# Speed vs Voltage, My Thoughts



## toddalin (Jan 4, 2008)

In another thread, it was asked how much faster a engine would go at a given voltage relative to another voltage (18 volts vs 14 volts in that case). This represents an increase of just 4 volts.

I proposed that the speed is directly proportional to the square of the difference in voltage.

Thus (1-(18^2/14^2)) x 100% = 65% and the engine would go ~65% faster.

Similarly, the difference between 22 and 18 volts (also a 4 volt difference) is (1-(22^2/168^2)) x 100 % = 49%.

An the difference between 14 and 10 volts is 1-(14^2/10^2)) x 100% = 96%.

There are those that may doubt this theory, (even with their own supporting data staring them in the face), but let me put it into terms that may be more familiar to some.

A dynamic loudspeaker is simply a linear motor that operates on a/c current. It uses a coil(s) and magnet(s) to transform electrical energy into mechanical energy just as our trains do. It does not use the split ring armature, but that is just a way to alternate the current to the windings to change the linear motion to rotational motion and this is not done on a loudspeaker. OTOH, the loudspeaker has to change directions X times per second.

The dynamics of a loudspeaker are such that if you apply twice the power (not voltage) to the coil, the speaker cone travels twice as far. Ideally, the travel is linear to the power. The power is equal to the voltage squared divided by the impedience/resistance of the coil as it moves through the gap.

So if 0.01 watt moves the cone 0.001", 0.1 watt moves it 0.01", 1 watt moves it 0.1", and 100 watts moves it 1" (long excursion surround). Like all things in nature, this represents an ideal situation and we must contend with such things as friction and slight changes impedience. (But, impedience is largely governed by frequency and NOT power levels.)

If we were to change the weight of the speaker cone (weight of the armature in a rotary motor), this would require a higher power level to accelerate the cone to the same distance, which is why I alluded to a power-to-weight ratio.

Now lets play a 100 Hz tone through our "linear motor." At 1 watt the cone moves in and out 0.1" and at 2 watts it moves 0.2". But even at 2 watts it is still doing it 100 times a second, as it does at 10 watts or 100 watts!

_*THEREFORE, with twice the power, the cone moves twice as far in the same time frame, and so it must travel twice as fast.*_ So doubling the power doubles the speed of the cone, just as it doubles the speed of the dynamic motors in our trains. And the power is based on the square of the voltage.


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## Totalwrecker (Feb 26, 2009)

Is speaker cone speed instantaneous or does it accelerate/decelerate?


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## toddalin (Jan 4, 2008)

It is almost under continual acceleration, both positive and negative. I imagine that like a piston in an engine, there may be a brief period (nanoseconds?) when it stops at the top and bottom of travel before changing direction.

Recognize that a rotary motor is also under continual acceleration, as is anything that moves in a rotary motion.


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## Greg Elmassian (Jan 3, 2008)

I agree with the speaker analysis, except that it is very nonlinear at the ends of the excursion.

Not sure about your motor theory, but maybe I missed your "square of the difference" term before, I swear you said linearly proportional.

Would be interesting to test the theory, but the power of 2 difference is definitely not linear and does roughly match what I have observed without actually going back and calculating.

I know you can find linear tables of motor speed vs voltage, but we are not talking unloaded motors, and if you look up torque vs. voltage vs. rpm you see that things are indeed not normally linear:









I really am less interested in the exact theory, since different motors perform differently, and there are variations in the pulse width frequencies used by modelers, just wanted to communicate to people that the difference in top speed between 14 volts and 18 volts can be a lot, and they cannot linearly estimate it. Until your theory is proven to me with a number of locos and good empirical data, it remains your theory in my mind. My position is real results, or a proven theory.

Regards, Greg


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## chuck n (Jan 2, 2008)

It's supposed to rain tomorrow, but as soon as I can I'll take some different engines out and measure speed and voltage. I'll test LGB Forney, Bachmann Shay, and an USAt SD70 MAC. If someone wants to do it first, go for it. I believe in data. Theory is fine, just ask the navigators before 1492.

Chuck


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## toddalin (Jan 4, 2008)

Thanks Chuck.


