updating from 2014 to 2015....
You can test your calculation of the planetary speed ratio against the calculator here:
Note that the numbers in the final column are only for a specific gearset. But the formulas in terms of R (teeth on the ring) and S (teeth on the sun) are general.
Stall: If you plot volts on the y axis and current on the x axis then the slope is just R.
K: similarly if you plot (V-iR) on the vertical axis and omega on the x axis... If omega = 0, what should (V-iR) be?
The '2' at end of line 2 is for the 2nd footnote, not to square the denominator.
You are using the R, K and T_f given in the problem statement. With these constants, you know everything about the motor. You can estimate the current, torque and speed for ANY voltage. So you can compute plots of efficiency versus speed or power versus speed, as was done for the final figures of the Motors notes (handed out in class and on Coursework).
For those students who are getting confused, here is what might be the most straightforward approach:
- You know that current is going to vary between a maximum i_stall and a minimum, i_nl, where i_stall and i_nl are computed
using V - i*R - k*omega = 0 and k*i - Tf = TL, with V =12 volts and k, R, Tf as above.
- Make a column in your spreadsheet (or a make a vector in Matlab) with a list of 10 or more currents spanning from i_stall to i_nl.
- Make a column of omegas in radians/second using the equation above for each corresponding value of i.
- Make a column of Power In = V i for each i
- Make a column of Power out = TL * omega where TL = ki-Tf and omega is from the previous column
- Make a column of efficiency = (Power Out)/(Power In)
If using Matlab, each column corresponds to another vector made by doing some operations on the "i" vector.
In the table below, zeros have been put in where we know the result should be zero.
If it's not exactly zero due to numerical roundoff, you can force it to zero. If it's not real close, you have an error.
|current (amps)||omega (rad/s)||Power In||TL (Nm)||Power Out||Efficiency||Omega rpm|
|i||= (V-iR)/k||= V*I||= k*i - Tf||= TL * omega||= Pout/Pin||rpm=omega*30/pi|
So how much torque does the motor need?
It would need 0.5Nm * (1/36) if the gearbox was perfect. But the gearbox is only 90% efficient. As explained in the Power Loss short video on OpenEdX, speed HAS to be exactly 1/36 of the input speed because the gears don't slip. So the 0.9 factor is in the torque: Tout = 0.9*36*T_L.
What will the drill motorís power and efficiency be at that speed?"
Notice that efficiency we're talking about here is the motor, not the gearbox. So you can get the motor efficiency at the particular speed directly from the plots above. We are talking about a particular torque requirement, TL, from the motor. This torque corresponds to a particular value of current, i. And therefore there is also a particular value of omega.