Cheap Torque Measurement?

[ChuckDickerson]

Thanks for the info! I have questions:

In your motor/generator pair:

(1) Make both the motor and generator the same type of motor.

Why is it important to use to of the same type motor instead of an "unkonwn" motor as the test and a known motor (one you have power curves for) for the load motor/generator?

(2) Wire a power resistor across the generator's electrical "outputs". Be careful to size the resistor properly.

What it the power resistor used for? Is this so that the generator now acts like a brake? Or am I missing the point and the point is not to actually load/brake the test motor. How do you determine the proper size power resistor? I am not an electronics person. I am more of a mechanical guy.

(3) Measure the voltage and current at the "input" and "output". Also measure the speed at the spindle/coupler.

(4) Adjust the input voltage to a known value (in our case it's almost always 12.0V). Note that you do not need a precision high current power supply... Just use an Exide SLA and let it slowly discharge as you run the MG and take your readings at 12V.

Note also that you can ignore (4) depending on exactly the type of data you are going for...

For a given load resistance, you will take your readings:

Input Power (W) = Input Voltage (V) * Input Current (A)
Output Power (W) = Output Voltage (V) * Output Current (A)
Mechanical Power at coupler (W) = [Input Power (W) + Output Power (W)] / 2
Torque at coupler (N*m) = Mechanical Power (W) / Spindle Speed (radians/sec)
Power Lost in Motor (W) = Input Power (W) - Mechanical Power at Coupler (W)

et cetera...

Easy enough! I can handle that!

Hope this helps...

A BUNCH! Thanks Mike!

[Al Skierkiewicz]

I was thinking along the same lines as Mike but by using a variable resistor on the output of the generator. (Or switched resistors works as well) I think if you throw in a fudge factor for efficiency of each of the motors (I am guessing between 90 and 95%) all of the available input power passes straight through to the output electrically. Measuring voltage and current in the load ought to give you the numbers you need to make a pretty good guess. By using two identical motors you only have to calculate the efficiency for one since they should be the same.
Please remember that trying to duplicate the published specs will get you frustrated. The power supply and instrumentation used by the manufacturers is high quality (read "expensive") so that all data can be tested and repeated.

[Mike Betts]

Originally Posted by Al Skierkiewicz

...I think if you throw in a fudge factor for efficiency of each of the motors (I am guessing between 90 and 95%)...

Al,

The efficiency of DC motors is dependent on speed and is usually much less. Take the CIM for example:

Stall Current = 114A
Stall Torque = 2.45 N*m
Free speed = 5342 RPM = 559 radians/sec

Efficiency at Max Power = (1.23 N*m * 280 rad/sec) / (12V * 57A) = 50%

Therefore, the mechanical power at the spindle will be 343W and I would expect the power dissipated in the generator’s brake resistor to be about 172W or so…

Mike

[Al Skierkiewicz]

Mike,
You got me again. I keep thinking of efficiencies at their peak and not at the operating point. I have got to get out of this rut.
Now what are you doing up at 2AM.?

[ChuckDickerson]

Originally Posted by Mike Betts

Efficiency at Max Power = (1.23 N*m * 280 rad/sec) / (12V * 57A) = 50%

Therefore, the mechanical power at the spindle will be 343W and I would expect the power dissipated in the generator’s brake resistor to be about 172W or so…

Mike,

What is your source for these numbers? The only source I have seen for the CIM motor specs is from the FIRST web site but some of the numbers don't quite match: http://www2.usfirst.org/2005comp/Specs/CIM.pdf

I am by no means trying to nit pick and I understand your numbers were just an example of the calculation but I am just wondering if there is another better source for the CIM motor specs that I should be using that I don't know about.

Using your example but taking the numbers for max power directly from the FIRST PDF spec sheet above I get that the efficiency is even worse (41%).

Torque = 171.7 Oz-In = 1.21 N*m
Speed = 2655 RPM = 278 rad/sec
Current = 67.9 A
Power = 337 W
Efficiency = 41% = (1.21 N*m * 278 rad/sec) / (12V * 67.9A)

The should be at least 337 W * 41 % = 138 W.

Thanks,
Chuck

[sanddrag]

Here's some specs that FIRST gave us on the CIM in 2002 if these are of any help.
Stall Torque: 2.22 N-m
Stall Current: 107 A
Free (no load) speed: 5,500 RPM
Free (no load) current: 2.3 A
Peak Power: 321 W

[Mike Betts]

Originally Posted by DeepWater

...What is your source for these numbers? ...

Chuck,

My source was: http://www.chiefdelphi.com/forums/sh...ad.php?t=32796

Note that even the pros will have variations in measurements from motor to motor due to manufacturing variations. Even the room temperature and relative humidity will affect performance.

The difference in numbers is no biggie. I'd expect that you would want to design the load/brake resistor for 200 or 225W continuous operation...

Mike

[Mike Betts]

Originally Posted by DeepWater

...Why is it important to use to of the same type motor instead of an "unkonwn" motor as the test and a known motor (one you have power curves for) for the load motor/generator?...

As I noted in my first post, we are sacrificing some accuracy for expediency here... The efficiency of a motor as a motor may not be exactly equal to a motor as a generator. However, for the purposes of this experiment we can assume that a motor is a perfect bidirectional conversion device.

As Al Skierkiewicz stated, your measuremnts will not be as accurate as a manufacturer's lab. By using two identical motors, you are signifigantly reducing errors caused by your instrumentation being exactly the same as a lab's equipment.

Using the same motor for both motor and generator makes the math a lot easier... You will find that losses in a motor are not linear and are dependent on speed, torque, voltage and current.

Lastly, by using the same motor, you are assured that you are using the generator within it's operating range (the max speed of a CIM is not the same as a FP, for example).

Originally Posted by DeepWater

...What it the power resistor used for? Is this so that the generator now acts like a brake?...

Exactly. You need both voltage and current to have power (P=V*I). Without a resistor, your output power is zero therefore your mechanical power is only the losses in the generator (fairly small in a good motor).

Mike

[ChuckDickerson]

Ahh! I think the light bulb is starting to glow! Thanks for all the great info guys!