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I'm working through a paper to better understand Differential Drive Mobile robots and I'm a bit stuck(fairly early in the paper) on how one of the constraints is derived.

The paper is here, https://www.hilarispublisher.com/open-access/dynamic-modelling-of-differentialdrive-mobile-robots-using-lagrange-and-newtoneuler-methodologies-a-unified-framework-2168-9695.1000107.pdf.

My question is how is equation 6 derived. Equation 5 defines a no slip lateral slip condition in the ydot robot frame, then states "Using the orthogonal rot matrix R(theta), the velocity in the in inertial frame give -xdot_a * sin(theta) + ydot_a * cos(theta) = 0.

Hmm, now that I think about it if I take the inverse of the Rotation matrix and multiply it by the inertial frame I get xdot_i * -s(theta) + ydot_i * c(theta) = ydot_a which equal 0. This is then defining that no slip condition in the robots y direction. That still doesn't explain why the paper has body frame and I have inertial frame.

Could anyone elaborate?

Thanks!

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  • $\begingroup$ Add some punctuation to give this a chance of being readable as this is just a wall of text. $\endgroup$
    – Solar Mike
    Jul 6, 2022 at 15:52

1 Answer 1

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Without thinking to much of the correctness of the method, the argument is something like this:
The velocity $\dot{y}_a^r=0$ is the $y$-component of the velocity of the point $A$ in the body-fixed frame. The same velocity represented in the inertial frame is $$\begin{bmatrix}\dot{x}_a \\ \dot{y}_a\end{bmatrix}=R(\theta)\begin{bmatrix}\dot{x}_a^r \\ \dot{y}_a^r\end{bmatrix}=R(\theta)\begin{bmatrix}\dot{x}_a^r\\0\end{bmatrix}$$ We can use this to write $$\begin{bmatrix}\dot{x}_a^r\\0\end{bmatrix}=R(\theta)^T\begin{bmatrix}\dot{x}_a\\\dot{y}_a\end{bmatrix} $$ which gives two equations where one is of the form$$0 = -\dot{x}_a\sin(\theta)+\dot{y}_a\cos(\theta)$$

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  • $\begingroup$ I appreciate it. Your explanation cleared things up for me. $\endgroup$
    – Lysford18
    Jul 6, 2022 at 16:21

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