In general, there are some kinds of mathematical descriptions for doing such things.
1st there are parametric surfaces. They are of the general form
S(u,v)
where S is a vector valued function (3 components since we are talking about 3D spaces, I think), and u,v are the parameters (2 components since they should describe a surface in 3D). These equations are fine to generate meshes.
Typical surfaces of this kind are spline patches. But also spheres, cylinders, tori, and others could be given this way, what typically yields in meshes like the one Thr33d showen above. As jyk has written above, using some trigonometrics with u and v makes the way.
2nd there are implicit functions where the locus of 0's defines the surface
f(x,y,z) = 0
where x,y,z are arbitrary space co-ordinates. The surface are just all those co-ordinates that fulfil the equation. Often these functions are algebraic, so that in fact an implicit algebraic surface is given then. These equations are less suitable to make meshes since they are implicit, but they are/maybe fine for other things like ray-tracing, collision detection, and such.
Typical surfaces of this kind are quadrics (i.e. where the function is a 2nd order polynomial): Spheres, ellipsoids, cylinders, some paraboloids, and some hyperboloids. Less often higher order degree polynomials could be found in 3D gfx, e.g. the "hyper-quadrics". Often there must be some additional restrictions active, e.g. to restrict the infinite quadric cylinder to a specified height.
Look e.g. at www.euclideanspace.com or wikipedia for some formulas.
[Edited by - haegarr on July 6, 2006 2:06:48 AM]
Render Spheres, torus and other complex objects by algorithm?
A good resource for procedural surface construction is Paul Bourke's surface pages.
Edit: Might be a good idea to check out his geometry pages aswell.
Edit: Might be a good idea to check out his geometry pages aswell.
darookie said: "Your sample image suffers from problem due to the singularity at the poles. For spheres, a sub-division approach seems better suitable, especially since mapping coordinates can be interpolated easily."
@darookie: I was trying to give a simple example because when I read,
"I realy want to get deep into this topic."
I mis-read... thought he *didn't* want to get deep into this topic, (the context supports this)
That's why my example was oversimplified.
@directNoob: you said: Ok, that´s easy.
r^2 = x^2 + y^2 + z^2 =>
z = +-sqrt(r^2 - (x^2 + y^2))
With this formula, I get every point on the sphere.
Did you ask for this?
Yeah... now just sample it in a way that you get a *finite* number of points and "stitch" (not a technical term) them together into triangles.
Others have mentioned using surface patches, splines etc. These methods will help to give a more uniform distribution than a, for instance, cylindrical approach.
A "simple" approach would be (from here)
Treat the sphere as a 2d grid (darookie mentioned why this *wasn't* a good idea already)
have a nested loop, find the x,y,z points for a regular grid wrapped over the surface of the sphere.
(Do you understand what I'm saying?)
Here, I'll get you started (that's what he asked for in his first post - I'm just trying to help him visualize a nieve approach to this)
Write some psuedo-code using u and v as x-axis and y-axis 2d space over the surface of the sphere. Find the x,y,z points for a given 2d UV coordinate
assume RADIUS is a declared variable, you have access to trig functions, and there exists some array to output x,y,z points
put it in the format of
---
(someone *please* clean this up and slap me around if my example is misleading)
I know it's rather stupid to use cylindrical coordinates for a sphere... but it's a good "first step" to visualizing the x,y,z positions mathematically.
-Michael g.
p.s. If you're "beyond" this already, ignore this, and check out the resources previously mentioned by others.
edit: formatting *ugh*
@darookie: I was trying to give a simple example because when I read,
"I realy want to get deep into this topic."
I mis-read... thought he *didn't* want to get deep into this topic, (the context supports this)
That's why my example was oversimplified.
@directNoob: you said: Ok, that´s easy.
r^2 = x^2 + y^2 + z^2 =>
z = +-sqrt(r^2 - (x^2 + y^2))
With this formula, I get every point on the sphere.
Did you ask for this?
Yeah... now just sample it in a way that you get a *finite* number of points and "stitch" (not a technical term) them together into triangles.
