Eastern Railroad Discussion > Power Factor Question

I understand that just a 1% uphill
grade triples the power requirements
for a train.

Some mainlines have 2% grades. What
is the power factor for such a grade?
Would the factor be six?

CSX106CL Wrote:
-------------------------------------------------------
> I understand that just a 1% uphill
> grade triples the power requirements
> for a train.
>
> Some mainlines have 2% grades. What
> is the power factor for such a grade?
> Would the factor be six?

To understand how this works, you first need to understand the difference between tractive effort and horsepower. I will give a simple example to demonstrate the difference between the two:

Imagine that you want to pick up a one ton object with a crane. The hook on the crane will see a force of 2000 lb when it first lifts the object off of the ground. If it can't develop that force, the object will remain on the ground.

This is analagous to the tractive effort from a locomotive consist, in that a minimum force on the coupler is necessary to get a train to start moving. If the locomotive can't generate that force, the train won't move.

Lifting the weight requires almost no horsepower. In fact, a person, which can develop less than one horsepower, can lift the one ton object with enough pulleys and gears, he just wouldn't lift it very quickly.

Where horsepower comes in is in how quickly the object can be lifted. If I want to lift the one ton object at say 5 feet per second, I would need to have a motor of at least 18 horsepower, ignoring frictional losses, to lift it that quickly. If I want twice that speed, then I would need twice the horsepower, or a motor that could put out 36 horsepower.

Note that in the example the force (equivalent to tractive effort) has not changed, since the object's weight hasn't changed, only the speed changed. That is what requires horsepower.

Now to your questions about trains. If you want to simply move a car at 10 mph, you will need a force of about 300 pounds on the front coupler of a car that weighs 100 tons. This overcomes the rolling resistance of the car, since air resistance is negligible at that speed. You will also need about 8 horsepower to move at 10 mph.

If you now want to move that same car at 60 mph, you might assume that you would need six times the horsepower, or 48 horsepower, but that only works if the pull on the car's coupler remained the same. It is more complicated than that.

The reality is that as speed increases, the resistance due to air also increases, so the force needed moves up as well, to about 1,000 lb. at 60 mph. This boosts the horsepower required to 160 hp. Therefore, six times the speed requires 20 times the horsepower. That ratio will vary depending on the type of car, and how difficult it is to pull through the air.

A grade adds additional force, proportional to the steepness of the grade. If the locomotive cannot provide at least the grade force, it will stall on the grade, just like the crane in the first example would not be able to lift the object off the ground. This force does not vary with speed, but is simply added to the rolling and air resistance of the car at whatever speed the car is moving.

In the case of a 1% grade, the force due to gradient alone on the front coupler of our 100 ton car would be 2000 lb. Adding the rolling and air resistance would give a total of 2,000 + 300 = 2300 lb. Note how much more this is than the 300 lb. required simply to get the car moving at 10 mph on flat ground. To move this car at 10 mph on a 1% grade would therefore require about 61 horsepower. To move the same car up the grade at 60 mph, the force to overcome would be 2,000 + 1,000 = 3,000 lb, and require 480 horsepower.

On a 2% grade, the grade force would be twice that on the 1% grade, or 4,000 lb. Working through all the numbers, to move the car at 10 mph on the grade would require about 115 hp. At 60 mph, it would require 800 horsepower.

The reality is that railroads often power trains to make track speed on flat land, and accept the lower speed on grades as a cost of doing business. Looking at it this way, if a train can maintain 60 mph on flat track, it will drop to about 20 mph on a 1% grade, and 12 mph on a 2% grade. Both of those speeds assume that the locomotives have enough weight on drivers (tractive effort) to pull the train without too much wheelslip.

Confused yet?

