What does power have to do with the speed calculations of a Train?

[Tim Leijten]

Hi there again,

I have been wondering why train simulators(and trains themself) have a kW specification, AND a maximum force indication.
If I would calculate the speed of a train using some formula, then where is the power value used for?
Does it have something to do?
Or is only the maximum force used for speed calculations?
And I also have another question, I understand trains have a tractive effort curve, where the amount of force at what speed is specified:
Is it possible to calculate such a curve, how?
I am trying to understand train physics, so I hope someone can help me,
Thank you very much!

 

[kuruman]

Here is an educated guess - I didn't even know that there is such a thing as "train physics."

Locomotives have to overcome the power that is dissipated by all sorts of non-conservative forces, friction in all moving parts, air resistance etc. in order to get the train moving and then to maintain a speed. They should also be able to overpower gravity when going uphill. There is probably a table or at last a rule-of-thumb somewhere that indicates what to expect in terms of these forces depending on how many cars and of what weight are attached to the train. So the power rating of the locomotive is useful when one has to decide how many locomotives one has to use to pull a train of a given length and weight.

 

[Tim Leijten]

Here is an educated guess - I didn't even know that there is such a thing as "train physics."

Locomotives have to overcome the power that is dissipated by all sorts of non-conservative forces, friction in all moving parts, air resistance etc. in order to get the train moving and then to maintain a speed. They should also be able to overpower gravity when going uphill. There is probably a table or at last a rule-of-thumb somewhere that indicates what to expect in terms of these forces depending on how many cars and of what weight are attached to the train. So the power rating of the locomotive is useful when one has to decide how many locomotives one has to use to pull a train of a given length and weight.

Thanks for your reply!
Well, I don't think there is something like Train Physics either, but it just seemed like a good term to use, so I did. ;)
I understand what you are saying, so Power(kW) isn't used in calculations(e.g. a game) but more used as an indication of what train could take what load.
So one could simply calculate the power required to pull wagons, and then take enough locomotives to get enough power.
But if I understand needed in speed calculations in games(e.g. a Train Simulator)?thanks

 

[jbriggs444]

I have been wondering why train simulators(and trains themself) have a kW specification, AND a maximum force indication.

At low speeds, the train will be traction-limited. Usable power <= force * velocity. Too much power and you just spin the wheels.
At high speeds, the train will be power-limited. Required power >= force * velocity. Too little power and you can't provide the force.

 

[Tim Leijten]

At low speeds, the train will be traction-limited. Usable power <= force * velocity. Too much power and you just spin the wheels.
At high speeds, the train will be power-limited. Required power >= force * velocity. Too little power and you can't provide the force.

Thanks for your reply, but I really still don't understand.
Could you maybe explain it in a way even a 7y old can understand?(Ofc not meant litteraly, just in a bit more easy way)Thanks!

 

[russ_watters]

I have been wondering why train simulators(and trains themself) have a kW specification, AND a maximum force indication.
If I would calculate the speed of a train using some formula, then where is the power value used for?
Does it have something to do?
Or is only the maximum force used for speed calculations?
And I also have another question, I understand trains have a tractive effort curve, where the amount of force at what speed is specified:
Is it possible to calculate such a curve, how?
I am trying to understand train physics, so I hope someone can help me,

The second part of your post is the answer to the first part!

Do you know the equation for translating torque to linear force?
Do you know the equation for rotational power?

Combine them!

 

[Tim Leijten]

The second part of your post is the answer to the first part!

Do you mean that the power is actually the force at a speed of the curve?(I am probably wrong)

Do you know the equation for translating torque to linear force?
Do you know the equation for rotational power?

I believe it is:
Torque = force * length (not sure)
And
T = r x f

Combine them!

But i don't really have a clue how i would combine them.

Could someone help me?

 

[Drakkith]

Thanks for your reply, but I really still don't understand.
Could you maybe explain it in a way even a 7y old can understand?(Ofc not meant litteraly, just in a bit more easy way)Thanks!

Imagine you have a sports car. If you are going slow, you can floor the gas pedal and the engine will provide so much torque to the wheels that they just start spinning and you burn/peel out. The traction of the wheels is not high enough to prevent this. The available traction can be expressed as a force (I'll get back to this down below). However, if you don't have a high-performance sports car, but a used 4-cylinder with fewer liters than a quart of milk, the engine might not be able to output enough torque to the wheels to overcome traction. Just like traction, we can express the final output of the engine as a force instead of a torque (again, more about this below).

