southa wrote: I think perspective means that the fading slices would be of constant z/w, i.e. all going through the origin in eye coordinates.
I think I see how the painting idea works. So if there were three small objects at (0,0,-1,-10), (0,0,0,-10) and (0,0,1,-10), would one object overpaint the others?
The Adanaxis renderer would blend the contributions from each object.
I don't think I can do a torus right now. Are we talking about the path swept out by a 3D sphere taken round in a circle, whilst being kept perpendicular to the direction of travel? The scripts can do a cube taken round in a circle.
southa wrote:I think we have a wire crossed here. I forgot to say that the observer is looking in the -w direction, so anything with negative w is in front of the observer. z is the hidden axis which doesn't map to a specific position or shrinking on the screen.
southa wrote:Ah, so you're talking about a double projection - 4D to 3D then 3D to 2D?
Does this hide things from the viewer that the tetronian can see?
In terms of this scheme, the Adanaxis 3D to 2D transformation is a non-perspective blend without depth hiding.
PS: I dont understand how the depth order makes a difference if you use HAC. Isnt then simply the sum/integral taken, which is independent of the order?
southa wrote:There's already a perspecive projection in place
In many ways discard-z is simple - it's like throwing away the z coordinate and viewing the resulting (x, y, w) coordinates as a normal 3D scene.
Depth sorting is done using w - effectively the distance between the viewer and the object. HAC applies to z, the hidden axis.
It is not the same to first parallel project (discarding one axis) and then perspective project, compared to first perspective project and then parallel project.
I thought that the brightness of each point on the screen is computed by the weighted sum of the brightness of the points on the corresponding ray to the eye (or in your case the ray perpendicular to the screen at that point).
southa wrote:Perspective then parallel:
(x,y,z,w) -> (x/w, y/w, z/w) -> (x/w, y/w)
Parallel then perspective:
(x, y, z, w) -> (x, y, w) -> (x/w, y/w)
Along the ray in 4D, normal occlusion operates, and that's what the depth sorting (approximately) does.
The HAC operates on the retina, and combines all of pixels on a line in the retina into one screen pixel.
The 3d world *is* the tetronians retina. The question is then about how to present this a 3d being. To do that there are established techniques available.
*gives up asking for the application of the rendering order, since it was 3 times not answered*
southa wrote:The 3d world *is* the tetronians retina. The question is then about how to present this a 3d being. To do that there are established techniques available.
So this looks like the nub of our disagreement. You appear to treat the retina as a 3D object in a 3D world that's viewed by an external observer. I won't do that in my application because I believe it would create a false impression of depth, i.e. one unrelated to the distance from 4D viewer to 4D object.
quickfur wrote:Projecting from 4D to 3D is the easy step. I think we all agree that we want perspective projection from 4D to 3D with occlusion (e.g., shouldn't see through walls). This projection is done from the 4D viewpoint.
What's really needed is to use transparent surfaces instead of lines, so that you can see the shapes of the projected volumes.
Why? Because if you force the second projection to happen orthogonally to the 4D viewpoint, you get a lot of illusions. For example, if you look down a cubical passage in 4D, the projected image has the 4 side walls as frustums (analogous to the side walls of the cube-within-a-cube projection of the 4-cube). But now you project to 2D parallel to the coordinate axes, and suddenly you can't see 2 of the frustums anymore because their opposite edges coincide. What you really want is to project to 2D using another viewpoint (which resides in 3D), that looks at the retina from an angle, so that you can see the 6 frustum volumes in the 3D retina separately. The cubical wall at the end of the corridor will be much more obviously a cube; if you had projected from an orthogonal viewpoint, it simply becomes a square, and you have absolutely no information about whether it's a square, a cube, or a cuboid in the 4D view.
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