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I am trying to grasp the concepts of forces applied to a rigid body resulting in a net change of the rotation of the object. This is not a home assignment and I read some resources on both the web and YouTube but haven't found the answer to this.

The problem

I was thinking of how I can calculate the resulting rotational matrix $R$, given a timestep $dt$, the force $F$ and the vector $r$ with the inertia matrix $I$ for the object in question.

  • $dt$: The timestep
  • $F$: is a general 3-dimensional force.
  • $r$: is the position where this force acts relative the object's center of rotation.
  • $I$: is the 3x3 inertia matrix (which also is assumed to be invertible).

We have torque $\tau = F \times r$.

We also have $\tau = I \cdot \alpha$, where $\alpha$ is the angular acceleration.

To calculate a net rotation, assuming $I$ is invertible, we have:

equation(1): $\alpha = I^{-1} \cdot \tau = I^{-1} \cdot (F \times r)$

My initial thought was to integrate over a time step $dt$ and calculate the angular velocity:

$\alpha \cdot dt$

Then from this calculate a normalized vector around where the rotation occurs and resulting rotation angle around this axis. Let's call this vector $w$ and the angle $a$

After this calculate the matrix from:

$K = \begin{bmatrix} 0 & -w_z & w_y \\ w_z & 0 & -w_x \\ -w_y & w_x & 0 \end{bmatrix}$

We then can use Rodrigues' rotation formula ($I$ in the formula below is the identity matrix):

$R = I + sin(a)\cdot K + (1-cos(a))\cdot K^2$

Question: Given the integrated angular velocity for a small timestep $\alpha \cdot dt$, where $\alpha$ comes from equation(1) above. How can I calculate the normalized vector $w$ and the angle $a$?

Alternatively can I calculate a generalized vector $w$ where the length $|w|$ is proportional to the angle $a$?

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  • $\begingroup$ A lot of your questions are answered with careful reading of Notes I of cs.cmu.edu/~baraff/sigcourse $\endgroup$ Nov 26, 2023 at 7:54
  • $\begingroup$ Please note that the rotational equations of motion are not $\tau = I \alpha$, but rather a bit more complex. See my answer below. $\endgroup$ Nov 26, 2023 at 8:39

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Nothing in Dynamics is simple

Your question can be split up into several steps

  1. Given a known orientation ${\rm R}$ and angular velocity $\vec{\omega}$ of a rigid body and the net torque $\vec{\tau}_C$ summed about the center of mass, find the angular acceleration of the body $\vec{\alpha}$.

    • Usually, the mass moment of inertia of a body is given on body-fixed coordinates $$\overline{I}_{\rm body} = \pmatrix{I_1 & & \\ & I_2 & \\ & & I_3}$$

    • Transform the MMOI matrix into the inertial orientation, summed about the center of mass with $$ \overline{I}_C = {\rm R}\, \overline{I}_{\rm body} {\rm R}^\intercal$$

    • Angular momentum vector of the body summed about the center of mass is $$ \vec{L}_C = \overline{I}_C \vec{\omega}$$

    • Rotational equations of motion for the body are $$ \vec{\tau}_C = \overline{I}_C \vec{\alpha} + \vec{\omega} \times \vec{L}_C $$

    • Rotational acceleration of the body is $$ \vec{\alpha} = \overline{I}^{-1}_C \left( \vec{\tau}_C - \vec{\omega} \times \vec{L}_C \right)$$

  2. Integrate angular acceleration $\vec{\alpha}$ and angular velocity $\vec{\omega}$ over a time step ${\rm d}t$ to find new values for the velocity and the rotation matrix ${\rm R}$.

    • Integration of rotational velocity is straightforward

    $$ \vec{\omega} \leftarrow \vec{\omega} + \int \vec{\alpha}\,{\rm d}t$$

    • Integration of the rotation matrix is not advised as it diverges from representing a rotation quickly. Note that at any instant $$ \tfrac{\rm d}{{\rm d}t} {\rm R} = \vec{\omega} \times {\rm R} $$ which can be done using linear algebra using the $K$ matrix in your question. So integration would be $$ {\rm R} \leftarrow {\rm R} + \int \vec{\omega}\times {\rm R} \,{\rm d}t$$ which quickly diverges from being a rotation matrix.

