Tons per horse power?

Depends on hills, etc. On the BCR we need 2.2 hp per ton to make it over the steeper hills between Squamish and Kelly Lake.That's a train with about 50 loads and 3 Dash 9's, or about 45 loads and 3 SD75I's as the EMD's can't pull as much. A train on the CN mainline can run with about 0.7 Hp/ton.

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.  On steep grades the company may add a bit for speed, I would expect about 2.5 HPPT vs the 2.2 that Tyler guotes for example.  Most lines will not go much below 1 HPPT becuase HPPT translates directly to speed on any given grade.  Again roughly if 1 HPPT gives you 12 MPH on 1%, you will get 24 on .5% and 48 on .25%.  Rember flange and journal friction is generally equavalent to .2% and it is always working.

If a train needs to make a particular schedule giving it more than the minimum HPPT improves its chances of making the schedule by providing the ability to make up time.  A hot van train may run with 3 or 4 HPPT for that reason.

On the other side bulk unit trains are not in a hurry and usually get no more power than is required.  The big advantage of AC traction is that the traction motors basically have no minimum continurous speed, as do DC motors.  This allows a reduction in the number of units, hence HPPT, with AC power holding train weight constant.  The fly in this ointment is that it could take hours for your coal train to climb up a long grade, say Cranberry on the CSX former B&O in West Virginina.  At some point very slow on the ruling grade becomes too slow to handle either the volume or the mix of traffic.

In short the HPPT actually assigned is a balancing act.  Also note that the power neither knows nor cares if cars are aluminum, steel, or kryptonite.  The only thing the power feels is gross trailing tons.  Aluminum cars are used to improve the ratio of payload to tare.  That too is an economic decision, not an engineering one.

Mac 

PNWRMNM

  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. 

I'd say most of the coal trains going to most midwest destinations are in the 0.5 to 0.6 (rounded to the nearest 10th) range.  There's a few that go to foriegn lines that have fewer cars and get to the 0.8 or 0.9 (again rounded off) range.  Those figures are what I see for most coal trains on the UP from North Platte eastward across Nebraska, Iowa and Illinois.  

The nice things about engines, is that they can easily add another one farther down the line.  So once they get out of my area, they may do just that.

Jeff