Steam & Excursion > Tracktive Effort

Please explain to me what is meant by tractive effort in a steam locomotive. Is it like horsepower in a Diesel engine?

Thanks,

Hal

Tractive effort is a theoretical calculation of the locomotives pulling ability, usually obtained by the formula

(k x P x (C squared) x S) / D

where

k is a constant, usually .85

P is boiler pressure, in psi

C is the diameter of the cylinders, in inches

S is the stroke, in inches

D is the driver diameter in inches

Drawbar pull is the amount of pulling ability, usually measured by a dynamometer car. It is essentially the tractive effort less the amount of effort it takes to move the engine itself.

In designing a steam locomotive, tractive effort was designed to be about 25% of the weight on drivers. A higher percentage would tend to make a locomotive slippery.

MTM, your last sentence....does that equate to "factor of adhesion?" I seem to remember a quote from Paul Keifer about shooting for 0.4 but I can't remember the details.

The figure I gave, expressed as a percentage, is the coefficient of adhesion.

The factor of adhesion is the inverse: That is, 1 divided by the coefficient of adhesion.

Thus a CoA of 25% is equivalent to FA of 4.0, which is probably what you're trying to remember.

Some representative examples

FA / Locomotive

3.63 T&P 2-10-4 610
4.67 B&O 0-6-6-0 2400 "Old Maude"
4.35 ATSF 4-8-4 3771
4.63 MILW 4-4-2 1 "Hiawatha"
3.54 GN 4-8-0 103
4.13 PRR 4-4-4-4
3.80 NW 2-6-6-4 1218
4.27 C&O 2-6-6-6 1605
4.91 GN 2-6-6-2 1800

I should further point out to Alaska that tractive effort is simply a force - the amount of effort available to push or pull.

Horsepower is a force moving through a distance in time. At 0 mph a locomotive delivers maximum tractive effort, since the engineer is applying full steam pressure to the cylinders. Horsepower is 0, since the force is not active through a distance.

As the train picks up speed, the engineer "hooks up" the valve gear, which reduces the mean effective pressure in the cylinder and thus reduces the tractive effort. Horsepower goes up, since the force is now moving the train a distance in time. At 40 mph or so, depending on the design of the engine, the force-distance-time relationship delivers maximum horsepower. Above that, tractive effort drops so drastically that the horsepower curve drops despite the increase in distance over time.

Tractive effort is what starts a train. Factor of Adhesion determines how well the engine can deliver its tractive effort to the rail. Horsepower determines how fast it can be moved, once started.

Thanks. Not only did I invert 4.0, but learned that FOA is the inverse of .......OK. Now, why was 4.0 an important number? Was it above 4 or below that made engines more susceptible to slip?

Any locomotive with a factor of adhesion of less than 4 was considered slippery.

The inverse of the factor of adhesion is the coefficient of friction. FofA of 4.0 = CofF of 25%. Whether a loco was slippery or not depended a lot more on rail conditions and the skill of the engineer than on a numeric ratio of any kind. MTM's comments are very well phrased here.

To clarify the distinction between tractive effort and horsepower a bit further:-

Tractive effort is the theoretical maximum force that a loco can put onto the rails at the wheels. For almost any type - steam, electric or diesel, this will be when starting, ie at close to zero speed, and may be limited by the adhesion. The greater the speed of the engine the more power is needed to produce a given tractive effort. So an engine producing a 50,000lbs t.e. at 60 mph will be developing twice as much power as one producing 50,000lbs at 30mph.

MTMEngineer Wrote:
-------------------------------------------------------
> Tractive effort is a theoretical calculation of
> the locomotives pulling ability, usually obtained
> by the formula
>
> (k x P x (C squared) x S) / D
>
> where
>
> k is a constant, usually .85
>
> P is boiler pressure, in psi
>
> C is the diameter of the cylinders, in inches
>
> S is the stroke, in inches
>
> D is the driver diameter in inches
>
> Drawbar pull is the amount of pulling ability,
> usually measured by a dynamometer car. It is
> essentially the tractive effort less the amount of
> effort it takes to move the engine itself.
>
> In designing a steam locomotive, tractive effort
> was designed to be about 25% of the weight on
> drivers. A higher percentage would tend to make a
> locomotive slippery.

Something seems missing in this equation. The radius of the drive rods needs to be considered, I would think. The torque applied to the wheel is not the same as the torque applied to the rail by the wheel. The larger the ratio of driver radius (or diameter) to wheel radius, the more this force is reduced. Is this accounted for by the size of the stroke. I would think that might do it as the stroke must be the same as the drive rod circle diameter and the expression can be rearranged to isolate S and D as a ratio.

rdsexton Wrote:
-------------------------------------------------------
> Something seems missing in this equation. The
> radius of the drive rods needs to be considered, I
> would think. The torque applied to the wheel is
> not the same as the torque applied to the rail by
> the wheel. The larger the ratio of driver radius
> (or diameter) to wheel radius, the more this force
> is reduced. Is this accounted for by the size of
> the stroke. I would think that might do it as the
> stroke must be the same as the drive rod circle
> diameter and the expression can be rearranged to
> isolate S and D as a ratio.

I think you answered your own question. The mechanical advantage of the lever formed by the axle, the radius of the crankpin, and the edge of the wheel is taken care of by dividing the stroke by the driver diameter.