Wednesday, 20 November 2019

Differentiation of kinematic differential equation for quaternion



We know the kinematic equation in terms of quaternion is the following:


$$\dot{q}(t) = \frac{1}{2}\Omega(t)q(t)$$ where $q(t)$ is the unit quaternion. Now if I want to differentiate the above equation w.r.t "$t$" again:


$$\ddot{q}(t) = \frac{1}{2}\dot\Omega(t)q(t)+\frac{1}{2}\Omega(t)\dot q(t)$$ where $$\dot\Omega(t) = \begin{bmatrix} 0 & \omega_3(t) & -\omega_2(t) & \omega_1(t) \\ -\omega_3(t) & 0 & \omega_1(t) & \omega_2(t) \\ \omega_2(t) & -\omega_1(t) & 0 & \omega_3(t) \\ -\omega_1(t) & -\omega_2(t) & -\omega_3(t) & 0\end{bmatrix}'\\ = \begin{bmatrix} 0 & \alpha_3(t) & -\alpha_2(t) & \alpha_1(t) \\ -\alpha_3(t) & 0 & \alpha_1(t) & \alpha_2(t) \\ \alpha_2(t) & -\alpha_1(t) & 0 & \alpha_3(t) \\ -\alpha_1(t) & -\alpha_2(t) & -\alpha_3(t) & 0\end{bmatrix}$$


where $\alpha(t)$ denotes the angular acceleration.



The derivation of the above is one step of my research, just sincerely make sure that if the above makes sense (mathematical and physical). I search a few papers, just by the title of my question; however, I have not seen "one step" of these papers related to the above.


References:


http://www.tu-berlin.de/fileadmin/fg169/miscellaneous/Quaternions.pdf


http://www.euclideanspace.com/physics/kinematics/angularvelocity/QuaternionDifferentiation2.pdf


http://ieeexplore.ieee.org/document/127239/


https://arxiv.org/abs/1510.02224




Actually I am reading a paper related to the soft-landing problem:


http://web.mit.edu/larsb/www/iee_tcst13.pdf


And focus on the state space representation of a system with the state $x = [q \ \ \dot q ]$. And include this to a optimization problem.



Hope this will gain a bit physical point of view.



Answer



You can confirm the results by comparing with the standard derivation of angular acceleration using euler angles.


For example if the three rotations have $i=1 \ldots 3$:



  • Rotation angle $\theta_i$

  • Local Rotation axis ${\bf z}_i$

  • 3×3 elemental rotation matrix is ${\rm R}_i = {\rm Rot}({\bf z}_i,\theta_i)$

  • Total rotation ${\rm R} = {\rm R}_1{\rm R}_2{\rm R}_3$



The rotational velocity of the body is


$$ {\boldsymbol \omega} = {\bf z}_1 \dot{\theta}_1 + {\rm R}_1 \left( {\bf z_2} \dot{\theta}_2 + {\rm R}_2 \left( {\bf z}_3 \dot{\theta}_3 \right) \right) $$


This is defined from $\dot{\rm R} = {\boldsymbol \omega} \times {\rm R}$


hint at: https://physics.stackexchange.com/a/73969/392


Collect the terms and form using a 3×3 coefficient matrix (called the Jacobian) and the vector of angle speeds.


$$ {\boldsymbol \omega} = \left| \matrix{{\bf z}_1 & {\rm R}_1 {\bf z_2} & {\rm R}_1 {\rm R}_2 {\bf z_3}} \right| \pmatrix{ \dot{\theta}_1 \\\dot{\theta}_2 \\\dot{\theta}_3 } $$


First derivative


$$ {\rm J} = \left| \matrix{{\bf z}_1 & {\rm R}_1 {\bf z_2} & {\rm R}_1 {\rm R}_2 {\bf z_3}} \right|$$ $$ \pmatrix{ \dot{\theta}_1 \\\dot{\theta}_2 \\\dot{\theta}_3 } = \left({\rm J}^{-1} \right) {\boldsymbol \omega} $$


NOTE: The 3×3 Jacobian matrix ${\rm J}$ is invertible unless there is gimbal lock condition.


This should be equivalent to



$$ \dot{q} = \frac{1}{2} \Omega q $$


Second Derivative


$$ \dot{ {\boldsymbol \omega} } = {\rm J} \pmatrix{ \ddot{\theta}_1 \\ \ddot{\theta}_2 \\ \ddot{\theta}_3 } +\dot{\rm J} \pmatrix{ \dot{\theta}_1 \\ \dot{\theta}_2 \\ \dot{\theta}_3 }$$


Note that $\dot{\bf z}_i =0$ since the local rotation axes are fixed.


$$\dot{ {\boldsymbol \omega} } = {\rm J} \pmatrix{ \ddot{\theta}_1 \\ \ddot{\theta}_2 \\ \ddot{\theta}_3 } + \left| \matrix{0 & \dot{\rm R}_1 {\bf z_2} & \left(\dot{\rm R}_1 {\rm R}_2 + {\rm R}_1 \dot{\rm R}_2\right) {\bf z_3}} \right| \pmatrix{ \dot{\theta}_1 \\\dot{\theta}_2 \\\dot{\theta}_3 }$$


We can expand $$\dot{\rm R}_1 = \left(\boldsymbol{z}_{1}\dot{\theta}_{1}\right)\times{\rm R}_{1} $$ $$\dot{\rm R}_2 = \left(\boldsymbol{z}_{2}\dot{\theta}_{2}\right)\times{\rm R}_{2} $$ $$\dot{\rm R}_1 {\rm R}_2 + {\rm R}_1 \dot{\rm R}_2 = \left(\boldsymbol{z}_{1}\dot{\theta}_{1}+{\rm R}_{1}\boldsymbol{z}_{2}\dot{\theta}_{2}\right)\times{\rm R}_{1}{\rm R}_{2} $$


$$\dot{ {\boldsymbol \omega} } = {\rm J} \pmatrix{ \ddot{\theta}_1 \\ \ddot{\theta}_2 \\ \ddot{\theta}_3 } + {\rm K} \pmatrix{ \dot{\theta}_1 \\\dot{\theta}_2 \\\dot{\theta}_3 }$$


with the non-invertible cross terms matrix ${\rm K}=\left| \matrix{0 & \left(\boldsymbol{z}_{1}\dot{\theta}_{1}\right)\times{\rm R}_{1} {\bf z_2} & \left(\boldsymbol{z}_{1}\dot{\theta}_{1}+{\rm R}_{1}\boldsymbol{z}_{2}\dot{\theta}_{2}\right)\times{\rm R}_{1}{\rm R}_{2} {\bf z_3}} \right|$


$$\pmatrix{ \ddot{\theta}_1 \\\ddot{\theta}_2 \\\ddot{\theta}_3 } = {\rm J}^{-1} \left( \dot{ {\boldsymbol \omega} } - {\rm K} \pmatrix{ \dot{\theta}_1 \\\dot{\theta}_2 \\\dot{\theta}_3 } \right) $$


This should be equivalent to



$$ \ddot{q} = \frac{1}{2} \dot{\Omega} q + \frac{1}{2} \Omega \dot{q}$$


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...