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loose nut
12-25-2009, 06:00 PM
I have a DC geared motor with speed controller with the following rating

115 volts
11 amps
48 in/lbs torque
117 RPM output (speed range of 1 to 117 rpm's)
20.4 / 1 ratio
Physical dimensions are about 4" Dia x 8" long.

With that I figure

1.that the motor has a max speed of over 2300 rpm's at full speed before the speed reducer.

2.the 48 in/lbs of torque is at the max geared down speed of 117 rpm's

So if I can remove the speed reducer or increase the speed in some manner back up to 2000 rpm's what kind of power will it have, is the torque going to fall off to a undesirable level (or not):confused: . It would seem that drawing 11 amps, it would be a powerful motor for it's size.

Anyone out there understand these things. Motor ratings aren't my bailiwick.

Black_Moons
12-25-2009, 06:03 PM
If you remove the speed reducer, you will have 20 times less torque
Well, technicaly you'll have maybe 15~19 times less torque, as removing gearing will remove 'losses'

Machinist-Guide
12-25-2009, 06:26 PM
I agree with black moon. you are getting your torque from the gear train. It's like a stick shift car. You get higher speeds in high gear but you don't have the torque to pull out. In low gear you have the torque to pull out but you run at a lower speed

mechanicalmagic
12-25-2009, 07:43 PM
With that I figure

2.the 48 in/lbs of torque is at the max geared down speed of 117 rpm's


Nope. Maximum torque is at zero rpm, it falls off to zero at top speed.

about 2/3 of the way down.
http://lancet.mit.edu/motors/motors3.html#torque

Also, your motor, without the gear train MIGHT have more torque than the 20 times the gear train provides. Some gear boxes are down rated because of the allowable stress on the gears.

EDIT: Looked up in "Pocket Ref" by Glover.
Your motor is over 1 hp. A 1 hp motor would have 504 inch pounds of torque at 125 rpm. (Assuming no losses). Your gear box is fragile, compared to the motor.

whitis
12-25-2009, 08:48 PM
Nope. Maximum torque is at zero rpm, it falls off to zero at top speed.

The linear torque/speed curve shown in you link applies to permanent magnet DC motors. Series motors, shunt motors, universal motors, and induction motors have different curves. Some have torque curves which are somewhat flat until you get near top speed. In permag DC motors, speed is limited by voltage which is reduced by back EMF as the motor runs. Ultimately, this reduces the power to the motor. In series/shunt motors, you also have a field winding, which complicates matters. For an induction motor, speed is limited by frequency.
http://www.reliance.com/prodserv/motgen/b7097_2.htm
http://www.reliance.com/mtr/mtrthrmn.htm

Speed controls can also complicate matters. In some cases, these produce nearly constant torque vs speed over a wide range. A VFD on an induction motor, for example. Or a DC motor control which can overvoltage the motor to maintain constant current. Or a DC series/shunt motor control where the field winding current can be reduced to reduce the back EMF.
A 117V control driving a DC motor wound for 60V can give constant torque over the full speed range without overloading the motor, if it is designed to do so.

Thus we don't have enough information to predict torque.

mechanicalmagic
12-25-2009, 10:17 PM
A 117V control driving a DC motor wound for 60V can give constant torque over the full speed range without overloading the motor, if it is designed to do so.
I will agree that a control that can over voltage the motor, and maintain a higher torque at higher rpms. However, I maintain that a motor rated for 60V (your example) and 11 Amps (from OP) cannot be continuously run at 117V and 11 Amps, without exceeding the operating limits. If the motor was designed for that current and voltage, it would be on the specs.

Thus we don't have enough information to predict torque.
Yes, since maximum torque is at 0 rpm, it cannot be predicted from the info available. Other motors might be very close, so an approximation could be done. But, without actually measuring the torque, it's probably a +-25% guess.