run8 Wrote:
-------------------------------------------------------
> CSX106CL Wrote:
>
> --------------------------------------------------
> -----
> > I understand that just a 1% uphill
> > grade triples the power requirements
> > for a train.
> >
> > Some mainlines have 2% grades. What
> > is the power factor for such a grade?
> > Would the factor be six?
>
> To understand how this works, you first need to
> understand the difference between tractive effort
> and horsepower. I will give a simple example to
> demonstrate the difference between the two:
>
> Imagine that you want to pick up a one ton
> object with a crane. The hook on the crane will
> see a force of 2000 lb when it first lifts the
> object off of the ground. If it can't develop that
> force, the object will remain on the ground.
>
> This is analagous to the tractive effort from a
> locomotive consist, in that a minimum force on the
> coupler is necessary to get a train to start
> moving. If the locomotive can't generate that
> force, the train won't move.
>
> Lifting the weight requires almost no
> horsepower. In fact, a person, which can develop
> less than one horsepower, can lift the one ton
> object with enough pulleys and gears, he just
> wouldn't lift it very quickly.
>
> Where horsepower comes in is in how quickly the
> object can be lifted. If I want to lift the one
> ton object at say 5 feet per second, I would need
> to have a motor of at least 18 horsepower,
> ignoring frictional losses, to lift it that
> quickly. If I want twice that speed, then I would
> need twice the horsepower, or a motor that could
> put out 36 horsepower.
>
> Note that in the example the force (equivalent
> to tractive effort) has not changed, since the
> object's weight hasn't changed, only the speed
> changed. That is what requires horsepower.
>
> Now to your questions about trains. If you want
> to simply move a car at 10 mph, you will need a
> force of about 300 pounds on the front coupler of
> a car that weighs 100 tons. This overcomes the
> rolling resistance of the car, since air
> resistance is negligible at that speed. You will
> also need about 8 horsepower to move at 10 mph.
>
>
> If you now want to move that same car at 60 mph,
> you might assume that you would need six times the
> horsepower, or 48 horsepower, but that only works
> if the pull on the car's coupler remained the
> same. It is more complicated than that.
>
> The reality is that as speed increases, the
> resistance due to air also increases, so the force
> needed moves up as well, to about 1,000 lb. at 60
> mph. This boosts the horsepower required to 160
> hp. Therefore, six times the speed requires 20
> times the horsepower. That ratio will vary
> depending on the type of car, and how difficult it
> is to pull through the air.
>
> A grade adds additional force, proportional to
> the steepness of the grade. If the locomotive
> cannot provide at least the grade force, it will
> stall on the grade, just like the crane in the
> first example would not be able to lift the object
> off the ground. This force does not vary with
> speed, but is simply added to the rolling and air
> resistance of the car at whatever speed the car is
> moving.
>
> In the case of a 1% grade, the force due to
> gradient alone on the front coupler of our 100 ton
> car would be 2000 lb. Adding the rolling and air
> resistance would give a total of 2,000 + 300 =
> 2300 lb. Note how much more this is than the 300
> lb. required simply to get the car moving at 10
> mph on flat ground. To move this car at 10 mph on
> a 1% grade would therefore require about 61
> horsepower. To move the same car up the grade at
> 60 mph, the force to overcome would be 2,000 +
> 1,000 = 3,000 lb, and require 480 horsepower.
>
> On a 2% grade, the grade force would be twice
> that on the 1% grade, or 4,000 lb. Working
> through all the numbers, to move the car at 10 mph
> on the grade would require about 115 hp. At 60
> mph, it would require 800 horsepower.
>
> The reality is that railroads often power trains
> to make track speed on flat land, and accept the
> lower speed on grades as a cost of doing business.
> Looking at it this way, if a train can maintain
> 60 mph on flat track, it will drop to about 20 mph
> on a 1% grade, and 12 mph on a 2% grade. Both of
> those speeds assume that the locomotives have
> enough weight on drivers (tractive effort) to pull
> the train without too much wheelslip.
>
> Confused yet?


This is why they need to buy some surplus mine hoists to pull trains up grade with a counterweight running down grade - or have a compressor car feeding air to air driven motors driving each axle to reduce the load on each loco and lash up. my two cents

run8 Wrote:
-------------------------------------------------------
> CSX106CL Wrote:
>
> --------------------------------------------------
> -----
> > I understand that just a 1% uphill
> > grade triples the power requirements
> > for a train.
> >
> > Some mainlines have 2% grades. What
> > is the power factor for such a grade?
> > Would the factor be six?
>
> To understand how this works, you first need to
> understand the difference between tractive effort
> and horsepower. I will give a simple example to
> demonstrate the difference between the two:
>
> Imagine that you want to pick up a one ton
> object with a crane. The hook on the crane will
> see a force of 2000 lb when it first lifts the
> object off of the ground. If it can't develop that
> force, the object will remain on the ground.
>

Snip Snip

Let's add a nice twisty track to that grade!!!