Now, for both types of cars, if you're already going at a high speed, then flooring the gas pedal may not do anything at all. In fact, if you're going fast enough, you can actually start to slow down if you go up a hill even if the pedal is floored. The engine simply cannot provide enough force to overcome wind resistance, rolling resistance, frictional losses, and gravity. Crucially, the power output of an engine is related to the amount of torque (and thus force) that is available to the vehicle. Obviously an engine with a higher maximum power output would be able to drive the vehicle at a higher top speed, but, more importantly, an engine with a higher maximum power will be better able to handle hills and slopes or changes that increase the required towing force (we don't usually run vehicles at their maximum speed, so that extra power can be used to overcome gravity or those changes).

Next, imagine that you hook up a trailer or two to your car. All of the various resistive and frictional losses in the trailers can be changed into a single value: the force required to pull them. And, if you've done any physics before, you'll remember that accelerating an object requires a force (f=ma) and that the required force increases as acceleration increases. So accelerating an object at twice the acceleration requires twice the force.

Remember what I said about traction and engine power being expressed as a force? This is where it comes in. We can combine both traction and engine output as a single value: tractive force. Tractive force is simply the lower of the two. The amount of tractive force determines how large of a load you can hook up and accelerate. Higher tractive force means a larger force available to tow with and a larger force means you can tow a larger load or accelerate the same load at a larger acceleration.

Here the wikipedia article on tractive force, and the paragraph about railed vehicles:

In order to start a train and accelerate it to a given speed, the locomotive(s) must develop sufficient tractive force to overcome the train's drag (resistance to motion), which is a combination of inertia, axle bearing friction, the friction of the wheels on the rails (which is substantially greater on curved track than on tangent track), and the force of gravity if on a grade. Once in motion, the train will develop additional drag as it accelerates due to aerodynamic forces, which increase with the square of the speed. Drag may also be produced at speed due to truck (bogie) hunting, which will increase the rolling friction between wheels and rails. If acceleration continues, the train will eventually attain a speed at which the available tractive force of the locomotive(s) will exactly offset the total drag, causing acceleration to cease. This top speed will be increased on a downgrade due to gravity assisting the motive power, and will be decreased on an upgrade due to gravity opposing the motive power.

Remember that tractive force takes into account both the amount of traction and the torque/force the engine is able to output. This means that tractive force generally decreases as speed increases, as the graph in the article shows.

 

[Tim Leijten]

Imagine you have a sports car. If you are going slow, you can floor the gas pedal and the engine will provide so much torque to the wheels that they just start spinning and you burn/peel out. The traction of the wheels is not high enough to prevent this. The available traction can be expressed as a force (I'll get back to this down below). However, if you don't have a high-performance sports car, but a used 4-cylinder with fewer liters than a quart of milk, the engine might not be able to output enough torque to the wheels to overcome traction. Just like traction, we can express the final output of the engine as a force instead of a torque (again, more about this below).

Now, for both types of cars, if you're already going at a high speed, then flooring the gas pedal may not do anything at all. In fact, if you're going fast enough, you can actually start to slow down if you go up a hill even if the pedal is floored. The engine simply cannot provide enough force to overcome wind resistance, rolling resistance, frictional losses, and gravity. Crucially, the power output of an engine is related to the amount of torque (and thus force) that is available to the vehicle. Obviously an engine with a higher maximum power output would be able to drive the vehicle at a higher top speed, but, more importantly, an engine with a higher maximum power will be better able to handle hills and slopes or changes that increase the required towing force (we don't usually run vehicles at their maximum speed, so that extra power can be used to overcome gravity or those changes).

Next, imagine that you hook up a trailer or two to your car. All of the various resistive and frictional losses in the trailers can be changed into a single value: the force required to pull them. And, if you've done any physics before, you'll remember that accelerating an object requires a force (f=ma) and that the required force increases as acceleration increases. So accelerating an object at twice the acceleration requires twice the force.

Remember what I said about traction and engine power being expressed as a force? This is where it comes in. We can combine both traction and engine output as a single value: tractive force. Tractive force is simply the lower of the two. The amount of tractive force determines how large of a load you can hook up and accelerate. Higher tractive force means a larger force available to tow with and a larger force means you can tow a larger load or accelerate the same load at a larger acceleration.

Here the wikipedia article on tractive force, and the paragraph about railed vehicles:Remember that tractive force takes into account both the amount of traction and the torque/force the engine is able to output. This means that tractive force generally decreases as speed increases, as the graph in the article shows.

Thank you! Thank you!
You really helped me a lot!
So just to verify, a tractive force is the amount of force the engine can output?
And thus like you said, if the car or train would be lighter, but has the same wheels and engine, it will go faster?
Thank you for your explanation, it really helped!
I now also understand what @russ_watters said about the traction curve being the answer.
So thanks to him too.

 

[russ_watters]

Do you mean that the power is actually the force at a speed of the curve?(I am probably wrong)

I believe it is:
Torque = force * length (not sure)
And
T = r x fBut i don't really have a clue how i would combine them.

