tractive effort

[Modelcar]

....Boyscouts Motto: Be Prepared.

[Mookie]

QUOTE: Originally posted by Modelcar

....Driver may have Tylenol available.....

yeah, he keeps it in my purse! [;)]

[Anonymous]

thanks all three:jamie, larry, peter.

[tree68]

QUOTE: Originally posted by cbt141

tree,
do i have the following correct?
in your explanation of lever...
radius from axel to rim is long arm of your lever
and from axel to point where connecting rod joins eccentric is short arm of lever.
thank you
cbt141

I was actually glossing over the "other end" of the lever, for the sake of simplicity. The relative power of the steam, pistons, rods, etc, factor into the virtual length of the end of the lever "you push down on," just as the weight of the person pushing down on the lever makes a difference on the weight that can be lifted by the other end.

Usually when we think of a person moving a large object with a lever, we think of the "long end" being where the person is pushing, while the "short end" is where the lifting is occuring. That provides the necessary leverage to move the heavy object.

As Peter alluded, the short end of the "lever" in my illustration is actually where the power is applied, while the long end of the lever does the "lifting". Think teeter-totter, with 240 lb me on one end and a 50 lb child on the other. I can be very close to the fulcrum and still lift the child off the ground. In the case of that 4-8-4 Peter mentions, you have a "short end" of 16" and a "long end" of over 70", both depending on the specific loco. With an engine with 80" drivers, that's a 5:1 ratio, just like me and that kid.

It's actually pretty amazing to consider the stresses on those drivers.

I also pretty much ignored weight on drivers, etc, since the question had to do with the diameter of the wheels. They are very definitely factors, which Jamie and Peter covered very well. Thanks, guys.[:D]

[M636C]

cbt141,

Yes, you are right about the two arms of the lever. Remember that the crank radius is set by the length of stroke of the cylinder. This was about 12 inches (24" stroke) on old 4-4-0 American types and up to 16 inches (32" stroke) on modern 4-8-4s.

As Jamie said, the weight on drivers is a consideration. A good generalisation was that you could rely on 25% adhesion, so the weight on drivers should be four times the tractive effort. On some locomotives like the Pennsylvania S-1 duplex, the tractive effort was too high for the weight on drivers. Not that the overall weight wasn't high in that case, but any locomotive with four driving axles that required six carrying axles in the trucks wasn't going to have good adhesion characteristics.

The Illinois Central rebuilt a 2-8-4 as a 4-6-4 with relatively small driving wheels for fast freights, but again, a loco with more carrying axles than driving axles didn't really work out in freight service. It was faster than it had been, owing to the larger wheels, but the tractive effort per axle increased, and the adhesion was reduced.

Peter

[jchnhtfd]

QUOTE: Originally posted by cbt141

tree,
do i have the following correct?
in your explanation of lever...
radius from axel to rim is long arm of your lever
and from axel to point where connecting rod joins eccentric is short arm of lever.
thank you
cbt141

Keep in mind a couple of things... tractive effort is also affected by piston diameter and steam pressure, but ultimately it's limited by the weight on the drivers.

[Anonymous]

tree,
do i have the following correct?
in your explanation of lever...
radius from axel to rim is long arm of your lever
and from axel to point where connecting rod joins eccentric is short arm of lever.
thank you
cbt141

[tree68]

QUOTE: Tree - Mookie was ready to have a headache after reading the question, but congrats - I got through the explanation with only a few twinges!

Mook

Will save headache for driver....

And the wheels were a-turnin' ... [swg]

[Modelcar]

....Driver may have Tylenol available.....

[Mookie]

Tree - Mookie was ready to have a headache after reading the question, but congrats - I got through the explanation with only a few twinges!

Mook

Will save headache for driver....

[tree68]

I'll assume you are talking about steam locomotives.

It all has to do with the size of the lever.

Think of a wheel as a lever. The fulcrum (pivot point) is the axle. If you have a short distance between the fulcrum and the load to be lifted, you can lift great amounts of weight. If you have a longer distance, you can lift less weight (although you can lift it higher). (We'll assume that you have a constant length for the end you're pushing down on.)

Thus, a small driver can move more weight, but not very fast. A large driver can't move as much weight, but it can move it faster, hence the 70-80" wheels on some speedster steam locomotives. Most steam yard goats were down in the 50" range, I believe. A speeder (with sufficient weight - obviously a limiting factor) could theoretically move a sizeable train, just not very fast.

There are a lot of other factors involved in determining tractive effort. For the sake of this question, though, we'll assume they are more or less equal.

As for diesels today, wheel size may play a part, but much less so. Dynamic augment was a factor on steam locos - not so on D/E, so the wheels can be made to spin as fast at their bearing will allow. Gearing is more significant, although the lever analogy is probably still relevant. If you understand the significance of the gear ratio in the differential on you car, you understand the gearing on modern locos. .

Dynamic Augment - the interaction of the drivers, counterweights, rods, etc. Balancing all those factors was an art. Some locos were great, others not so great. What might be OK at one speed would nearly throw the loco off the track at another. It often manifested itself in vertical motion and was hell on the track.

[Anonymous]

why is tractive effort inversely proportional to driver diameter?