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## Totalwrecker (Feb 26, 2009)

toddalin said:


> It is almost under continual acceleration, both positive and negative. I imagine that like a piston in an engine, there may be a brief period (nanoseconds?) when it stops at the top and bottom of travel before changing direction.
> 
> Recognize that a rotary motor is also under continual acceleration, as is anything that moves in a rotary motion.


No sorry, the piston does not work as an example, a piston moves at a variable speed because it's motion is controlled by the circular motion of the crankshaft. At the top and bottom horizontal travel controls the piston's vertical motion, slowing it compared to travel in the middle of the stroke.
Traveling as you say, I fear the piston would self destruct.

Must be a better example somewhere.... carry on.

Are Stepper motors accelerating? .... can't realize that.

John


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## toddalin (Jan 4, 2008)

Totalwrecker said:


> No sorry, the piston does not work as an example, a piston moves at a variable speed because it's motion is controlled by the circular motion of the crankshaft. At the top and bottom horizontal travel controls the piston's vertical motion, slowing it compared to travel in the middle of the stroke.
> Traveling as you say, I fear the piston would self destruct.
> 
> Must be a better example somewhere.... carry on.
> ...


As I noted, anything that rotates is under acceleration because the vector direction is continually changing.


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## chuck n (Jan 2, 2008)

Todd and Greg:

I just finished my data collection. I'd like to thank you both for this discussion. I learned a lot.

First, I learned that my multimeter was broken. After getting a replacement I ran the engines at 2 volt increments.

There are some grades on my layout so the running voltages are higher than the starting voltages. I'd still be out there waiting for the Forney to go around. It just barely made the crest at 18v.
I used a Bridgewerks powersupply.

LGB Forney (0-4-4)

started moving at about 13v, needed 18v for complete circuit.
amps 0.7

Bachmann Shay (2 truck)

started moving at about 10v needed 14v for complete circuit.
amps 0.7

USAt SD70MAC

started moving at about 6.7v, started measurements at 10v.
amps 1.4.


Sorry the data table didn't export with proper layout. You are stuck with the graph.











The time is how long it took for one revolution of the layout (87').

I was surprised by the voltage needed to move the Forney and the small amount needed for the USAt. 

It is definitely not a linear relationship between speed and voltage. I guess that for having atotal lack of knowledge on the subject I had assumed a linear relationship. So much for my assumptions.

Chuck


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## toddalin (Jan 4, 2008)

Certainly is an exponential curve. The SD gives a nice representation. I'll have to look at the data.

Thanks


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## chuck n (Jan 2, 2008)

Todd, if you want the table to play with, PM me with a nonMLS email address and I'll send it as an attachment.

Chuck


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## Totalwrecker (Feb 26, 2009)

toddalin said:


> As I noted, anything that rotates is under acceleration because the vector direction is continually changing.


Pistons don't rotate. Their velocity varies.
The sky is blue.

John


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## toddalin (Jan 4, 2008)

Chuck,

I think that the data for your SD pretty well fits the curve I've predicted and you can in fact use one speed and voltage within reason until you start to reach the limitations of the motor (highest voltage). Like a speaker, at the highest reaches, things heat up and you undergo dynamic compression, as well as other anomalies like increased friction due to thermal expansion.

OTOH, it is also possible that at the higher voltages you were get more wheel slippage (especially with grades), so the motor and wheels are actually running faster than the engine speed would indicate. Probably a combination of factors.

I just pulled it off the chart as close as I could so the values may be off a bit, which could equate to a couple mph. 

So, if it is 87 feet around and takes 88 seconds, the engine is traveling 87 feet / 88 sec = 0.9886 feet/sec / 5,280 feet/mi x 60 sec/min x 60 min/hr x 29 (scale) /1.

19.5 mph at 10 volts.

For 12 volts, (12^2/10^2) x 19.5 mph = 28 mph. You read 65 sec = 26.5 mph.

For 14 volts (14^2/10^2) x 19.5 mph = 38.2 mph. You read 48 sec = 35.8 mph.

For 16 volts (16^2/10^2) x 19.5 mph = 49.9 mph. You read 35 seconds = 49.1 mph.

