I need some way of generating points on the surface of a sphere (actually, vectors, but if I have points on the surface I have vectors) that are uniformly distributed on a "circular section" of the sphere's surface. This diagram should explain what I mean: The diagram is in 2d, but the same applies in 3d. The blue line is the vector given. Theta is the "threshold angle" - the randomly generated vectors are allowed to form an angle of theta or less with the given vector. The red part is the surface of the sphere (in the image, the arc of the circle) that the points are allowed to be on, and must be uniformly distributed on that area. I have one current solution, but it only works for the vector (0,0,1):

theta = fRand( 2*M_PI );
tMin = cos( randomAngle * M_PI_180 ); - randomAngle is the "threshold angle"
t = fRand( tMin, 1 );
phi = acos( t );
sinPhi = sin( phi );
return Vector3( sinPhi * cos( theta ), sinPhi * sin( theta ), cos( phi ) );

Also, it uses a lot of trig functions. Is there any more efficient ways of doing this that work with any vector? Thanks.

Gumgo,

I'm not sure what you mean by uniformly distributed, but I'm assuming you mean that the point on the surface of the sphere that is randomly generated is the same distance in longitude as it is latitude away from the starting vector.

In either event, I think the best way to go about solving this is to use spherical coordinates.

Spherical coordinates represent a point on a sphere by the ordered pair:
(r, theta, phi)

Where r is the radius, theta is the angle on the YZ-plane (latitude), and phi is the angle on the XY-plane (longitude). To solve this you only need to convert from cartesian coordinates to spherical, add in your random angle, and then convert back into cartesian coordinates.

The equations to convert from cartesian to spherical as follows:

r = sqrt( x^2 + y^2 + z^2)
theta = arccos( z / r )
phi = atan2(y,x)

Now, it's relatively easy to normalize your vectors first, however as you can see above it's not necessary, as the calculations are part of the conversion. But, in the event you're working with a unit circle, r = 1, which means theta is just the arccos(z).

So simplified:

r = 1
theta = arccos( z )
phi = atan2( y, x )

This gives you theta and phi, which is the position of your existing vector in spherical coordinates. Now, compute your random angle with something like

angle = random(-threshold, threshold)

This'll give you a random angle that lies in between the threshold. Now, just add the random angle to your existing spherical coordinates.

theta1 = theta + angle
phi1 = phi + angle

Now convert back to cartesian coordinates using the following:

x = r * sin(theta1) * cos(phi1)
y = r * sin(theta1) * sin(phi1)
z = r * cos(theta1)

As you can see, this is again greatly simplified if you're using a unit circle, as r = 1, however either way it works, and you can cut down on the number of computations by caching intermediate calculations. So the total method would look something like.

float angle = fRand(-threshold, threshold);
float r = sqrt( x^2 + y^2 + z^2);
float theta = acos( z / r ) + angle;
float phi = atan2(y, x) + angle;
float rSinTheta = r * sin(theta);
x = rSintheta * cos(phi);
y = rSinTheta * sin(phi);
z = r * cos(theta);

This requires:
1 square root
1 acos
1 atan2
2 cos
2 sin

If you're using a unit circle (r=1), the above is reduced to:

float angle = fRand(-threshold, threshold);
float theta = acos( z ) + angle;
float phi = atan2(y, x) + angle;
float sinTheta = sin(theta);
x = sinTheta * cos(phi);
y = sinTheta * sin(phi);
z = cos(theta);

Either way this method uses 2x cos, 2x sin, 1x acos, and 1x atan2.

Well, I hope this has been helpful. It's not really any more efficient than your method, but it's straight forward, easy to read, and should work on all 3 axis, regardless of the starting vector.

Cheers and Good Luck!