Calculating Motor Engine torque/power using accelerometer data

[FFog]

Hi all, got a little physics problem that I wonder if anyone could help me with.

I'm using an accelerometer to record the rate of acceleration of a vehicle over time (1500kg). With this data I am calculating speed and distance traveled over time. I'm now trying to figure out a way to calculate either the torque or the power produced by the engine over time (I can calculate the other using recorded rpm @ the time). Drive train losses, rolling resistance and aerodynamic drag will all play their part, but I shall ignore these presently till I figure out this first bit.

Thanks much for reading.

 

[Bob S]

Hi FFog-
Power(watts) = torque(Neutons) times speed (in radians per second).

Average power (watts) = change in kinetic energy (joules) divided by elapsed time (seconds).
So a 1500 Kg car going fom 0 to 30 meters per sec in 5 seconds is

(1/2)Mv²/time = (1/2) 1500 30²/ 5 sec = 135,000 joules per sec = 181 HP

 

[FFog]

Thanks for the reply. The problem is that I want power over time, not the average power, so it can be plotted on a graph over time along with the speed/acceleration/
Would (1/2)Mdeltav²/deltatime work? my data looks something like this in my spreadsheet

Time(s),Accel(m/s2),Speed(m/s),Distance(m),Power (hp)
0,0.00,0.00,0.00,?
0.222,5.77,1.28,0.14,?
0.416,8.52,2.93,0.55,?
0.633,8.24,4.72,1.38,?
0.861,6.96,6.31,2.64,?
1.089,6.40,7.77,4.24,?
1.319,5.35,9.00,6.17,?
1.55,5.18,10.19,8.39,?
1.781,5.37,11.44,10.89,?
2.01,5.73,12.75,13.66,?
2.241,5.08,13.92,16.74,?
2.453,5.19,15.02,19.80,?
2.665,4.88,16.06,23.10,?
2.898,5.42,17.32,26.99,?
3.139,7.29,19.07,31.37,?
3.379,3.41,19.89,36.05,?
3.614,4.34,20.91,40.84,?
3.83,4.09,21.80,45.45,?
4.054,4.16,22.73,50.44,?
4.262,3.95,23.55,55.25,?
4.483,3.76,24.38,60.55,?
4.705,3.66,25.19,66.05,?
4.923,3.39,25.93,71.63,?
5.138,3.35,26.65,77.28,?

Thanks again

 

[Bob S]

Hi FFog-
"Would (1/2)Mdeltav2/deltatime work? my data looks something like this in my spreadsheet"
Yes. To get power (watts) from HP, multiply by 746

to get torque (Neuton-meters) from RPM, multiply power (watts) by 60, and divide by 2 pi times RPM:

T = 60 P/(2 pi RPM)

Bob S

 

[FFog]

Thanks, this works for me.

 

[nidhi.monika]

hi
I really need help. I want to calculate speed from a tri axial acclerometer.
I tried calculating the speed from data which i mentioned below
Calculation of velocity from discrete acceleration data:

Integral of acceleration will give velocity value. Since data given for acceleration is discrete so we have calculated velocity value at x direction (Vx) through integrating under discrete acceleration data at x direction by using MATLAB programme. Similarly Vy and Vz was calculated. Total velocity value will be sqrt(Vx*Vx + Vy*Vy + Vz*Vz)

but the speed which i am getting through this method is 30m/s for one subject which is impossible.
I didnot process the data before calculation (which could be the reason behind inaccuracy).
Please. somebody help me in processing my data and correcting the method of calculating the speed.
I am using tri axial acceleromter (glink), and i attached it on sacrum during run. The subject ran on 400m track which is not bumpy . distance was 50m on plane surface.
Please some body help

 

[rcgldr]

power = force time speed. To calcuate watts.

power/watts = force (Newtons) x speed (m/s)

This eliminates the need to calculate any torque related values, so gearing isn't an issue.

The early part of the launch will be traction limited.

Your input is acceleration versus time. You might want to smooth this data with a curve fit, then use that curve fit function to generate more sample points.

For a linear approximation of velocity, use the average acceleration for each time interval:

v_n+1 = 1/2 (a_n+1 + a_n) (t_n+1 - t_n) + v_n
v_n+1 = 1/2 (a_n+1 + a_n) Δt + v_n

A quadratic approximation results in the same equation because it also assumes acceleration changes linearly with time during each interval:

v_n+1 = c Δt² + a_n Δt + v_n
a_n+1 = 2 c Δt + a_n
c = (a_n+1 - a_n) / (2 Δt)
v_n+1 = ((a_n+1 - a_n) / (2 Δt)) Δt² + a_n Δt + v_n
v_n+1 = 1/2 (a_n+1 + a_n) Δt + v_n

The quadratic approximation could be used in this form to calculate intermediate values for velocity versus time, and integrating it could be used to better calculate distance.

v = 1/2 ((a_n+1 - a_n) / (t_n+1-t_n)) t² + a_n t + v_n

d = 1/6 ((a_n+1 - a_n) / (t_n+1-t_n)) t³ + 1/2 a_n t² + v_n t + d_n

 

[nidhi.monika]

hey thanks. Let me try your solution and will get back to u ASAP.. thanks a tonn