# Help with pulse dc motor calculations- torque, eff, hp

1. Jul 13, 2012

### mnsman

Hello all,

I have a few questions about calculating efficiency, torque and horsepower for a small pulse dc motor. The only parameters that I have are input voltage = 39Vdc, input current = 2A and rpm = 90,000 rpm. Are the above values obtainable with the given information?

This is not a homework problem per se. It is homework in the sense that I am a hobbyist and I've built the motor with the values I've provided at home.

Thank you in advance for any help provided.

2. Jul 13, 2012

### vk6kro

That figure of RPM seems very high. Are you sure?

All you can say is that the motor is using 2 amps at 39 volts or 78 watts.

If the motor was 100 % efficient the output power would be 78 watts / 746 watts per horsepower, or 0.105 horse power. The actual horsepower would be something less than this.

You can't really work out the efficiency or torque.

3. Jul 14, 2012

### mnsman

Yes that is an accurate rpm measurement. The rpm has varied and that is the maximum recorded. It was measured with a tachometer and oscilloscope connected directly to the hall sensor input of the circuit. The rpm was also closely confirmed by a spectrum analysis of the audio output of the motor.

But that isn't the issue. I should have simply asked if it were possible to determine efficiency, horsepower and torque given power input (E x I) and rpm.

Just for clarity I will add that this motor has one drive coil and one diametrically magnetized magnet attached to a ~6" stainless steel shaft.

I've run in circles trying to figure this one out. From my research it seems you need one of the missing values to calculate the others. That is why I'm here. It's above my pay grade so to speak.

It is safe to assume that efficiency is not 100% given the friction loss of the bearings alone. Assuming an efficiency value doesn't help matters.

If I provide the mass of the rotor does that help? If I'm not mistaken I can weigh the rotor and convert that to mass. Or are there any other values that I may use? I don't know what they could possibly be but I guess I should ask.

I don't want to further complicate matters but to properly analyze this system's efficiency it really should be considered a motor generator. There is recovered energy that is stored in a capacitor bank and can be used to power a load. This energy is the combination of the recovered inductive collapse from the motor's drive coil and the ac waveform generated by the rotor magnet passing over the drive coil during the off cycle of the pulse motor. BTW It also has a 120db audio output so wear your ear protection while operating. I added that bit only because I've successfully converted some of that audio output to usable electrical energy by way of an acoustical energy harvesting technique I devised. Obviously that adds to the overall efficiency of the system. For the time being I would like to keep it simple and consider this system as only a motor. I mention this now only because I would like to add this to the equation in the future if only this first step of analysis is successful.

4. Jul 14, 2012

### vk6kro

It is safe to assume that efficiency is not 100% given the friction loss of the bearings alone. Assuming an efficiency value doesn't help matters.

Yes it does. If we adopt a 100% efficiency, we can set an upper limit on the output horsepower. The output power can't be higher than the input.

Other than that, you don't know the load on the motor, or the torque, so you can't work out the output power or the efficiency.

5. Jul 14, 2012

### mnsman

Okay you got me on that one. Agreed it sets up the best case scenario but it is still not what I'm after since I know I don't have 100% efficiency. At least I know I can solve for best case.

I thought mass may come into play here and it could help solve the problem.

So in summary you say it can't be done without a one of the missing values I am after.

Tell me what you think of this, I just found it on another forum:

"Stall Torque could be measured with a spring scale (Fish scale)" by mounting a bar on the
motor shaft with spring scale hanging off the bar at a known radius and secured to something immovable for
this purpose, then calculating the torque (Lb/Ft, Oz/In, etc) from the measured force and radius. Just start
stepping up the motor current while monitoring the force. "

"P = I * V = Torque * shaft frequency ==> Torque = (I * V) / shaft frequency "

In response to that someone wrote:
"Be carefull. If I recall correctly, the mechanical power in the load is P=T*omega, where omega is angular frequency. So you need to transform your rpm to angular frequency. "

I'm not sure if any of that helps or not.

6. Jul 14, 2012

### vk6kro

It is often useful to start off with a perfect situation when you are dealing with motors or transformers and then start adding the losses if you have to.

Often you can see that something won't work, or will work regardless of losses, so you can proceed even if you don't know everything.

