Projecting an ellipse defined on a sphere onto the XY plane

Question

I want to find the equation for and the area of the ellipse defined by $\Theta_H$ and $\Psi_H$ when it gets projected onto the XY plane.

The following diagram shows these ellipses.

Let us suppose we know the radius $r$. I know that I can approximate the major and minor axes of the ellipse via:

$$ a = 2 r \text{tan}\left(\frac{\Theta_H}{2}\right) $$ $$ b = 2 r \text{tan}\left(\frac{\Psi_H}{2}\right) $$

If I could find the vector equation of the ellipse, then I could just set the z component to zero to project it onto the XY Plane.

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Questions:

1. What is the vector equation for the ellipse? I.e. the 3D xyz equation.
2. Is there another way to do this projection rather than approximating the curve with a ellipse in a plane, finding the equation for that plane, and then setting the z component to zero?

Answer 1

Let : $a:=\frac12 \Theta_H$ and $b:=\frac12 \Psi_H$.

Let $(\psi_0,\theta_0)$ be the spherical coordinates of center of the curve.

A first parameterization is

$$\begin{cases}\psi&=&\psi_0+a \cos(t)&(i)\\\theta&=&\theta_0+b \sin(t)&(ii)\end{cases}\tag{1}$$

i.e., a "true ellipse" wrt coordinates $(\psi,\theta)$ parameterized by a 'non-physical" parameter $t$.

In a second step, we plug these expressions in the classical cartesian/spherical formulas:

$$\begin{cases}x&=&\cos \psi \cos \theta&(i)\\ y&=&\sin \psi \cos \theta&(ii)\\ z&=&\sin \theta&(iii) \end{cases}\tag{2}$$

One gets the projection onto the $xy$ plane, indeed, by plainly suppressing parameter $z$ in (2).

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Edit (2023-04-16): Let us go a step further.

Let us show how to get a polar equation for this projected curve.

First of all, as the projection can be written :

$$\binom{x}{y}=\cos\theta\binom{\cos \psi}{\sin \psi},$$

it means that the polar equation of this curve is

$$r=\cos(\theta) \ \ \iff \ \ r=\cos(\theta(\psi))\tag{3}.$$

The second form in (3) stresses the fact that we desire to express $\theta$ as a function of $\psi$ in order to get finaly the polar equation under the forme $r=F(\psi)$ for a certain function $F$.

Using first (1)(ii), then (1)(i), we have :

$$r=\cos(\theta_0+b \sin(t))=\cos(\theta_0+b \sin(\cos^{-1}(\tfrac{\psi-\psi_0}{a})))$$

Otherwise said :

$$r=\cos\left(\theta_0+b \sqrt{1-(\tfrac{\psi-\psi_0}{a})^2}\right)$$

which is the function $r=F(\psi)$ we were looking for.

Calculating (in a numerical way here) the area enclosed by a polar curve is easy : it is

$$\frac12 \int_{\psi=0}^{2 \pi} r^2 d\psi$$