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The usual measure is horsepower per ton. For tonnage trains in the range of 1% to 2.2% ruling grade with DC power the lowest possible ratio to get over the ruling grade at mimimum continuous speed happens to be about the same as the ruling grade. . . . Rember flange and journal friction is generally equavalent to .2% and it is always working. [snipped; emphasis added - PDN]
That's a really good answer in a short space from Mac. I'd not seen the underlined rule of thumb before, but it makes sense. In brief (by ignoring friction and other losses): 1 HP = 550 ft.-lbs. per second (that means lifting 550 lbs. at a rate of 1 ft. per second, or 1 lb. at 550 ft. per second, or any other equivalent combination). Since we're dealing with 1 ton = 2,000 lbs., 1 HP would lift 1 ton at the rate of 550 / 2,000 = 0.275 ft. per second. On a 1.00% grade, each 1 foot forward raises the train 0.01 ft., so to lift the train at the rate of 0.275 ft. per second it would have to go forward at a speed of 27.5 ft. per second, which is about 18.7 MPH. Knock off 20% to allow for the 0.2% friction, and you get 15.0 MPH, which is a very common speed for grades - see Krug's essay below for more on that. And as Mac said, double the HP Per Ton and you'll roughly double the speed - but then there's more friction and also additional air turbulence resistance, etc.
At the other extreme, Amtrak's AEM-7's have 7,000 HP for trains of roughly 70 to 75-ton passenger cars, so a 10-car train would have about 10 HP/ ton - on a basically flat level line !
For lots more detail and explanations, see Al Krug's "Tractive Effort vs Horsepower' webpage at: http://www.alkrug.vcn.com/rrfacts/hp_te.htm
And see also his "Train Forces Calculator" at: http://www.alkrug.vcn.com/rrfacts/RRForcesCalc.html , which allows you to play around with this a little bit without having to do all the math.
You didn't ask, but here's another formula that's often helpful: MPH x TE / 375* = HP
(*375 for "no friction or other losses"; a percentage of that, such as 318 [85%] or 338 [90%] are sometimes used instead to allow for mechanical losses in the locomotive.)
For 1 ton on a 1.00% grade, the TE is (2,000 lbs. x 1.00% ) = 20 lbs., so the formula becomes:
MPH / 18.75* = HP (per ton); the HP requirement will increase roughly proportionally to the grade and speed. (*Note that this is the same figure as the 18.7 MPH in the paragraph above.)
There's a bunch of limitations and exceptions to all this for tractive effort limited by adhesion, short-time motor overload ratings, etc., but I'm leaving those out here for space and time reasons.
- Paul North.
"This Fascinating Railroad Business"
(title of 1943 book by Robert Selph Henry of the AAR)