma)&-\sin(\gamma)\\ 0&\cos(\gamma)&\sin(\gamma) \end{bmatrix}\begin{bmatrix} r_x\cos(\theta)\\ r_y\sin(\theta)\\0 \end{bmatrix}=\begin{bmatrix} r_x\cos(\theta)\\ r_y\sin(\theta)\cos(\gamma)\\ r_y\sin(\theta)\cos(\gamma) \end{bmatrix}.$$

A rotation by angle $\alpha$ around the $y$ axis can be performed by the following matrix: $$\begin{bmatrix} \cos(\alpha)&0&\sin(\alpha)\\ 0&1&0\\ -\sin(\alpha)&0&\cos(\alpha) \end{bmatrix}.$$

Multiplying the vector by this matrix we get $$\begin{bmatrix} \cos(\alpha)&0&\sin(\alpha)\\ 0&1&0\\ -\sin(\alpha)&0&\cos(\alpha) \end{bmatrix}\begin{bmatrix} r_x\cos(\theta)\\ r_y\sin(\theta)\cos(\gamma)\\ r_y\sin(\theta)\cos(\gamma) \end{bmatrix}=$$ $$=\begin{bmatrix} r_x\cos(\theta)\cos(\alpha)+r_y\sin(\theta)\cos(\gamma)\sin(\alpha)\\ r_y\sin(\theta)\cos(\gamma)\\ -r_x\cos(\theta)\sin(\alpha)+r_y\sin(\theta)\cos(\gamma)\cos(\alpha) \end{bmatrix}.$$

We are interested only in the $z$ coordinate of the twice rotated ellipse: $$z(\theta)=-r_x\cos(\theta)\sin(\alpha)+r_y\sin(\theta)\cos(\gamma)\cos(\alpha)$$ and we are interested in the lowest point. So, we need to find the absolute minimum of $z(\theta)$ over the interval $0\le \theta <2\pi$.

We have to solve the following equation:

$$\frac{dz}{d\theta}=r_x\sin(\theta)\sin(\alpha)+r_y\cos(\theta)\cos(\gamma)\cos(\alpha)=0.$$

Let $A=r_x\sin(\alpha), B=r_y\cos(\gamma)\cos(\alpha), \text{ and }x=\sin(\theta)$. With this we have two equations

$$\begin{cases} i.&Ax+B\sqrt{1-x^2}=0, \text{ if } 0\le \theta <\frac{\pi}{2} \text{ or }\frac{3\pi}{2}\le \theta <2\pi\\ ii.&Ax-B\sqrt{1-x^2}=0, \text{ if } \frac{\pi}{2}\le \theta <\frac{3\pi}{2}\end{cases}.$$

These equations are easy to solve. However, the final result gets complicated...