RobbieKnobbie
12-25-2009, 10:37 PM
Lovely theoreticals aside, you're probably lookig at a 2500 rpm, 1 hp motor.

gearing that down to 2000 rpm, ignorong losses, puts you at about 2.6ish foot pound. I don't know what you're thryng to drive or what requirements you have, I'd say that's a lot less than what you started with.

Out of curiosity, what exactly are you motivating with this setup?

darryl
12-26-2009, 01:57 AM
Never mind. I think my math was bad. I had some calculations done, but I don't think I had the formula right. Probably from too much turkey and rum. I'll do it over tomorrow and see if it comes out the same.

Forrest Addy
12-26-2009, 03:02 AM
I might add the motor may develop 48 in lb at some overload. More realistically, the figure would be about 25 in lb but that's only an estimate based on the sketchy info provided. If the reducer has a ratio of 20.4 to 1 the torque goes up about 20.4 time less losses. My guess is the output torque of the reducer shaft is about 40 ft lb.

Abense of factory specs or test data reduces this whole discussion to best guesses and speculation. If you wish for more reiable data you may wish to run the motor loaded with a brake and intrument it for volt, amp motor temp, torque load and RPM. Runthe motor increasing the load until the motor temp get to the max temp rie for which the motor nsulation is rated. Not a difficult test for a modetly equipped home shop experimentor.

If you don't want the hassle of an elaborate test and think the motor will serve the purpose you wish to put it to, get a cooking thermometer and attach it to the motor so its internal temp registers. Run the motor for a time under actual operating conditions until you can determine whether the motor has to be load limtated or not and set limits accordingly.

I agree the motor is probably 1 HP. Ordinary HS physics and shop math works well for rough problems. If the motor's RPM works out to 2300 RPM at full voltage and no load so be it. That's close enough to work pulley and gear ratio. Expect 5% to 8% RPM reduction at full load from an unregulated drive.

EVguru
12-26-2009, 04:12 AM
I will agree that a control that can over voltage the motor, and maintain a higher torque at higher rpms. However, I maintain that a motor rated for 60V (your example) and 11 Amps (from OP) cannot be continuously run at 117V and 11 Amps, without exceeding the operating limits. If the motor was designed for that current and voltage, it would be on the specs.

A motor is simple an energy conversion device, the sated ratings are just one possible operating point.

The major losses in the motor are current squared time the resistance. Since the current is 11 amp in both cases then the losses and the thermal loading are the same.

As long an the rpm stays within the safe limit and the commutation is good there should be no problem. I've run a 72 volt motor at over 200 volt and three times its original rated power. Because of the internal fan running at much higher rppm, the motor ran cooler.

Black_Moons
12-26-2009, 08:44 AM
Evguru makes an intresting point. motor losses are primarly resistive losses, followed id assume by eddy losses and some other magnetic losses.

Many motors (see steppers, servos) ARE actualy driven at WAYY over 'rated' voltage but constant current so that they draw full design current over a wide range of RPM's, insted of having a current that drops off with RPM's resulting in low torque as RPM's incress.

IE: while your motor may be rated 11A and 70V DC at rated RPM's and torque load, at 2x the RPM's it might draw 7A at 70V DC, Much less produced torque unless you incress the voltage to 110V, then you achive 11A again, the motors current limit.

The current limiting also has the added benifit that an *overloaded* (See: loaded untill RPM's fall below rated speed, ie more current at the same voltage) motor will not overheat, At least, assumeing its cooling is not also hampered by being at low RPM's.

IE: a stalled motor no longer dims the shop lights and blows the braker. :)

loose nut
12-26-2009, 10:30 AM
I don't have anymore info on the motor other than what has been stated all ready, there isn't any info on the type of motor design, it's out of a mig welder wire feeder and is made to run all day long without any problem. Very quiet when running.

Any testing is out of the question. All I'm trying to do is see if the motor speed can be increased to a usable level, 117 Rpm's is to low for most applications, and still have the power to do useful work. Variable speed drive on a small lathe for example. A few other things come to mind also.

