Projecting an arc through two points about a sphere - c++

Question

I'm trying to solve a problem where I need to essentially project an arc from P0 some distance through P1 about a sphere of a known radius, however I have a fairly limited knowledge solving this sort of problem outside of the standard Cartesian context.
To simplify, if I were doing this on a plane I would simply do the following
float distance = 30;
Vector3 direction = (P1 - P0).Normalize();
Vector3 endPoint = P0 + (direction * distance);
I'm essentially trying to see if there is a way I can solve that same sort of problem, only on a sphere instead of on a plane. This get's conceptually difficult as I need to account for the fact that P0 and P1 can (and will) be at arbitrarily different altitudes on the sphere.
I have a feeling that I will potentially need to convert from Cartesian to spherical space, but finding a jump off point here has proven very difficult.
Thoughts?

Answer 1

Let P0 and P1 be points, and consider a sphere of radius r.
First, say that P0 and P1 lie on the sphere, and that we want to travel a distance d. First note that we need P0 ≠ P1, just like in the linear algebra case. (otherwise, P0 = P1 and so there is no unique arc through P0 and P1.) Similarly, we need P0 ≠ -P1, as otherwise there is no unique arc through P0 and P1.
Now the arc from P0 to P1 lies on what is known as a great circle of the sphere. Basically, this is an intersection of the sphere with some plane through O = (0,0,0). More precisely, this is the unique plane through the points O, P1 and P2.
Some basic linear algebra will show you that this is the plane perpendicular to the vector P0 × P1, the cross product of P0 and P1. Note that the order of P0 and P1 is important in this product: in general, P0 × P1 ≠ P1 × P0, and if we took P1 × P0 instead then the direction of motion would be reversed. (i.e. we would be going from P1 to P0 instead.) Therefore the "direction" of P0 to P1 is given by a rotation about P0 × P1.
Now consider general points P0, P1. As melak47 points out, we can first project P0 and P1 onto the sphere, so we can get the rotation above in the "direction" P0 to P1. To project these onto the sphere, we further need that both P0 and P1 are non-zero, i.e. are not the origin. We also run into a problem when P0 and P1 lie on the same line through the origin, because then they project to the same point (or antipodal points) on the sphere and so don't define a unique arc: so in general, we need to check that P0 and P1 are linearly independent. Now note that P0 lies on the sphere of radius ||P0||, where ||P0|| is the norm of the vector P0.
Note that the radius r introduced above hasn't actually been used anywhere yet, and so we can set r = ||P0||. To travel a total distance of d on the arc P0 to P1', where P1' is P1 projected onto the sphere of radius ||P0||, the angle becomes d/||P0||, as above. The good news is that P0 × P1 is still a valid axis of rotation.
So in summary, you want to rotate P0 about P0 × P1 by d/||P0|| radians. In pseudo-code,
float d;
Vector3 P0, P1;
float r = norm(P0);
float angle = d / r;
Vector3 axis = crossProduct(P0, P1);
Matrix3 rotationMatrix = RotationMatrix(angle, axis);
Vector3 endPoint = rotationMatrix * P0;
Checking that P0 and P1 are linearly independent is the same as checking that P0 × P1 is non-zero, so in general you may want to check that axis != Vector3(0,0,0) before constructing the rotation matrix.
If you instead want to travel in the direction P0 to P1 on the arc P0' to P1, where P0' is P0 projected onto the sphere of radius ||P1||, we first need to replace r = norm(P0) by r = norm(P1) (obviously). Further, P0' = ||P1||/||P0|| * P0, and so we also need to replace rotationMatrix * P0 by rotationMatrix * (r / norm(P0) * P0), or rotationMatrix * r * normalize(P0) if you prefer.

Answer 2

IMO, your question doesn't have a single answer. It seems to be a 2D problem that occurs in the plane formed by the two given points and the center of the sphere.
But you need to give more constraints on the trajectory. It could be a straight line, an Archimedean spiral, a logarithmic spiral, a spline... or whatever you like.