The following is a concrete example of what happens when you concatenate two vector-quaternion "frames" (rigid body transformations). For the purpose of making this comprehensible, I'm leaving out a lot of the gritty details of quaternion multiplication.
Frame notation: <Rotation operator, Translation operator>.
Input frames:
A = <Rot(90, y), Trans(10, 0, 0)>
B = <Rot(90, x), Trans(0, 0, 5)>
And we want to calculate (and verify) the product:
C = A * B
As a starting point, we expand the frames into operators and vectors:
C * v =
<Rot(90, y), Trans(10, 0, 0)> * <Rot(90, x), Trans(0, 0, 5)> v =
<Rot(90, y), Trans(10, 0, 0)> * (Rot(90, x) * v + (0, 0, 5)) =
Rot(90, y) * (Rot(90, x) * v + (0, 0, 5)) + (10, 0, 0)
Quaternions for A and B
(x, y, z in the below are "unit x", "unit y", "unit z")
A, Rot(90, y): p = exp((pi/4)*y) = [cos(pi/4), sin(pi/4)*y] = (sqrt(2)/2)[0, y]
B, Rot(90, x): q = exp((pi/4)*x) = [cos(pi/4), sin(pi/4)*x] = (sqrt(2)/2)[0, x]
Equivalent quaternion-vector expression for C * v
p * (q * v * q' + (0, 0, 5)) * p' + (10, 0, 0)
Simplify the first Quaternion product:
q * v * q' = (vx, -vz, vy)
This is Rot(90, x). The intuition is that a rotation around the x axis leaves x fixed. The rotation itself can be interpreted as a rotation of the basis. If the x axis points out of the screen:
y y'--x
| |
z--x => z'
New basis:
x' = x
y' = z
z' = -y
[x z -y] * (vx, vy, vz) =
[1 0 0][vx] [ vx]
[0 0 -1][vy] = [-vz]
[0 1 0][vz] [ vy]
Which leaves:
p * ((vx, -vz, vy) + (0, 0, 5)) * p' + (10, 0, 0) =
p * (vx, -vz, vy + 5) * p' + (10, 0, 0)
Simplify the second quaternion product for some input vector u (makes the calculations nicer):
p * u * p' = (uz, uy, -ux)
Intuitive picture with y pointing out of the screen:
y--x x'
| |
z => y--z'
x' = -z
y' = y
z' = x
[-z y x] * (ux, uy, uz) =
[ 0 0 1][ux] [ uz]
[ 0 1 0][uy] = [ uy]
[-1 0 0][uz] [-ux]
Substitute u = (vx, -vz, vy + 5):
p * (vx, -vz, vy + 5) * p' = (vy + 5, -vz, -vx)
Original equation is now:
(vy + 5, -vz, -vx) + (10, 0, 0) =
(vy + 15, -vz, -vx) =
(vy, -vz, -vx) + (15, 0, 0)
Decomposing, this is the net-effect of:
<Rot(90, y) * Rot(90, x), Trans(15, 0, 0)>
If you turn the above rotations into matrices and multiply, it is easy to verify that the (vy, -vz, -vx) coming from the quaternion multiplication matches:
Rot(90, y) * Rot(90, x) * v =
[ 0 0 1][1 0 0][vx]
[ 0 1 0][0 0 -1][vy] =
[-1 0 0][0 1 0][vz]
[0 1 0][vx] [ vy]
[0 0 -1][vy] = [-vz]
[-1 0 0][vz] [-vx]