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What would the equations be for the robot's angular and linear velocity at P and also P2? I think I'm doing it wrong...

WL = left wheels angular velocity WR = right wheels angular velocity

For P I had for example the linear velocity = (1/3)rWL + (2/3)2rWR

Am I on right track?

  • $\begingroup$ Can you explain that diagram? If I'm reading it properly, it looks like the boxes on the outside are supposed to be the wheels, and the r and 2r labels are radii (even though drawn as diameters). Is that correct? $\endgroup$
    – Ian
    Feb 24 '14 at 16:50
  • $\begingroup$ diameter. The below has examples of how this works, i just cant figure it out. rose-hulman.edu/~berry123/Courses/ECE425/Spring07_files/… $\endgroup$
    – user3893
    Feb 25 '14 at 21:41

The linear velocity $v$ of a differential drive robot is the average of the velocities of its wheels [^]. In mathematical terms:

$$ v = \frac{v_L + v_R}{2} $$

Where $v_L$ and $v_R$ are respectively the linear velocity of the left and right wheels. These are functions of their angular velocities and radii, such that:

$$ v_L = r_L \omega_L $$

$$ v_R = r_R \omega_R $$

Where $r_L$ and $r_R$ are respectively the radii (not the diameters!) of the left and right wheels, and $\omega_L$ and $\omega_R$ likewise the angular velocities.

The angular velocity $\omega$, on the other hand, is the average of the opposing contributions of left and right wheels: the left wheel contributes clockwise ("negative") motion, whereas the right wheel contributes counter-clockwise ("positive") motion. In mathematical terms:

$$ \omega = \frac{\phi_R - \phi_L}{2} $$

Where the wheel contributions $\phi_L$ and $\phi_R$ are given by:

$$ \phi_L = \frac{r_L \omega_L}{l_L} $$

$$ \phi_R = \frac{r_R \omega_R}{l_R} $$

And $l_L$ and $l_R$ are respectively the left and right shaft lengths between the center of rotation and the left and right wheels.


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