Others have mentioned using surface patches, splines etc. These methods will help to give a more uniform distribution than a, for instance, cylindrical approach.
A "simple" approach would be (from here)
Treat the sphere as a 2d grid (darookie mentioned why this *wasn't* a good idea already)
have a nested loop, find the x,y,z points for a regular grid wrapped over the surface of the sphere.
(Do you understand what I'm saying?)
Here, I'll get you started (that's what he asked for in his first post - I'm just trying to help him visualize a nieve approach to this)
Write some psuedo-code using u and v as x-axis and y-axis 2d space over the surface of the sphere. Find the x,y,z points for a given 2d UV coordinate
assume RADIUS is a declared variable, you have access to trig functions, and there exists some array to output x,y,z points
put it in the format of
//psuedo-code//for simplicity's sake, assume cylindrical coordinatesfor (theta = 0; theta < TWO_PI; theta+=...){ for (z = -1; z < 1; z+=...) { //find the x, y, and z, values here }}---
(someone *please* clean this up and slap me around if my example is misleading)
I know it's rather stupid to use cylindrical coordinates for a sphere... but it's a good "first step" to visualizing the x,y,z positions mathematically.
-Michael g.
p.s. If you're "beyond" this already, ignore this, and check out the resources previously mentioned by others.
edit: formatting *ugh*
Author
Hello Thr33d.
First of all, thanks for your help by providing as much text.
Before I can proceed, can you explain this text a bit more precisely, please.
I don´t understand the part with the "2d space over the surface of the sphere".
Alex
First of all, thanks for your help by providing as much text.
Before I can proceed, can you explain this text a bit more precisely, please.
Quote:
Write some psuedo-code using u and v as x-axis and y-axis 2d space over the surface of the sphere
I don´t understand the part with the "2d space over the surface of the sphere".
Alex
Again, we're simplifing this down to cylindrical coordinates.
Here's how to visualize it
Take a piece of 8.5" x 11" paper. Orient it this way
_____
| |
|_____|
Draw say, 6 dots along the top and bottom edge.
then draw six dots horizontally along the middle.
You can connect the dots with vertical lines if you'd like as well (and if you connect the dots horizontally, you get a kind of grid paper)
Now take this and wrap it, dots on the outside, into a cylinder
See this page for a visual of a cylinder and it unwrapped.
Say you have a ball that this wraps nicely around.
You now have a ball with a cylindrical piece of paper wrapped around it.
The dots down the middle of the paper should be along the 'equator' of the ball.
Now, crumply in the top and bottom so the paper *wraps* against the ball.
(Think about wrapping those oddly shaped gifts for Christmas)
The dots along the top and bottom edges have converged to 2 single points. The points (or, if you've connected the dots with lines, their intersections) represent vertices. If you've drawn lines, these represent triangle edges (except you'll notice they're quadlaterials, you have to divide them each in two, no problem).
Here's a picture of what that (a sphere) formed in this way would look similar to.
here
Since it's kinda silly to have the vertex placement on fixed z-axis spacing, one could modify this to break it up based off of spherical coordinates instead and step along a constant angle interval.
In the end I believe geodesic spheres give the most bang for the vertex :) (Involves starting with a "balanced" convex polyhedron and subdividing the sucker!)
Please heed what the people above have already posted, they'll help you over the hurdle of parametric surfaces, which I didn't bother trying to cover - what I've written is mainly to help you have a simple example.
-Michael g.
Here's how to visualize it
Take a piece of 8.5" x 11" paper. Orient it this way
_____
| |
|_____|
Draw say, 6 dots along the top and bottom edge.
then draw six dots horizontally along the middle.
You can connect the dots with vertical lines if you'd like as well (and if you connect the dots horizontally, you get a kind of grid paper)
Now take this and wrap it, dots on the outside, into a cylinder
See this page for a visual of a cylinder and it unwrapped.
Say you have a ball that this wraps nicely around.
You now have a ball with a cylindrical piece of paper wrapped around it.
The dots down the middle of the paper should be along the 'equator' of the ball.
Now, crumply in the top and bottom so the paper *wraps* against the ball.
