Twisted Oak

Quaternions part 2

posted by harrison on August 31, 2012

This will be a short post explaining why conjugation is rotation, and why the axis-angle formula works. If you didn’t see it, here’s part 1.

Why is conjugation rotation?

To see why conjugation is rotation, it is helpful to expand the multiplication (assuming q is a unit quaternion):

$q = cos(\frac{\theta}{2}) + sin(\frac{\theta}{2})\hat{v}$

$quq^{-1} = (cos(\frac{\theta}{2}) + sin(\frac{\theta}{2})\hat{v}) u (cos(\frac{\theta}{2}) - sin(\frac{\theta}{2})\hat{v})$

$quq^{-1} = cos^2(\frac{\theta}{2})u - cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) u \hat{v} + cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) \hat{v} u - sin^2(\frac{\theta}{2})\hat{v} u \hat{v}$

$quq^{-1} = cos^2(\frac{\theta}{2})u + cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v}u - u \hat{v}) - sin^2(\frac{\theta}{2})\hat{v} u \hat{v}$

When multiplying vectors as quaternions, notice that since the real part is 0, $uv = 0 - u \cdot v + 0v + 0u + u \times v = -u \cdot v + u \times v$ .

$quq^{-1} = cos^2(\frac{\theta}{2})u + cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) ((\hat{v} \cdot u + \hat{v} \times u)- (u \cdot \hat{v} + u \times \hat{v})) - sin^2(\frac{\theta}{2})\hat{v} u \hat{v}$

$quq^{-1} = cos^2(\frac{\theta}{2})u + cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u - u \times \hat{v}) - sin^2(\frac{\theta}{2})\hat{v} u \hat{v}$

$quq^{-1} = cos^2(\frac{\theta}{2})u + 2cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u) - sin^2(\frac{\theta}{2})\hat{v} u \hat{v}$

$quq^{-1} = cos^2(\frac{\theta}{2})u + 2cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u) - sin^2(\frac{\theta}{2})(-\hat{v} \cdot u + \hat{v} \times u)\hat{v}$

$quq^{-1} = cos^2(\frac{\theta}{2})u + 2cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u) - sin^2(\frac{\theta}{2})(-(\hat{v} \cdot u)\hat{v} - (\hat{v} \times u) \cdot \hat{v} + (\hat{v} \times u) \times \hat{v})$

$quq^{-1} = cos^2(\frac{\theta}{2})u + 2cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u) - sin^2(\frac{\theta}{2})(-(\hat{v} \cdot u)\hat{v} + (\hat{v} \times u) \times \hat{v})$

To finish, we need to talk about parallel and perpendicular components of $u$ to $\hat{v}$ . $u_\parallel = (\hat{v} \cdot u) \hat{v}$ and $u_\perp = (\hat{v} \times u) \times \hat{v}$ . Notice that $u_\parallel + u_\perp = u$ .

$quq^{-1} = cos^2(\frac{\theta}{2})u + 2cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u) - sin^2(\frac{\theta}{2})(-u_\parallel + u_\perp)$

$quq^{-1} = cos^2(\frac{\theta}{2})u + 2cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u) - sin^2(\frac{\theta}{2})(-2 u_\parallel + u )$

$quq^{-1} = (cos^2(\frac{\theta}{2}) - sin^2(\frac{\theta}{2}))u + 2cos(\frac{\theta}{2}) sin(\frac{\theta}{2}) (\hat{v} \times u) + sin^2(\frac{\theta}{2})(2 u_\parallel)$

And now we can start using trig to get rid of those half angles:

$quq^{-1} = (\frac{1+cos(\theta)}{2} - \frac{1-cos(\theta)}{2})u + sin(\theta) (\hat{v} \times u) + (1-cos(\theta)) u_\parallel$

$quq^{-1} = cos(\theta)}u + sin(\theta) (\hat{v} \times u) + (1-cos(\theta)) u_\parallel$

$quq^{-1} = cos(\theta)}(u - u_\parallel) + sin(\theta) (\hat{v} \times u) + u_\parallel$

$quq^{-1} = cos(\theta)}u_\perp + sin(\theta) (\hat{v} \times u) + u_\parallel$

From here, you can see that the parallel component remains unchanged, as expected. Also, notice that $u_\perp$ is just $\hat{v} \times u$ rotated $-\frac{\pi}{2}$ around $\hat{v}$ , So you can see that the perpendicular part is being rotated around $\hat{v}$ by $\theta$ .

next time: exponentials, logs, and slerp

Comments are closed.

Twisted Oak Studios offers consulting and development on high-tech interactive projects. Check out our portfolio, or Give us a shout if you have anything you think some really rad engineers should help you with.

Quaternions part 2

Why is conjugation rotation?

Archive

More interesting posts (26 of 33 articles)