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Unread 12-06-2005, 23:56
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Al Skierkiewicz Al Skierkiewicz is offline
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Re: Cheap Torque Measurement?

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.
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Unread 13-06-2005, 03:41
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Re: Cheap Torque Measurement?

Quote:
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
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As easy as 355/113...
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Unread 13-06-2005, 07:10
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Re: Cheap Torque Measurement?

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.?
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Unread 13-06-2005, 12:34
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Re: Cheap Torque Measurement?

Quote:
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
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Unread 13-06-2005, 13:05
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Re: Cheap Torque Measurement?

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
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Unread 13-06-2005, 13:57
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Re: Cheap Torque Measurement?

Quote:
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
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As easy as 355/113...
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