(Think about wrapping those oddly shaped gifts for Christmas)
The dots along the top and bottom edges have converged to 2 single points. The points (or, if you've connected the dots with lines, their intersections) represent vertices. If you've drawn lines, these represent triangle edges (except you'll notice they're quadlaterials, you have to divide them each in two, no problem).
Here's a picture of what that (a sphere) formed in this way would look similar to.
here
Since it's kinda silly to have the vertex placement on fixed z-axis spacing, one could modify this to break it up based off of spherical coordinates instead and step along a constant angle interval.
In the end I believe geodesic spheres give the most bang for the vertex :) (Involves starting with a "balanced" convex polyhedron and subdividing the sucker!)
Please heed what the people above have already posted, they'll help you over the hurdle of parametric surfaces, which I didn't bother trying to cover - what I've written is mainly to help you have a simple example.
-Michael g.
Author
Thanks again.
But do you have any suggestions where to start? I mean, and you just said it, parametric surfaces.
I feel like a little ant underneath the whole thing and don´t know what to do!?!?
I just need something, from where I can proceed myself.
Again, where do you think, I should start with??? Is it the sphere?
All I want is to render my own procedurally created objects and no bad downloaded objects from the net or any tut. It is not worth to go ahead like I do...
Just begging for little hint once more!
Greetings
Alex
P.S.
@Thr33d
Mybe this is the requested code you asked me for?! I didn´t try it, so I don´t know if it works. This shoud be a point list of the structure of a sphere...
But now, it needs to be indexed.
[Edited by - directNoob on July 6, 2006 8:50:34 PM]
But do you have any suggestions where to start? I mean, and you just said it, parametric surfaces.
I feel like a little ant underneath the whole thing and don´t know what to do!?!?
I just need something, from where I can proceed myself.
Again, where do you think, I should start with??? Is it the sphere?
All I want is to render my own procedurally created objects and no bad downloaded objects from the net or any tut. It is not worth to go ahead like I do...
Just begging for little hint once more!
Greetings
Alex
P.S.
@Thr33d
//l = longitude//c = colatitude//r = radius//vb = vertexbuffer//max = max vertices stored in vbi = 0 // for the vb count// colatitude needs just be calculated from 0 to Pi// longitude needs a full circle, hence 0 to 2*Pi// this leads to the followingfor(c = 0; x < PI; c += 1/6*PI){ for(l = 0; l < 2*PI; l += 1/6*PI){ x = r * cos(l) * sin(c); y = r * sin(l) * sin(c); z = r * cos(c); // the top and bottom point is needed just once, // so proceed with the next angle if(c == 0 || c == PI) c += 1/6*PI; // store the calculated point in the locked vertex buffer // for example: #define FVF D3DFVF_XYZ vb = VERTEX( x , y , z ); i++; }}max = i;Mybe this is the requested code you asked me for?! I didn´t try it, so I don´t know if it works. This shoud be a point list of the structure of a sphere...
But now, it needs to be indexed.
[Edited by - directNoob on July 6, 2006 8:50:34 PM]
The code looks good to me... try it out to check for sure (loop through the list, drawing a point at that 3d location in space, make sure the sphere equation is all good)
About indexing, check out the d3d chm file. Basically though, you'll either use an indexed triagle list, where you feed it an array of indexes, or you feed it the indexes in a "special" way (read up on triagle strips and triangle fans)
For your purposes, just start out with an indexed triangle list.
Then set up the 3 loops to create the sphere's index list
Recognize there are 3 sections: the top (1) and bottom (2) are special (they connect many triangles hingeing off of one point.
The middle is the general case - you have a grid of points, set it up to index these to form triangles.
(I'm pretty sure you understand what I mean, and that you can come up with the code to do this)
Some things to note, (I can't remember the default vertex order CW or CCW)
Always order the indexes for each triangle, with the same rotation order. This is because we generally use closed surfaces, so we can make the triangles single sided - using back face culling. In order to choose a "front" and "back" side, you choose a standard.
1
2 3
CCW "front"
1
3 2
CW "front"
(you probably know that 1-2-3 is the same as 2-3-1 is the same as 3-1-2, so don't worry about absolute positioning of the verticies for index choice, just relative positions)
....
