Rotations about an arbitrary axis

[Spencer W. Thomas]

In article <1990May31.171937.14296 at watcgl.waterloo.edu> jdchrist at watcgl.waterloo.edu (Dan Christensen) writes:

me> In article <SPENCER.90May22144823 at spline.eecs.umich.edu> spencer at eecs.umich.edu (Spencer W. Thomas) writes:
me>	You want to rotate the vector V1 into the vector V2.  We can
me>do this in two steps: (1) rotate V1 to the X axis and then (2) rotate
me>the X axis to V2.  (Why do this, you say?  Because each step is easy.)

> But there is more than one rotation mapping a vector V1 to a vector V2.
> The final result can be oriented in any way about V2.  The original
> poster indicated that he wanted the one that rotates about V1xV2,
> ie. the most "direct" rotation.  I don't think that your solution 
> does this. 

> For example, suppose in a right handed coordinate system the poster
> wants to rotate the y axis to the z axis (by rotating about y cross z,
> ie. the x axis).  This will map y -> z, z -> -y and x -> x.  I believe
> that your solution will map y -> z, z -> -x and x -> -y which is
> probably not what is wanted.  (I may have the details wrong, but I
> think the idea is right.)

I agree that the problem, as I stated it, is ill-defined.  However, my
solution definitely maps the normal vector onto itself, by
construction.  The first matrix takes v1->x, n->z (and m->y), the
second matrix takes (x->v2, z->n (and y->p)).  Thus, the normal vector
(rotation axis) remains fixed.  Since each matrix is carefully
constructed to be a pure rotation, the result must be a pure rotation.

Let's try it with your example.

1.	n = v1 x v2 = (0 1 0) x (0 0 1) = (1 0 0)
2.	m = n x v1 = (1 0 0) x (0 1 0) = (0 0 1)
3.	R1 = (v1' m' n') = (0 0 1)
			   (1 0 0)
			   (0 1 0)
4.	p = n x v2 = (1 0 0) x (0 0 1) = (0 -1 0)
5.	R2 = (v2) = (0  0 1)
	     ( p)   (0 -1 0)
	     ( n)   (1  0 0)
6.	R1 R2 = (1  0  0)
		(0  0  1)
		(0 -1  0)

So the final result takes x -> x, y -> z, and z -> -y, as desired.

--
=Spencer (spencer at eecs.umich.edu)