The consensuses seems to be that it is 1+ HP, I believe that is before the speed is slowed by the gearbox, correct me if I'm wrong.

mechanicalmagic
12-26-2009, 11:40 AM
The consensuses seems to be that it is 1+ HP, I believe that is before the speed is slowed by the gearbox, correct me if I'm wrong.
Yes, it's a 1+ hp motor.
117 rpm @ 48 in pounds is about 1/10 hp.

EVguru
12-26-2009, 04:06 PM
I don't have anymore info on the motor other than what has been stated all ready, there isn't any info on the type of motor design, it's out of a mig welder wire feeder.


The consensuses seems to be that it is 1+ HP


Yes, it's a 1+ hp motor.
117 rpm @ 48 in pounds is about 1/10 hp.

If it's out of a MIG welder, then it may well be only 1/10th hp.

The shaft power out of a gearbox is equivilent to the input power minus the losses in the gearbox. Worm reduction drives can get pretty lossy, but not 90%.

Whilst the torque output of the gearbox might be below that produced by the motor and multiplied by the reduction, you wouldn't want to do that without limiting the input torque by limiting the motor current. You generally don't over-rate a motor by very much, it's too costly.

The size originally quoted for the motor would be very large for 1/10 hp, or perhaps marginal for 1hp given the rpm (117 * 20.4 = 2387). Are the 8" by 4" dimenstions for the motor, or the motor and reduction?

mechanicalmagic
12-26-2009, 04:26 PM
The size originally quoted for the motor would be very large for 1/10 hp, or perhaps marginal for 1hp given the rpm (117 * 20.4 = 2387). Are the 8" by 4" dimenstions for the motor, or the motor and reduction?

I'm starting to think the motor must be mis-stamped. I'll bet it's a 1.1 amp motor. No reason for a 1 hp in that application.

whitis
12-26-2009, 05:40 PM
I will agree that a control that can over voltage the motor, and maintain a higher torque at higher rpms. However, I maintain that a motor rated for 60V (your example) and 11 Amps (from OP) cannot be continuously run at 117V and 11 Amps, without exceeding the operating limits. If the motor was designed for that current and voltage, it would be on the specs.


While your argument superficially appears to be correct, it is fallacious based on oversimplification. You have neglected that at full speed, the back EMF of the motor is 60V and thus the 120VDC suppllied by the controller at that speed would be reduced to 60V for power calculations. The equivalent circuit is a 120V battery in series with an ideal inductor and a resistor (winding resistance) and a 60V battery then back to the other side of the 120V battery. Still a little oversimplified, but sufficient for these purposes.

Another way of looking at it is that the resistance of the winding is constant (at a given temperature) and thus at 11A the power (P=I^2*R) is the same no matter what voltage is required to induce that current. Thus the motor does not overheat as long as it is continuous duty and its specs were not fiction and it is run in its intended environment. The motor is more likely to overheat at stall because the fan, if any, isn't running, the power isn't spread over multiple winding segments, and you don't have rotation induced convection currents inside the motor to move heat from the rotor to the stator.

The back EMF of 60V is the reason you had to raise the voltage to 120V in the first place.

The insulation on the motor windings should be able to withstand the 120V and sees much worse in normal operation.

Specs on motor nameplates are seriously oversimplified for electricians.

darryl
12-26-2009, 06:37 PM
I have to agree that for that application, a motor of that rating would be way overkill. Its stated size though and type suggests that it would be capable of one horse or so. Seems a mis-match. With a 20 to 1 step down, there would be gobs of torque to pull a wire off a spool. There's something missing in this whole thing-

By the way, as I re-do my torque calculations for 1 hp and 2400 rpm, I come up with 2.2 ft/lbs. That makes it 26.4 in/lbs. I extrapolated this from the formula 1 hp = 550 ft/lbs/sec.

