How to Make a Robot Car Go in a Straight Line

After messing around with my micropython-based esp8266 chip, I’ve finally got to the point where I can make the robot go in a straight line.

The Hardware

My robot car is a dual-motored car. This means that unavoidably I’ll need to correct the tiny differences between the two motors so as to make it go in a straight line. To achieve this goal, I bought an MPU9250 motion sensor which can detect its acceleration $\mathbf{a}$, angular velocity $\omega$ and the earth’s magnetic field $\mathbf{B}$.

The Original Algorithm

My original algorithm is to increaase the speed of one motor and decrease the speed of the other when I detect the robot is turning to one direction. As the motors are controlled by the voltages supplied, when I apply a voltage difference between the two motors, the motors will operate at different speeds.

The following is a theoretical analysis of my original algorithm.

1. The scalar angular velocity $\omega$ can be determined by projecting the vector onto the gravitational unit vector:

   $$\omega =\overrightarrow{\omega }\cdot \frac{\overrightarrow{g}}{|\overrightarrow{g}|}$$

2. The angular velocity can be approximated as a linear function of $V$, the voltage difference between the two motors I apply:

   $$\omega (V)=a+bV$$

3. The voltage difference is assumed to have a linear relationship with respect to he angle $\theta$ and the angular velocity $\omega$:

   $$V(\theta ,\omega )=p\theta +q\omega$$

4. In step 3, $\theta$ is determined by numerically integrating $\omega$:

   $$\theta ={\int }_0^t\omega (t)dt$$

5. Then a differential equation can be set up and solved with separation of variables:

   $$\begin{array}{r}\begin{array}{rl}\dot{\theta }& =a+b(p\theta +q\dot{\theta })\\ (1-bq)\frac{d\theta }{dt}& =a+bp\theta \\ {\int }_0^{\theta }\frac{d\theta }{a+bp\theta }& ={\int }_0^t\frac{1}{1-bq}dt\\ \frac{1}{bp}\ln \left|\frac{a+bp\theta }{a}\right|& =\frac{1}{1-bq}t\\ 1+\frac{bq}{a}\theta & =\exp \left(\frac{bp}{1-bq}t\right)\\ \theta & =\frac{a}{bq}(\exp \left(\frac{bp}{1-bq}t\right)-1)\end{array}\end{array}$$

This implies that my original algorithm should never be able to achieve a perfect straight line.

Nevertheless, I tried it out.

Note

The units of $p$ and $q$ were all messed up because I was using degrees and the voltages I used were not in volts either. I omitted them because they are irrevelant here.

$p$	$q$	$\mathrm{\Delta }t$	Result
10	20	0.2	oscillates
10	10	0.2	tilts about 25 degrees
50	10	0.2	oscillates
20	10	0.2	tilts about 25 degrees
20	12	0.2	tilts about 15 degrees
20	10	0.6	tilts about 5 degrees

($\mathrm{\Delta }t$ is the period with which I update the voltages)

As you can see, the results are far from satisfactory. Oscillation is not shown in the theoretical derivation but it was probably a result of latency. Another significant problem was that none of them works on relatively smooth surfaces. (It oscillates violently)

The Solution

I eventually solved this problem by using accumulative voltage difference with weight decay.

Instead of using a fixed formula for $V$, I update the value of $V$ with this recursive formula:

$$V_n=\alpha V_{n-1}+(p\theta +q\omega )$$

where $\alpha$ is a constant in the interval $(0,1)$. This way, more recent measurements of $\theta$ and $\omega$ constitute a larger portion of $V$ while older ones constitute a smaller portion.

This not only smooths out the curve for $V$ but also implies that I will be able to reduce $\mathrm{\Delta }t$ to a smaller value and thus gain more accurate control.

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