For 18 volts (18^2/10^2) x 19.5 mph = 63.1 mph. You read 30 seconds = 57.3 mph.

For 20 volts (20^2/10^2) x 19.5 mph = 78 mph. You read 23 seconds = 74.8 mph.

For 22 volts (22^2/10^2) x 19.5 mph = 95.7 mph. You read 21 seconds = 81.9 mph.

For 24 volts (24^2/10^2) x 19.5 mph = 113.9 mph. You read 18 seconds = 95.6 mph.


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## toddalin (Jan 4, 2008)

Totalwrecker said:


> Pistons don't rotate. Their velocity varies.
> The sky is blue.
> 
> John


The reference was to the stepper motors that do rotate.

A piston accelerates from a stop to an certain speed then slows to a stop (negative acceleration), changes directions, and repeats the process. The piston continually changes its speed though the stroke so is under continual acceleration. 

Any change of speed or direction denotes an acceleration.


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## Greg Elmassian (Jan 3, 2008)

The original point was that at the top end of the motor voltage that the voltage vs. speed is nonlinear.... seems you now say that too.

Speakers go nonlinear at max or min excursion due to the non linearity of the spring (the surround) at these excursions and also that the voice coil is leaving the magnetic flux at one end or the other.

The inefficiency at this point is also paired with higher drive currents, and that causes heating.

So is it now ok to say that the difference in speed of 14v vs 18 or higher can make a larger, non linear difference in speed?

Is this not where we started?

Greg


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## toddalin (Jan 4, 2008)

Greg Elmassian said:


> The original point was that at the top end of the motor voltage that the voltage vs. speed is nonlinear.... seems you now say that too.
> 
> Speakers go nonlinear at max or min excursion due to the non linearity of the spring (the surround) at these excursions and also that the voice coil is leaving the magnetic flux at one end or the other.
> 
> ...


 
You are unbelievable!

The original point is that the speed is proportional to the power and the OP used 14 to 18 volts. I've (Chuck) demonstrated that the equation would certainly seem to hold true in that range.

Again, you love to twist stuff.

We don't know that Chuck's nonlinearity at the high end wasn't due to wheel slippage or any number of other things that could have affected his testing. Remember that his loop is only 87 feet around. While the rolling resistance through the straights is directly proportional to speed, the friction through these curves increases at some exponential factor with speed. The additional friction slows the trains, and of course the equation is based solely on motor speed so doesn't consider this. Certainly the predicted top speed is right in line with what you've measured on your engines.

I came up with this based on the Scientific method (i.e., make observations, fit a hypothesis to those observations based on real world experience/research, test the hypothesis with real data, and modify it as necessary). 

It seems to hold true and you can't handle that, so need to cast aspersions rather that providing your own data to support your views.


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## chuck n (Jan 2, 2008)

There was no wheel slip.

Chuck


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## Randy Stone (Jan 2, 2008)

Knowing the different speeds by voltage is great but I'd like to know the lbs of pull at the coupler per the different voltages.


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## Dr Rivet (Jan 5, 2008)

Randy

Go buy an inexpensive fish scale [if you don't already have one]. Attach it to something rigid. Attach the scale to the locomotive coupler. Read the pounds / ounces on the scale as you raise the voltage until you get wheel slip. Record your results and post on this thread.


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## chuck n (Jan 2, 2008)

Randy:

Jim's suggestion is a good one, but having done several parts of the exercise, I don't think that you will find much difference with tractive effort and voltage.

When I measured the tractive effort of a number of my engines with a fisherman's scale I found that the TE went up rather quickly and stayed there until wheel slip.

Based on my experience, I think that it will take two people: one to read and adjust the voltage and a second to read the scale.

My thought is that adding weight to a train will only increase the voltage needed to start it moving. I don't think that a static test, pulling against a spring, will show that much.

Chuck


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## SD90WLMT (Feb 16, 2010)

Actually Chuck the static test is the most common method used to determine pulling effort..or tractive effort..and is typical of the tests performed in the loco reviews we read in GR...

Because I was running tires..and did not want them ruined in a slipping condition..I developed a tractive effort test car that can move/ be pulled behind a loco and perform the test in a moving state...the drag is adjustable to increase the load a loco pulls against...