I found a formula- torque = hp x 5252, divided by rpm. That's pretty easy to use. In this case, it resolves to 2.188 ft/lbs for a 1 horse motor turning at 2400 rpm.

mechanicalmagic
12-26-2009, 07:07 PM
While your argument superficially appears to be correct, it is fallacious based on oversimplification. You have neglected that at full speed, the back EMF of the motor is 60V and thus the 120VDC suppllied by the controller at that speed would be reduced to 60V for power calculations.
I believe at full speed, the back EMF equals the input voltage, minus that energy used to heat the windings (from resistance), air resistance, bearing and brush friction. If there was a load, we would not be at full speed.


The equivalent circuit is a 120V battery in series with an ideal inductor and a resistor (winding resistance) and a 60V battery then back to the other side of the 120V battery. Still a little oversimplified, but sufficient for these purposes.
Oversimplified to the point where there is no provision for the load. The winding resistance does not change. So the back EMF MUST change with a load. At zero rpm, there is no back EMF, since the rotor is not moving, there is only winding resistance. It turns out that the actual motor current is determined by the difference between the applied voltage and the back EMF divided by the parasitic resistance.

I agree there is no way to accurately determine the parameters for this motor.

darryl
12-26-2009, 08:02 PM
Going back to the original post by loose nut, the motor is 115v, 11a, 48 in/lbs torque- using those figures we have a motor drawing 1265 watts, which is 1.7 horse. The amp figure could be the max continuous, or could be the stall current, or could be the current at the rpm where the 48 in/lbs are produced.

Going back to using 1 hp as a reasonable guess for the hp capability, 48 in/lbs corresponds to a motor speed of 1320 rpm. Through the speed reducer, that becomes 65 rpm. The stated rpm is 117, so something is fishy with the assumptions.

I think the 11 amp figure must be stall current, and in no way relates to the power capability of the motor. Much more likely then that the stated torque figure is for the output shaft. It makes a lot more sense anyway, because the loading is relatively light- drawing a wire off a spool at a slow enough speed to be useful in a wire feed welder.

So- more calculating- 48 in/lbs and 117 rpm- same formula- power becomes about 1/11 hp. Seems way more reasonable. The motor would be somewhat more powerful than that to account for losses, so we could say about double that. That makes it about a 150 watt motor. Since I've accounted for loss of efficiency already, we can put that figure directly into an equation and say that motor draws 1.3 amps under rated output speed and torque. In that case, an 11 amp draw at stall could be about right. Consider the case of the (beaten to death) vacuum cleaner motor specs.

A lot depends on what type of motor this actually is. I suspect this last outline of mine comes a lot closer to the reality of the situation.

whitis
12-27-2009, 12:06 AM
I believe at full speed, the back EMF equals the input voltage, minus that energy used to heat the windings (from resistance), air resistance, bearing and brush friction. If there was a load, we would not be at full speed.


No, the back EMF does not equal the input voltage at full speed when using a controller that is capable of supplying higher than motor nameplate voltage. Unless you are at no load, in which case the controller will lower the average input voltage to conditions similar to a controllerless motor. Your assumption that full speed equals no load is not necessarily true in the presence of a controller.

For a motor connected to a simple power supply at nameplate voltage back EMF equals the input voltage at full speed under true no load conditions - without deductions for the winding resistance. However, the effects of internal friction and the small voltage drop across the winding resistance when supplying just enough current to overcome the internal friction can result in the nameplate speed being adjusted downward slightly from the theoretical value where back EMF equals nameplate voltage.

To add to the confusion, of course, we don't have a nameplate for the motor but one for the combination of motor, gearbox, and possibly speed control.

But without a controller, load reduces speed which causes the back EMF to drop which causes the current to increase which causes torque to increase. If load is increased sufficiently, you reach 0 rpm, no back EMF, and maximum torque.



