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I'm trying to find the required Newton or Nm of torque to move 6000 kg up a grade at 100 km/h.

I'm trying to work out the closest suited engine platform to do the job while been efficient, durable,cost effective and most of all required torque.

  • Mass: 6000 kg
  • Speed: 27.77 m/s = 100 km/h
  • Grade: 7° = 12%
  • Wheel size: 830 mm
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    $\begingroup$ Is this a homework problem? $\endgroup$
    – Daniel K
    Jan 14 '18 at 0:08
  • $\begingroup$ No it is not, general question. I am looking repowering or upgrading my vehicle and my calculations seem to unlikely at best. $\endgroup$ Jan 14 '18 at 6:30
  • $\begingroup$ Hi @Q.stratford it would be great if you could supply us with a simple sketch. I am not sure if I understand the setup. Are you assuming that friction can be neglected? And do you have any specifications for the rate of acceleration of the 6000kg mass? $\endgroup$
    – rul30
    Jan 14 '18 at 10:54
  • $\begingroup$ if you have calculations, show what you've done so far and we can check. Beyond that, my suggestion is to start with energy, not torque. Work out how much horesepower you need to lift the given mass at the given vertical rate. After that, you can calculate torque from the power and the wheel RPM. $\endgroup$
    – Daniel K
    Jan 14 '18 at 12:07
  • $\begingroup$ The acceleration required is a key point - and is it able to have a run-up on the flat to reach the required speed? The speed itself is not connected to the torque required to pull a mass up a slope, unless it affects the friction/air resistance, which we know nothing about. More information required! $\endgroup$ Jan 14 '18 at 13:25
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To overcome hill resistance, you need 2977 Nm or 2195 ft*lbs of torque at the wheels.
Assumed that the gravitional acceleration in your situation is $9.81m/s^2$, as it approximtely is everywhere on Earth, your weight is:

$F_g=m*g=6000kg*9.81=58860N$

The hill resistance which drags you backwards is :

$F_h=sin(a)*F_g=sin(7deg)*58860=7173N$

To keep any given speed, the force must be the same at the outer side of your wheel. So any speed that is achieved, is maintained. (theoretically, no friction included)

Assumed you meant that the diameter(not radius) of your wheels is 830mm, this translates in a torque at the wheels of

$T_w=F*r=7173*(0.830/2)=2977Nm$

To drive your vehicle to the speed of 27.78m/s up a hill of 7 degrees, you need

$P=F*v=7173 * 27.78 = 199.257 kW$ (or roughly 270hp)

of power, when there wouldn't be any friction.

But there will be friction, so you'll need more power and torque. This isn't calculable though, since there isn't enough information supplied in the question. At a normal asphalt road with an expectable loaded truck of 6 tonnes, i'd guess that you need roughly 50% more power, and therefore 50% more torque at the same speed. You also need a margin to accelerate to that speed in not-too-much time. NB: this is just a wild guess, don't rely on it for any important decisions.

Don't forget that the transmission in a vehicle will lower the torque demand at the motor, at the price of the demand for a higher angular speed of the engine, relative to that of the wheels.

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It seems you are looking for power, not torque. Torque (together with RPM) is just one of the factors for calculating power and can be amplified or reduced by using gears. It is not possible to just give a needed torque figure for an engine.

All in all there are too many variables missing anyway. For all i know we could be talking about a moving vehicle or a stationary setup like a cable elevator, which could lead to a different solution.

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  • $\begingroup$ Welcome to Engineering.SE, a Question and Answer site. This is not an answer, it is a request for clarification. As such, it should have been posted as a comment and not an Answer. Once you get more rep you'll be able to post comments. $\endgroup$
    – AndyT
    Jan 15 '18 at 16:14

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