I don't think stall torque would help much as it is an abnormal situation and doesn't relate to the normal behaviour of the motor.

Strictly, you should apply a known braking force to the motor until the speed just starts to slow down.
This would be pretty hard to do with such a low powered motor.

Another way is to get two identical motors and have one drive the other as a generator. Then measure the electrical output and assume the losses are shared between the two devices.

I'm still intrigued by the 90000 RPM. Did you get the revs/second and multiply by 60 or by 3600?
3600 gives revs per hour.

7. Jul 16, 2012

### mnsman

The 90k rpm figure was derived with a tachometer that reads directly from a hall sensor output. The hall gen senses the position of the rotor magnet and is used for timing the drive coil output. My rotor magnet is a diametrically magnetized ring magnet. Therefore there is one symmetrical clock cycle per revolution of the rotor. The tachometer reading is confirmed by an oscilloscope which is directly connected to the hall sensor output. In the case of 90k rpm the scope read 1500hz. 1500hz x 60 = 90k. Yes it is scary fast. 90k rpm was the fastest it's ever run and so far was a single event. It has ran 70k+ on many occasions. Setting the coil and hall sensor positions relative to the magnet is critical to performance. When the record was set both were probably nearly perfectly positioned and I've chased that speed ever since. As a side note, due to an error on my part the coil that set the record has been destroyed. Ironically the motor was idle at the time. I had finished a test run after software changes to the arduino controller and mistakenly left the motor on. The coil fried in a hurry.

What I'm really after is the true efficiency calculation/measurement. Horsepower and torque can be derived from that and would be nice to have.

Being an amateur, I thought that it made sense that the amount of work performed (power out) could be calculated from rpm and mass alone. I have the power in ( E x I). A physical measurement, in the literal sense, like measuring torque is very tricky and may turn out to be inaccurate. As you've said it is even more difficult with such a small low power / torque motor. So I'm ruling that out for the time being.

Building a second identical motor is out of the question. This is a homegrown custom motor.

I can and have hooked up a second coil as a generator coil albeit in a poor manner. As a matter of fact I did it again last night only in a proper manner. I've been meaning to build a platform to mount a second coil on the top side of the magnet to be used either as an additional drive coil or generator coil. I finally bit the bullet and did it yesterday. It was all in effort to test a new coil design that was too large to fit on the lower side of the magnet where I normally mount my coils. Last night I had it running only for a few minutes because it was late when I finished the platform. Preliminary results: 20k rpm @ 12V 750mA input, generator coil produced ~72Vdc (bridge rectified to cap). The output was not loaded. Tonight I will load it and raise the power input up to 39Vdc.

As another side note: The coil I primarily use is a set of thirteen 2" diameter stacked pancake coils connected in series. It's a quite unique design and is the main reason for the speed of the motor. As a result of burning up my original coil, I have developed a method of winding perfect pancake coils. It was time consuming to do it, but I wound nearly one hundred of them, both single wire and bifilar. Now I'm testing different combinations in conjunction with other coil designs.

8. Jul 27, 2012

### mnsman

I will figure this out myself. Here is your clue as to how. What is work? Can anyone on this forum answer that question? Physics forum?

9. Jul 27, 2012

### Carl Pugh

Get a pony brake and put it on your motor.
You now have the RPM and the torque and it is a simple matter to calculate the motor output horsepower.

10. Jul 28, 2012

### jim hardy

force X distance
energy

nope. Power is a rate of performing work , amount per unit time.

if you know 'moment of inertia' of motor's rotating parts you might make some approximations by measuring its rates of acceleration when power is applied and deceleration when power is removed.

but as several folks have said before you really need to measure torque and it's not difficult look up dynamometer.

From "work is force X distance" it's an easy step to:
horsepower = 2pi X torque X rpm / 33,000

11. Jul 28, 2012

### jim hardy

was there any mechanical load on the shaft?
if not, then can say that your internal losses (friction, windage, magnetic, and electrical) total 78 watts , as vk6 mentioned .

12. Jul 28, 2012

### Staff: Mentor

And therefore, if the test condition is with the motor unloaded, there is no output power and the efficiency is zero.

