Linear accelerometer on a rotating platform

Question

I want to measure the linear speed of a vehicle putting the accelerometer on its wheel. Assume the vehicle is moving at a velocity 70 KMPH and its wheel has a radius (R = Radius) 30 CM. I made this analogy, If the accelerometer will be placed on the wheel at position R/2 whose Z axis is now orthogonal to the wheel and let us assume there is no tilt in X and Y axis.

The Accelerometer reading in X and Y should give the net accelerometer which is a_translational + a_rotational + TIGA = (R/2)alpha + (R/2) omega ^2 + TIGA where aplha is the angular accelartion and omega is the angular velocity and TIGA is the Tilt induced gravitational accelarion.

If my above observation is correct, how can i calculate TIGA factor? Will it change with the rotation degrees? How to balance it?

For calculating the speed (to correlate with 70 KMPH) i should integrate the accelration value in that period. In this case? Which accelration value from the accelrometer data should be considered? Should the a_rotational and TIGA factor must be removed from the data before integration?

Answer 1

This is what you are going to read from your accelerometer: $$ \begin{align*} a_{r}= & \frac{R}{2}\left ( 2\cos\vartheta \: \ddot{\vartheta } - \dot{\vartheta}^{2}\right )& +g \sin \vartheta\\ a_{t}= & \frac{R}{2}\left ( \ddot{\vartheta } - 2\sin\vartheta \: \ddot{\vartheta } \right ) & +g \cos\vartheta \end{align*} $$ and you want to get $$ v=R\dot{\vartheta} $$ If your speed is high enough (\$v=R\dot{\vartheta }\gg \sqrt{2gR} \approx 8.7km/h\$) you can neglect the other two factors and assume \$v=\sqrt{2a_{r}/R}\$ (this is the case for a normal travel by car).

Anyway the radial component can be easily filtered to obtain just \$\dot{\vartheta }\$, as the frequency of \$\vartheta\$ is way higher than the frequency of \$v\$.

I suggest you to simulate these equations and try to find a good strategy for your needs before going further. The following is a matlab code to simulate an experimental speed profile.

v = [20*ones(1,50*1e4) 20*2.^(-(1:(50*1e4))/(50*1e4)) 10+20/(50*1e4)*(1:(50*1e4)) 30*ones(1,50*1e4)];
plot(v)
disp('Velocity 72kph exponentially decreasing to 36kph then linearly increasing to 144kph')
pause

x = cumsum(v)/1e4;
theta = x / R;
xs = x + R/2*cos(theta);
ys = R/2*sin(theta);
xsd = diff(xs)*10000;
xsdd = diff(xsd)*10000;
ysd = diff(ys)*10000;
ysdd = diff(ysd)*10000+9.81;
ar = xsdd .* cos(theta(2:end-1)) + ysdd .* sin(theta(2:end-1));

plot(ar)
disp('Radial acceleration')
pause

plot(R*sqrt(-ar/R*2))
disp('Velocity recovered from radial acceleration')
pause

Answer 2

This is actually not a straightforward problem. The chain rule will bite you where you don't expect it to. Believe it or not, I'd low pass filter for the centripetal acceleration, use that and the tire circumference to calculate angular rate of the tire, and use that for speed.

Answer 3

You get a nice component of the acceleration due to the rotation. I would just measure that frequency and the velocity follows directly from the wheel circumference. This has the massive advantage that no calibration of the sensor position and orientation is needed.

To do this, calculate the mean value, subtract it from the signal, and find the zero crossings of the result. The time between successive positive crossings is a good measure of the time for one revolution. You may need to apply a small amount of low pass filtering to ensure you get just one crossing per revolution.

Or do an FFT of the signal and the largest frequency component is what you want (smoothing then comes for free).

Answer 4

Velocity is equal to the rotational velocity multiplied by radius, I suggest using this formula instead for your application, so you just have to do the math instead of using more electronics. So, use the formula: acceleration = (alpha)(radius) + (omega)^2(radius) and solve for omega.