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zix99

OpenGL Drawing 3D "manually"...

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Hi all I'm currently working on a project (along with a few others) to draw some 3D stuff, which will hopefully turn into a rendering engine eventually. I searched for a bit on these forums, and google with no luck. What i would like to do is draw 3D points on the 2D screen, with a depth buffer. I understand that the camera is a 4x4 matrix, along with the point, and i know all rotation matricies ant stuff like that. What is getting me is, given the projection (is it a matrix? in OGL it's gluPerspective(50,1,1,100) ) then how to I get the 2D point on the screen and how far away it is for Depth?? I've worked with OpenGL for a while now, with a little DirectX, and i think that will help me.. Thanks for anything (sorry if this topic has been brought up before) ~zix~ [Edited by - zix99 on November 22, 2004 3:05:15 PM]

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Yes, projection is a matrix. The DX SDK help files detail how the projection matrix is created.

What then occurs is: vertexposition (with an assumed w of 1...[x,y,z,1]) * world * view * projection.

The resulting vector is the divided by W. Now you're in clip space. In DX thats -1<x<1, -1<y<1, and 0<z<1. I think I've seen somewhere that in OpenGL z goes from -1 to 1 too... it's a full cube.

From here, you discard and/or clip the triangles to fit within those boundaries. You can also do this before dividing by W, where -w<x<w, -w<y<w, etc..

Then you apply viewport scaling. Basically scale and offset the x and y into pixel coordinates. For a 640x480 screen that would be x*320+320 and y*240+240. Then, rasterize your result.

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Yeah, look up the D3DXMatrixPerspectiveLH function in the DirectX SDK help file, it shows you exactly what the perspective matrix looks like.

It's this:


w 0 0 0
0 h 0 0
0 0 zf/(zf-zn) 1
0 0 -zn*zf/(zf-zn) 0
where:
h is the view space height. It is calculated from h = cot(fovY/2);
w is the view space width. It is calculated from w = h * aspect (where aspect = screenWidth/screenHeight)
zf = farthest visible distance, zn = nearest visible distance


So, this works with homogenous coordinates. for a vertex:


[x, y, z, 1]


you get the following: output vertex


[x*w, y*h, (z-zn)*zf / (zf-zn), z]


which becomes (After the homogenous divide by w)


[x*w/z, y*h/z, (1-zn/z)*zf / (zf-zn), 1]


In the case of D3D's matrices, you end up with x and y in the range [-1, 1] and z in the range [0, 1], 1 being farthest from the camera (the z-far plane) and 0 being nearest (z-near plane).

Hope that helps :)

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Drillan, thank you very much for that response, quite helpful. Now, sense I have one subject down, all that's left is drawing filled triangles (dont forget i'm using depth so if it is translated to 2D then drawn, i still need depth). Also texturing, but this can be covered later if needed.

One question for namethatnobodyelsetook
Quote:
What then occurs is: vertexposition (with an assumed w of 1...[x,y,z,1]) * world * view * projection.


So, if I am correct it would look something like (if camera was at 0,0,0 and point was at 5,5,0):

1 0 0 5 1 0 0 0 w 0 0 0
0 1 0 5 * WORLD? * 0 1 0 0 * 0 h 0 0
0 0 1 0 0 0 1 0 0 0 zf/(zf-zn) 1
0 0 0 1 0 0 0 1 0 0 -zn*zf/(zf-zn) 0

I figure rotating the camera is simply just multiplying the viewport coordinates by the rotation matrix?

Thanks again
~zix~

PS: I rated u guys up.. but it's not doing anything.. is the rating system broken??

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To draw a triangle, start by drawing the lines of the triangle. Look into Bresenham's line algorithm for details. When you draw the lines to the screen you only need to keep track of the smallest x value, and the greatest x value of the line.

So you could keep two array's of min x and max x.

MIN X MAX X
[0] [0]
[1] [1]
. .
. .
[480] [480]

The array would be the height of the screen, then just iterate through the array and connect the lines horizontally. In array position [0] you could have a value of 10, and array 2 position [0] you could have a value of 100, that means that left most outer edge of the traingle 0 pixels down on the screen is at 10, and the right most edge of the triangle 0 pixels down is at 100. A simple for loop from 10 to 100, filling in each pixel with an interpolated color or a sampled UV coord will give you a colored/textured triangle.

- Hope thats clear.

