tractive effort equation? draw bar pull equation?

[Maddhatter]

I will admitt I know nothing about the railroad system. However, I have always been interested in it.

As some of you know I am a college studing writing a paper on diesel locomotives and I have had the hardest time trying to understand Tractive Effort, I have yet to find a good equation for it. Same thing for drawbar pull, and rolling resistance of a train. I have also seen an equation for gradeablity, but it uses rolling resistance which is still confusing. I know the coef. of friction is really small on the rails (0.0015).

Given a diesel made of four 1,000hp electric motors at a speed of 12mph, and a 4 to 1 gear ratio, and 40 inch wheels. whats the equations for the tractive effort and drawbar pull?

Thank you all for the help!

[Allen Hazen]

Tractive effort isjust the FORCE the locomotive puts into pulling the train. So, at constant POWER, inversely proportional to VELOCITY.
So, if the locomotive has 4000 hp (the conventional rating for U.S. diesel locomotives is the power put into the main generator by the engine -- "raw" engine power is a bit more since it includes power for driving the radiator fans etc etc etc) and the transmission efficiency is 85% (reasonable for modern locomotives), the locomotive is putting 3400 hp into train-pulling. Or (33,000 ft-lbs/min) 112,200,000 foot-pounds per minute into it. At 12mph, that's a fifth of a mile, or 1,048 feet per minute. So a bit over 105,000 pounds of train-pulling force (TRACTIVE EFFORT).
--
(Theoretically modern locomotives can exert something like a third their own weight in tractive effort before their wheels lose traction and spin, but I suspect most railroads would prefer a margin of error. So, if the idea was to operate for any significant length of time at 12 mph, a six-axle -- and so six-motor -- 4,000 hp locomotive would be preferred to a 4 axle: contemporary locomotives for mainline service in North America tend to weigh in the 67,000 to 72,000 pound per axle range.)
---
Any chance your college library has back numbers of "Trains"? There was a very good multipart conceptual introduction to locomotives in the April/May/June/July 1979 issues. If you are still going to be working on this in a week or two, I could snailmail you a photocopy (I don't have a scanner, so I can't e-mail it).

[txbritt]

Al Krug, a BNSF engineer operating out of the Montana/Wyoming area maintains a website with Pictorial stories of some of his runs. He also has a technical area with several useful articles, one of which being an explanation of how Tractive effort works.

This is the link.

http://www.alkrug.vcn.com/rrfacts/rrfacts.htm

Hope this helps

TxBritt

[timz]

Given a diesel made of four 1,000hp electric motors at a speed of 12mph, and a 4 to 1 gear ratio, and 40 inch wheels. whats the equations for the tractive effort and drawbar pull?

Like they said, Al Krug's site should be a big help.

The one fact you have to keep in mind, is that power by definition is force times speed. So if the four electric motors could actually produce 1000 hp apiece at 12 mph, then the tractive effort would be 125,000 lb minus whatever is lost in friction between the motor pinion and the rail. The gear ratio and wheel size would be irrelevant.

Of course, in reality the gear ratio and wheel size are very relevant, because the motor can only produce 1000 hp over a certain range of speeds. Even if you geared it to the rail the locomotive wouldn't produce 1,500,000 lb at 1 mph.

[Typewriters]

Tractive effort is the force, measured in pounds, exerted in a direction tangential to the circumference of a locomotive's driving wheels, and in a direction to move the train along the rail. It is limited by the coefficient of friction between the wheel and rail; this means that the highest absolute value is determined by the force of the weight of the locomotive.

Starting tractive effort is a theoretical value based upon a simple percentage of a locomotive's weight. This high value of force is not achievable continuously, and cannot be achieved in less than optimal conditions (rain, snow, ice, oil on track, etc.) For old units, the value was normally 25% of the weight on drivers. For a 200,000 lb switching locomotive, then, the assigned 'rated' STE would be 50,000 lbs.

Years ago, the common tractive effort equation was (horsepower x 308) divided by speed = tractive effort.

Naturally, also (horsepower x 308) divided by tractive effort = speed.

Again, everyone is preparing to hit their "reply" buttons, and rightly so as this equation is extremely generalized. To start with, you must be sure that "horsepower" is measured in a relevant sense -- ie, that it is horsepower input to the generator and not brake horsepower.

Much more important is the fact that "308" is highly approximate. This conversion factor takes a number of things into account. If you rearrange the equation and call the conversion factor not 308, but call it "x" and then plug in various known ratings for locomotives at their minimum continuous speeds (also the maximum continuous tractive effort, it would follow) then you will find that the number isn't always 308, and in fact it isn't often 308. You can find conversions that come out in the high 290's and the low 310's.

But for older units (like, 1930's - 1940's - 1950's) the equation can provide rough curves with which you can calculate (again, rough) haulage capacities at speeds.

Drawbar pull is the force left to move the train, measured in pounds, after having accounted for the force necessary to move the locomotives. I seem to recall reading somewhere that a good rough figure (again, older) would be that you needed 35 lbs of tractive effort to start one ton of train weight in motion. If that holds, then you take the starting tractive effort of the locomotive in lbs, subtract the value ("loco weight in tons" x 35 lbs) and you'll have, again roughly, the force exerted to start the train.

Hope this helps.

-Will Davis

[timz]

Years ago, the common tractive effort equation was (horsepower x 308) divided by speed = tractive effort.

If speed is in miles/hr and TE in pounds. This just restates the definition of horsepower and assumes that 308/375 (i.e 82%) of the locomotive's rated power appears at the wheel rim.

I seem to recall reading somewhere that a good rough figure (again, older) would be that you needed 35 lbs of tractive effort to start one ton of train weight in motion.

So starting a 5000-ton train required 175,000 lb?

As you see, it's always been a mystery just what starting resistance amounts to.

[mxdata]

All excellent replies gentlemen! To put this in a typical product application scenario, on units which have static adhesion wheelslip systems (which Will describes as "older" units) that do not employ controlled rates of wheel creep, we normally do not like to assume much more than 18% dispatchable adhesion for figuring ratings, even though they will produce 25% under ideal conditions. The primary reason for being so conservative is that these units take a big hit in the autumn when you may have leaves falling on the track, and if you cut your margin too close you can get into trouble.

[yousouf leghari]

dear,
i am connected with transportation department in pakistan railways.
while going through your discussions,it seems that i am totally ignorant about basic of railways design.though tractive effort curves are provided by manufacturer but if i want to calculate tractive effort required for hauling of 2300ton,for calculating horse power of diesel electric locomotives.what are general factors affect on tractive effort.what are standards gear ratio for freight and passenger operation.i shall be very obliged if some one hlep me in this regard.
muhammad yousouf leghari