Now, go ahead and write the 3 sections of code mentioned above :)
(After you make it through this exercise, you should have a much better working knowledge of direct3d's architechture. Moving to parametric surfaces with this base of how to draw stuff on the gpu will be most benefical.)
(questions?)
-Michael g.
About indexing, check out the d3d chm file. Basically though, you'll either use an indexed triagle list, where you feed it an array of indexes, or you feed it the indexes in a "special" way (read up on triagle strips and triangle fans)
For your purposes, just start out with an indexed triangle list.
Then set up the 3 loops to create the sphere's index list
Recognize there are 3 sections: the top (1) and bottom (2) are special (they connect many triangles hingeing off of one point.
The middle is the general case - you have a grid of points, set it up to index these to form triangles.
(I'm pretty sure you understand what I mean, and that you can come up with the code to do this)
Some things to note, (I can't remember the default vertex order CW or CCW)
Always order the indexes for each triangle, with the same rotation order. This is because we generally use closed surfaces, so we can make the triangles single sided - using back face culling. In order to choose a "front" and "back" side, you choose a standard.
1
2 3
CCW "front"
1
3 2
CW "front"
(you probably know that 1-2-3 is the same as 2-3-1 is the same as 3-1-2, so don't worry about absolute positioning of the verticies for index choice, just relative positions)
....
Now, go ahead and write the 3 sections of code mentioned above :)
(After you make it through this exercise, you should have a much better working knowledge of direct3d's architechture. Moving to parametric surfaces with this base of how to draw stuff on the gpu will be most benefical.)
(questions?)
-Michael g.
Hi there,
here is a tutorial describing how to procedually generate a sphere. Maybe it is helpful: http://in4k.untergrund.net/html_articles/hugi_27_-_coding_corner_polaris_sphere_tessellation_101.htm
There is also another way of doing it, the result will be a so-called geosphere. You start off by creating a simple cube (hard-coded) and then you start to subdivide each triangle of the cube into two smaller ones, and then subdivide these two smaller ones again, etc... The more often you subdivide the cube's triangles, the more "smooth" the sphere will look. If you finally got a highly subdivided cube you loop through each vertex of it, "shoot" a ray from the center of the cube through the vertex and move the vertex to the point where the ray hits an imaginary sphere around the cube.
For the procedural generation of more complex objects than a sphere (for example meta-balls), you can look at the marching-cubes algorithm which basically extracts a polygonal mesh from an iso-surface.
http://en.wikipedia.org/wiki/Marching_cubes
http://www.essi.fr/~lingrand/MarchingCubes/accueil.html
here is a tutorial describing how to procedually generate a sphere. Maybe it is helpful: http://in4k.untergrund.net/html_articles/hugi_27_-_coding_corner_polaris_sphere_tessellation_101.htm
There is also another way of doing it, the result will be a so-called geosphere. You start off by creating a simple cube (hard-coded) and then you start to subdivide each triangle of the cube into two smaller ones, and then subdivide these two smaller ones again, etc... The more often you subdivide the cube's triangles, the more "smooth" the sphere will look. If you finally got a highly subdivided cube you loop through each vertex of it, "shoot" a ray from the center of the cube through the vertex and move the vertex to the point where the ray hits an imaginary sphere around the cube.
For the procedural generation of more complex objects than a sphere (for example meta-balls), you can look at the marching-cubes algorithm which basically extracts a polygonal mesh from an iso-surface.
http://en.wikipedia.org/wiki/Marching_cubes
http://www.essi.fr/~lingrand/MarchingCubes/accueil.html
Author
Hi!
Sorry that it took so long, but I had to do something for the shool...
However.
I made it, but the sphere is not a sphere. Its just the half of a sphere.
It´s exactly the same when I use this equation:
Here it is the problem that you gain
out of the equation after taking the square root(or how must it be written in english?)
I hoped that with the arc function i yield the absolute values without the +- thing.
Forget this! it works. it is just a matter of " how do i calculate the number of the max vertices?"
