I know DPU is being widely used (relatively speaking) and is a time tested principle. But when pushing from behind, does it cause the cars to want to zig-zag (opposite of stringling)? When pushing from the rear all slack and space will be used inorder for the rear loco to get as close to the front of the train as its power and terrain will allow. This would mean that cars would want to do this; ///// istead of this —. Although it may only be a fraction of an inch or so in each car. Does this cause increased wear on the rail head and flanges of the wheels? Or cause the wheels to pick a defect in the rail much easier? I read the article in Trains a few months back about the physics behind DPU but it made no mention of this. Just curious if I am making a mountain from a mole hill or if there is a concern. It obviously is not that big of concern or that large of a cost or the railroads would not be doing this.
I personally prefer Southern’s way of controlling the DPUs with the load cell in the front coupler of the mid train helper. All cars were being pulled this way. Of course I am not a loco engineer, civil engineer, or track maintainer to justify which way is truely “better”. Just my .02
I would suggest you read the TRAINS magazine issue about DPU from a few months back as it goes into detail about track dynamics including the issues you raise. Southern experimented with the drawbar sensor system you refer to but ultimately chose to use the Harris Locotrol RC system instead…
In theory, yes - at least on tangent track. But consider the following:
For each car, the push is on the near end’s coupler==>draft gear==>center sill or frame==>other end’s draft gear==>coupler; there is no push / compression/ “buff” forces applied to the car trucks or wheels. They’re just held in place underneath the car by their center pins, and will push outward only to the extent that the center sill/ frame pushes them to do that.
Which it will, though, because the center-line of the couplers is a couple of inches off the center-line of the car’s center-sill/ frame, and there’s also an inch or so of free ‘play’ between the flanges and the gage face of the rails. But even adding all that up, the sideways force vector would still be pretty small. Say that the couplers are 3" off center plus 1" for the flange space = 4" total over a car-length of say 50 ft. at the couplers - that’s 4" / 50’ = 1 in 150 or 0.00667 = 0.667%. So even at a maximum push force of 400,00 lbs. or so, the sideways force would be 0.667% x 400,000 lbs. = 2,670 lbs. or so sideways force at each truck, so that’s even spread out over 2 wheels = 1,335 lbs. per wheel, which are each already typically carrying from 9,000 lbs. (empty) to 35,000 l
Thanks Paul! That is the kind of answer I was looking for. I would have though there would have been more sideways force than that, but the numbers don’t lie.
Say the RR runs compass east-west. Put an 80-foot infinitely-rigid bar between the rails with its west end fixed to the south rail and its east end resting against the north rail. Then put another such bar next to it, with its west end contacting the east end of the west bar, next to the north rail, and its east end against the south rail. Apply a westward 400000-lb force to the east end of the east bar. What’s the lateral force on the north rail?
Good explanation, Paul, and I think you’re exactly right. Something additional to point out:
Too much tractive effort pushing from behind could, in theory, become a liability for the exact reason stated, just as too much force pulling from ahead can cause stringlining. With a DPU setup, you have both forces at work, generally speaking. But keep in mind these forces are the opposite of one another, and effectively cancel each other out. So by moving the train through a combination of pushing and pulling, you’re actually moderating these forces.
But this is the reason Union Pacific rosters “CTE” versions of the AC4400CW and ES44, the CTE being “controlled tractive effort.” These units are designed for helper service, and deliberately scale back tractive effort so as not to pop a car off the track while negotiating a curve. So apparently this is an issue to some degree.
There was an incident on the Tehachapi grade just a few years ago where the rear helper on a Santa Fe train shoved several cars off the track. Don’t recall if it was due to a wheel slip on the head end or just too much lateral force against empties on one of the many tight curves.
That would be in a shallow upside down ‘V’-shape if we’re looking at it with North being upwards - kind of like this: /\
The answer is twice the lateral component from each bar - 1 from each. To be more specific, I’d need to know the width of the bar so as to be able to calculate its angle relative to the track. Interestingly, the postulated 80 ft. length of the bars is about the length of a typical longest car, except for some TTX flats, auto-racks, passenger cars, and the like. At the worst if the bars are infinitely thin - essentially a line - then the ratio would be (4’-8-1/2" = 4.71’) / 80 ft. = a little more than 1-in-20 lateral / longitudinal - approx. 0.0509 = 5.09%. So the lateral force from each bar would be 5.09% x 400,000 lbs. = 20,360 lbs. or so sideways force at each bar. And since there are 2 bars at the north rail, the total force on it would be approx. 40,720 lbs.
Fortunately for me, I was addressing only the lateral force from one of the trucks at the end of each car - in these terms, the force on the end of 1 bar only, not the total force on the rail or