Vector rotation question

I *think* but am not totally sure that it would be:

RotationVector = (v1 + v2) / sqrt(v1^2+v2^2)
Angle = (d1 + d2)

I just added the vectors together to get the resultant, then normalized it because that''s what you should always do, and just added the angles together. I''m pretty sure this is right.

Shadow12345 - It's not.

Counter example:
v1 = [1, 0, 0] d1 = 90°
v2 = [0, 1, 0] d2 = 90°
Result:
v = [sqrt(2)/2, sqrt(2)/2, 0]
d = 90°

lwinkenb - That's exactly the problem quaternions solve.

Assume |v1| = |v2| = 1

Let A = [cos(d1/2), v1x.sin(d1/2), v1y.sin(d1/2), v1z.sin(d1/2)] = [Aw,Ax,Ay,Az]
Let B = [cos(d2/2), v2x.sin(d2/2), v2y.sin(d2/2), v2z.sin(d2/2)] = [Bw,Bx,By,Bz]

Q'= A*B = [Qw,Qx,Qy,Qz]
  = [ Aw.Bw - Ax.Bx - Ay.By - Az.Bz,
      Aw.Bx + Ax.Bw + Ay.Bz - Az.By,
      Aw.By - Ax.Bz + Ay.Bw + Az.Bx,
      Aw.Bz + Ax.By - Ay.Bx + Az.Bw ]

    cos(d'/2) = Aw.Bw - Ax.Bx - Ay.By - Az.Bz
v'x.sin(d'/2) = Aw.Bx + Ax.Bw + Ay.Bz - Az.By
v'y.sin(d'/2) = Aw.By - Ax.Bz + Ay.Bw + Az.Bx
v'z.sin(d'/2) = Aw.Bz + Ax.By - Ay.Bx + Az.Bw

|(Qx,Qy,Qz)| = |sin(d'/2)| 
             = sqrt(1-cos(d'/2)^2)
             = sqrt(1-Qw)

v' = (Qx,Qy,Qz) / sqrt(1-Qw)
d' = acos(Qw)*2  

[edited by - Fruny on May 3, 2004 12:51:08 PM]

does it work unless you rotate one axis onto another though?