# SLERP velocity and acceleration (3d Orientation interpolation)

Hey guys I am quite new to quaternions and I am a bit unsure as to how to use them. How are the velocities and accelerations trajectories determined using quaternions?

I have been using slerp to interpolate the orientations.

\begin{align} Slerp(Q_0, Q_1,t) = Q_0(Q_0^{-1}Q_1)^t \end{align}

Where $$t$$ is the interpolation parameter from $$[0;1]$$. I read on wikipedia that:

*The derivative of $$Slerp(q_0, q_1; t)$$ with respect to t, assuming the ends are fixed, is $$log(q_1 q_0^{−1})$$ times the function value, where the quaternion natural logarithm in this case yields half the 3D angular velocity vector.*

This is further supported by a document I found on the topic of quaternion interpolation in animation (Dam, E. B., Koch, M., & Lillholm, M., (1998)), where in equation: (6.12), they show the first and second derivative (using their notation):

\begin{align} \frac{d}{dt}Slerp(p,q,h) &= \frac{d}{dt} p (p^* q)^h \\ &= p(p^* q)^h log(p^* q)\\ &= Slerp(p,q,h) log(p^*q)\\\\ \frac{d^2}{dh^2}Slerp(p,q,h) &= p(p^* q)^h log(p^*q)^2\\ &= Slerp(p,q,h) log(p^* q)^2 \end{align}

In other words, to get the second derivative of $$Slerp$$ you need to times the function value with $$log(Q_1, Q_0^{-1})^2$$.

Is this legitimate? And can I use these resulting quaternions for:

\begin{align} X_d = \begin{bmatrix}p_d\\ Q_d \end{bmatrix}, \quad \dot{X}_d = \begin{bmatrix}\dot{p}_d\\ \dot{Q}_d \end{bmatrix} \quad \ddot{X}_d = \begin{bmatrix}\ddot{p}_d\\ \ddot{Q}_d \end{bmatrix} \end{align}

where $$p = [x,y,z]^\intercal$$ and $$Q = \{\eta, \epsilon \}$$, and the dots $$\dot{X}$$ and $$\ddot{X}$$ are the respective velocities and accelerations.

Because I am not sure what to make of the results. Maybe because I don't really understand quaternions. For context here's an example:

Python:

import numpy as np
import quaternion
import matplotlib.pyplot as plt
import matplotlib as mpl
mpl.style.use('seaborn')
plt.rcParams['figure.figsize'] = [10, 15]

t = np.linspace(0,1,100)
Q = np.zeros((4,100))
Qd = np.zeros((4,100))
Qdd = np.zeros((4,100))

rot0 = np.array([[ 1.0,  0.0,  0.0],
[ 0.0, -1.0,  0.0],
[ 0.0,  0.0, -1.0]])

rotf = np.array([[ 0.0, -1.0,  0.0],
[-1.0,  0.0,  0.0],
[ 0.0,  0.0, -1.0]])

q0 = quaternion.from_rotation_matrix(rot0)
qf = quaternion.from_rotation_matrix(rotf)

for i in range(100):
slerp = q0 * (q0.inverse() * qf)**t[i]
slerp_d = slerp * np.log(qf * q0.inverse())
slerp_dd = slerp * np.log(qf * q0.inverse())**2

Q[:,i] = quaternion.as_float_array(slerp)
Qd[:,i] = quaternion.as_float_array(slerp_d)
Qdd[:,i] = quaternion.as_float_array(slerp_dd)

fig, (ax1, ax2, ax3) = plt.subplots(3, 1)
ax1.plot(Q.T)
ax1.set_title("Slerp")
ax1.legend(['w','i','j','k'])
ax2.plot(Qd.T)
ax2.set_title("Slerp_d")
ax2.legend(['w','i','j','k'])
ax3.plot(Qdd.T)
ax3.set_title("Slerp_dd")
ax3.legend(['w','i','j','k'])
plt.show() ## 2 Answers

Derivative of $$\sin$$ is $$\cos$$, and the derivative of $$\cos$$ is $$-sin$$.

Given a quaternion definition of:

$$q = \cos{a} + \mathbf{r}\sin{a}$$

$$\mathbf{r}^2 = -1$$

I would expect to see what is effectively a phase shift at every derivative level, and that's what I'm seeing in your curves. I noticed your magnitude seems to be growing, but that may be because you need to normalize at every step.

• It was the magnitude that was throwing me off. Yup, thanks for pointing out that I forgot to normalize according to the timestep. I think this resolves it. – Spaceman Jun 2 '20 at 12:48

I think you may have a mistake in python code. That is:

slerp_d = slerp * np.log(qf * q0.inverse())

I think the correct one is:

slerp_d = slerp * np.log(q0.inverse()*qf)