There are a couple of ways we could do this. One would be to explicitly compute the matrix for
T R^{–1}
and find its trace, the sum of its diagonal elements. Any rotation matrix in 3D by an angle β has a trace of:
1 + 2 cos β
So if we solve:
trace(T R^{–1}) = 1 + 2 cos 2πp/N
for α, for any integer p, that value for α will do the job.
But if computing those matrices sounds like too much work, there’s an even simpler method. We can combine the two rotations, T and R^{–1}, using the “mirror trick”.
Any rotation by an angle β is equivalent to two successive reflections, by a pair of mirrors whose planes meet along the axis of rotation, and are separated by an angle of β/2. We are free to rotate the two mirrors around the axis in any way we like, so long as we keep the correct angle between them.
This means that if we want to combine two rotations around two different axes, we can choose the second mirror for the first rotation, and the first mirror for the second rotation, to lie in exactly the same plane: the plane that contains both axes of rotation.
But two successive reflections in the same plane cancel out. That means the result of performing the two rotations we’re combining is now equivalent to reflections in the first mirror of the first rotation and the second mirror of the second rotation, and the angle of rotation is just twice the angle between those mirrors!
=> More informations about this toot | View the thread | More toots from gregeganSF@mathstodon.xyz
text/gemini
This content has been proxied by September (ba2dc).