How to calculate movement from 2 local velocities at an object with one axis

Question

Given a movable object with two wheels (connected through an axis) of which each wheel has a velocity pointing always in the direction the wheel rolls (so we assume perfect friction, no drifting is possible).

An top-down illustration is given below

The black object is the object in question in State $S_1$ at time point $t_1$ while the gray object is the object in question in State $S_2$ at time point $t_2=t_1 + \Delta t$.

How can I calculate $p_{1t2}$ and $p_{2t2}$, given $p_{1t1}$, $p_{2t1}$, $v_{1t1}$, $v_{2t1}$ and $\Delta t$ (where $p$ denotes an absolute position in space)?

Note that all calculations are done in 2D space.

To clarify things, I'm talking about this kind of movement

Answer 1

The velocity of the midpoint of the axis is the average of the velocities of the two end points. So $v_x = \frac{v_1+v_2}2$. There are many ways to see this, but perhaps the easiest is to transform to a reference frame moving at $\frac{v_1+v_2}2$. In this frame, the two wheels have the same speed, but in opposite directions, so the axis is spinning in place, and the middle is stationary.

As for the rate of rotation of the axis, again it can be beneficial to change to the reference frame from the above paragraph. In that reference frame, each wheel is moving at a speed of $\frac{|v_1-v_2|}{2}$, and they're moving in a circle of radius half the axis. So if the axis is $a$ long, the time it takes to do one full revolution is $$ \frac{2\pi\cdot \frac a2}{\frac{|v_1-v_2|}{2}} = \frac{2\pi a}{|v_1-v_2|} $$ Converting this to angular velocity for convenience, we get $$ \frac{|v_1-v_2|\cdot 180^\circ}{\pi a} $$

Answer 2

I finally figured it out myself. I'm giving a detailed solution in case someone else needs this.

Solution

First, we calculate the theoretical position $\vec q_1$ and $\vec q_2$ of each wheel by adding the theoretically traveled distance $$\vec q_1=\vec p_{1t1}+\vec v_{1t1}*\Delta t$$ $$\vec q_2=\vec p_{2t1}+\vec v_{2t1}*\Delta t$$

Second, we construct the two direction vectors of both states $$\vec v_1=\vec p_{2t1}-\vec p_{1t1}$$ $$\vec v_2=\vec q_2-\vec q_1$$

Next, find the intersection of the two lines $$ l_1 = \vec p_{1t1} +a*\vec v_1$$ $$ l_2 = \vec q_1 +b*\vec v_2$$ using Cramer's Rule (like here) which gives us the center $\vec c$ of the arc we want to move along (the turning point).

After that, we get the radius $r$ and arc length $s$ of the arc $$r=||\vec c - \vec p_{1t1}||$$ $$s=||\vec q_1 - \vec p_{1t1}||$$

from which we can calculate the angle $\theta$ we have to rotate around the center $\vec c$ in order to travel the full distance $$\theta = \frac{s}{r} \mbox{sign}( \mbox{atan2}(y_{\vec v2}, x_{\vec v2})-\mbox{atan2}(y_{\vec v1}, x_{\vec v1})) $$

the sign function is needed in order to determine the correct sign for $\theta$ (as it would always be positive otherwise).

Finally, we rotate our original points $\vec p_{1t1}$ and $\vec p_{2t1}$ by $\theta$ radians around our arc center $\vec c$ which gives us the actual destination points $\vec p_{1t2}$ and $\vec p_{2t2}$ that we tried to find.

Examples

Here are some examples.

• The blue points denote the origin points $\vec p_{1t1}$ and $\vec p_{2t1}$.
• The orange points denote the destination points $\vec p_{1t2}$ and $\vec p_{2t2}$.
• The brown cross marks our arc center $\vec c$.
• The red, dotted lines show the theoretical track (the end of both lines describe $\vec q_1$ and $\vec q_2$).
• The green line shows the correct track that we want to move.