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Quote:
Original post by zix99
So, if I am correct it would look something like (if camera was at 0,0,0 and point was at 5,5,0):

1 0 0 5 1 0 0 0 w 0 0 0
0 1 0 5 * WORLD? * 0 1 0 0 * 0 h 0 0
0 0 1 0 0 0 1 0 0 0 zf/(zf-zn) 1
0 0 0 1 0 0 0 1 0 0 -zn*zf/(zf-zn) 0

I figure rotating the camera is simply just multiplying the viewport coordinates by the rotation matrix?


The correct order of operations is world*view*proj.

I don't understand what you mean by point was at 5,5,0 do you mean the look at point for the camera? I'm not sure what you mean but, for camera, every thing is inverted, because the world is transformed to move the camera back to 0,0,0. So if the camera was translated 5,5,0 forward the matrix would have -5,-5,0 in the translation portions.

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Maybe this is clearer?

(5,5,0) (0,0,0)
Vertex Pos World Camera Projection
1 0 0 5 1 0 0 0 w 0 0 0
0 1 0 5 * WORLD? * 0 1 0 0 * 0 h 0 0
0 0 1 0 0 0 1 0 0 0 zf/(zf-zn) 1
0 0 0 1 0 0 0 1 0 0 -zn*zf/(zf-zn) 0



According to namethatnooneelsetook it is
Quote:
vertexposition (with an assumed w of 1...[x,y,z,1]) * world * view * projection.

And i'm a little confused on what the world matrix vs view matrix is. I'm guessing the view matrix is the same as the camera (inverted); in that case what is the world?

Thanks
~Zix

PS: Thanks for the triangle algo, once you have the 3 positions of the triangle, then you simple interpoliate the UV coordinates and draw accordingly (relativly simple).. correct? I'll have to do some thinking on this (along with the aspect of depth buffer), but i think i may have it.

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OpenGL doesn't necessarily have a world, you have a modelview matrix which is both world and view together (I could be wrong, I've never used OpenGL).

In DirectX the World matrix is the matrix that transforms a vertex to world space. ie: It positions and rotates a mesh.

The view matrix is the inverse of the camera's world matrix. If your camera was at 1,2,3, the inverse of that would be a translation matrix of -1,-2,-3. Instead of actually moving a camera through the world, the world is dragged (and rotated) to the camera.

So, lets say we have a vertex at the origin. Lets say our mesh is at 0,0,10. Lets say the camera is at 0,0,2.

vertex pos = 0,0,0
*worldmatrix = 0,0,10
*viewmatrix = 0,0,8
*projection = ...

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Thanks everyone.. I'll test all these ideas out and let you know how it goes.

I'll also rate you guys up as soon as it lets me.
EDIT: It just wont let me rate Drillan... is it because he's a GDNet member? o well, i guess i'll report to bugs

~zix~

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Quote:
Original post by Namethatnobodyelsetook
What then occurs is: vertexposition (with an assumed w of 1...[x,y,z,1]) * world * view * projection.

The resulting vector is the divided by W. Now you're in clip space. In DX thats -1<x<1, -1<y<1, and 0<z<1.


I have a question on this. What happens if W is now 0? I've been working on a project like this for a class at school and I am having that problem. Where I have a unit pyramid with the top point at (0, .5, 0, 1) and after I do all my transformations and go to do perspective division W is now 0 so it screws it all up. BTW, I have my world axes with X->right, Z->up, Y->in (because our teacher wants it that way).

So maybe I'm doing something wrong with the transformations. I don't know. But if you could just answer that question for me I'd be grateful. :)

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If your w is 0 when it comes to perspective division, you probably screwed up your transformation matrix. IIRC w should come out equal to the distance between the point and the camera...

For the most part it seems like you can treat w as a binary flag. If it's 1, translations will be applied to the vector. If it's 0, they won't.

Quote:
Original post by zix99
I'll also rate you guys up as soon as it lets me.
EDIT: It just wont let me rate Drillan... is it because he's a GDNet member? o well, i guess i'll report to bugs
Don't worry, your rating's been recorded. It just doesn't have any effect at the moment because you and he both have the same score. I, however, can rate him up, and change his score. *rates*

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Quote:

I have a question on this. What happens if W is now 0? I've been working on a project like this for a class at school and I am having that problem. Where I have a unit pyramid with the top point at (0, .5, 0, 1) and after I do all my transformations and go to do perspective division W is now 0 so it screws it all up. BTW, I have my world axes with X->right, Z->up, Y->in (because our teacher wants it that way).

So maybe I'm doing something wrong with the transformations. I don't know. But if you could just answer that question for me I'd be grateful. :)


W=0 implies that the point is 'at infinity'. If you imagine a line passing through your eye (from in front of you, through your head and out the back), the perspective transform kind of turns space inside out (it's hard to visualize).

