# Power? Voltage versus amps



## Randy Stone (Jan 2, 2008)

OK, I think I finally got my head around this battery voltage/Amps equals Torque question in my mind.

Voltage gives the motor speed.
Current/Amps give the motor torque.

So, when picking a battery pack to run your train, you need to think about :
1.	What speed do you want out of the loco?
2.	What torque will you need out of the motor? (How many cars are you going to pull)
3.	What other accessories will be powered off the same battery pack? (multiple locomotives, multiple motors, Sound, smoke, Lights, Lights in passenger cars, etc)

Battery packs? On Cordless Renovations website’s HC series batteries are as listed below.

14.8 Volt 3000mah
14.8 Volt 6000mah
14.8 Volt 9000mah

18.5 Volt 3000mah
18.5 Volt 6000mah
18.5 Volt 9000mah
18.5 Volt 12000mah

22.2 Volt 3000mah
22.2 Volt 6000mah
22.2 Volt 9000mah

Now all along, all I heard was, (All you need is 14.8 Volts to power your train) Higher voltage only makes the train go faster.)

Well this is only true to a point.

A Shay, pulling 10 loaded logging cars up a 4% grade might bogged down and not be able to pull the load up the hill with a 3000mah battery pack, but install a 14.8 V 6000mah or 14.8 V 9000mah battery pack and the Shay could then pull the train up the hill.

So, the same Shay could have the 22.2 V 3000mah battery pack and still not pull the train up the hill.

Now, take the Shay and put it in the barn and hook an Aristo Craft Dash 9 to those same 10 loaded log cars and with the same 14.8 Volt 6000 or 9000mah battery pack and the Dash 9 might bog out and fail to pull the train up the same 4% hill the Shay pulled. Why, well the Dash 9 has (4 motors) compared to the Shays (1 motor). 

Simply put, if your locomotive will not break the drive wheels loose and spin them when pulling too heavy of a load up a hill, (Rubber tired drive wheels not included) then your battery pack doesn’t have enough amperage to provide enough current for the motor to create enough torque and may need more voltage.


----------



## Dan Pierce (Jan 2, 2008)

Also remember that the ratings are for a specified time.. if a battery is rated at 2000mah, that is 2 amps for 1 hour, or 1 amp for 2 hours. So if an engine is maxing out a batteries current, run time will be minimal!!!!

SO, it is voltage, current and TIME.


----------



## SD90WLMT (Feb 16, 2010)

Some how a controller device is in play here..a decoder..or track throttle..
You can turn the knob all day long..as much as you want..yet..you cannot force the depicted overloaded loco to work harder..regardless of volts or amp behind the given supply..
The motor or motors will over load..get hot...draw more current..read that as a higher load..till it all stalls and stops..forcing more volts may not deliver more..but will create more heat..increase the amps draw as a result...even cook the motors if left in such a condition..fuses blowing if on board...

The mah rating on batteries is like an air compressor..it may only have a 5 or 8 gallon tank..running out quickly..or a 50 or 120 gallon storage tank..lasting far longer before refill/recharge..


----------



## Randy Stone (Jan 2, 2008)

Yes, we have used the mah of a battery pack to determine length of time the train will run. But ultimately, it is the load being applied that determines the time provided.


----------



## East Broad Top (Dec 29, 2007)

> ...A Shay, pulling 10 loaded logging cars up a 4% grade might bogged down and not be able to pull the load up the hill with a 3000mah battery pack, but install a 14.8 V 6000mah or 14.8 V 9000mah battery pack and the Shay could then pull the train up the hill....


No. Not the case. The capacity of the pack has nothing to do with how much a train can pull. It _only_ impacts the length of time the train can pull it. The Shay is going to draw the same current pulling the train regardless of what's powering it. If your Shay is bogging down pulling 10 loaded log cars up a grade, the only way to get it not to bog down is to give it more voltage. Upgrade from 14 volts to 18 volts if that's the case. 

Think of it in terms of a flashlight. If you use a AA battery, it will make the bulb glow at a given brightness. Installing a C or D-sized battery won't make the bulb burn any brighter, it'll just allow it to burn longer. 

