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What I know of it (which isn't much) is, if you have a camera at -10*z
looking at the origin (<0,0,0>) using 'angle' 22.6 you get 4 POV units
wide by 3 POV units high approximately.
And if using 'direction' instead of angle for the "lens zoom" setting
you would need 'direction' 10/3 (ten divided by 3) to approximate a
'angle' 22.6 (the old POV-Ray camera default).
So 'direction' 10/3 is about the same area as angle 22.6.
In your example to get that -30*x object to show at the far left when
the camera is located at -30*z would require 'angle 90' or so. A very
wide lens setting. Keep in mind this is the default 'perspective'
camera.
Anyhow, dividing 30 by 10 (distance) you get 3 times the area. These
aren't exact numbers at all remember, only close.
Hope this is a start of help about this.
I Chuprov wrote:
>
> Ok, say I set up a scene, and place the camera at <0, 0, -30> and look_at is set
> to <0,0,0>
>
> How many units are across the "screen" (so to speak)?
> If I put something at <-30,0,0>, will it be on the far left of the screen?
>
> Thanks,
> Igor
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