    To solve this problem, people usually use quaternions $q$ which encode the same information as ${\rm R}$ but with 4 values instead of 9 in the rotation matrix. In fact any quaternion is a combination of a scalar value and a vector $$q = (q_s |\, \vec{q}_v)$$ These are operations you need to work with quaternions

    • Construct a quaternion $q$ from a rotation axis $\hat{z}$ and angle $\theta$ $$q = ( \cos \left( \tfrac{\theta}{2} \right) | \, \hat{z}\sin \left( \tfrac{\theta}{2} \right) )$$

    • A sequence of two rotations is represented by the quaternion product $$ a b = ( a_s b_s - \vec{a}_v \cdot \vec{b}_v | a_s \vec{b}_v + b_s \vec{a}_v + \vec{a}_v \times \vec{b}_v)$$

      Where $\cdot$ and $\times$ are the vector dot and vector cross product respectively.

    • The rotation matrix of quaternion $q = (q_s |\, \vec{q}_v)$ is extracted by $$ {\rm R} = {\bf 1} + 2 q_s [\vec{q}_v\times] + 2 [\vec{q}_v\times] [\vec{q}_v\times]$$ where ${\bf 1}$ is the identity matrix, and $[\vec{q}_v\times]$ is the 3×3 skew-symmetrix matrix similar to the $K$ matrix above representing the cross product as matrix multiplication. The inverse rotation matrix is calculated similarly but with a negative sign in front of $q_s$.

    • The rate of change of quaternion based on a body rotational velocity of $\vec{\omega}$ is evaluated with the following quaternion product $$\begin{aligned}\tfrac{{\rm d}}{{\rm d}t}(q_{s}|\,\vec{q}_{v}) & =\tfrac{1}{2}(0|\,\vec{\omega})(q_{s}|\,\vec{q}_{v})\\ & =\tfrac{1}{2}(-\vec{\omega}\cdot\vec{q}_{v}|\,q_{s}\vec{\omega}+\vec{\omega}\times\vec{q}_{v}) \end{aligned} $$

    • The integration of the orientation of the body is done with $$q \leftarrow q + \int \tfrac{1}{2} ( 0 |\, \vec{\omega}) \,q\,{\rm d}t$$ which surprisingly does not diverge away from a rotation very quickly. In practice a renormalization process of $$ q \leftarrow ( \frac{q_s}{\sqrt{q_s^2 + \vec{q}_v \cdot \vec{q}_v}} | \, \frac{\vec{q}_v}{\sqrt{q_s^2 + \vec{q}_v \cdot \vec{q}_v}})$$ is needed every few dozen integration steps.

    • Since there isn't an analytical solution to the equations above, a numerical solver for ODEs is needed to carry out the time steps.


I wanted to state that in practice, the above process is never done in terms of linear and angular velocity. What is preferred is to keep track of linear $\vec{p}$ and angular $\vec{L}_C$ momentum and perform the integrations just as follows

$$ \vec{p} \leftarrow \vec{p} + \int \vec{F} {\rm d}t $$

$$ \vec{L}_C \leftarrow \vec{L}_C + \int \vec{\tau}_C {\rm d}t $$

where $\vec{F}$ and $\vec{\tau}_C$ are the applied forces and torques.

To extract the motion from momentum use

$$ \vec{v}_C = \tfrac{1}{m} \vec{p} $$ $$ \vec{\omega} = \overline{I}_C^{-1} \vec{L}_C $$

The above process is far simpler than trying to calculate $\vec{\alpha}$ and then integrating it.