13. Jul 28, 2012

### I_am_learning

You could try measuring the resistance of the Motor.
Say its R.
Then, Power lost in resistance = I^2*R = 2.0^2 * R = 4R.
Total Power input = V*I = 39*2 = 78watts
Hence, Mechanical Power Output = Power Input - Power Lost in Resistance.
Of course, whole of this mechanical Power is lost in Windage and Bearings, so there is no useful power out, hence 0 efficiency as russ_watters pointed out.
However, we can assume mechanical Power output as our actual Power out.
In this case, efficiency is simply,
Efficiency = Mech_Power_Out / Power_in
Horse_Power = Mech_Power_out /746
Torque = Mech_Power_out / w
where w is the angular velocity in radian per second = RPM * 2*PI / 60.
Please note that the Horse_Power and Torque Calculated here isn't the maximum obtainable from the motor, but are the values for the no load condition.
Maximum Mech_Power is obtained when Efficiency is 50%.

14. Aug 15, 2012

### mnsman

It's late and I decided to drop by for some unknown reason so I haven't given this much thought.

First of all I'd like to say that I appreciate people still tackling this question.

My naive thinking is this... If I were a worker is paid by food (energy) to turn a wheel on a shaft and bearings, wouldn't that be considered work? It sure would seem like it to me. Of course, the employer would want to pay me the least amount of food (calories) to turn the wheel. So he would want me to be efficient. Best case scenario I am 100% efficient. That is I use all of the calories he pays me just to turn the wheel and nothing else. If it takes me x amount of calories to turn the wheel y times. Shouldn't we be able to determine how efficient I am by knowing my energy input and output (work performed)? The amount of work performed would be measured by the amount of times the wheel is turned in a given amount of time. The mass of the wheel and shaft and the friction of the bearings would determine how many times it could be turned in z amount of time given x amount of calories.

Thanks again. I will look over the comments when it's not so late.

15. Aug 15, 2012

### jim hardy

only if the wheel were driving some machine, maybe a pump to lift water or make compressed air for a pipe organ.

or if the purpose of turning the wheel was to warm the room from the heat generated in its bearings.

In either case you'd have to measure the output of the wheel, either the amount of fluid pumped(water or air) or heat added to the room.

not that today's economy doesn't have plenty of people working hard at producing nothing
but that's a societal problem not a physics problem. Google "Parkinson's Law" .

old jim

16. Aug 16, 2012

### mnsman

I must have a fundamental misunderstanding of the formula W=Fd. I don't see anything about the output has to be meaningful as in the case of my example. Doesn't the spinning wheel, in this case, represent kinetic energy and that in itself is a result of work performed? In my real life case we substitute wheel with rotor magnet and shaft.

My main points are: 1) Whether or not the work performed is purposeful or not it should still be considered work. 2) The output of my motor is the rotation of a mass (rotor and shaft) and should be considered kinetic energy that is a result of transformation of electrical energy to electromagnetic energy and then to energy of motion. Therefore my thinking is that I should be able to calculate how efficient my motor is simply based on it's input power (electrical) and kinetic energy output (rotating mass). Does this make sense?

I thank you for your time, input and patience. I hope that I'm not coming across as argumentative. Ignorant- good possibility. I'll take that chance.

Update: my motor is now a 97k rpm motor.

17. Aug 17, 2012

### jim hardy

no you're okay. Just in physics we usualy consider work as useful output of a machine or process, in some units of energy.

In your example, you would be the machine doing work on the wheel, for whatever reason.

I was thinking of the wheel as the machine with you as its energy source, and looking for useful output from the wheel.

If you consider your 97K motor as the machine,
and you figure(correctly) that its only output as heat from friction&windage, and noise, and you consider those useful,
then your efficiency is 100%. There's no other output. All the electrical energy that goes in comes out as one of those.

If instead you use your motor to drive a shaft, say to turn fan blades, then it has a useful output and your efficiency would be however much work it does on the fan blades divided by electrical input. Clearly that's less than 100% because of the noise etc that dont help turn the fan blades..

Think about a computer. Its only outputs are heat, light and frustration none of which i consider useful. Programmers feel differently though.

Big part of our economic problem today is society has lost ability to distinguish between useful and non-useful work. Basic Thermo should be taught in high school.

old jim