What is the area of a cinema screen when seen from an angle?

Question

Let's say that you are going to the cinema. You are choosing where to sit. You know that if you are sitting too close to the cinema screen, the screen will look distorted. If you are sitting too far to the sides of the screen, it will look distorted. The question is, how do you go about finding the apparent area of the rectangular screen (assuming you know the area) given where you are seated?

I assume that the viewer's eye instinctively tracks the centre of the screen. I would think that the area scales by $\cos{\theta} $ (with $\theta$ being the angle of offset). But that would only apply if the viewer was orthogonal to the screen on one of the axes. But, when you are watching a cinema screen, you are looking at it from a deviation on both the vertical and horizontal axis.

Answer 1

Consider two intersecting planes $P_1$ and $P_0$ in the Euclidean space $R^3$ and note by $\theta\in [0,\pi/2]$ their angle. If you are not sure of the definition of $\theta$ consider two vectors $u_0$ and $u_1$ orthogonal to $P_0$ and $P_1$ with length one. Then $|\langle u_0,u_1\rangle|=\sin \theta.$ Now consider the orthogonal projection $x\mapsto p(x)$ from $P_1$ onto $P_0$. Finally if $A\subset P_1$ has area $|A|$ then $$|p(A)|=|A|\cos \theta.$$ Application: $P_1$ is the screen, $u_0$ is the direction of your eye, $A$ is the screen.

Answer 2

If $\vec{A}$ is the area vector of the screen, (i.e, $\left|\vec{A}\right|$ is the area of the screen and $\hat{A}$ the normal vector pointing outwards from the screen) and let $\hat{u}$ denote the unit vector along which you are facing the screen, then $$ A'=\left|\vec{A}\cdot\hat{u}\right| $$ denotes the projected area onto your vision.

This is only applicable if the screen in its entirety is always visible. Otherwise we would have to exclude the area which wouldn't be inside the field of view.