After the perspective transform you are in clip space and it is perfectly possible for primitives to wrap around infinity and come back (e.g. a line like above). You MUST clip to your view frustum (which is actually a 'parallelpiped' or box in clip space). After clipping you will only get W=0 for a point if you don't have an offset in your z (in other words Zn=0).

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Yeah, w=0 implies that the point is at infinity. Another way to look at it is that, if w=0, the x,y, and z coordinates are the direction TO that point (so if the coordinate is [0,1,0,0] then the point is at infinity down the +Y axis).

You definitely shouldn't end up with w = 0 after the perspective matrix is applied.

Really, you want to treat your vectors as a 4x1 matrix, so it'd be:


(5,5,0) (0,0,0)
Vertex Pos World Camera Projection
1 0 0 0 w 0 0 0
5 5 0 1 * WORLD * 0 1 0 0 * 0 h 0 0
0 0 1 0 0 0 zf/(zf-zn) 1
0 0 0 1 0 0 -zn*zf/(zf-zn) 0


The transform you're doing won't come out correctly because a vector is not the same thing as a transform matrix :)

Hope that helps!

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Thanks all again for the information.. and good thing i kept up on this otherwise i would've never treated the vector as a 4x1 matrix, i would've treated it as a 4x4.

Thanks again Drillan, seems you have some background in this.

And thanks superpig for taking care of the rating

~zix~

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No, you can perfectly well have W=0 because after the perspective transform W is proportional to the original Z coordinate.

In fact, taking Drillians example:

(I'll discard the World and camera transform, presume we're in camera space and this point happens to fall at Z=0)

v' = v * P

x' = (5 * w) + (5 * 0) + (0 * 0) + (1 * 0)
y' = (5 * 0) + (5 * h) + (0 * 0) + (1 * 0)
z' = (5 * 0) + (5 * 0) + (0 * ...) + (-zn*zf/(zf-zn))
w' = (5 * 0) + (5 * 0) + (0 * 1) + (1 * 0)

So this point does have W=0!

The point is that this point will be clipped away since z' < Zn.

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Ah! You're correct. Forgot about the ol' "z = 0" case. Yeah, because the perspective is, at its most basic level, a divide by z (which is why you set the w coordinate to z: because in order to get the normalized coordinate you divide by the w parameter [x y z w] -> [x/w, y/w, z/w, 1]). But when z = 0, the perspective divide becomes a divide by zero, which effectively puts the coordinate at infinity in post-perspective space.

But JuNC is right, that point gets clipped away due to the near Z value, so in practice you should never actually see this.

Good catch!

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This seems interesting. I might give it a try and program my own software renderer.

My only question is, what does the projected point look like? Are the coordinates from (-1,-1) to (1,1), where (0,0) is the center of the screen, and outside that range is outside of the screen?

And whats the standard algorithm for, say, a color filled polygon? You project the 3 vertices, draw the edges to find min and max values in terms of rows, and then fill each row?

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Quote:
Original post by Max_Payne
My only question is, what does the projected point look like? Are the coordinates from (-1,-1) to (1,1), where (0,0) is the center of the screen, and outside that range is outside of the screen?

To do simple project to -1,-1 to 1,1 you use this matrix.

1 0 0 0
0 1 0 0
0 0 1 1/D
0 0 0 0

The X, Y, Z get multiplied by the 1 across the diagnal, and the Z get's multipled by the recipical of D. So the Z value is stored in the W value of the vertex you multiply it by. This is useful when you actually do projection and divide (x,y,z,w) by w, to get (x/w, y/w, z/w, 1) you can discard the z and w and get your projected coordinates (x/w, y/w).

Those numbers are in-dependent on your viewport or field of view. The viewport is factored into the w(x) and h(y) values of the above matrix. In directX the width can be calculated using field of view or viewport. If you want to depend on the view port you would do this.

w = 2 * z-near / ViewportWidth;
h = 2 * z-near / ViewportHeight;

so your matrix is
w 0 0 0
0 h 0 0
0 0 1 1
0 0 0 0

Also, you probally are going to want to factor z in for the near and far plane, to do that use
Q = Zfar / ( zfar - znear )

to get the matrix
w 0 0 0
0 h 0 0
0 0 q 1
0 0 0 0

That will account for a viewport that can be set rather then the standard -1,-1, to 1,1.

Quote:
And whats the standard algorithm for, say, a color filled polygon? You project the 3 vertices, draw the edges to find min and max values in terms of rows, and then fill each row?


Read my post above, but basically what you said sums it up.

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