# Help Needed from E&M and Motor Gurus!

1. Oct 16, 2006

### marcusl

Some years ago I helped a talented young boy do a project for his elementary school science fair. He wanted to build an electric motor and “see how fast it would go with different batteries.” After building the usual wobbly paper-clip-and-toothpick type motor that needed careful jiggling to get it to move, he suggested using Lego Technics, with which he had hours of experience building cool vehicles and devices with winches and little pneumatic drives. Lego blocks are solid and precise, the fluted axles spin true and fast in their plastic bearings with a little oil, and with an 8-32 steel screw wrapped with wire and mounted through a Lego T-piece slipped onto the axle, 2 round refrigerator magnets and 2 brushes cut from a flat bronze spring, the result was a high quality DC motor.

To measure “how fast it would go,” I had him tie a small fishing weight to one end of a string and tie the other end onto the axle. He started a stopwatch upon connecting the D-cell battery, stopping it when all the string wound up on the axle and the weight hit the top. Because of the small diameter of the axle and the long string (about 15 or 20 inches), it took a while to raise the weight so his time measurements were pretty accurate. I came across a photocopy of his hand-drawn chart yesterday, and have replotted and attached it. The y axis is the speed or, more precisely, the length of string in inches divided by the number of seconds it took the weight to travel from bottom to top. He was able to present “how fast it would go with different batteries” and won a ribbon from the judges (teachers).

He didn’t realize that, because of how I had him measure “speed”, he was actually measuring power; indeed horsepower was originally measured by seeing how long it took a horse to raise a known weight a known distance. I naively expected to see something like a quadratic curve, since we are accustomed to thinking that P is proportional to V^2 in common situations. I was surprised to see that the power of this motor depends linearly on voltage!

I've thought about why this might be, but I’m curious to hear from any experts out there who can explain it.

#### Attached Files:

• ###### Motor power.pdf
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2. Oct 16, 2006

### Danger

Are you taking into account that as the string winds up, it's changing the diameter of the shaft and therefore the 'gear ratio'?

3. Oct 16, 2006

### marcusl

Since that occurs the same way every time, I'd be surprised if it matters.

4. Oct 16, 2006

### Danger

Oops... I misinterpreted the experiment. I was thinking that you varied the voltage as the thing was winding up. I take it then that you did a complete run on each voltage setting.

5. Oct 16, 2006

### marcusl

Yes that's right.

6. Oct 16, 2006

### leright

well, you WERE measuring speed, not power. You could divide the length of the string by the time and get the linear velocity of the rod (at the surface of the rod). Then you could divide that linear velocity by the radius of the rod to get the angular velocity. divide the angular velocity by 2pi to get the revolutions per second (or minute or whatever).

Now, why does the SPEED vary linearly with voltage? Well, the current through the armature varies linearly with applied voltage, right (and the current is what matters, since it's the current that is producing the Lorentz force)? We know that the force on the ring (or solenoid if you will) due to the external field is F = ILB (since the B field and current direction are perpendicular. However, there are two forces acting on the ring (or solenoid) that each for a couple, producing a torque. The torque is T=2*F*(radius of current ring), and we can say this because it is reasonable to approximate by saying the force is always perpendicular to the vector pointing in the direction of the radius. So, T=2*I*L*B*(radius of current ring). However, the B-field is constant, L is constant, the radius of the ring is constant, and I we vary. I is related to V by I = V/R (r is the resistance of the current ring or solenoid). Plugging I = V/R into the torque equation we get T = 2*(V/R)*L*B*(radius of ring). Everything is a constant except V and we can group it all into a constant K. So, T = KV. Torque is proportional to voltage! BUT, eventually the back EMF (due to the solenoid cutting the magnetic flux as it rotates) will oppose the direction of the current caused by the applied voltage and the ring (or solenoid) current will become zero, and no torques will be applied to the ring. However, it will continue to rotate at a constant speed because of its angular momentum....so, speed varies linearly with applied voltage.

