Re: 2D in a 3D world

On 8 August 2010 04:21, Jason Hughes <[email protected]> wrote:

> - @Colin: I did read that previously, which is why I attempted a half-pixel
> offset with the projection and/or view matrices.  This did not seem to have
> the desired effect of correcting a solid gray to black and white.  I was a
> bit surprised at that.  No value I could put in there seemed to do more than
> 25% correction, at best, which leads me to believe it's a texture issue.
>

I see where you said that now: I was a little bit too anxious to stick my
oar in. Apologies!


> I guess the real question I had was, how is the best way to correct for
> pixel/texel mismatches?  Do most people adjust the view matrix or the
> projection matrix, or do you modify the vertices on the quads you generate,
> or do you trick it with texture matrix modifications or generate the UVs
> differently?  Lots of options.  The easiest seemed to me to be the view
> matrix, but when it didn't work, I started looking for other things that
> could affect the calculation, but didn't find any culprits.
>

I like to think of it not so much as an adjustment of those matrices, but as
another transform applied after the model/view/projection. If you consider
just the x-coordinate, the post-projection value of the sides of your
fullscreen quad are -1 and 1. The viewport transform is:

x * 0.5 * vpw + (vpx + 0.5 * vpw)

Which (assuming vpx is 0) gives you:

-1: -0.5 * vpw + 0.5 * vpw = 0
1: 0.5 * vpw + 0.5 * vpw = vpw

Where you actually want to be, accounting for the position offset, is -0.5
and vpw - 0.5. So you solve for x and you get (leaving out vpx for
simplicity):

x * 0.5 * vpw + 0.5 * vpw = -0.5 -> x = -(1 + vpw) / vpw
x * 0.5 * vpw + 0.5 * vpw = vpw - 0.5 -> x = (-1 + vpw) / vpw

Subtract -1 and 1 respectively and you get a constant offset of -1/vpw. This
is essentially the same thing that Jon said already, except I'm working in
post projection space so my offset is double his ( at least, I hope that's
why :-) ).

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