I'm wondering the same thing. You seem to consistently miss the point he is making and instead try to make it sound as if he's implying stuff he is not. You're also resorting to kindergarden-level...

Right, took me a few moments, but I get your approach. You're assuming the current point is part of a level set and compute the gradient of the point, which is of course normal to the level set.
...

You're computing the gradient there, which can indeed be the zero vector, and which should in general not be confused with the normal vector of a surface.

For a heightfield h(x, y) you find the...

Have you tried a wider pencil?


Seriously, if you want any sensible replies you will have to give a lot more detailed information on how you're currently drawing things.

The second edition of PBRT is due this summer, and will include info on bidirectional path tracing IIRC.

In the meantime you should read Veach's thesis, which explains it in great detail, it is...

When considering a workstation from Dell, it was either an intel thing or a Quadro-line card. Nope, couldn't get anything in between, then we had to go for a non-workstation model (which was poor in...

My gut feeling is that games are still the primary driver of GPU sales. However most gamers can play their games at max quality using "only" a low to midrange card.

The lack of need for a high-end...

Are you certain that glGetUniformLocation() is successful and returns a valid location? Your problems could indicate that it does not.

Someone might spot your error. However it's usually easier to get replies if you know where the issue is.

In order to determine the location of the problem I would start with determining if the...

You're welcome. Hope it works out alright :)

Regular linear interpolation should be used, so if you drop the division by two there it should work :)

Well, if all normals point in the Z direction (I'm going to assume the same Z direction, since the mesh should be "nice"), then there's nothing to interpolate. If one of the normals does not point...

I must admit I've only used the barycentric version of this method. The way I interpret that passage is that you pick one of the normals consistently for all points P within the quad. Which normal...

The normal approach would be to use Phong normal interpolation. It's essentially using linear interpolation of the vertex normals, which is then normalized. You can read up on it here:...

I suggest you read the specifications, it's section 4.1.7 and 6.1.12 in the OpenGL 2.1 specs. To quote the relevant part:



Also check out the extension text, it contains some more useful info...

Couldn't you use occlusion queries for this? Start an occlusion query, render the object, get the number of samples (which, afaik, equals to fragments if multi/super sampling is off). Then re-render...

Unless you're looking for ultimate speed, it's not that hard, a double loop should do... see http://mathworld.wolfram.com/MatrixMultiplication.html (equations 3-12 should give you enough to work...

Quoting chapter 3 in the spec's (2.1):



From this one can induce that the area covered by a fragment will always be 1 in grid coordinates (ie pixels).

This does not change with MSAA as MSAA...

So in other words floats aren't accurate enough.

The article I mentioned describes how to compute the sum and dot products up to K times the working precision, using only working precision...

Perhaps you can use some techniques mentioned in this paper:

"Accurate Sum and Dot Product" http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.2.1547

See the "Overview" section of http://www.opengl.org/registry/specs/ARB/texture_rectangle.txt

A regular NPOT texture behaves like a normal POT texture. A texture rect has non-normalized texture...

Excellent, well at least that you figured it out :D

I suspect that internally the driver would have converted to float anyway, but perhaps I'm wrong.

Have you tried using floats instead of ints, and three coords instead of two? Just to see if there's another nice bug in the ATI drivers :)

Also, is this (in essence) the only code you've got? If...

I don't see "glEnableClientState(GL_VERTEX_ARRAY);" in that code, did you omit it?

Perhaps you could do your own "two-pass" technique. First use a somewhat large step, and find the last empty position before you hit the surface. Also find the first empty position on the back, which...