Could someone help me?

The 1st equation is correct.

I'm tying to help you help yourself - it will be better for you in the long run.

It looks to me like you posted the same equation twice, once with words and once with symbols. Do you know what the word "power" means? If you don't know the answer to these questions off the top of your head, please don't wait for us to spoon feed you the answers. You can literally copy and paste pieces of the questions into Google and Google will give you the answers.

 

[Drakkith]

So just to verify, a tractive force is the amount of force the engine can output?

Kind of. The engine actually outputs torque, not linear force, but the two are related. Also, don't forget traction. If you look at the graph on the wiki page I linked, you'll see that from 0 m/s to about 23 m/s the graph is just a straight line. I believe this is because the engine is able to output enough torque to overcome traction, making traction the limiting factor in how much force the locomotive has available to tow with at that range of speeds. Once the locomotive exceeds about 23 m/s, the engine can no longer overcome traction, so the engine output becomes the limiting factor.

And thus like you said, if the car or train would be lighter, but has the same wheels and engine, it will go faster?

A lighter car will accelerate faster, but its top speed doesn't usually change much. However that's for a car, not a train. I don't know what proportion of losses are due to weight-related causes for a train towing heavy loads.

 

[Tim Leijten]

Kind of. The engine actually outputs torque, not linear force, but the two are related. Also, don't forget traction. If you look at the graph on the wiki page I linked, you'll see that from 0 m/s to about 23 m/s the graph is just a straight line. I believe this is because the engine is able to output enough torque to overcome traction, making traction the limiting factor in how much force the locomotive has available to tow with at that range of speeds. Once the locomotive exceeds about 23 m/s, the engine can no longer overcome traction, so the engine output becomes the limiting factor.

Thanks, i mostly understand, I only have 3 more questions:
Could you maybe explain the difference between torque and traction?
And so the power value is needed to plot the tractive effort curve, but what more is it needed for?
And could you maybe tell me a bit what power means in train terms?(I really don't understand)

Thank you very much!

 

[cosmik debris]

Power is also used in velocity calculations to do with air resistance. If you double the speed of the train the air resistance goes up with the square of the velocity, and the power goes up with the cube of the velocity.

Cheers

 

[Tim Leijten]

Power is also used in velocity calculations to do with air resistance. If you double the speed of the train the air resistance goes up with the square of the velocity, and the power goes up with the cube of the velocity.

Cheers

Thanks for your reply, but the power of what?
I think the train, but if the power of the train goes up, that doesn't have anything to do with the speed of the train right?(Only force, right?)Thanks

 

[A.T.]

Thanks for your reply, but the power of what? I think the train, but if the power of the train goes up, that doesn't have anything to do with the speed of the train right?(Only force, right?)

https://en.wikipedia.org/wiki/Power_(physics)#Mechanical_power

 

[Tim Leijten]

https://en.wikipedia.org/wiki/Power_(physics)#Mechanical_power

I really still don't understand where you need force for.
I should probably find a book or something.
Maybe someone can provide a good explanation for not so smart people ;) (if not too much trouble)

 

[cosmik debris]

Thanks for your reply, but the power of what?
I think the train, but if the power of the train goes up, that doesn't have anything to do with the speed of the train right?(Only force, right?)Thanks

If it takes 100HP to keep the train traveling at 50mph against air resistance it will take 800HP to keep the train traveling at 100mph. The air resistance goes up by 2 squared and the HP by 2 cubed.

Cheers

 

[jbriggs444]

I really still don't understand where you need force for.

You need force for acceleration. That's Newton's second law. Take total force and divide by total mass and you have acceleration.

You also need force to compensate for frictional resistance. That's still Newton's second law. You need total force not to be negative. So your driving force had better be at least as much as friction.

Power is the rate at which an engine can provide energy. So many units of energy per second. As it turns out, the faster you are going, the more energy it takes to maintain a given force. That can be thought of as a matter of gear ratio...

If you are pedalling your bicycle uphill in low gear, you are pushing with your feet with a low force through a large distance. If you are pedalling your bicycle uphill in a high gear, you are pushing with your feet with a high force through a smaller distance. Either way, the rate at which you are climbing the hill and the rate at which your muscles are providing energy to the bike can be the same.

As a train goes faster and faster, it takes more and more power to maintain a given force.

 

[Tim Leijten]

Ah, thanks for the explanations.
But what is power used for?
Why does it exist?
And if i would want to calculate the speed of a train using all values and such, do i need to really know power, or do i just need the tractive effort values?
I btw. will go to the library tomorrow, see if they have some book about physics or power and force or something.Thanks

 

[Drakkith]

Could you maybe explain the difference between torque and traction?