I have run comparison tests hand holding a scale behind a running loco. ...and recorded higher readings than I can get from the test car... 

I will be doing some strictly bench only tests for this in the near future...

Dirk


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## chuck n (Jan 2, 2008)

I agree Dirk! My comment about pulling weight was directed at the possibility that starting voltage may vary with load. I don't see how you can do that with a static test.

One other comment. As I remember when I did my static TE tests. I didn't have to increase the voltage very much until I encountered wheel slip. Certainly, nowhere near full power (>20v).

Chuck


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## SD90WLMT (Feb 16, 2010)

Well I guess there are some varibles here....

Starting .....

Against a spring scale...
Against a train with loose couplers...
Against a train with taunt couplers...

All different conditions...

Measuring tractive effort is in its own category I think....generally measuring the maximum pulling abilities of a given loco...

Dirk


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## SD90WLMT (Feb 16, 2010)

I guess a point I did not clarify well..

A short train with tight couplers may "present" a greater load to move..initially...then a longer train set with loose couplers..yet once up and running the longer train may still generate overall a higher load to the power applied..

Dirk


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## chuck n (Jan 2, 2008)

Dirk:

All good points, which is why I think Randy's request will be hard to accomplish. Perhaps it would be a good science fair project for someone's child.

Chuck


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## East Broad Top (Dec 29, 2007)

I'll keep it in mind for when Suzi's in middle school. 

Later,

K


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## chuck n (Jan 2, 2008)

Randy:

The wheel slip starts almost immediately. This morning I took out the USAt SD70MAC with my scale and multimeter. The engine started to creep at a little over 4v. If I put the scale on the coupler, while it was moving, it would immediately go into wheel slip.

If I started it with the spring on, and gradually increased the voltage, I got wheel slip at about 6.4v with a TE of 2.5 lbs.

At 6 volts the engine is barely moving.

Chuck


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## SD90WLMT (Feb 16, 2010)

From your former chart for T.E. Chuck, what did you previously record for your SD70Mac..?

ThX

Dirk


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## chuck n (Jan 2, 2008)

It previous reading was 4.5 pounds. I better go out and try again.

It was a different engine the first time. That was the Rio Grande (silver and yellow) fallen flag version. This was a Black Rio Grande.

Chuck

Just tried the other engine. About the same, between 2.5 and 3 pounds TE. When I did the original test, the silver and yellow engine was almost new. Both of these engines have been run a lot. I wonder if the running, polished the wheels?


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## Randy Stone (Jan 2, 2008)

So maybe tractive effort measured from a standstill isn't the same as load pulling ability for a said loco on say a 2% grade. What I am getting at is, will a loco with a 18.5 volt battery pull more frt cars up a 2% grade than it would pull with a 14.8 volt battery? This assuming the loco is weighted heavy enough to not slip the driving wheels.


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## chuck n (Jan 2, 2008)

Randy:

Tractive effort is tractive effort. Any grade will reduce the effective TE. I doubt if the difference in voltage will affect the TE. It will affect the speed.

There is a reason it is measured on level track. That is the maximum. Any grade will reduce that value.

Get yourself a fisherman's scale and pull a string of cars on the level and then on your 2% grade. See the difference. The train on the grade will have a larger drag.

Chuck


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## toddalin (Jan 4, 2008)

Randy Stone said:


> So maybe tractive effort measured from a standstill isn't the same as load pulling ability for a said loco on say a 2% grade. What I am getting at is, will a loco with a 18.5 volt battery pull more frt cars up a 2% grade than it would pull with a 14.8 volt battery? This assuming the loco is weighted heavy enough to not slip the driving wheels.


 
It's ~56% more power, so yes.


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## chuck n (Jan 2, 2008)

If wheels slip at 14v they will still slip at 18 v.

Until wheel slip the only difference will be speed.
Chuck


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## SD90WLMT (Feb 16, 2010)

From experience with meters on board...
Pulling 70 cars up a 2.6% grade, was so heavely loaded that ..yes the speed was reduced...and the loco could only "see" 12- 14 volts...while the amps went ballistic. ...I tried to increase throttle input...which at that point only increased the voltages...with out any effective change to the locos output...put continued to increase current draw...

