Calculate path of vehicle with two wheels parallel to each other

Question

There's a vehicle that starts at $P(0|0)$ oriented in positive y direction. This vehicle has two wheels that are a certain distance $d$ apart, have a diameter of $\frac{10}\pi cm$, are parallel to each other and can move at different speeds ($v_r$ and $v_l$) backwards and forwards. $d$, $v_l$ and $v_r$ are known constants.

I'm wondering if it is possible to calculate the vehicles coordinates at any point in time.

A concrete example would be for $d=8cm; \; v_l=-1\frac{rev}{s}; v_r=3\frac{rev}{s}$

Here's a visualization of the above description:

Note that this is only a model representation and the scale might not be perfectly correct.

Edit: The black dots in the first and in the second figure represent the center of the vehicle and not the point $P(0|0)$. I just marked the point in the first figure to clarify, that the vehicle's center is starting at $(0|0)$.

I've noticed that there always seems to be a point where both wheels are rotating around. If this is true finding this point could help also finding the location of the vehicle but I'm not 100% sure

Answer 1

With the help of this sketch:

We know that the length of an arc is $l=\beta R$

Let the center of rotation be O, located in the x-axis, at distance s from the y-axis

In a time $dt$ the right wheel goes $l_1=v_1dt$, while the left wheel moves $l_2=v_2dt$

So, $l_1=v_1dt=d\beta (\frac{L}{2}+s)$ and $l_2=-v_2dt=-d\beta (\frac{L}{2}-s)$
Notice the "-" sign for $v_2$ as it goes as "-y".

The angle $d\beta$ is the same for both wheels, so:

$\frac{v_1dt}{\frac{L}{2}+s } \; =\;\frac{-v_2dt}{\frac{L}{2}-s }$

$v_1\frac{L}{2} - v_1s \;=\; -v_2\frac{L}{2} - v_2s$

$(v_1+v_2)\frac{L}{2} \;=\; (v_1-v_2)s$

$s= \frac {v_1+v_2}{v_1-v_2}\frac{L}{2}$

That formula works for all values of $v_1$ and $v_2$, except for the case $v_1=v_2$ when the vehicle draws a straight (radius=$\infty$) and not a curve.
Also notice that if both wheels spin in the same direction then $s>\frac{L}{2}$, out of AB zone.

You see, the distance "s" is the same over time, which means that the path of every point in the wheel-axis is a circumference, centered at point O.

Then, the path for the middle point of the wheel-axis (your black box) is also a circumference centered at $(-s,0)$ and radius $R=s$

$(x+s)²+y²=s²$

Expressed with parametrics and time:

$x=s·cos (\frac{v_1·t}{\frac{L}{2}+s}) - s\;\;\;$ and $\;\;\; y=s·sin (\frac{v_1·t}{\frac{L}{2} +s})\;\;\;$ if $v_1 \neq 0$

or

$x=s·cos (\frac{v_2·t}{\frac{L}{2}-s}) - s\;\;\;$ and $\;\;\; y=s·sin (\frac{v_2·t}{\frac{L}{2} -s})\;\;\;$ if $v_2 \neq 0$

In the above formulas the velocities are expressed in linear units per time. If you have rev/s you need the radius a of the wheel to find $v_i= v_{rev}·a$

Answer 2

The following simple situation as I understood the problem.... There is no forward motion but pure rotation. The wheel velocities in proper units should be cms/sec.

$$ v_{right}=3 \text{ cm/sec}, \;v_{left}= 1 \text { cm/sec}\;$$

but should not be given in revs/sec. Disregarding wheel size since you call it as velocity. The wheel size does not matter as long as it is the velocity that is given. The velocity diagram should be drawn so that the center of stationary rotation is located dividing axle distance in the ratio

$$ 3:1= 6:2 $$

Else the axle would break. The wheels rotate around a stationary vertical axis at $O$, there would be no forward motion of the single axle Segway sort of vehicle. The angular velocity

$$\omega= \dfrac{3}{6}=\dfrac{1}{2}\text{ = 0.5 radians/sec}$$

We have here pure circular motion:

Left/Right wheels motion

$$ (x_{left},y_{left})=\{2,6\} (\cos \omega t,\sin \omega t)\text{units} $$

EDIT1:

The above is repeated to include same direction of inner wheel velocity, for generality to include wheel directions parallel and anti-parallel.

When steering a four wheeled automobile and considering the front pair of wheels the inner wheel has lower speed and radius of curvature compared to the outer..considered in design of steering mechanisms (like the Ackermann's).

We consider velocities in same direction as plus, in opposite direction as minus.

$$ L = 8 \text{cm},v_r=3, v_l= \mp 1 \text{ cm/s}$$

$e$ is distance offset from the common origin to locate center of rotation. from triangle similarity

$$\dfrac{v_l}{v_r}=\dfrac{e}{e+L}\to e=\dfrac{L v_l}{v_r-v_l}; \omega= \dfrac{ v_r}{L+e}= \dfrac{v_r- v_l}{L}; $$

The last can be directly written by consideration of relative velocity of axle ends.

Same direction (not in question) $$ (e,\omega)=(-2,\dfrac14) $$

Opposite sense direction $$ (e,\omega)=(4,\dfrac12) $$

The centers of rotation in these cases are $(-2,0), (4,0); $

The calculations are shown in the following vector diagram for $-1 $ cm/s opposite sense of direction:

$$ \omega= \dfrac12 ; \;(x_{left},y_{left})= (2 \cos \omega t,2\sin \omega t +4)$$

Radius of curvature of inner turning wheel:

$$= \dfrac{v_l=-1}{\omega= \dfrac12}= -2\; cm. $$

i.e., 2 cm for the CW rotating circle.

If $v_r= v_s $ the Segway goes straight and when $v_r<v_l $ it turns left as can be shown in these equations and the associated vector/ locus diagram.

Answer 3

You should resolve the wheel rotation rates into common rotation and differential rotation. Common rotation will make the vehicle move forward or reverse in the direction it is currently pointing. Differential rotation will make the vehicle rotate around its center. In your example the common rotation is $+1 \frac {rev}s$ and the differential rotation is $+2 \frac {rev}s$ where positive common rotation is forward, the direction that starts pointing to $+y$ and positive differential rotation causes the vehicle to rotate counterclockwise.

The linear speed of the wheels caused by differential rotation is $10\frac {\text{cm}}{\text{rev/sec}}$, so here is $20 \frac {\text{cm}}{\text{sec}}$. One rotation of the vehicle requires $8\pi \text{ cm}$ of motion of each wheel, so the vehicle will rotate $\frac {20\ \text{rev}}{8\pi\ \text{sec}}$ or $5\frac {\text{rad}}{\text{sec}}$. At the same time it will be moving at a constant rate of $10 \frac {\text{cm}}{\text{sec}}$ in the direction the body is pointing.

If we measure time in seconds, distances in cm, and angles in radians from the starting position we have $$\theta=5t \\ \dot y=10 \cos (5t)\\ \dot x=-10 \sin (5t)\\ y=2 \sin(5t)\\ x=2 \cos(5t)-2$$