MtArarat Wrote:

> This is why they need to buy some surplus mine
> hoists to pull trains up grade with a
> counterweight running down grade -

You would need a steel cable of about 3 inches in diameter to be able to handle the pulling power of one six axle locomotive. It would be an interesting exercise to figure out how to attach it to a train, and how to protect the crew if the cable broke and whipped back at the train. Not to mention the horsepower needed for the winding engine, and the logistics of randomly handling uphill and downhill trains. No thanks. It's simpler to just put a helper locomotive on the train.

> or have a compressor car feeding air to air driven motors
> driving each axle to reduce the load on each loco
> and lash up.

Yikes. Compressors are very inefficient ways of providing power, not to mention the motor losses and the size of pipe needed to transmit the power down the train, plus the need for inspection and maintenance. Charles Law on gas expansion would be a real constraint. There's a reason why compressed air is rarely used for high power applications.

jonnycando Wrote:
-------------------------------------------------------
>
> Let's add a nice twisty track to that grade!!!

The usual way to consider curves is to convert the resistance they provide to an equivalent grade, and add it to all the other forces. A moderate 6 degree curve is considered to be about the same as a 1/2 percent grade, which is a fair addition.

One interesting thing about curves is that since they always resist train movement, they benefit downhill trains in that they lessen the effective grade under the train, making braking easier.

run8 Wrote:
-------------------------------------------------------
> jonnycando Wrote:
>
> --------------------------------------------------
> -----
> >
> > Let's add a nice twisty track to that
> grade!!!
>
> The usual way to consider curves is to convert
> the resistance they provide to an equivalent
> grade, and add it to all the other forces. A
> moderate 6 degree curve is considered to be about
> the same as a 1/2 percent grade, which is a fair
> addition.
>
> One interesting thing about curves is that since
> they always resist train movement, they benefit
> downhill trains in that they lessen the effective
> grade under the train, making braking easier.
>

Ah yes, and observed personally to be true. Afton Mountain on C&O North Mountain Sub has compound curves and exceeds 1 percent westbound on the east slope. Speed uphill is limited to 25 to prevent stringlining. Going up can be mightily difficult if underpowered, at times impossible. Coming down these same curves, dynamic braking is usually sufficient as the curves hang the train up. Not certain of the arc of these curves but most quite sharp. Classic mainline mountain railroading to be sure.

Greetings from the Eastern Wilderness, As I previously mentioned in my earlier post on the Western Board regarding traction and slippage it is very easy to set up a system using a tugger hoist with a counterweight to haul things up and down slopes. and it uses very little energy because you are helping reduce the load by increasing the speed of the train while the train and hoist are under power.

Great physics lessons; effect of friction in curves is especially interesting. Is it possible that a long, light train like baretables might actually require tractive effort to DESCEND a moderate grade?

Freddie Wrote:
-------------------------------------------------------
> Great physics lessons; effect of friction in
> curves is especially interesting. Is it possible
> that a long, light train like baretables might
> actually require tractive effort to DESCEND a
> moderate grade?

It is possible yes. I have on rare occasion PULLED a train down a grade. Though most often this is the result of using too much air brake. Since you don't want to release the brakes on the grade and risk having to reapply them before they've charged up fully, you might pull against them to keep from stopping. In all the effect is the same a your scenario of a long light train hanging up in curves. When pulling such a train through curves you have to be careful, too much power and it will try to straighten itself out, turning the rail over as it does. This is called stringlining.

Freddie Wrote:
-------------------------------------------------------
> Is it possible that a long, light train like baretables might
> actually require tractive effort to DESCEND a moderate grade?

To add to the comments in the previous post, a grade doesn't have to be very steep for cars to roll away. Something less than 0.01 percent will do the trick.

However, if there are curves, the resistance of the curves will slow the train down, and you might have to use power. As an example, if the train was moving through a 10 degree curve, like Horseshoe curve near San Luis Obispo, the effect of the curve is like a 0.8 percent grade. This only applies to the portion of the train in the curve. Therefore, if the curve were on a 5% grade, and the entire train was on the curve, the net effect to a downhill train would be like having to pull up a 0.3% grade. Tractive effort would be required to keep the train moving, even though it was moving downhill.

The other interesting thing about grades is that the effect does not vary with the weight of the cars in the train. A train of heavy cars would be affected just the same way as a train of empty cars. That is, the resistance due to the curve will increase as the weight of each car increases.