# B Electric motor: power and RPMs

1. May 17, 2017

### serpentine009

I am helping a friend with a project, and am wondering if this statement is true.
"To double the RPMs of a central drive shaft motor, you must quadruple the input."

In other words, if an electric motor being supplied with 1 Watt spins at 500 RPMs, one would need 4 watts to spin at 1,000 RPMs, and 16 watts to spin at 2,000 RPMs, and so on...

Thanks

2. May 17, 2017

### Staff: Mentor

Friction usually increases with the square of RPM.

But the speed of a motor depends on the mechanical load connected to it.

So, what load is your motor driving?

3. May 17, 2017

### serpentine009

You already lost me.... I don't know how to answer your question regarding load. It's been a very long time since I studied or had to apply anything remotely close to this.

I am helping with potential marketing for a concept of "over unity". I know, I know it's snake oil...but I want to be able to intelligently confirm or deny some base concepts that have been presented to me. One was simply the phrase that I posted at the top. I guess the concept is that in this particular motor's design, the ratio of input power to output RPMs is linear as opposed to exponential.

Thanks

4. May 17, 2017

### Staff: Mentor

Discussion of pseudoscience is against PF rules, even if it is to debunk it.

However, so long as you keep it to a discussion of actual motors, this is ok.

5. May 17, 2017

### serpentine009

Yikes, my apologies. For my motor specific question, I guess let's start with no load. Just simply turning a drive shaft @ 500 RPMs. If we assume that 1 watt is required to accomplish that, is it true that 4 watts would be required to achieve 1,000 RPMs?

6. May 17, 2017

### Staff: Mentor

But turning an unloaded motor shaft is not very productive. What would the load be in real life? Just a flywheel? Or is the motor doing some useful work?

7. May 17, 2017

### serpentine009

It would be providing electricity for other devices, motors etc... Maybe like a backup generator.

8. May 17, 2017

### Staff: Mentor

To repeat and reword:
For example, if your motor is spinning a fan, the input power varies as a cube function of rpm. If moving a conveyor, it is a direct proportion.

Input power vs rpm when unloaded isn't very useful and I doubt very predictable.

9. May 17, 2017

### Staff: Mentor

So you're running the motor backwards like a generator? Then the mechanical load presented to the shaft would vary with the output power being drawn...

10. May 17, 2017

### Dr.D

Has the type of motor ever been specified? Is it an AC induction motor? An AC synchronous motor? A DC shunt motor? A DC series motor? IF so, I must have missed it, but I don't really see how this can be discussed without specifying the motor type.

11. May 18, 2017

### Tom.G

Yes IF...The load is purely frictional AND the motor is ideal.

Good luck meeting either of the prerequisites.

12. May 18, 2017

### Staff: Mentor

Sorry, but I disagree.

Students should first be introduced to the concept that a motor or generator is merely an energy conversion device. X amount of energy of type A goes in, and X of type B comes out. A might be electrical and B mechanical, or visa versa. That is the primary concept.

Efficiency, motor type, AC/DC, voltage, current, speed, everything else is secondary.

The OP does not understand the primary concept (indeed since he mentioned "over unity" he thinks he can get more out than he put in), so it is no help to introduce secondary considerations.

13. May 18, 2017

### Dr.D

Per DrClaude's condition, we are not discussing over unity devices.

The OP speaks of "the input" but does not define what he means. If "the input" is voltage, the effect of doubling input voltage on a DC machine is quite different from that on an AC machine. If "the input" is frequency, the the whole concept is only meaningful for an AC machine, but meaningless for a DC machine. And so on ...

To say that a motor or generator is "just an energy conversion device" is simplistic and less than meaningful in my opinion.

14. May 18, 2017

### CWatters

Given the "background" to this thread you should also think about the efficiency of the set up. The efficiency is the useful power produced divided by the input power, or in short.

Efficiency (%)= 100 * Power out/Power in

If there is no load (eg the motor is turning but isn't connected to anything) then the useful power out is zero so the efficiency is also zero. All the input power is wasted as heat etc.

For most motors it's possible to plot a graph of load vs efficiency and this will usually have a peak. Peak efficiency for some motors can be as high as 97%.

If you and your friends are building you own motors or machines then measuring the efficiency and comparing them with existing motors or machines would be a reasonable thing to do.

One way to measure output power is to measure the output torque and angular velocity (calculated from the rpm). In the right units the output power is just these two multiplied together. You can sometimes measure the torque by mounting the motor so that the casing can rotate slightly. Then attach a lever arm to the casing and either hang weights from it or have it press on a weighing scale. However this method includes any power needed to overcome friction so isn't very accurate.

If you don't measure the output power and calculate the efficiency it's very easy to fool yourself into thinking you have a wonderful new invention. Best do this before posting on YouTube :-)

15. May 18, 2017

### sandy stone

I'm going to go out on a limb here and assume that the OP is concerned with the relationship between torque, shaft RPM, and power. So, power is proportional to torque x RPM. Assuming the motor is providing the same torque at twice the RPM (not a very realistic situation, I know), then power has doubled, not quadrupled. Hope this is what the OP was getting at.

16. May 23, 2017

### Shane Kennedy

Power is proportional to RPM squared, multiplied by torque. if you can measure the current, voltage, and RPM, you can derive the torque (ignoring friction and other inefficiencies).

17. May 23, 2017

### jbriggs444

RPM is not squared. That would be dimensionally inconsistent. Power is proportional to RPM multiplied by torque.

With an appropriate choice of units, it is equal to angular velocity (e.g. in radians per second) multiplied by torque (e.g. in newton-meters), yielding power (e.g. in watts).

If the torque were proportional to the rotation rate then you could square the rotation rate. But then you would not properly multiply by torque. Instead you would multiply by torque per unit rotation rate. That would be dimensionally consistent.

18. May 23, 2017

### Shane Kennedy

I sit corrected