And when they measured power using the method you described back in the old days they multiplied the weight of the weight (lol) by the distance the weight moves and then divide by the time it takes the weight to move. This is what they were calculating.

Now, of course you'd have a factor of N in there for a solenoid, but I did the calculations for a RING with only one turn. The point is still the same.

So bottom line is, you aren't measuring power. Hell, your units on the y-axis are in inches/second, which is of course not power. Power is Energy/second, not distance/sec.

Last edited: Oct 16, 2006
7. Oct 16, 2006

### Claude Bile

So at 1.5 Volts, the motor speed was 0? Do you remember if this was actually the case? i.e. did you have to increase the voltage to a particular value before there was movement?

Claude.

Last edited: Oct 16, 2006
8. Oct 16, 2006

### marcusl

Wait. If I measure how long it takes to expend the same amount of work (U=m*g*h) every time, aren't I measuring power?

Same thing here.

Last edited: Oct 16, 2006
9. Oct 16, 2006

### marcusl

Yes, the motor wouldn't lift the weight at 1.5V.

10. Oct 16, 2006

### leright

so, clear as mud? I remember you saying you're a physicist, so it's probably clearer than mud. :p

yes, you'd be measuring power, but (string length)/(time) isn't power, which is what you (or the kid) measured.

power would be (weight of mass)*(distance traveled by mass)/(time of travel)

Last edited: Oct 16, 2006
11. Oct 16, 2006

### leright

There's going to always be a deadzone due to friction. The curve is definitely going to be linear.

Last edited: Oct 16, 2006
12. Oct 16, 2006

### marcusl

Easy there! You are working up too much of a lather. If you multiply the y axis by weight, you get power.
Weight=m*g,
unchanging during the experiment, and the work done during each run is
W = m*g*h.
Power is
P=Work/time = m*g*h/t,
and the y axis is showing exactly h/t. You may not like the normalized units I used, but this motor displays a linear relationship of POWER to voltage.
No, wrong. If it did, then the power would be proportional to V^2, or P=V*I=V^2/constant, as I said in my original post. You've kind of missed the whole point.

Last edited: Oct 16, 2006
13. Oct 17, 2006

### leright

good point....I didn't realize at the time that power was proportional to speed, but I suppose it should have been obvious. If you want to look at it that way then I think the back (or counter) EMF might be the answer. As the solenoid is rotated due to the lorentz force the solenoid is cutting the magnetic flux, which translates into a change in the flux. This produce an EMF that causes a current that opposed the current caused by the battery. This might explain the conservation of energy issue, which is essentially the issue you're presenting.

Here's a bunch of different lego motors that all show linear voltage/speed (or, voltage/power if you will) graphs.

http://www.philohome.com/motors/motorcomp.htm

I will think about this a little more and respond later. However, the analysis I provided earlier still works....there's no debating that the current in the ring is V/R and therefore the power transferred by the battery is V^2/R. Now, why are we not getting that much power out? Well, there is energy stored in the magnetic field produced by the magnet which counteracts the power by producing counter EMF....not to mention air resistance, which does work in the form of friction, but I suspect this plays a negligible role here.

EDIT (further explanation): To add, you can account for the back EMF in the equation I provided for torque on the ring, which was T = 2*(V/R)*L*B*(radius of ring)....back EMF is proportional to linear velocity (or angular velocity...take your pick). so, Vback = K*omega, where omega is angular velocity. The current that is produced by the back EMF opposes the current produced by the battery and is obviously Iback = (K*omega)/R. The net current is of course the difference between the current caused by the battery voltage and the current caused by the back EMF due to flux cutting. So, Inet = (V/R)-[(K*omega)/R]. We can substitute this back into the Torque equation to get the following: T = 2*[(V-K*omega)/R]*L*B*(radius of ring). Notice that omega is a function of time and so is torque. As soon as the battery voltage is applied omega is zero and the net torque will be a maximum and the motor will accelerate. Eventually, omega will grow and the battery voltage and counter EMF will cancel, producing no current, but the motor will continue to spin at the speed it reached due to its inertia, right? When it begins to decellerate due to friction or an external force or whatever, the counter EMF will decrease and there will again be a net current in the ring and thus a torque (since the battery current overpowers the counter EMF) and the motor will continue to maintain its speed.

