Coordinate Transformation of Torque

Let's look at this example:

The Euler equations of motion given in components of a body fixed frame are:

$$I,vec{dot{omega}}_B+vec{omega}_Btimes (I,vec{{omega}}_B)=vec{tau}_Btag 1$$

assume that you know the components of the external torque in inertial frame $vec{tau}_I$, so to solve equation (1), you have to transfer the torque components from inertial frame to the body frame.

$$vec{tau}_B=R,vec{tau}_I$$

where $R$ is the rotation matrix between body frame and inertial frame that depend on the Euler angles $R=R(alpha,beta,gamma)$

thus eq. (1)

$$I,vec{dot{omega}}_B+vec{omega}_Btimes (I,vec{{omega}}_B)=R(alpha,beta,gamma),vec{tau}_Itag 2$$

equation (2) are 3 equations for 6 unknowns ,$quadvec{omega}quad$ and $quad alpha,beta,gamma$

the additional 3 equations are the angular velocity equations those are function of the Euler angles .

$$vec{omega}=A(vec{phi}),vec{dot{phi}}tag 3$$

where the matrix $A$ is a $(3times 3)$ matrix, and$quad vec{phi}=left[alpha,beta,gammaright]^T$

equations (2) and (3) are 6 equation for 6 unknowns $vec{omega},,vec{phi}$

Numerical simulation:

for numerical simulation we must transfer equation (1) and (2) to the first order differential equations $vec{dot{y}}=vec{f}(vec{y},t)$

with :

$$vec{y}=begin{bmatrix} vec{omega} vec{phi} end{bmatrix}$$

we get with eq. (2) and (3)

$$vec{dot{y}}=begin{bmatrix} I^{-1}left[R(vec{phi}),vec{tau}_I-vec{omega}_Btimes (I,vec{{omega}}_B)right] A^{-1}(vec{phi}),vec{omega} end{bmatrix}=vec{f}(vec{y})$$

notiz that the equation has singularity in one of the Euler angles

Let's say you have a 3DOF rotational joint, like a gimbal, with three rotations, about three local axes $boldsymbol{hat{z}}_1$, $boldsymbol{hat{z}}_2$ and $boldsymbol{hat{z}}_3$ with angles $theta_1$, $theta_2$ and $theta_3$ in sequence.

The resulting rotational velocity is

$$ boldsymbol{omega} = boldsymbol{hat{z}}_1 dottheta_1 + mathrm{rot}(boldsymbol{hat{z}}_1,theta_1) left( boldsymbol{hat{z}}_2 dot{theta}_2 + mathrm{rot}(boldsymbol{hat{z}}_2,theta_2) boldsymbol{hat{z}}_3 dot{theta}_3 right) tag{1}$$

From that you extract the 3×3 Jacobian matrix

$$ boldsymbol{omega} = mathbf{J} pmatrix{ dot theta_1 dot theta_2 dot theta_2 } tag{2} $$

where $mathbf{J}$ is defined by three column vectors

$$ mathbf{J} = left( begin{array}{c|c|c} boldsymbol{hat{z}}_1 & mathrm{rot}(boldsymbol{hat{z}}_1,theta_1)boldsymbol{hat{z}}_2 & mathrm{rot}(boldsymbol{hat{z}}_1,theta_1)mathrm{rot}(boldsymbol{hat{z}}_2,theta_2)boldsymbol{hat{z}}_3 end{array} right) tag{3} $$

Now suppose each axis has a motor attached to it supplying torques $Q_1$, $Q_2$ and $Q_3$ such as the total power of the system is $$ P = Q_1 dot theta_1 + Q_2 dot theta_2 + Q_3 dot theta_3 tag{4}$$

The question is, how are the individual torques add up to the total torque on the base of the gimbal $boldsymbol{tau}$?

If the end effector 3×3 mass moment of inertia matrix is $mathbf{I}$ then the answer is

$$ boldsymbol{tau} = underbrace{ mathbf{I}, mathbf{J} left( mathbf{J}^top mathbf{I} ,mathbf{J} right)^{-1} }_{mathbf{J}^star} pmatrix{Q_1 Q_2 Q_3} + ldots text{(velocity related terms)}tag{5} $$

Where the 3×3 matrix $mathbf{J}^star = mathbf{I}, mathbf{J} left( mathbf{J}^top mathbf{I} ,mathbf{J} right)^{-1}$ is the Jacobian pseudo inverse (transposed), since $mathbf{J}^top mathbf{J}^star = mathbf{1}$.

You can easily prove the above by evaluating power as follows and comparing it to (4)

$$ P = boldsymbol{omega} cdot boldsymbol{tau} = (mathbf{J} dot{q})^top (mathbf{J}^star Q) = dot{q}^top ( mathbf{J}^top mathbf{J}^star ) Q = dot{q}^top Q tag{6} $$

where $dot{q} = pmatrix{ dot theta_1 dot theta_2 dot theta_3}$ are the joint speeds and $Q = pmatrix{Q_1 Q_2 Q_3}$ the joint torques.

If you expand out (6) you will find (4).

The reverse operation is rather simple. Going from the torque vector $boldsymbol{tau}$ to the individual motor torques is done by

$$ Q = mathbf{J}^top boldsymbol{tau} tag{7} $$

Appendix

The full expression for torque (5) including the velocity related terms is

$$ boldsymbol{tau} = mathbf{J}^star Q + left( mathbf{1} - mathbf{J}^star mathbf{J}^top right) left( mathbf{I} ,dot{mathbf{J}} dot{q} + boldsymbol{omega} times mathbf{I} boldsymbol{omega} right) $$

where $mathbf{1}$ is the 3×3 identity matrix, and $dot{mathbf{J}}$ is the time rate of the jacobian. It can be found from the derivative of (3) as

$$ dot{mathbf{J}}dot{q}=mathbf{R}_{1}left(left(boldsymbol{hat{z}}_{1}timesboldsymbol{hat{z}}_{2}right)dot{theta}_{2}dot{theta}_{1}+left(boldsymbol{hat{z}}_{1}timesmathbf{R}_{2}boldsymbol{hat{z}}_{3}right)dot{theta}_{3}dot{theta}_{1}+left(mathbf{R}_{2}left(boldsymbol{hat{z}}_{2}timesboldsymbol{hat{z}}_{3}right)right)dot{theta}_{3}dot{theta}_{2}right) $$

and with the rotation matrices $mathbf{R}_{1} = mathrm{rot}(boldsymbol{hat{z}}_1,theta_1)$ and $mathbf{R}_{2} = mathrm{rot}(boldsymbol{hat{z}}_2,theta_2)$.

You can also prove (7) by

$$ Q = mathbf{J}^top boldsymbol{tau} = mathbf{J}^top mathbf{J}^star Q + mathbf{J}^top left( mathbf{1} - mathbf{J}^star mathbf{J}^top right) left( mathbf{I} ,dot{mathbf{J}} dot{q} + boldsymbol{omega} times mathbf{I} boldsymbol{omega} right)$$ $$require{cancel} = cancelto{1}{(mathbf{J}^top mathbf{J}^star)} Q + left( mathbf{J}^top - cancelto{1}{(mathbf{J}^top mathbf{J}^star)} mathbf{J}^top right) left( mathbf{I} ,dot{mathbf{J}} dot{q} + boldsymbol{omega} times mathbf{I} boldsymbol{omega} right)= Q $$