Question

Spherical texture mapping - rotation problem

Asked by: InteractiveMind

My algorithm involves filling a circle with the centre at (0,0). For each pixel in the circle, I colour it with the colour returned by my colorAt(x,y,dt,dp) function, below. The x and y values are the coordinates on-screen, and dt and dp are the offset angles for the t and p coordinates.

When I call colorAt(x,y,0,dt), the output is good:
http://www.youtube.com/watch?v=ExpsS6TnB64&feature=channel

But if I call colorAt(x,y,pi/2,dt), the output is undesired (but somewhat expected):
http://www.youtube.com/watch?v=fiQQkVljWT4&feature=channel

Also, colorAt(x,y,dp,0) is slightly odd too:
http://www.youtube.com/watch?v=iU6CfZLEeDY&feature=channel

I think I'm doing the rotation all wrong. How should I do this? Thanks

public int colorAt(double x, double y, double dt, double dp)
{
    double z=sqrt(r*r-x*x-y*y);
    double t=atan2(z,x) + dt;
    double p=acos(y/r) + dp;
    
    while(t<0)t+=2*pi;
    while(p<0)p+=2*pi;
    
    p%=2*pi;
    if(p>pi)
    {
        p=2*pi-p;
        t+=pi;
    }
    t%=2*pi;
    
    double u=t/(2*pi);
    double v=p/pi;
    
    return texture[(int)(TEX_WIDTH*u),(int)(TEX_HEIGHT*v)];
}

                                  
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Asked On
2009-09-16 at 11:47:10ID24737435
Topic

Math & Science

Participating Experts
2
Points
500
Comments
12

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Answers

 

by: ozoPosted on 2009-09-16 at 11:56:51ID: 25348838

For rotations on other than the vertical axis. you may find it easier to convert to cartesian coordinates.

 

by: InteractiveMindPosted on 2009-09-16 at 14:06:12ID: 25350151

Do you mean transform (x,y,z) before projection?

I'm beginning to think I might need to do something like this, as I'm currently only displacing the UV coordinates which explains the effect. But I expect that transforming every vertex will be expensive.

I'll give it a go though, but would prefer the fastest method you (or anyone else) can think of. Thanks

 

by: InteractiveMindPosted on 2009-09-16 at 14:35:44ID: 25350452

>Do you mean transform (x,y,z) before projection?

It works nicely, but as predicted is very slow

 

by: InteractiveMindPosted on 2009-09-16 at 14:59:37ID: 25350662

Would I be best to use a different approach, like gnomonic projection?

 

by: ozoPosted on 2009-09-16 at 15:45:19ID: 25351079

 

by: InteractiveMindPosted on 2009-09-17 at 03:10:12ID: 25354363

I've already tried applying the Euler rotation matrices to each (x,y,z) point before projection, if that's what you're hinting at..
However, a typical sphere that I'm rendering can occupy more than 80,000 pixels (vertices), so it's too slow..

 

by: InteractiveMindPosted on 2009-09-17 at 03:20:29ID: 25354425

Actually, this makes no sense to me...

The following works:

  (x,y,z) -> rotate with matrices -> (new x, new y, new z) -> transform to spherical coordinates (r,t,p) -> (u,v)

but this does not:

  (x,y,z) -> transform to spherical coordinates (r,t,p) -> rotate by adding on angles -> (r,t+dt,p+dp) -> (u,v)

But, aren't these two the same thing?!?

 

by: NovaDenizenPosted on 2009-09-17 at 09:26:52ID: 25357976

Here are two problems.  I don't know if they are your only problems, but these are definitely two of them. :)

> p%=2*pi;
> t%=2*pi;

This doesn't work how you expect it.  In C, '%' only works on integers.  It will truncate p to an integer, convert 2*pi to 6, perform the remainder (it doesn't calcluate the modulus), then convert this integer back to a floating point number and store it in p.

Instead you want to use the fmod function.  It's in the standard C math library.  http://linux.die.net/man/3/fmod

p = fmod(p, 2*pi);
t = fmod(t, 2*pi);

But this still isn't enough, because p and t could have a value between -2pi and 0.  Also, pi is defined in the standard headers as M_PI, so you might as well use it.

p = fmod(p, 2*M_PI);
if (p < 0) p += 2*M_PI;

t = fmod(t, 2*M_PI);
if (t < 0) t += 2*M_PI;

 

by: NovaDenizenPosted on 2009-09-17 at 09:50:13ID: 25358264

This might be a better way of doing it.  There is no branch here, which can be a big performance win.  There will be an accuracy problem for x of large magnitude which causes a return value slightly smaller than 0 or slightly larger than 2*pi.  This can be fixed in the pixel-index-calculating expression.

// converts an angle to 0 <= x < 2*PI
// but there is an accuracy hazard.  For large x, result may be <0 or >2*pi.
private double normalize_angle(double x) {
    double n = floor(x / (2*M_PI));
    return x - (n * (2 * M_PI));
}
 
public int colorAt(double x, double y, double dt, double dp)
{
    double z=sqrt(r*r-x*x-y*y);
    double t=atan2(z,x) + dt;
    double p=acos(y/r) + dp;
 
    p = normalize_angle(p);    
 
    if(p>pi)
    {
        p=2*pi-p;
        t+=pi;
    }
    t = normalize_angle(t);
    
    double u=t/(2*pi);
    double v=p/pi;
    
    int tx = (int)(TEX_WIDTH*u);
    int ty = (int)(TEX_HEIGHT*u);
 
    // to cover for the accuracy problem in normalize_angle
    tx = (tx + TEX_WIDTH) % TEX_WIDTH;
    ty = (ty + TEX_HEIGHT) % TEX_HEIGHT;
    
    return texture[tx, ty];
}

                                              
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by: InteractiveMindPosted on 2009-09-17 at 12:13:18ID: 25359803

Good thinking, but unfortunately it's the same result...

 

by: ozoPosted on 2009-09-18 at 03:46:20ID: 25364502

ir I understand your situation, you should be abls to so
> (x,y,z) -> rotate with matrices -> (new x, new y, new z)
just once for each point
and then do
> (new x, new y, new z) -> transform to spherical coordinates (r,t+dt,p) -> (u,v)
to rotate

 

by: InteractiveMindPosted on 2009-09-29 at 04:19:06ID: 25447648

I haven't abandoned this question, I just have no internet at home currently. I THINK I know what you mean, ozo, I shall give it a go and get back to this ASAP. Thanks

20120131-EE-VQP-002

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