# Calculating the angle of a disc to provide given ellipse

Consider a disc viewed from an angle, to give the illusion of an ellipse shape. How can the angle of the disc to the "camera" be calculated from a given ellipse? (For example, 0 degrees = the ellipse shape is a straight line, 90 degrees = the shape is a perfect circle) I suppose it is possible to calculate by measuring the minor and major axis of the ellipse, but how?

• How is the disc positioned? If positioned vertically, then it’s fairly easy to observe that $a=r|\sin\theta|$, $b=r$ – L Parker Feb 21 '18 at 7:52
• The disc is positioned at an angle of approximately 45 degrees to the camera, as measured by measuring the angle between the long axis of the ellipse and neutral plane. I am not quite sure what your abbreviations stand for, is a=short axis and r=long axis? – Tomas Nilsson Feb 21 '18 at 10:24
• By "positioning of disc" I mean this: Is there an angle $\neq 0$ between the normal vector of the disc and the horizontal ground? If the disc stands on the horizontal ground perfectly without tilting, then you should observe the elliptic projection as something like this: one of the axis $b$ does not change (hence equals the radius of disc $r$), whereas the remaining axis $a$ changes periodically with the angular displacement of the camera $\theta$. Absolute sign in my first comment is to make $a \geq 0$ at all times. – L Parker Feb 21 '18 at 10:30

Set up a Cartesian coordinate. Let the radius of disc be $r$.

Let the camera take the $xy$-plane.

Case (1): the plane of disc is parallel to the $xy$ plane

This is essentially an ellipse with one of the axes $b=r$, whereas the remaining axis $a$ changes periodically with the angular displacement of the camera $\theta$. $a=r|\sin\theta|$, in equation:

$$\frac{x^2}{(r\sin\theta)^2}+\frac{y^2}{r^2}=1$$

Case (2): the plane of disc is tilted w.r.t. to the $xy$-plane by an angle of $\phi$ ($0\leq\phi\leq \pi/2$). (angle of "tilting", like Tower of Pisa)

By Lambert's cosine law, we can first interpret the projection radius on the $xy$-plane $r'=r\cos\phi$. This "corrected" disc then rotates like Case (1).

Hence, in equation:

$$\frac{x^2}{(r'\sin\theta)^2}+\frac{y^2}{r'^2}=1$$

$$\frac{x^2}{(r\sin\theta)^2}+\frac{y^2}{r^2}=\cos^2\phi$$

Notice $\phi=0$ corresponds to Case (1).

How to find the angles? You measure three things, $r,a,b$,

$b$ being the elliptical axis that remains unchanged during rotation of the disc.

Then, find $\phi$ using the relation $b=r\cos\phi$

Find $\theta$ (your angle in prompt) using the relation $a=r|\sin\theta|\cos\phi$

$$\phi=\arccos\frac{b}{r} \quad \theta=\arcsin\frac{a}{b}$$