Since the motor stabilizes to a point where the net torque becomes zero, the speed of the motor is the speed that causes the current due to the back EMF to exactly cancel the current due to the battery. From the above equation, (V-K*omega)/R = 0 must be true when the torque is zero of the motor speed stabilizes to a certain value. Solving for omega yields omega = V/K. Or, to make you happy, Power = mgV/K So, power is proportional to battery voltage.

Last edited: Oct 17, 2006
14. Oct 17, 2006

### leright

Oh, and another thing about DC motors I'd like to add, since it is kinda interesting and important (especially to control engineering). If you look at the torque expression, for a given battery voltage V you can group things into constants and express torque as a function of angular velocity. You get Torque = K1 - K2*omega, where K1 and K2 are constants. K1 = (2*V*L*B*(ring radius))/R and K2 = (2*K*L*B*(radius of ring))/R. So, torque varies inversely with speed for a given voltage and the y-intercept (value of torque when speed is zero) is obviously K1, which is intuitively pleasing, since I said as speed increases the counter emf will increase which decreases the current flowing through the armature resulting in a lesser torque. This just shows that they are linearly related. When the motor starts at rest when the speed is zero the torque is K1. It is also obvious that the peak speed (speed obtained when torque becomes zero due to counter EMF) is given by the ratio K1/K2.

So, for a given input voltage, at low speeds you have high torque but at high speeds you have very low torque, and eventually there will be a max speed with no torque at all. However, when there is a load applied to the motor that decreases the speed there will suddenly be a torque and the speed will be maintained.

Sorry....I just find motors to be quite interesting. :tongue2:

Last edited: Oct 17, 2006
15. Oct 17, 2006

### marcusl

I worked it out last night also. We agree that the back EMF caused by sweeping the rotor coil through the static field is what generates the linear relation and also, as you point out, the minimum required voltage. (It's not due to friction as suggested in an earlier post.) You got, I think, the same results I did (I haven't read them carefully), but certainly beat me to the post! Thanks for your comments, I had fun with this!

There is a second source of back emf, the usual -L dI/dt in the coil, that is negligible for this motor but I think can be important for fast high-inductance motors like power tools or vacuum cleaners. Fast here also reflects the many armature windings used in commercial motors that greatly reduce the time that any one coil is energized.

For what it's worth, I've attached my derivation as a pdf file.

#### Attached Files:

• ###### motor eqns.pdf
File size:
22.1 KB
Views:
109
16. Oct 17, 2006

### leright

hmm...if you're interested I will write a "derivation" of my thought process also, if I can find the time.

17. Oct 18, 2006

### quinn

Correct me if I am wrong,...

Horsepower = energy / time

Voltage = energy /charge

therfore,

horsepower = voltage/ (time *charge)

18. Oct 18, 2006

### leright

um, no. power = (voltage*charge)/time, which is (voltage)*(current)

power is energy divided by time and voltage is energy/charge. This is correct but you messed your algebra up. :tongue: energy is voltage*charge. power is therefore voltage*charge/time.

horsepower is a measure of power.

19. Oct 18, 2006

### Astronuc

Staff Emeritus
In SI,

1 volt = 1 Joule/Coulomb (J/C) and 1 Amp = 1 Coulomb/second (C/s),

so Voltage * Current = Joule/second (J/s) = 1 W = Power

20. Oct 18, 2006

### quinn

oops,.. it was late,.. thanks for correcting me