Here is a summary of each time step in a rigid body simulation

Simulation steps for rigid body motion Formulas used
rotation matrix ${\rm R}$ from quaternion ${(q_{s}|\,\vec{q}_{v})}$ $${\rm R}={\bf 1}+2q_{s}[\vec{q}_{v}\times]+2[\vec{q}_{v}\times][\vec{q}_{v}\times]$$
mass moment of inertia tensor ${\overline{I}_{C}}$ at CG $$\overline{I}_{C}={\rm R}\overline{I}_{{\rm body}}{\rm R}^{-1}$$
integrate momentum ${\vec{p}}$ from force ${\vec{F}}$ $$\Delta\vec{p}=\int\vec{F}\,{\rm d}t$$
integrate ang. momentum ${\vec{L}_{C}}$ from torque ${\vec{\tau}_{C}}$ $$\Delta\vec{L}_{C}=\int\vec{\tau}_{C}\,{\rm d}t$$
calculation of velocity vector ${\vec{v}_{C}}$ at CG. $$\vec{v}_{C}=\tfrac{1}{m}\vec{p}$$
calculation of body rotational vector ${\vec{\omega}}$ $$\vec{\omega}=\overline{I}_{C}^{\,-1}\vec{L}_{C}$$
integrate position ${\vec{r}_{C}}$ from velocity $$\Delta\vec{r}_{C}=\int\vec{v}_{C}{\rm d}t$$
integration of quaternion $q$ from rotational velocity $\vec{\omega}$ $$(\Delta q_{s}|\,\Delta\vec{q}_{v})=\int \tfrac{1}{2}(-\vec{\omega}\cdot\vec{q}_{v}|\,q_{s}\vec{\omega}+\vec{\omega}\times\vec{q}_{v})\,{\rm d}t$$
renormalization of quaternion $$ \begin{array}{c} q_{{\rm mag}}=\sqrt{q_{s}^{2}+\vec{q}_{v}\cdot\vec{q}_{v}}\\ q=(\frac{q_{s}}{q_{{\rm mag}}}|\,\frac{\vec{q}_{v}}{q_{{\rm mag}}}) \end{array}$$
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  • $\begingroup$ So in practice you integrate the force into the momentum and then the torque into the angular momentum, as you noted in the last 4 formulas, and then you're done? $\endgroup$
    – Decaf Sux
    Nov 26, 2023 at 15:00
  • $\begingroup$ Force integrates into momentum, which is converted into velocity and integrated into position. Torque integrates into angular momentum, which is converted into rotational velocity and integrated into a quaternion orientation. $\endgroup$ Nov 26, 2023 at 17:02
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    $\begingroup$ I know this comment is going to be removed, but thanks for your effort to explain. I understand this much better now. Thanks :-) $\endgroup$
    – Decaf Sux
    Nov 26, 2023 at 17:16
  • $\begingroup$ @DecafSux - see edit on post for a cheatsheet needed to do rigid body simulations. $\endgroup$ Nov 26, 2023 at 17:27
  • $\begingroup$ @DecafSux - also read cs.cmu.edu/~baraff/sigcourse/notesd1.pdf to get a lot more details on how to do this. $\endgroup$ Nov 26, 2023 at 17:28
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Euler equation in body fixed frame \begin{align*} & I\,\dot{\omega}+\omega\times\,I\,\omega=\tau\tag 1 \end{align*} from the rotation matrix $~R(\phi_1~,\phi_2~,\phi_2~)~$ you obtain \begin{align*} &\omega=J_R\,\dot\phi\quad \phi=\begin{bmatrix} \phi_1 \\ \phi_2 \\ \phi_3 \\ \end{bmatrix}\\\ \end{align*} $\Rightarrow$ \begin{align*} &\dot{\phi}=[~J_R~]^{-1}\,\omega\tag 2 \end{align*} for the numeric simulation you have to solve equation (1) and (2). with the solution $~\phi_i(t)~$ you obtain the transformation matrix $~R(\phi_i)~$

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  • $\begingroup$ The jacobian is needed when using Euler angles, but it is not needed when using quaternions. In any case $$I\,\dot{\omega}+\omega\times\,I\,\omega=\tau$$ is valid in any basis vector orientation, either body aligned, or inertial aligned. Either way you either need to transform the torque/motion vectors to body frame, or transform MMOI to inertial frame (See Eq. (2.39) - Physically Based Modeling) $\endgroup$ Nov 27, 2023 at 23:53
  • $\begingroup$ For vehicle dynamics simulation, you don’t need the quaternions at all, because the roll and pitch angles are small, so you can use my solution, which is very simple compared to yours . $\endgroup$
    – Eli
    Nov 28, 2023 at 7:20

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