Math question- tractive effort to torque

[Spencer M]

I think this is the most fitting place to post this so here it goes. I am trying to find the torque produced by the 353. Since torque is a unit of measurement of force over a distance (pound-feet) I would assume that this can be worked backwards to find the "force" side of the equation.

Given the tractive effort of the 353, which is 31,200 pounds and
Given the wheel diameter of the same, 51 inches (4.25 feet)
By finding the circumference of the wheel with a diameter of 51 inches (4.25 feet) and
multiplying the circumference by the force in pounds would produce pound-feet

First we must convert inches to feet 51/12= 4.25 feet

then we find circumference C as per the equation C=pi*Diameter

3.14*4.25=13.345

so to find the torque of the 353 over one complete revolution, we would take the tractive effort, 31200 lbs and multiply by the circumference of the drivers, 13.345 to get 416346 foot pounds

31200*13.345=416346



Am I using the correct distance though? is the circumference the right unit for distance or do I have to use the diameter or radius of the wheels?

If I take the diameter of the drivers 4.25 feet and multiply by 31,200 I get 132,600 foot pound force

If I take the radius of the drivers 2.125 feet and multiply by 31,200 I get 66,300 foot pound force.

If I take the circumference of the drivers, 13.345 and multiply by 31,200 I get 416,364 foot pound force.

Which, if any is the most correct?

[Jerry Christiansen]

Hi Spencer,

Good idea. You have a very common error in you calculations. Torque is the product of an applied force and the length of the lever arm it acts on. Work is the product force and the distance it acts. Both are measured in ft-lb, but mean very different things.

Since tractive effort is measured in pounds, are we correct in assuming the 353 can pull with a force of 31,200lb? If so, then the following numbers can be fugured.

In your calculations you multiplied the tractive force times the circumference of the driver. Your number (416,364ft-lb) is the Work done during one revolution of the wheel.

The Torque is 66,300ft-lb [31,200lb times 2.125ft (the radius of the wheel)]

The Torque and Work calculated above are the maximum values for the 353. These values would be measured when the throtle is wide open as the engine is moving. I suspect that doesn't happen very often. We may have a method to measure the force the 353 pulls with. Check with me this weekend (May 31 - June 1) and we can discuss this.

Later,
Jerry Christiansen

[Spencer M]

Thanks Jerry that cleared it up a lot! When I was a teachers assistant I had to correct science tests and the students would often have the right numbers but the units would be incorrect. I think that using the right units and converting them is the most difficult part of math for a majority of people.

The torque figure is really a moot point because the locomotive isn't pushed to its limits in normal operation but it allow things to be put in perspective.

Thanks for your help again

[Spencer M]

Jerry you mentioned that a scheme could be devised to reckon the horsepower of the locomotive. What did you have in mind?

About how fast does the locomotive go up the grade? I'm not good with speeds.

[Tim Moen]

One must also consider upon how the builders arrived at the 31,200 number. There are many times that we pull the grade and spin! We have just exceeded the Tractive Effort of 353. Very interesting question Spencer. Tim

[Jerry Christiansen]

Spencer,

We have a device on the grounds that can be used to measure drawbar force. If we got that rigged up on the 353, then we could measure a distance along the track and time the 353. With those numbers we could calculate horsepower.

We may not get a real big horsepower number because the 353 usually doesn't go very fast. The force measured would be the tractive force. As Tim mentioned, when the 353 spins, thats all the force you can get. I am interested to know whrere the 31,200lb number was determined.

Later,
Jerry Christiansen

[Spencer M]

It would really be interesting to test this on the locomotive! It would be like our own dynamometer car. The rule of thumb for steam locomotives is that they can pull any train they can start, and the opposite is generally true for diesel-electrics. (The torque produced by a DC motor at low speeds is immense but so is the current, and if the locomotive doesn't accelerate the train after a certain period of time the motors will burn out.) The only thing that burns out on a steam locomotive would be the fireman :P

Tractive effort= Constant*Boiler Pressure*(bore*bore)*stroke all divided by Driver diameter

The constant is .85 and this takes into account various losses. It was standardized by the Association of American Railroads

in the case of the 353 the boiler pressure=180 PSI, the bore=20 inches, the stroke is 26 inches and the driver diameter is 51 inches

the result is 31200 lbs, the same as stated by the poster in the roundhouse.

[P.D.]

The waters aren't muddy enough as yet so I'd like to submit a thing or two for consideration.
I would suggest it isn't the tractive effort but the adhesion factor that has been exceeded when the drivers spin. Normally the train is still moving or at least the drivers are moving so there is more energy than can be applied to the rails. I would think if the load exceeded the tractive effort you'd be at a dead stall.

Horsepower numbers may surprise you if the peak energy expended in starting the train can be quantified. The friction bearings on the cars require many times the effort to get them rolling than it takes to keep them rolling. One of the greatest advantages of the roller bearings is the low starting effort needed compared to brasses......PD

[Spencer M]

Yeah I agree the locomotive spins because it loses grip, not because it is giving up