Selecting locos for a consist
If you want a little more than a rule of thumb, consider this.The following is a summary of info from Al Krug's site at
http://www.alkrug.vcn.com/rrfacts/hp_te.htm which I recommend you read.
Sizing the locomotive power required for a consist requires several steps and is best explained by an example.
Assume a consist weighing 15,000 tons is to be moved over a route that has a max incline of 1%. The math in this example is done for you and the notes at the bottom shows how it was done.
If the consist is stopped on that 1% incline, gravity would produce 20 lbs of force acting back down the track for each ton of weight (1.) for a total of 300,000 lbs (2.). To keep the consist from rolling back down the incline requires a locomotive with a TE (tractive effort) of 300,000 lbs. TE of a locomotive depends on the weight of the loco and the adhesion of the wheels on the track. Assuming an adhesion of 25% the required loco weight is 1,200,000 lbs (3.) . The max allowable weight on each steel wheel on a steel rail is 35,000lbs or 70,000lbs per axle. The required number of axles on the set of locos is therefore 1,200,000/70,000 = 17.14 or 18. The minimum number of locos for this consist is then three 6 wheel diesels provided that the total weight of the three locos is 1,200,000lbs. The SD70MAC weighs 420,000lbs and has 6 axles which fits this requirement.
Notice that horsepower (HP) hasn't even been mentioned yet. HP output of a loco depends on the throttle setting. The total HP required is the TE of the loco times the maximum speed desired. Assume the consist is to be pulled up the 1% incline at 15 MPH. The total HP required is 11,455 HP (4.). The SD70MAC is rated at 4000 HP so 3 locos will provide the required HP with a little to spare.
Notes:
1. (1 ton=2,000 lbs X .01=20 lbs per ton)
2. (15,000 tons X 20 = 300,000 lbs.)
3. (300,000/0.25 = 1,200,000 lbs.)
4. (300,000 lbs. X 1.4 ft/sec per mph X 15 mph/550 ft-lbs/sec = 11,455 HP.)