grades and hauling force


gregc

Apprentice Modeler
i had read that the average friction of a rail car is about 7 lbs / ton. this would be the force an engine needs to produce just to keep a train in motion on flat ground.

using a little trigonometry, i was curious about what grade would be equal to this friction force and found it to be 0.35%. this would mean the load on the engine is double. at 0.7 it's 3x, 1.05 4x, and so on ...

i was surprised to see that a 1% grade quadruples the hauling load on an engine, and can now appreciate why a 1% grade is so significant.

do these number sound correct?
 
Greg, I think your numbers are pretty close. The only additional consideration for grades is curves and how this affects rolling friction. I don't know who to calcualte it but the sharper the curve, the nore flanges against the curved rail add to rolling friction. For our typical model railraod curves, I'd guess that it has the effect of increasing the grade by at least 50%.
 
Rule of thumb is that a train needs the % grade in hp/tt to make it up the grade without stalling. So if its a 5000 ton train and a 2.5% grade, it would need 2.5 x 5000 = 12500 hp, so 3 SD60's, 3 SD70's or 4 SD40's would work.
 
That's a little over kill, don't you think.

That would mean that on a 2% grade you would need 16x the hp you would need on level ground. So if a SD40 can pull 5000 tons on level ground, by the doubling rule you would need 2 SD40's to pull 5000 tons on a .5% grade, 4 SD40's on 1% grade, 8 SD40's on 1.5% grade and 16 SD40's on 2% grade (32 SD40's on a 2.5% grade and 64 SD40's on 3% grade).

The "same hp/tt as the grade rule" is more reasonable. The 5000 ton train would need 10,000 hp on a 2 % grade, 3 or 4 SD40's.

By the way, all these rules tend to become less accurate with AC engines. They can go down to about 1/2 to 1/3 hp/tt per percent of grade because they have way better slow speed traction. Where a regular DC powered train might have 1.5 hp/tt, an AC powered train might get away with .8 hp/tt.
 



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