Mapping between generalized forces and external torque of a rigid body whose rotation is described by quaternion is not unique(?)

Question

In this paper the mapping between generalized forces and external torques for a rigid body (when the rotation is described by a quaternion) is derived:

$$\textit{F}_Q = 2\textbf{G}^TT'$$

where $\textit{F}_Q$ is a vector of generalized forces (quaternion elements as generalized coordinates) and $T'$ is a vector of external torque and $\textbf{G}$ is the matrix that maps between those two

$$G = \begin{bmatrix}-Q_1 & Q_0 & Q_3 & -Q_2\\ -Q_2&-Q_3& Q_0& Q_1 \\ -Q_3& Q_2& -Q_1& Q_0\end{bmatrix}.$$

I have derived a vector of generalized forces, I mapped it to a vector of external torques, which works correctly. However, when I mapped the external torques back to generalized forces, the generalized forces were different. Does that mean that this mapping is not unique and different generalized forces in terms of quaternion can rotation represent the same external torques applied to a rigid body?

Edit:

I am creating a biomechanical model. My Equations of motion (for 1 joint) are:

$$\textbf{M}(Q)\dot{q}=f(q)$$

where $q = (Q_0,Q_1,Q_2,Q_3,\omega_x,\omega_y,\omega_z)^T$, $\textbf{M}(Q)$ is joint-space mass matrix and $f(q)$ are the external forces ($\omega_x$, $\omega_y$, $\omega_z$ are angular velocities). To add a contribution of muscle force to the equations of motion, I calculate the muscle length $l_m = f(Q_0,Q_1,Q_2,Q_3)$, then I calculate the $4 \times 1$ jacobian $J_Q$ w.r.t. the quaternion coordinates

$$J_Q = \frac{\partial l_m(\textbf{Q})}{\partial Q_i} $$

from this I calculate $3 \times 1$ vector of external torques

$$\tau_E = \frac{1}{2} G J_Q F_m$$

where $F_m$ is a scalar muscle force. Then I can finish my equations of motion:

$$ \textbf{M}(Q)\dot{q}=f(q)+ \begin{pmatrix} \textbf{0}^{4 \times 1} \\ \tau_E^{3 \times 1} \end{pmatrix} $$

and it works perfectly. For some reason I need to map the $3 \times 1$ external torque back to $4 \times 1$ generalized forces (if I can call it this way). And this mapping back doesn't give the same forces as the one calculated by using the jacobian of muscle lengths.

Answer 1

$\def \b {\mathbf}$ Starting with the Euler equations $$ \b I\,\b{\dot{\omega}}+\omega\times\,\b I\,\omega=\b\tau_{\rm{E}}\tag 1$$

with

$$\omega=2\,\b Q\,\b{\dot{z}}\tag 2$$ where $$\b Q= \left[ \begin {array}{cccc} -b&a&-d&c\\ -c&d&a&-b \\ -d&-c&b&a\end {array} \right] \quad \b{\dot{z}}= \begin{bmatrix} \dot{a} \\ \dot{b} \\ \dot{c} \\ \dot{d} \\ \end{bmatrix}$$

the constraint equation for the quaternions is:

$$ a^2+b^2+c^2+d^2=1\quad, a\,\dot a+b\,\dot b+c\,\dot c+d\,\dot d=0$$

from here you obtain

$$ \b{\dot{z}}= \begin{bmatrix} \dot{a} \\ \dot{b} \\ \dot{c} \\ \dot{d} \\ \end{bmatrix}= \underbrace{\left[ \begin {array}{ccc} -{\frac {b}{a}}&-{\frac {c}{a}}&-{\frac {d }{a}}\\1&0&0\\ 0&1&0 \\0&0&1\end {array} \right] }_{\b T} \,\underbrace{\begin{bmatrix} \dot{b} \\ \dot{c} \\ \dot{d} \\ \end{bmatrix}}_{\b{\dot{q}}} $$ hence equation (2)

$$\omega=\underbrace{2\,\b Q\,\b T}_{\b J (3\times 3)}\,\b{\dot{q}}$$

the generalize torque is:

$$\b\tau_G=\b J^T\b\tau_E\quad, \b\tau_E=\left(\b J^T\right)^{-1}\b\tau_G$$

---

the determinate of $~\b J~$ is equal to $~\frac 8a~$

how to simulate

from the constraint equation

\begin{align*} &\dot{a}=-\frac{1}{a}\,(b+c+d) \end{align*} and with $~\omega=\b J\,\b{\dot{q}}~$ you obtain

\begin{align*} &\begin{bmatrix} \dot{b} \\ \dot{c} \\ \dot{d} \\ \end{bmatrix} =\frac 12\,\begin{bmatrix} a & d & -c \\ -d & a & b \\ c & -b & a \\ \end{bmatrix}\,\begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z \\ \end{bmatrix} \end{align*}

those are 4 first order differential equations for the quaternions.

additional form equation (1) you obtain 3 first order differential equations for the angular velocity $~\b \omega~$ so altogether 7 first order differential equation. notice that the initial conditions must fellfield the constraint equation $~a(0)^2+b(0)^2+c(0)^2+d(0)^2=1~$ and also you have to check it during the simulation

$~a(t)^2+b(t)^2+c(t)^2+d(t)^2=1~$

the rotation matrix between body system and inertial system is a function of the quaternions

\begin{align*} &\b S= \left[ \begin {array}{ccc} {a}^{2}+{b}^{2}-{c}^{2}-{d}^{2}&2\,bc+2\,a d&2\,bd-2\,ac\\ 2\,bc-2\,ad&{a}^{2}+{c}^{2}-{d}^{2}- {b}^{2}&2\,cd+2\,ab\\ 2\,bd+2\,ac&2\,cd-2\,ab&{a}^{2 }+{d}^{2}-{b}^{2}-{c}^{2}\end {array} \right] \end{align*}