Allow me to present the code to you:
I really hope it is readable!
The vertices are badly calculated!
Do you have any suggestions?
Here is a nice picture:
Even better image:
When the vertices are correctly initiated, i can start indexing it.
Greetings
Alex
[Edited by - directNoob on July 10, 2006 4:35:26 PM]
Sorry that it took so long, but I had to do something for the shool...
However.
I made it, but the sphere is not a sphere. Its just the half of a sphere.
It´s exactly the same when I use this equation:
r^2 = x^2 + y^2 + z^2Here it is the problem that you gain
+-rout of the equation after taking the square root(or how must it be written in english?)
I hoped that with the arc function i yield the absolute values without the +- thing.
Forget this! it works. it is just a matter of " how do i calculate the number of the max vertices?"
Allow me to present the code to you:
void Setup(){ // Longi -> Longitude // Colati -> Colatitude float radius = 1.0f; // Degree measure int iLongiDeg, iColatiDeg; iLongiDeg = iColatiDeg = 5; // Angle of 5 degrees // Radian measure float fLongiRad, fColatiRad; fLongiRad = ((float)iLongiDeg * D3DX_PI)/180.0f; fColatiRad = ((float)iColatiDeg * D3DX_PI)/180.0f; // figure out, how many vertices are needed: // horizontally int iHorzVertices = 360 / iColatiDeg ; // vertically int iVertVertices = 180 / iLongiDeg ; // this must be improeved, I can´t figure out how much vertices are used!!! bad!! int iMaxVertices = iHorzVertices * (iVertVertices) * 10 ; g_iMaxVertices = iMaxVertices; // now, calc the indices. int iIndices = 0; // On the top and on the bottom are triangle strips // top iIndices += 1 + iHorzVertices; // bottom iIndices += 1 + iHorzVertices; // the left vertices must be organized in a triangle fan iIndices += iMaxVertices; VERTEX *v; WORD *w; g_pDev->CreateVertexBuffer(iMaxVertices * sizeof(VERTEX), D3DUSAGE_WRITEONLY, FVF, D3DPOOL_MANAGED, &g_pvb, NULL) ; g_pDev->CreateIndexBuffer(iIndices * sizeof(WORD), D3DUSAGE_WRITEONLY, D3DFMT_INDEX16, D3DPOOL_MANAGED, &g_pib, NULL); g_pvb->Lock(0, 0, (void**) &v, 0); float x, y, z; int topbottom = 1 + iHorzVertices; int iCounti = 0; int iCountj = 0; // Calc the vertices on the colatitue for(int i= 0; i < 180 / iColatiDeg; i++) { // Calc the vertices on the longitude for(int j = 0; j<360 / iLongiDeg; j++) { int index = (iCounti * (360 / iLongiDeg)) + iCountj; x = radius * cosf(fLongiRad * j) * sinf(fColatiRad * i) ; y = radius * sinf(fLongiRad * j) * sinf(fColatiRad * i) ; z = radius * cosf(fColatiRad * i); v[index] = VERTEX(x,y,z, 0.0f,0.0f,0.0f, 0.0f,0.0f); if(i * fColatiRad == 0.0f || fabs((D3DX_PI - i * fColatiRad)) <= 0.00001f ) { i++; j = 0; } iCountj++; } iCounti++; } g_pvb->Unlock();}I really hope it is readable!
The vertices are badly calculated!
Do you have any suggestions?
Here is a nice picture:
Even better image:
When the vertices are correctly initiated, i can start indexing it.
Greetings
Alex
[Edited by - directNoob on July 10, 2006 4:35:26 PM]
Author
Alright, here is my approach of indexing this sphere.
Has anyone some hint for me, how to index this sphere?
This thing is absolutly jagged. And by the way, I get a maximum of vertex count
of 88200 vertices. Is that possible with a longitude and a colatitude of 5 degrees?
Alex
Has anyone some hint for me, how to index this sphere?
This thing is absolutly jagged. And by the way, I get a maximum of vertex count
of 88200 vertices. Is that possible with a longitude and a colatitude of 5 degrees?
Alex
This topic is closed to new replies.
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