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d123
04-11-2004, 05:59 AM
Hi guys,
I was wondering if I can get some help about
how hardware is implementing the zbias.
If i'm using a clasic zbuffer not wbuffer how is
the zbias applyed? Is it in homogenouse space
before w divide or in screen space after w divide?
Is it 1/(w + offset) or 1/w + offset ? I'm
considering that the offset is constant per pixel.
If the offset is not constant per pixel inside a
primitive what is the offset formula ?
Regards.

jonasmr
04-11-2004, 07:27 AM
Hi.

The offset is done in screenspace, and it is constant for each polygon. See the spec, sec. 3.5.5.

Regards, Jonas

d123
04-11-2004, 11:58 AM
Hello,
Using the following formulas for calculating the
offset: offset=m*sloapFactor + r * unitsFactor, where m is the maximum z sloap and the r for the an integer zbuffer is one whats the sense for sloapfactor ?

Regards.

d123
04-11-2004, 12:22 PM
I mean if you render a sphere at resolution 640x480 and take a zbuffer picture and after
you render the sphere at 800x600 and take z pictures the zvalues for the 2nd sphere will be
greater than the first sphere (the sphere remains
at the same position in camera space) considering that the sloapScale >= 1. Thats because the m will be grater in the 2nd case since m = max(abs (dz/dx), abs(dz/dy)) (2nd sphere dx is greater than 1st sphere dx).

jonasmr
04-11-2004, 10:54 PM
m for the second sphere(at 800x600) should be smaller, since dx is larger and the denominator :)

Your observation, is as I see it, correct. The offset does depend on the resolution.

To see why this is desireable, consider shadow mapping - nvidia has a presentation about it, which explains the details. (Its a bit hard to explain, and can be tricky to understand)

dorbie
04-13-2004, 06:52 AM
Slope factor is there because rendering has rounding errors and these impact what interpolated fragment depth you end up with. The steeper the derivative of z for a polygon the greater the potential z sampling error is due to transformation precision, subpixel rounding, interpolation precision, eye z precision etc. A fixed offset cannot compensate for this on a polygon with varying orientation. Basically the steeper the slope in z the more offset you need, but you don't want excessive offset when the derivative of z is low.

The offset gives you two values, one is a constant offset for when the polygon is orthogonal to the viewer, the other is an additional offset as a multiplier on the derivative of z to increase the offset as z interpolation increments increase and z rounding and interpolation errors increase.

d123
04-20-2004, 01:00 AM
Thanks a lot guys, that makes sense.