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Motor Quiz
What voltage would be required to operate a CIM at 60 oz-in at 3700 rpm? |
Re: Motor Quiz
9 volts
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Re: Motor Quiz
Spoiler for Work:
10.1 V |
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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. |
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:
10.43V |
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? |
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? |
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 |
Re: Motor Quiz
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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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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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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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