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Unread 22-12-2011, 23:44
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Motor Quiz



What voltage would be required to operate a CIM at 60 oz-in at 3700 rpm?


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Unread 22-12-2011, 23:46
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Re: Motor Quiz

9 volts
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Unread 22-12-2011, 23:53
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Re: Motor Quiz

Quote:
Originally Posted by xSAWxBLADEx View Post
9 volts
Wild guess?


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Unread 22-12-2011, 23:57
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Re: Motor Quiz

Spoiler for Work:
60 oz-in (or 0.42 Nm) is about 1/6 (0.17) of stall torque for a CIM (2.43 Nm), at which a DC motor runs at about 5/6 of free speed. 3700 rpm is 0.83 (5/6) of 4473 rpm, the free speed at our mystery voltage. Free speed scales with voltage, so if a CIM's free speed is 5310 rpm at 12V, the mystery voltage is 12V*4473/5310 =


10.1 V
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Unread 23-12-2011, 00:01
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Re: Motor Quiz

Quote:
Originally Posted by Ether View Post
Wild guess?


yep never did like quizes, multiple guess tests are better.
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Unread 23-12-2011, 00:15
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Re: Motor Quiz


Quote:
Originally Posted by compwiztobe View Post
10.1 V
At 10.1 volts the free speed is (10.1/12)*5310 = 4469 rpm

The stall torque is (10.1/12)*343.4 = 289 oz-in.

So at 3700 rpm the torque would be 49.7 oz-in, not 60 oz-in.


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Unread 23-12-2011, 00:40
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Re: Motor Quiz

I see, so stall torque also scales in the same way. A convenient way to can all this then would be with giving the speed vs. torque curve as speed/free speed + torque/stall torque = voltage/spec voltage.

For another shot then:
Spoiler for Let's try that again:
Free speed = 5310rpm
Stall torque = 2.43Nm
Spec voltage = 12V

Speed = 3700rpm
Load = 60oz-in = 0.42 Nm

So voltage = 12 * (3700/5310 + 0.42/2.43) =

10.43V
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Unread 23-12-2011, 00:42
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Re: Motor Quiz

10.5 Volts, give or take.

I arrived at my answer by a very different (and probably incorrect) method from the poster above. I figured motor constants for RPM/Volt and Oz-In/Amp, found how much current it would take to hold 60 oz-in at stall, considered the resistance of the motor's windings, then figured out how many volts it would need to hold 60 oz-in at stall, then I figured how many volts it would need to spin 3700 RPM at no load, then added those two voltages to arrive at my answer. I did have some rounding error, but perhaps my method is entirely flawed and just happened to be close by coincidence? I am on holiday here....

EDIT: Revising my answer with less rounding, I get 10.458 Volts by the above method.

I'm confused by the above poster's method of adding ratios of speed and torque. Why does that work?
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Last edited by sanddrag : 23-12-2011 at 00:57.
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Unread 23-12-2011, 01:03
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Re: Motor Quiz

My work:

CIM Motor Constants
442.5 RPM/Volt
2.581954 oz-in / A
Resistance of motor windings is 0.0902255639 Ohms by Ohm's Law

Thus:
60 oz-in requires 23.238214 A to stall, from a 2.09668 V source

Also, it takes 8.3615819 V to spin 3700 RPM under no load

Add the voltages to arrive at 10.458 V needed to spin 60 oz-in load at 3700 RPM

Check: At this voltage:

Free speed is 4627.65 RPM
Stall Torque is 299.2731 oz-in

Operating parameters are at:
20.0486% of stall
79.9514% of free speed = 3699.87 RPM

Less than 0.4% Calculation error.

Can someone tell me if this is the proper way to do this, or if my hairbrained idea just happened upon a lucky coincidence?
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Unread 23-12-2011, 06:29
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Re: Motor Quiz

Come on engineers, significant figures.
2 digits of data yielding 8 digits of precision?
Use 10.5 or 11 volts and call it a day.
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Unread 23-12-2011, 10:07
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Re: Motor Quiz


Quote:
Originally Posted by sanddrag View Post
I'm confused by the above poster's method of adding ratios of speed and torque. Why does that work?
See attached speed vs torque drawing.

E is the desired operating point with speed C and torque D

Line AB is the 12V speed vs torque curve (given)

If we can find G, then the "mystery" voltage is 12*(G/A).

Proceed as follows:

From similar triangles, (G-C)/D = A/B

Multiply both sides by D/A to get (G-C)/A = D/B

Expand the left side to get G/A - C/A = D/B

Move C/A to the other side to get G/A = C/A + D/B

So V = 12*(G/A) = 12*(C/A + D/B) which is what Aren did.


Of course, this ignores the 2.7 amp free current. What role should that play in this analysis?



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Unread 23-12-2011, 10:19
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Re: Motor Quiz


Quote:
Originally Posted by MooreteP View Post
Come on engineers, significant figures.
2 digits of data yielding 8 digits of precision?
Use 10.5 or 11 volts and call it a day.
The whole concept using "significant figures" this way, although still taught in high schools, is deeply flawed.


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Unread 23-12-2011, 11:25
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Re: Motor Quiz


Quote:
Originally Posted by sanddrag View Post
I arrived at my answer by a very different (and probably incorrect) method from the poster above. I figured motor constants for RPM/Volt and Oz-In/Amp, found how much current it would take to hold 60 oz-in at stall, considered the resistance of the motor's windings, then figured out how many volts it would need to hold 60 oz-in at stall, then I figured how many volts it would need to spin 3700 RPM at no load, then added those two voltages to arrive at my answer.
The above method is arithmetically identical to what Aren did:

current to hold "D" oz-in at stall = (D/B)*Istall

volts to hold "D" oz-in at stall = (D/B)*Istall*R = (D/B)*Istall*(12/Istall) = 12*(D/B)

volts to free-spin at "C" rpm = 12*(C/A)

adding these 2 gives:

"mystery volts" = 12*(C/A+D/B)



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Unread 23-12-2011, 12:07
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Re: Motor Quiz

Quote:
Originally Posted by Ether View Post

The whole concept using "significant figures" this way, although still taught in high schools, is deeply flawed.
It's also at least touched on at the college level. I'm curious why you say it's flawed--besides the fact it confused me every time I tried to use it! (I typically hold about a 2-3 decimal digit precision in the final answer, but I'll often go to 4-5 decimal digits or more during calculations.)
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Unread 23-12-2011, 12:17
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Re: Motor Quiz

Yes, I know my significant figures and decimal places are a complete mess. I was rushing. I too would like to hear Ether's explanation of the flaws in the significant figure method taught in schools.

The 2.7A free-running current I believe plays a role in determining the motor's torque constant. Rather than being 343.4 Oz-In / 133A it would be 343.4 oz-in / (133-2.7 A)

I'm actually rather surprised I did this correctly, albeit in a somewhat roundabout method. This was fun.
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