When figuring out which battery you need for your train, get an old-fashioned analog throttle and hook it up to your railroad. Run the locomotive with a typical train. If the loco bogs down, what's the first thing you do? Increase the throttle until it gets moving again. That's increasing the _voltage_. Measure the voltage going to the tracks with the train running at that speed. That's the minimum voltage you're going to need in your battery pack. You can add more, but that's the minimum. 

Now, look at the locomotive and where you're going to be putting the battery. How much space do you have available. Find the pack that's closest to the voltage you need without going under, and look at the sizes of the packs of the different capacities. Pick the biggest one that will fit the space. You will have a battery that will provide the voltage needed to get your trains up and over your hills, and as much capacity as you can physically fit in the locomotive so it can do it for the longest possible amount of time.

Later,

K


----------



## Randy Stone (Jan 2, 2008)

East Broad Top said:


> No. If your Shay is bogging down pulling 10 loaded log cars up a grade, the only way to get it not to bog down is to *give it more voltage*. Upgrade from 14 volts to 18 volts if that's the case.
> 
> 
> 
> ...


WHAT????? NOW, you're telling me it takes more voltage to pull a heavier load? How many times you and others have said increasing the voltage only increases the speed of the loco.


----------



## SD90WLMT (Feb 16, 2010)

Increasing voltage does not mean a train will run faster...esp if it has reached that point of motor overload vs. power available. Again..a case where in forcing more volts as increased feed power will not generate results expected or wanted.


----------



## Semper Vaporo (Jan 2, 2008)

You have to remember the linear relationship of Volts, Amps, and Resistance. You can have a million Amps available, but you can't make them go anywhere unless you have some Volts behind them pushing. It takes 1 Volt to push 1 Amp through a resistance of 1 Ohm... 2 Volts to push 1 Amp through 2 Ohms... or 1 Volt to push 2 Amps through 1/2 Ohm. 

"E=I*R" (where E is Electromotive Force in Volts, I is current in Amps, and R is Resistance in Ohms).

Speed, Torque and burning up your motor are also all interrelated, but not by any sort of common formula, it depends on who made the motor.


----------



## Totalwrecker (Feb 26, 2009)

Randy Stone said:


> WHAT????? NOW, you're telling me it takes more voltage to pull a heavier load? How many times you and others have said increasing the voltage only increases the speed of the loco.



Think of driving your car up a hill, by your self acceleration up hill is easy, add a full car of adults and you'll probably need to give it more gas to achieve the same results. While on a flat, that would increase speed, the hill converts it to power to overcome the resistance. See?
More gas = more volts. A bigger gas tank has no effect on the motor, just time.
John


----------



## chuck n (Jan 2, 2008)

Randy

Let us say your engine will pull a 10 car train on the level at a reasonable speed at 12 volts. When the train gets on the grade it stops. The 12 v to the motor can't turn it against the added drag of the cars. You are now adding some of the weight of the cars to the overall friction of the trucks. Increasing the voltage will get the train moving as long as the drag isn't too much. If the drag exceeds the drawbar pull of the engine, when increasing the voltage, the wheels will start to spin and the train just stays in place.

This is what I did when I measured the draw bar pull of some engines. I held the engine back with a fisherman's scale and raised the voltage until the wheels started to spin. I then read the weight on the scale to determine the drawbar pull.

It would be interesting to look at the amps as the voltage is increased and what happens to the current draw when the wheels start to slip. My guess is that the current draw would increase until wheel slip and then it would fall off. I'm not an electrical engineer, so I really don't know what would happen, just a guess.

I am using Cordless Revolutions 18.5 v/5600mah batteries. I use them on a Bachmann K-27, AristoCraft Mikado and Mallet, and my Delton Doozie/goose. I get over two hours of run time. I've never run them long enough, in a single session to run out of power. Occasionally, at a later session the engines will stop. I've never timed the total time for the battery to die.

I also run a USAT Hudson with that battery. Pulling 8 heavyweights or 8 USAT streamliners it goes a little slower than I'd like it to. I'm thinking about getting a 22v battery for it.

I'm using one of CR's 14v batteries for my Accucraft Goose #2. Top speed for that is about 20 scale mph, just about right.

I support the recommendation of some of the others. Get hold of an analog power supply, make up a typical train for your engine and then run it at a speed that you like. Then measure the voltage at that speed. If it were me, I'd add a couple of volts for good measure and then look at batteries that are close to that total.

As was said earlier, MAH, is just the size of the fuel tank. The bigger the number the longer the run time.

