That's an interesting method of calculating power... I'm not entirely sure it's correct. Probably because I gave an incomplete explanation compared to kinetic energy.

PE=mgh only requires a difference in height.

g is always equal to gravity (vertical movement) and not the speed at which something moves laterally (generally speaking, velocity).

Say, if 200kg of mass were to climb 1m above ground, you would get this:

m = 200kg

g = 9.8m/s^2

hA = 0m

hB = 1m

= mg(hA-hB)

= 200 * 9.8 * (0-1)

= -1959 kg*m^2/s^2

Now, because energy is never lost and changes from one form to another, Law of Conservation of Energy, when an object is on the ground (height = 0), it has zero potential energy and all of it is converted into kinetic energy.

Formula for KE: KE = (1/2)mv^2

Since we know the (m) mass to be 200kg and we know that energy conserves itself, or changes forms, KE = PE, we can work out this:

KE = 1959 kg*m^2/s^2

m = 200kg

v = ?

Using the above formula, but rearranged to solve for (v)

v = sqrt (2KE/m)

= sqrt (2*1959/200)

= sqrt (19.59)

= 4.43 m/s

So that would mean, assuming constant velocity, that you lose 4.43m/s of kinetic energy going up a hill and gain 1959 Joules of potential energy.

Conversely, that means you gain 4.43m/s of kinetic energy at the base of the hill by trading it with 1959 Joules of potential energy.

I think that I have managed to work this out - I take the weight to be 3,500kg and the speed to be 22.35 metres per second. The delta height per metre is 1/894 of a metre, and, taking your constant for g at 9.8 m/s^2, we get 38.37 joules per metre by multiplying 3,500kg by 9.8 and (1/894), which, multiplied by 22.35 meters per second gets us 857.5 Jules per second - in other words, Watts.

Rounding this up, this requires deducting only one kilowatt from the calculated power, giving a total power of 37Kw. Thank you very much for your assistance!

Delta height should be total height difference, not a ratio. The gradient ratios are not useful unless we know the horizontal distance to work out total height difference.

using what you have

m = 3500kg

g = 9.8m/s^2

vA = ? (possibly 0?)

vB = 22.35m/s

hA = ? (peak of the hill)

hB = ? (base of the hill)

Hopefully, these would explain it better than I:

http://science.howstuffworks.com/engineering/structural/roller-coaster3.htmhttp://www.physicsclassroom.com/mmedia/energy/ce.cfmA key formula, would be this combined one

KE initial + PE initial = KE final + PE final

This one, which is possibly what you want, is slightly beyond what I have learned or can remember

KE initial + PE initial + W external = KE final + PE final

W = Work done, which would be the locomotive accelerating, or putting additional kinetic energy into the total. I think it should be something like joules * time.

PE final, would always be zero in this case, since we can safely assume the train only reaches top speed at the base of the hill.