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The one thing that I think is questionable is your estimate of losses at 37%. The losses will be a combination of: • aerodynamic drag • rolling resistance (RR) – friction in rail contact and rolling losses in bearings. For a Japanese passenger train running at high speed, aerodynamic losses will be relatively high, but starting and stopping will also be important, depending on the drive cycle. Every time the train stops, you have to put back all the KE to accelerate it back up to speed. At steady speed on the flat, 100% of the power from the engine goes into rolling losses (aero + RR)… so it is the stopping and starting (S&S) that reduces this to 37% of the total for a typical passenger train drive cycle! ie RR/(RR+S&S) =0.37. The mining train's drive cycle is completely different: almost no S&S. So 37% is not the right number to use in your calcs.... I think it should be much less. Alternative Calc: An alternative to assuming a constant percentage energy loss would be to calculate the work done against the resistance… The total drag force will be: F= ½.r.V2 CD A + mgCR • CR is the rolling resistance coefficient of the train: about 0.002. (See Table here: https://en.wikipedia.org/wiki/Rolling_resistance ) m is the unladen mass; Dm is the payload. • CD is the drag coeff, and A is the frontal area... If the vehicle was of ‘truck’ length, CD would be about 0.3-0.5. But, for a 2km long train, CD must depend on the length because of the skin friction and hundreds of gaps between wagons with ‘mini-wakes’. (On a quick Google, I haven’t been able to find any suitable drag coeff for a mining train… needs a more thorough search.) On the other hand, for a mining train running at 60km/h, aerodynamic losses will probably be small compared with the rolling resistance, so you could most likely neglect it. The work done against resistance F will be FX (assuming steady speed), where X is the distance rolled (in metres). Downhill: The PE gain is: DPE = (m+Dm)gDh The energy loss into RR (ignoring aero) is: (m+ Dm)gCRX Dh = 600m CR = 0.002 X = 140km = 140E3 m CRX = 0.002(140E3) = 280… same order of magnitude as Dh=600. % Energy loss = Energy loss / DPE = CRX/Dh = 280/600 = 0.47 è 47% Uphill: PE Gain is: mgDh Rolling resistance is mgCRX Total energy required to pull the vehicle up the hill is mg(Dh+CRX) % Energy loss = RR / Total energy = CRX/(Dh+CRX) = 280/(600+280) = 0.32 è 32% So, I take it back… your assumed 37% was not too far off! D