Chuck


----------



## Randy Stone (Jan 2, 2008)

Guys, I agree with all of you. BUT, for years on here, all I've heard was 14.8 volts was all you need. all a 18.5 or 22.2 volt pack is going to do is give you more speed.


----------



## chuck n (Jan 2, 2008)

Randy

Yes, more volts increases speed. 

In the analog DC world a train that draws 2 amps, if powered by a 1 amp power supply, will run until the power supply over heats and throws the thermal circuit breaker or fries the components. I'm sure something similar happens with batteries, but that is beyond my knowledge. 

Chuck


----------



## Greg Elmassian (Jan 3, 2008)

To Kevin's "rebuttal", what he says is true, EXCEPT if you cannot get enough amperage that the loco "desires".

There is resistance in the circuit, the wires, the motor, the controller, etc.

So, sometimes you DO need a higher amperage pack if you are running a load that requires a lot of amperage.

Also, as another person mentioned, your battery will not perform well if you are trying to pull a large amount of current. Rule of thumb, you want to be pulling no more than 1/2 C (the mah capacity is C) under load.

Yes (and the airplane guys need NOT point up the aircraft usage here) you can run 2 amps from a 2 amp hour battery, theoretically, but often the battery really does not perform well at that load, and most times the output voltage drops significantly, like any other system in the world, running at max load.

So stay from the extremes of the maximum specs on the motor, controller and the battery and all will be well.

Greg


----------



## toddalin (Jan 4, 2008)

Randy Stone said:


> Guys, I agree with all of you. BUT, for years on here, all I've heard was 14.8 volts was all you need. all a 18.5 or 22.2 volt pack is going to do is give you more speed.




And you believed them? 

In mechanical terms, power is work over time.

For the same power, you can move a load at a certain speed, twice as fast with half the load, or twice the load, but half as fast. Except for things like friction and heat that come into play and the limitations of the wiring/circuit to carry the necessary current, it's a linear relationship.

In electrical terms, power is the square of the voltage divided by the resistance (V^2/R).

So a certain voltage will move a load at a certain speed, and half the power (based on square of the voltage) will move it at half the speed and twice the power (based on square of the voltage) will move it at twice the speed. Alternatively, the same power will move half the load twice as fast, or twice the load half as fast. Again, except for..., a linear relationship.

You could do the same exercize using amps where you increase/reduce the resistance thereby changing the power. But, while we can easily change the power by changing the voltage to the track or motor, it is relatively impossible to change the power by varying the resistance, unless you are controlling your trains with an old fashion rheostat with a constant voltage supply.

In the past, I had presented a discussion on the relationship of voltage to speed and why a little difference in voltage makes a fast train so much faster, but it takes a lot of difference in voltage to make a slow train much faster. I had also proposed a new method for speed control using these principles with the throttle based on the inverse square of the voltage rather than ramping up the voltage in a linear fashion.

BTW, 18.5 volts will move a train 56% faster than 14.8 volts. Alternativey, you can pull 56% more load at the same speed.

22.2 volts will move the trains 125% faster or pull 125% more load at the same speed as 14.8 volts.


----------



## East Broad Top (Dec 29, 2007)

Greg Elmassian said:


> So, sometimes you DO need a higher amperage pack if you are running a load that requires a lot of amperage.


That would be the case if the output of the battery is somehow governed at levels below what the locomotive would draw. I have heard that some rechargable NiMH cells are limited to a half-amp maximum draw, though I never experienced that with the NiMH cells I used. Many Li-Ion packs you can purchase commercially will tell you in the specs what their maximum current draw is. (The 2600mAh packs I use have a continuous draw current limit of 5 amps. The 4800mAh packs are governed at 6 amps.) The NiCad and NiMH packs designed for racing R/C cars I used to use? No limit there. The high-end car motors draw 10 - 20 amps, and the batteries are designed to provide it. 

Later,

K


----------



## TonyWalsham (Jan 2, 2008)

So far no one has made any mention of loco gearing.

A low geared loco will always require more voltage to move any sort of load at the same speed as a higher geared loco will require.

Bachmann locos will generally perform adequately on 14.8 volts whilst Big locos like a Dash9 or SD45 will require higher voltage because they are lower geared so they can pull greater loads.
Those facts screw up generalisations like nobodies business. There is no one voltage that suits every loco.