You can not "force" something into something where it will not fit....

You can not force more work out of a loco ..once it has reached it's limit of abilities....

Adding a higher voltage battery under such loaded conditions will not generate greater results!!

Go try it...in real world terms...use on board meters to know precisely what is occuring while it takes place....

Oh..don't ferget to have loads of Fun while playing trains....

Thanks for your efforts again Chuck!!! I'll add more as I can here...

Dirk


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## Randy Stone (Jan 2, 2008)

So, if a loco is designed to run on 22 volts, you're saying it would not do better pulling 10 cars up a grade on 18 volts when it would not pull them up the hill with a 14 volt battery.


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## SD90WLMT (Feb 16, 2010)

Really only way to know is to repeat conditions ..using an on board volt meter..
What you really need to know is what it might be running at under your load of 10 cars..
If it runs right at the 14 volts, then an increase may help..but after changing the supply voltage you need to re-run again and see what or if it is using any of the available increased battery supply..

Too many of these type conversations are broadly generic...and readily could be solved with the proper use of test equipment ..rather than lots of subjective "I think so's"...that never readily address the issue...

I subscribe to try ..but test...it is not a monthly magazine tho!!

I will continue to test here as I develope more ideas!!!

Dirk


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## Greg Elmassian (Jan 3, 2008)

Todd: I'm unbelievable? You must be senile!

Here are the 2 first lines of the first post by YOU

In another thread, it was asked how much faster a engine would go at a given voltage relative to another voltage (18 volts vs 14 volts in that case). This represents an increase of just 4 volts.

I proposed that the speed is directly proportional to the square of the difference in voltage.

Where is the word power in these statements? You only say voltage.

Get a checkup.

Greg



toddalin said:


> You are unbelievable!
> 
> The original point is that the speed is proportional to the power and the OP used 14 to 18 volts. I've (Chuck) demonstrated that the equation would certainly seem to hold true in that range.
> 
> >> excess verbiage snipped


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## toddalin (Jan 4, 2008)

Greg, I guess this is just all beyond your comprehension level and I'll leave it at that.


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## chuck n (Jan 2, 2008)

Randy, I think that you might be mixing tractive effort with power. In my mind tractive effort is the weight, drag, an engine can pull before the wheels start to slip. Power is the ability of the motor in the engine to move it forward at a given voltage. In my test the Forney started moving at 12v. But it could not clear the hill until 18v. It easily made the grade at 20v. This had nothing to do with tractive effort. It was all related to the ability of the motor to turn the wheels against the grade and weight of the engine and pull the engine up at a given voltage. If the motor can't turn the wheels, it just stops, the wheels don't spin.

The more volts you have the more your engine can pull, until you add enough cars to get wheel slip which is when tractive effort raises it's ugly head.

I maybe all wet in this thinking, I'm sure some of our engineering friends will chime in.

Chuck


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## Greg Elmassian (Jan 3, 2008)

Sounds correct to me Chuck, one small addition, the more volts you have, the more current will flow, and thus the motor can generate more horsepower/torque.

Regards, Greg


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## SD90WLMT (Feb 16, 2010)

Ah yes..mmm...but...

Each motor is only capable of producing horsepower..up to its limit..

Randomly implying that continually increasing voltage to supply more current ..thus producing more power has limits also...

Once a motor has reached its limit..increases in voltage really only result in higher currents...which due to a horsepower limit ..the motor now is no longer able to produce even more power...all the current only turns into "heat".. and is unable to generate more rpm or power...

Some how I think that the realities of increased friction..increased drag..and mounting heat have been left out of this conversation...they do effect a motors performance as rpm and power are increased....
We can ask more of the motor...but it can't give everything we ask of it....

Ponder on...

D


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## Randy Stone (Jan 2, 2008)

Chuck, your comment is exactly what I believed all along. Unless the motor is only designed for 14.8 volts, putting an 18.5 or higher voltage battery in will not only increase speed, it increases power to pull larger loads.


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## Greg Elmassian (Jan 3, 2008)

Once moving, friction/drag is relatively constant. (linear with velocity)

Greg


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