Estimating moment of inertia for Vex 269 and 393 motors

Jpearman has done some great lab testing with Vex motors.  This post shows the motor current response for both 269 and 393 motors to a step response.   The steady state is achieved in around 100ms for both motors.

Good data.   It appears that both motors have a time constant tau = 20 ms based upon the initial slope or using 100ms as a 5*tau value.

We can use this data to calculate the approximate moment of inertias of the motors.    Since the motor time constant  tau = max speed/max accel  and the max speed is the same for both motors … we know that the motor max accelerations are identical for both motors.    Since max speed = 10.4 rad/s   and max accel = stall torque/Izz   we can calculate Izz for a 269  as

Izz_269  = tau*stall torque/max speed  = 2.8 in_lb*.113 nm/in_lb*tau/10.4  = 6.085*E-4 kgm^2

Using the equality of motor max accelerations gives a  ratio to get  Izz_393  from Izz_269  and the motor torques.  So

Izz_393 = Izz_269*Istall_393/Istall_269 = 6.085E-4*14.7/2.8 =  3.19E-3 kgm^2

So we now have enough data to do a pretty good motor model.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: