I'm not sure if this is the right forum for this question, and it must be already answered somewhere so you can just point me to the answer.

If there is a disc of radius $r$ and an observer is moving away from it at a constant speed while staying "aligned" with the center (on an orthogonal line that passes through the center), how does the ratio $\frac{\text{area of the disc}}{\text{total area of visible things in the plane of the disc}}$ evolve?

Let's say the observer can see anything that doesn't require him to move his eyeball (he's a cyclope w.l.o.g) by more than a certain angle $\theta\in (0,\pi)$. Let's call $L$ the distance between the eye and the center of the disc. If $L=\frac{r}{\tan\theta}$ then the ratio occupies all of the visible area and our ratio is $1$.

I found that the ratio is $\frac{\pi r^2}{\pi (L\tan\theta)^2}=\frac{r^2}{L^2\tan^2\theta}$ so the evolution of the space taken by the disc in the field of vision is $\frac{\partial}{\partial L} \frac{r^2}{L^2\tan^2\theta}=\frac{r^2}{\tan^2\theta}\frac{-2}{L^3}$ does that make sense?


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