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View Full Version : What is the max value for FOV ?



transnept
02-08-2002, 10:21 AM
Hi folks !

What is the maximum value for a field of view.
Can i go to 180 ? 360 ?

Knowing that the rendering would be done in a normal 1280x768 viewport.

Thanks !

Bob
02-09-2002, 01:57 AM
There's no maximum value of the FOV, only a threshold where you, as a user, find the projected image too distorted.

I, personally, think a vertical FOV of about 60 degrees for a fullscreen window is enough. Anything larger results in too much distortion.

Tron
02-09-2002, 08:53 AM
@Bob:
normal FOV is 90 degree, so 60 degree is pretty distorted and zoomed

@transnept:
the limit is < 180, cause the matrix is generated with tan(FOV / 2) and tan(90) is undefiend

Bob
02-09-2002, 01:36 PM
90 degress, huh? I think a scene is pretty distorted then. As I said, it's the user that defines the limit, and every use has a different limits. Saying 90 degrees is normal, is as wrong (and correct) as saying 60 is normal. i said what I think is normal.

And how can you say something is normal in 90 degrees but distorted in 60? Heck, place a cube in a corner in both 60 and 90 degree FOV. With which FOV does the cube look most like a cube (all corners closest to 90 degrees). That's what I mean with distortion. In 90 degree FOV, a cube in the corner of a view doens't look like a cube, but a pyramid or something similar.

Tron
02-09-2002, 09:04 PM
this is a simple perspective projection matrix:
A 0 0 0
0 B 0 0
0 0 C D
0 0 1 0

A = 1.0 / Aspect
B = 1.0
C = (Far + Near) / (Far - Near)
D = -2.0 * Far * Near / (Far - Near)


this is a perspective projection matrix with FOV:
A 0 0 0
0 B 0 0
0 0 C D
0 0 1 0

A = 1.0 / tan(Angle / 2.0 / 180.0 * M_PI) / Aspect
B = 1.0 / tan(Angle / 2.0 / 180.0 * M_PI)
C = (Far + Near) / (Far - Near)
D = -2.0 * Far * Near / (Far - Near)

they're only equal for the case FOV == 90
take pen and paper and build a projection matrix yourself and you'll see...


btw: in Quake the normal FOV is 90 too and i think JC know's much about projection and FOV.
and with every other value things look zommen and distorted.

02-09-2002, 09:57 PM
looking at this, im wondering if it works on half the angle basis ie straight ahead is 0 and 45 degrees is the viewing angle, aka 90, this looks like it could be the case but i cant say for certain.
45 0 45
\ | /
\|/

Bob
02-10-2002, 02:43 AM
Tron, I can see that the two matrices are equal then the FOV is 90 degrees, but what does that has to do with the fact that objects looks distorted or not? When playing any FPS with 90 degree FOV, I can clearly see how the objects are distorted when they reach the edge of the screen. In my oppinion, it's quite annoying.

My monitor is about 13 inch high, and I have my eyes about 18 inch away. Using some trigonometry and geometry, I get the vertical FOV to about 40 degrees. That's the true FOV I see when looking at a fullscreen window on my monitor. An I have tried it, and it doens't look bad at all. The fact that you don't see much of the surrounding is another problem. But that's the amount of environment I would see if I was looking through a rectangle of that size, at that distance, in real life.