----------



## Randy Stone (Jan 2, 2008)

Exactly Tony. PLUS, the Dash 9 has 4 motors. You have to consider how many motors and accessories will be pulling power from the power source.


----------



## TonyWalsham (Jan 2, 2008)

Yes, accessories have to be considered, but in general it doesn't matter really how many motors a loco has. They will still only consume the amps they need, to shift a given load.


----------



## Greg Elmassian (Jan 3, 2008)

and enough voltage to run at a prototype speed.


----------



## SD90WLMT (Feb 16, 2010)

Randy...the motor counts do not add to each motor..
1 motor..14 volts
2 motors 14 volts
6 motors. 14 volts..

The load across any given motor amount is related to and because of the the train load..how many cars...get pulled!
Accessories add to the load total running amps...


----------



## CliffyJ (Apr 29, 2009)

Some batteries (like LiPo's) have max discharge rates (in mA). Maybe that was affecting things with your 3,000mah battery Randy?


----------



## Randy Stone (Jan 2, 2008)

SD90WLMT said:


> Randy...the motor counts do not add to each motor..
> 1 motor..14 volts
> 2 motors 14 volts
> 6 motors. 14 volts..
> ...


Yes, the volts are the same until you start pulling on them. Then the amps go up and the volts go down. This is why you need more volts. Why else did the automotive industry go to 12 volts? Even with 12 volts, when you hit the starter on your car, the amps go up and the volts go down.


----------



## Greg Elmassian (Jan 3, 2008)

Randy, you got it. One thing people are forgetting, that under heavy loads, more voltage is lost in the wiring, it's not like the motor is 1000 ohms and your wiring is 1 ohm so voltage lost is miniscule, the resistance of the motor gets closer to the resistance of the wiring.

Even a 1 volt drop to a motor provides a noticeable (not huge but noticeable) drop in speed.

This is indeed why automotive companies want to go to 48 volts, thinner wire, less % of voltage lost in the wiring.

Greg


----------



## SD90WLMT (Feb 16, 2010)

Ahh..yes..correct Randy...its the pulling that cannot be forced however..
Once the tipping point...or balance point is reached and exceeded...the volts will drop...no matter how much you have to push into a motor...

Load creates higher amp draw
Higher amp draw creates heat
Heat increases resistance..which

Creates more amp draw..

A vicious cycle is exceeded...the load has become the master..
Not increasing the throttle..thus more volts...

Most don't run at this level of demand...at least I hope not..


----------



## toddalin (Jan 4, 2008)

Greg Elmassian said:


> Even a 1 volt drop to a motor provides a noticeable (not huge but noticeable) drop in speed.
> 
> 
> Greg


~12% difference from 17 to 18 volts.

~10% difference from 21 to 22 volts.


----------



## toddalin (Jan 4, 2008)

SD90WLMT said:


> Load creates higher amp draw
> Higher amp draw creates heat
> Heat increases resistance..which
> 
> ...



A higher load requires more voltage to do the work at the same/expected rate.

Amps = Volts/Resistance.

As voltage goes up, and resistance remains fairly constant (within the range of intended use), the amps go up as a consequence.

Simply adding load should not require more amps if you let the train run at its "comparable" slower speed (i.e., work/time).


----------



## chuck n (Jan 2, 2008)

Gentlemen

I have tried to follow this discussion, but my knowledge of electricity is minimal. I can usually figure out how to do something simple, but I had a hard time with some of the written comments. I understand that speed increases with voltage, I had no concept of how that effected the current.

Yesterday afternoon between rain showers, I took out a Bachmann 2-truck Shay and a USAt SD70Mac. I measured the speed and amps as I increased the voltage. The speed was measured in seconds needed for one complete lap around my mainline (87'). I tried to get a reading on the amps as the engines went up grade and down grade. 

I have put the data on the following plots. The numbers in parentheses are the amps. The first number in each group is the down grade amp reading and the second is the up grade reading. For simplicity I only show the values for the lowest and highest voltage. 









When it stops threatening storms, I want to repeat this with a load. I think I will take out 4 Accucraft San Juan coaches. They are heavy and roll easily. This should illustrate the effect of weight on the up and down grade, I hope. 

This exercise has helped me visualize what is going on, maybe it will help others.

