# Size of an object (the space it occupies in your whole field of vision) as you move away from it

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?