I just read the article on distributed power in the September issue of Trains. I think there are a couple errors in the diagram on page 26. The third row says “64 cars stretched”, but is should be 66. Also, the drawbar force on the single locomotive should be 1/4 of the conventional head end or 3/8 of the same train head end, but not “1/3 drawbar force of head end.” That loco DOES provided 1/3 of the power to the train, but about a quarter of that power goes to the cars in front if it, so the load on its drawbar for 25% of the train should be 25% of the conventional tension. Or am I missing another factor?
First - [#welcome] to the Forum !
Next - good catch ! At least I think so, on most of your comments/ questions, as follows:
The 64 vs. 66 cars is easy enough - with 9 cars being pushed ahead of the 75th car, the correct matching number is 66 cars being stretched.
I also agree with this comment, although the portion “about a quarter of that [single/ rear DPU loco’s] power goes to the cars in front if it” is a little cryptic. Where that comes from is that the DPU has 1/3 = 33.3 % of the train’s power, but only 1/4 = 25 % of the train is trailing it, so - using the ‘least common denominator’ - the difference (4/12 - 3/12 = 1/12 or 8.33 %) can be used to push the cars ahead - maybe 8 cars instead of the 9 labeled in that 3rd diagram for In-Train Distributed Power on page 26. The 1/12 of the power being used to push is of course a quarter of 4/12, the same as 8.33 % pushing is a quarter of 33.3 % from that loco, so that does hold out when fully labeled and so understood.
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Thanks, Paul. You did lay out the derivation of the ‘one quarter’ and the ‘3/8’ very clearly.
I sometimes suffer from the math teacher habit of waving my hands and saying, “we can see that…” when really we can not see anythying of the sort.
Hey, you’re welcome. Sometimes I belabor the obvious, but I do so simply so that everone can follow the process . . .
A couple aspects not stated here or in the article is that we’re assuming all of the cars in these trains are the same or uniformly similar in weight, rolling friction, etc., and that all of the locomotives have identical power and tractive effort outputs in the situations that are being considered. That’s usually true for unit train operations - coal, grain, fertilizer, multi-level automobiles, etc. - but not so for mixed freight and maybe some intermodal moves, too.
Next - a physics or vector aspect, which can get a little complicated: As I read it, the article seemd to claim that the DPUs reduce wear on the inner rails of curves because the DPUs reduce the pulling forces - also known as ‘draft’, tension, or “stretching” in the article - which tend to ‘string-line’ the cars into a straighter line on a shorter path across the curve, which is resisted by the inner rails. At the anecdotal level, I’ve never seen a rail that’s been curve-worn from being on the inside of a curve - the rail heads there are usually ‘head-mashed’ and displaying heavy overflow on both sides of the rail head from all the weight on the low or inner rail from slow-moving trains; the only time I’ve seen curve-worn rail there is when the rail was transposed or rel
I have been mulling some of these things since reading the article but soon realized this is a non-trivial problem, meaning it would require a pencil and paper. Add to the factors you mentioned—I think the bearing surfaces on the wheels have a slight conic angle to them that keeps them centered on the straight and helps them turn (rotate) in a curve. If a single, detached truck entered a curve, it would tend to the outside because inertia would keep it moving in a straight line. As the truck moved to the outside of the curve, the outer wheel will make contact closer to the wheel flange where the diameter is larger and the opposite for the inside wheel. They are fixed on the same axle, so the turn at the same rpm and the outer wheel travels farther in one revolution than the inside one (because of the difference in diameter contacting the rail) causing the truck to turn. This is a turning effect that does not depend on the flanges, so would have to be subtracted. Some railroad guy somewhere must have already figured all this stuff out…I wonder if I can google it?
For most curves - certainly the sharp ones we’re discussing here where ‘string-lining’ would be a concern - the conicity effect is too small to really affect the results.
I believe that the standard tread taper of freight car wheels is 1:20, and that there is about 1’’ of total free distance between the flanges and the sides of the rail head. So if the wheelset is centered in the track approaching a curve, it will have offset at most 1/2’’ when the flange contacts the rail head. The outer wheel will be riding on a slightly larger diameter, and the inner wheel on a slightly smaller diameter. Altogether, there will have been a 1’’ lateral shift, and so the change in wheel radius from one wheel to the other will be 1/20 of that, or 0.050’‘. Since the circumference is 2 x pi x R, the change in circumference is 2 x 3.14 x 0.05’’ = 0.314’', over about 5.0 ft. between the centers of the rail heads.
Later on I’ll try to remember - yet again - how to relate that to what radius track curve will also produce that differential in distance between the rail heads. That too will take pencil and paper - maybe even a slide rule, too - at least for me . . .
Did anyone besides me find the “Force against Rail on Curve” graphic on Page 30 confusing?
Yes.
Other things being equal, the lateral forces are going to be essentially the resultant force of the vectors on the two couplers minus the friction losses. Thus the closer the car is to the engine, the larger the component vectors will be and the larger the lateral force will be. In a similar fashion, the outward forces would be greatest immediately in front of the pusher engine and taper off as the compression diminishes farther ahead.
Thank you, KJ. Sometimes a few well-chosen words can beat the stuffings out of a graphic.