Chuck

Something that surprised me was that the USA diesel started moving at a lower voltage that the Bachmann Shay.


----------



## toddalin (Jan 4, 2008)

chuck n said:


> Gentlemen
> 
> I have tried to follow this discussion, but my knowledge of electricity is minimal. I can usually figure out how to do something simple, but I had a hard time with some of the written comments. I understand that speed increases with voltage, I had no concept of how that effected the current.
> 
> ...


Your test bear out what I've said all along, the speed is proportional to the square of the voltage, and the amps follow along accordingly.

Note those nice exponential curves. 

~10.5 volts @ ~76 seconds.

So, at ~21 volts:

21^2 / 10.5^2 = ~4 times as fast

76 seconds / 20 sec = ~3.8 times as fast


----------



## Randy Stone (Jan 2, 2008)

toddalin said:


> Your test bear out what I've said all along, the speed is proportional to the square of the voltage, and the amps follow along accordingly.


Your getting back to speed. This thread has nothing to do with speed. It's about "TORQUE". More voltage means more torque. 
I created this thread to show that you need more voltage to provide more torque to pull more load. This has been verified by several responders, yet some of you are still stuck on only wanting to say more voltage only increases the speed.


----------



## toddalin (Jan 4, 2008)

Randy Stone said:


> Your getting back to speed. This thread has nothing to do with speed. It's about "TORQUE". More voltage means more torque.
> I created this thread to show that you need more voltage to provide more torque to pull more load. This has been verified by several responders, yet some of you are still stuck on only wanting to say more voltage only increases the speed.


No Randy, you are missing the correlation.

It is not about speed. It is about _power_. More voltage means more _power_. Based on the same load, proportionally, more _power_ creates proportionally more speed. And this _power_ is based on difference in the square of the voltage.

But it could just as well be applied the other way and you can use that increase in _power_ to haul a proportionally larger load at the same speed or some proportionally intermediate load at some proportionally intermediate speed.

Also, based on Ohm's law, we can calculate the resistance of the motor (but the amp data is a bit broad).

Still, if we assume that the motor pulls 2 amps at 20 volts, 20 volts / 2amps = 10 ohms). And this is confirmed at 10 volts and 1 amp (10v / 1 amp = 10 ohms).

So again, as I had contended before, over the working range of the motor the resistance remains fairly constant.

One would expect to see a slight slowing at greater speeds from the predictions based on voltage when one tries to use anything other than a straight line for testing because curves introduce non-linear forces into the equation. For example angular momentum through a curve involves the square of the velocity introducing a non-linearity into the equation.


----------



## chuck n (Jan 2, 2008)

I completed my experiments this afternoon. I brought out four brass Accucraft J&S San Juan coaches and pulled them with the Shay and SD70MAC. Each car weighed was 10.5 pounds, so the total weight the engines were pulling 42 pounds. The cars have ball-bearing wheels, so the rolling resistance is very small. It is just the work of pulling that weight around.

The added weight had a significant effect on the Shay, not so much on the diesel.

The diesel would move at 10 and 12 volts, but stopped on the grades. I had to stop the diesel at 20 volts because it was going too fast for the coaches.

Here is the final graph.










This answered some of my questions, maybe it answered some of yours.

Chuck


----------



## Greg Elmassian (Jan 3, 2008)

it would be neat to have the volts and amps on both of those curves, but you have done a lot already Chuck. Interesting the much higher voltage needed to have the same speed, since most of the work is lifting in your case, due to the ball bearings, there's not much load keeping it rolling.

Greg


----------



## toddalin (Jan 4, 2008)

chuck n said:


> I completed my experiments this afternoon. I brought out four brass Accucraft J&S San Juan coaches and pulled them with the Shay and SD70MAC. Each car weighed was 10.5 pounds, so the total weight the engines were pulling 42 pounds. The cars have ball-bearing wheels, so the rolling resistance is very small. It is just the work of pulling that weight around.
> 
> The added weight had a significant effect on the Shay, not so much on the diesel.
> 
> ...


Totally expected.

That's because the added cars represent a bigger proportion of the total load for the shay.

With the Mac, 14 volts versus 12 volts represents the addition of 36% to the load at the ~ same speed (i.e., 14^2 /12^2).

With the Shay, the load represents a 92% addition at ~ the same speed (i.e., 18^2 / 13^2).


----------