Torque is simply rotational force. It can be converted to linear force in various ways, such as by using a wheel.
Traction is just the friction with the ground

But what is power used for?
Why does it exist?

It exists because we created it and defined it to be something. Power is just a quantity that we use to say how much energy is used or generated per unit of time.

And if i would want to calculate the speed of a train using all values and such, do i need to really know power, or do i just need the tractive effort values?

You could probably do it using just the provided tractive effort, but you'd need to know a lot of other stuff too.

 

[kuruman]

But what is power used for?
Why does it exist?

What is length used for? Why does length exist?
Power, like length, is used to quantify a physical entity. Power is the rate at which energy is generated or dissipated. Imagine me having to move 5 boxes of books each weighing 50 lbs up a flight of stairs. I will spend the same amount of energy if I move all 5 boxes together and take 2 minutes to do it as if I move them one at a time and take 10 minutes to do it. So why would I choose to move the boxes one at a time? Answer: Because I am incapable of generating the energy needed for the task in 2 minutes but I can generate the same energy over 10 minutes. My metabolism limits the rate at which I produce energy which is another way of saying that my power has an upper limit and moving 250 lbs up a flight of stairs exceeds that limit.

 

[sophiecentaur]

But what is power used for?
Why does it exist?

Many answers to this question already!
Power and a lot of other quantities were identified as suitable ways of describing Scientific processes. If you go back in the history of Science, there was a lot of verbal attempts at explaining things and very often there was confusion about how things were defined and categorise the various quantities. We now tend to take it for granted that we have no chance of understanding things if we can't use Maths with properly specified quantities. Power was recognised as a suitable way to describe how fast Energy (another of those quantities) is being transferred (time was also involved, and that was yet another).
As with the rest of Scientific Maths, we can ask the question whether the Maths was always at work in natural processes or is it just the way we choose to describe them. Maths is a really weird thing, as a matter of fact. Best not think too deeply about it if you want to sleep tonight.

 

[GoatGuy]

Tim… some basic physics will help.

Nº 1: E = ½ mv²
Nº 2: E = ∫ P(t) dt
Nº 3: P = mva (very useful for moving bodies)
Nº 4: d = ½ at² (sometimes very useful, but has initial conditions to work)
Nº 4a: d = ½ at² + v₀t (including initial velocity)

E is ENERGY. P is POWER. Big difference: kilowatts versus kilowatt-hours. Or in physics, watts (power) versus joules (energy). Equation Nº 2 relates to Nº 1 exactly, showing the equivalence. If you've not run into it before "∫ x(t) dt" is a physicist's / mathematician's way of saying, integrate the area under the curve of function x(t) for all changes of t. In the case of Nº 2 … E = ∫ P(t) dt … means "energy (in joules) is the integration of all power added or subtracted over time".

For trains, if you have an engine that (say) puts out 10,000,000 W of power (which might be called 10 megawatts, or 10,000 kilowatts, or in English measure 13,410 horsepower) and you have a train of 100 cars weighing a total of 10,000,000 kg. (10,000 tons). At low velocities near zero, {P = mva} has a 'v' term near zero. Using a bit of algebraic rearrangement:

Nº 3: P = mva … a = P/mv

You maybe can see that the closer 'v' is to zero, the higher the acceleration for a given P (maximum output of the engine) and m (mass of the whole train). But you can also see that the acceleration also changes with the change in velocity of the train as it picks up speed. Right?

In this way trains are no different than cars, no different than big flywheels (or any size). As the speed of “the thing” picks up, an input of constant power to accelerate the thing results in less, and less acceleration. And, in a way, that too should make sense: If…

E = ½mv²

And we're putting constant power INTO the acceleration of the train, then it wouldn't make sense for the train to pick up a linearly unvarying amount of velocity.. If it were to, then it'd pick up kinetic ("moving") energy FASTER than the input power itself.

You're welcome,
GoatGuy

 

[Dr.D]

Most of the answers above are based on balancing the power required from the engine to supply the losses due to rolling resistance and windage. In a real train situation, there is also a limit imposed by the dynamics of the wheel rail interaction. If the train becomes unstable (begins to hunt), it is not safe to move to a higher speed.

 

[Tim Leijten]

Most of the answers above are based on balancing the power required from the engine to supply the losses due to rolling resistance and windage. In a real train situation, there is also a limit imposed by the dynamics of the wheel rail interaction. If the train becomes unstable (begins to hunt), it is not safe to move to a higher speed.

Thanks, but yes, I am aware of that.
But you mean by dynamic of the wheel rail interaction, something like the contact between the rails and wheel and stability of that?Thanks

 

[Dr.D]

There is a tendency for the wheel sets to oscillate laterally (across the direction of travel); this is called hunting.

 

[gary350]

Some math problems include unneeded information to see if your smart enough to know better.