What voltage would be required to operate a CIM at 60 oz-in at 3700 rpm?
9 volts
Wild guess?
Work
[spoiler]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 = [/spoiler]
10.1 V
yep 
never did like quizes, multiple guess tests are better.
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.
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:
Let’s try that again
[spoiler]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) = [/spoiler]
10.43V
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?
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?
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.
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?
*
The whole concept using “significant figures” this way, although still taught in high schools, is deeply flawed.
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)IstallR = (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)
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.)
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.
Are you saying that there’s more than 4 significant digits in the ‘best’ answer?
Me too, because I’m not sure in which way you find flaw.
The statement I use is “Just because your calculator shows 8 digits doesn’t mean they are all important”.
On the other hand, it is very important to know just how many digits you really have to work with, and a good idea of physical reality to check for reasonableness.
In real life, there are no multiple choice questions.
Usually…
Several parallel thoughts come to mind:
1) for a detailed (and humorous) treatment, see this link.
2) rounding the answer to 11 volts as suggested here gives a profoundly wrong answer*. At 11 volts, the torque at 3700 rpm is 75.5 oz-in
3) with modern computers and calculators, there is no reason to drop any digits of precision during intermediate calculations. carry all the precision through, then round the final answer. the number of digits to keep is not a simple matter of looking number of digits in the datum with the least digits. in the specific case of this problem, tenths of a volt is a reasonable (and necessary) precision to get a reasonably correct answer:
61 oz-in @ 3710 rpm requires 10.52 volts
59 oz-in @ 3690 rpm requires 10.40 volts
… so if you want to round to 10.5 volts that would be reasonable, but not 11 volts*
*we are talking about correct answers to the problem as stated. I know that manufacturing tolerances of the motor could cause wide variation.
No. But I am saying using only 2 digits (11 volts), as suggested, is not a correct answer.
Me too, because I’m not sure in which way you find flaw.
More detail posted here.
The statement I use is “Just because your calculator shows 8 digits doesn’t mean they are all important”
Agreed. But a student should learn how to use the calculator in such a way that all those digits get used through the intermediate calculations.
On the other hand, it is very important to know just how many digits you really have to work with, and a good idea of physical reality to check for reasonableness.
I agree completely.
I agree that the whole “sig figs” deal is flawed. I had a science teacher in high school who extolled it, and perhaps it was his inability to do basic algebra that led me to skepticism of that as well. My beef with it is that it is so dependent on the base in which you are representing the numbers. If you’re going to worry about such things, actually do the stats with how the initial standard deviations or uncertainties propagate through the calculations. If you don’t care that much, just round your final answer at the end to something that’s not way than you care about, but still gives detail relevant to the problem and its context. My answer 10.43 was perhaps too many digits: 10.4V would perhaps have been more appropriate given how much control we actually have over (and how much we actually about) the voltages in question. But it doesn’t really matter that much. And as Ether has pointed out, 11V is just plain wrong. And one should never round in the intermediate steps. The best way to go about these calculations is to solve for the answer symbolically and then just plug in your parameters, since the general solution in infinitely more useful and this avoids any intermediate rounding errors. I do a lot of calculations in excel and keep all the digits (the first 30 some anyway) around. However, when reporting these intermediate values, I will often round to keep it uncluttered.
Also (to sanddrag mostly), my method of solving was based on the fact that stall torque and free speed scaled with voltage, so I whipped a quick mathematical model that did just that. Your (sanddrag’s) approach went through the physics of it from the start, which is equally valid (in fact, that’s where we get the linear dependence of stall torque and free speed on voltage, as well the linear performance curve between them).