Oversimplified to the point where there is no provision for the load. The winding resistance does not change. So the back EMF MUST change with a load. At zero rpm, there is no back EMF, since the rotor is not moving, there is only winding resistance.



The simplification was omitting details that were not pertinent, not omiting important things like external load. I omitted capacitance and inductive effects due to commutation as they just confuse the issue. And internal mechanical losses which should be a small effect even without a speed control and irrelevant with a decent speed control. And then I made a point of advising the reader that I was leaving out minor details that had little bearing on the explanation lest they make the mistake of applying this information in a drastically different context. The simplifications I made, incidentally, are ones used on a daily basis when actually designing motor controls. When making detailed computer models, however, one should include the minor effects usually ignored.

Load does not affect back EMF except in that load can slow down the motor in the absence of a speed control. Back EMF is a function of speed, not some combination of speed and load. In the speed control case I was discussing, the controller could maintain speed by adjusting current and voltage as needed to adjust for load and back EMF. Thus we can be at full (nameplate) speed and full load with a suitable controller - that is more or less the point of using a decent speed controller - to make speed independent of load.

Winding resistance does change slightly with motor/winding temperature; under the constant power conditions I was describing, however, temperature would be fairly constant once the motor warms up (although there can be a loss of cooling at slow speeds). You can ignore the change in winding resistance most of the time.

The other curve ball with a speed control is that the max speed could be hardwired to a value less than the motor is capable of.

Since this is a wire feeder from a welder, though, the speed control might not be very sophisticated and not deserve the name speed control. It may amount to a variable voltage supply (or PWM equivalent) relying on back EMF to control speed with an oversized motor so load doesn't have much effect. Too much motor and not enough electronics. In that case, friction of the gear train must be considered also. I.E. the bare motor's no load speed may be higher than the gearbox output speed and gear ratio would suggest as the motor never reaches no load due to the friction of the gear train. With such a primitive control, however, you throw away torque at low speeds due to depriving the motor of power, lose torque at high speeds due to back emf, and have no real speed regulation over varying loads.



It turns out that the actual motor current is determined by the difference between the applied voltage and the back EMF divided by the parasitic resistance.

Slight oversimplification but one commonly used by pros. When dealing with simple DC machines, one usually ignores the other parasitics (except when solving different problems or considering instantaneous currents rather than average under adverse conditions). However, for stepper motors with lots of poles, winding inductance is the killer. The effect is somewhat analogous to back EMF but when dealing with steppers, you tend to ignore back EMF and think in terms of winding inductance. The effect of inductance is present whether the motor is turning or not, unlike back EMF.


I agree there is no way to accurately determine the parameters for this motor.

yep. Might be able to narrow it down a bit though based on motor RPM and electrical horsepower and an estimated efficiency. I.E. with some calculations one may be able to set an upper bound on the torque at any given speed based on the maximum power rating from the AC line. Actual torque may be less especially without a motor controller or with a primitive one.

loose nut
12-27-2009, 11:34 AM
I'm starting to think the motor must be mis-stamped. I'll bet it's a 1.1 amp motor. No reason for a 1 hp in that application.

Sorry, no. I've got two of these and they are both marked the same way.

darryl
12-27-2009, 04:32 PM
Loose Nut- I'm just curious- is that a permanent magnet motor, or is there a field winding, or is there a field winding and no brushes-

EVguru
12-27-2009, 05:35 PM
What's the resistance accross the terminals?

Measure with the motor in several positions and average out if there's much variation.

loose nut
12-28-2009, 04:19 PM
What's the resistance accross the terminals?

Measure with the motor in several positions and average out if there's much variation.

Say what????? "me no understand motors"

loose nut
12-28-2009, 04:23 PM
Darryl, I just went out and looked at the motor again and it does have brushes, that might have been a important bit I omitted before.

Actual dimension of motor are 3 x 6 1/2". The only other info on the label are the manufacturer name and model and that it was manufactured in New York State (couldn't make out the city, Waterford maybe).