Calculate distance after rotation? I'll start off by saying that I suck at math.
I'm trying to calculate the distance between a circle and the center of the screen after rotating an image that contains that circle by $45^\circ$ in $3D$,
(The $y$ distance of the object changes as the image rotates)

I hope I made myself clear, thanks in advance
 A: Let's suppose that the projection point is on the line perpendicular to the middle of the screen (at a distance $z$) then the vertical tangent at the view point will change from $\displaystyle\frac yz$ to $\displaystyle\frac {y\,\cos(\alpha)}{z+y\,\sin(\alpha)}$ after a vertical rotation of angle $\alpha$. 
This implies that $\,y\,$ will become $\ \displaystyle h(\alpha):=\frac {y\,z\,\cos(\alpha)}{z+y\,\sin(\alpha)}$.
For $\,\alpha=\frac{\pi}4\,$ you should get : $$y':=h\left(\frac{\pi}4\right)=\frac {y\,z}{y+\sqrt{2}z}$$

After rotation of $(oy)$ of an angle $\alpha$ we get following picture (from the side) :

Now observe the triangle obtained and notice that :
$$\frac{y\,\cos(\alpha)}{z+y\,\sin(\alpha)}=\frac{y'}z$$
A: I thought of a more naiive way of doing this calculation that I hope illustrates how images are really taken of things, and how that means you have to manipulate the camera to manipulate the image.
In step 1, the perspective point F and either of the other points (the centers of the planes) can be chosen arbitrarily.
Something to think about: images in our vision are really upside-down, but our brains correct them.

