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Ether 22-12-2011 23:44

Motor Quiz
 


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



xSAWxBLADEx 22-12-2011 23:46

Re: Motor Quiz
 
9 volts

Ether 22-12-2011 23:53

Re: Motor Quiz
 
Quote:

Originally Posted by xSAWxBLADEx (Post 1093279)
9 volts

Wild guess?



Aren Siekmeier 22-12-2011 23:57

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

xSAWxBLADEx 23-12-2011 00:01

Re: Motor Quiz
 
Quote:

Originally Posted by Ether (Post 1093283)
Wild guess?



yep :) never did like quizes, multiple guess tests are better.

Ether 23-12-2011 00:15

Re: Motor Quiz
 

Quote:

Originally Posted by compwiztobe (Post 1093285)
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.



Aren Siekmeier 23-12-2011 00:40

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

sanddrag 23-12-2011 00:42

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?

sanddrag 23-12-2011 01:03

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?

MooreteP 23-12-2011 06:29

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.:D

Ether 23-12-2011 10:07

Re: Motor Quiz
 
1 Attachment(s)

Quote:

Originally Posted by sanddrag (Post 1093296)
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?




Ether 23-12-2011 10:19

Re: Motor Quiz
 

Quote:

Originally Posted by MooreteP (Post 1093315)
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.:D

The whole concept using "significant figures" this way, although still taught in high schools, is deeply flawed.



Ether 23-12-2011 11:25

Re: Motor Quiz
 

Quote:

Originally Posted by sanddrag (Post 1093296)
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)




EricH 23-12-2011 12:07

Re: Motor Quiz
 
Quote:

Originally Posted by Ether (Post 1093336)

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.)

sanddrag 23-12-2011 12:17

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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