Derivation of Newton-Euler Inverse Dynamics

Question

I'm currently studying from the book Modern Robotics - Mechanics, Planning, and Control, and I'm having a hard time following the derivation of the Newton-Euler inverse dynamics algorithm, particularly the expression for the acceleration of rigid link {i}.

For clarity, the notation is defined as follows:

• $\mathcal{A}_i$: Screw axis of link frame {i}, expressed in frame-{i} coordinates
• $\mathcal{V}_i$: Twist of link frame {i}, expressed in frame-{i} coordinates
• $T_{i-1,i}(\theta)$: Configuration of frame {i} relative to {i-1} given the joint variable $\theta_i$
• $[a]_\times$: Skew-symmetric representation of $a\in\mathbb{R}^3$ $$[a]_\times = \begin{bmatrix}0&-a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0\end{bmatrix}$$
• $[Ad_{T_{i,i-1}}]$: Matrix version of adjoint representation of $T_{i,i-1}$ $$[Ad_{T_{i,i-1}}]= \begin{bmatrix}R_{i,i-1}&0 \\ [p]_{\times}R_{i,i-1}&R_{i,i-1}\end{bmatrix}$$

Now, the expression for $\mathcal{V}_i$ can be written as: $$\mathcal{V_i} = \mathcal{A}_i\dot{\theta}_i + [Ad_{T_{i,i-1}}]\mathcal{V}_{i-1}$$

Taking the derivative $\mathcal{V_i}$, we obtain... $$\dot{\mathcal{V_i}} = \mathcal{A}_i\ddot{\theta}_i + [Ad_{T_{i,i-1}}]\dot{\mathcal{V}}_{i-1} + \frac{d}{dt}\left([Ad_{T_{i,i-1}}]\right)\mathcal{V}_{i-1}$$ To simplify the derivative expression, let us focus on the final term... $$ \begin{align} \frac{d}{dt}\left([Ad_{T_{i,i-1}}]\right)\mathcal{V}_{i-1} &= \frac{d}{dt}\left(\begin{bmatrix}R_{i,i-1}&0 \\ [p]_{\times}R_{i,i-1}&R_{i,i-1}\end{bmatrix}\right)\mathcal{V}_{i-1} \\ &= \left(\begin{bmatrix}\frac{d}{dt}(R_{i,i-1}) & 0 \\ \frac{d}{dt}([p]_{\times})R_{i,i-1} + [p]_{\times}\frac{d}{dt}(R_{i,i-1}) & \frac{d}{dt}(R_{i,i-1})\end{bmatrix}\right)\mathcal{V}_{i-1} \end{align} $$

At this point of the derivation, Lynch & Park make the following claims:

• $\frac{d}{dt}(R_{i,i-1}) = -[\omega\dot{\theta}_i]_{\times}R_{i,i-1}$
• $\frac{d}{dt}([p]_{\times})R_{i,i-1} + [p]_{\times}\frac{d}{dt}(R_{i,i-1}) = -[v\dot{\theta}_i]_{\times}R_{i,i-1}-[\omega\dot{\theta}_i]_{\times}[p]_{\times}R_{i,i-1}$

where $\mathcal{A}_i = \begin{bmatrix} \omega \\ v \end{bmatrix}$

Now, I understand that the negative signs arise as the screw axis is defined in frame {i} while the adjoint representation is for the configuration of {i-1} relative to {i}. The part I'm struggling to understand is how Lynch & Park flipped the order of matrix multiplication in the expression for $\frac{d}{dt}([p]_{\times})R_{i,i-1} + [p]_{\times}\frac{d}{dt}(R_{i,i-1})$.

The expression I was expecting was: $$\begin{align} \frac{d}{dt}([p]_{\times})R_{i,i-1} + [p]_{\times}\frac{d}{dt}(R_{i,i-1}) &= (-[v\dot{\theta}_i]_{\times})R_{i,i-1} + [p]_{\times}(-[\omega\dot{\theta}_i]_{\times}R_{i,i-1}) \\ &= -[v\dot{\theta}_i]_{\times}R_{i,i-1} - [p]_{\times}[\omega\dot{\theta}_i]_{\times}R_{i,i-1}\end{align}$$

I tried to see if my expression is equal to the expression given in the book to no avail. I feel like I'm missing something relatively obvious, but I can't seem to figure it out. If anyone can give me some advice or direction, I would greatly appreciate it.

Answer 1

The issue is showing up as part of the simplification step. You are correct that the second term comes out to $[p]_\times \frac{d}{dt}(R_{i,i-1}) = [p]_\times (-[\omega \dot \theta_i]_\times R_{i,i-1})$. However, the issue is showing up because $\frac{d}{dt}([p]_\times) = [\dot p]_\times$ which is not the same as $-[\upsilon \dot \theta_i]_\times$. Therefore, the interim equation should actually be $$\frac{d}{dt}([p]_×)R_{i,i-1} + [p]_\times \frac{d}{dt}(R_{i,i-1}) = [\dot p]_\times R_{i,i-1} + [p]_\times (-[\omega \dot \theta_i]_\times R_{i,i-1})$$

Since the equation for $\dot p$ is actually $\dot p = (-\omega \dot \theta_i) \times p - \upsilon \dot \theta_i$, if we want to replace the reference to $\dot p$ with $\upsilon$ to make use of the twist directly, then the first term becomes $$\frac{d}{dt}([p]_×)R_{i,i-1} = [\dot p]_\times R_{i,i-1} = -\left[p \times (-\omega \dot \theta_i)\right]_\times R_{i,i-1} - [\upsilon \dot \theta_i]_\times R_{i,i-1}$$

If we use the Jacobi Identity, defined as $(a \times b) \times c = a \times (b \times c) - b \times (a \times c)$, on the second term, we can modify its form as follows (by assigning $a=p$, $b=-\omega \dot \theta_i$, and $c=R_{i,i-1}$ and rearranging the identity): $$[p]_\times ([-\omega \dot \theta_i]_\times R_{i,i-1}) = \left[p \times (-\omega \dot \theta_i)\right]_\times R_{i,i-1} - [\omega \dot \theta_i]_\times [p]_\times R_{i,i-1}$$

By dropping these two equations for each term into the interim equation from above and cancelling out the common term, we end up with Lynch & Park's result.