Re: [Algorithms] Hemisphere of points ?

[Peter-Pike S. <pp...@wi...>]

The differential solid angle is the same as it is for cube maps - you
are just stretch out the plane:

=20

1/( (u^2 + v^2 + 1)^(3/2) )

=20

This is assuming the "Z" is 1, and u/v are the XY axis (so a cube map
would just evaluate this expression from [-1,1], if you want a bigger
FOV just make u/v go outside of [-1,1]...)

=20

No numerical computation is necessary for this term.

=20

-Peter-Pike Sloan

=20

________________________________

From: gda...@li...
[mailto:gda...@li...] On Behalf Of
Jon Watte
Sent: Monday, March 05, 2007 3:48 PM
To: Game Development Algorithms
Subject: Re: [Algorithms] Hemisphere of points ?

=20



Paul at Home wrote:=20

Well, ok. I'll be honest, I don't really have the ability either. When I

read
=20
 =20

		You will have to compensate for the anisotropy of each
		pixel in the render target, but that is a pre-process
weighting matrix
		you can calculate once (and fudge if you want :-)
		     =20

=20
I start to dribble! Do you think rhombic pixels would have a big enough=20
effect to worry about ?
=20
 =20


My guess is that it's only the area that's a problem. You will have a
larger area area (in steradians) of a pixel at the corner, as opposed to
a pixel in the center, because of the planar projection. This is really
quite simple trig; turning one delta-degree into delta-length at the far
edge of a triangle (whose English name I've never learned). You can
calculate it with finite differencing, even, subtracting tan(A) from
tan(A+delta).

You can also plot the X, Y locations of the different directions onto a
flat grid, to see the distribution for yourself. You could even measure
(using a ruler) the difference in area, and fudge the compensation (it's
going to be the same for all pixels), which is why I said it's
fudge-able :-)

I've wanted to do something like this for quite a while, but never had
the time, so my ulterior motive is for you to do it and post the code,
btw :-)

Cheers,

          / h+