Torque calculation of an vehicle

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anoop
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hi,
i have 8000 lb machine it has 4 wheels (just like car), i want it to move on 2 ft/sec speed on cast iron wheels with steel track at flat surface. radius of every wheel in 5 inch. what will be the torque required to move this machine?? and what will be the torque on each wheel?? i want to put two motors on two drive wheels... and which type of motor is suitable in 3 phase motors
please help...
thanx in advance
 
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Hello anoop,

torque is needed for acceleration and to overcome friction. Suppose the friction can be neglected for the moment. Then the required torque depends on how fast you want to accelerate the thing to the desired 2 ft/s ! You'll need to look up some formulas to link angular acceleration and linear acceleration.
 
The size of your motor is determined by the amount of power needed.

The power needed to move your vehicle is the sum of the resistance forces (which equal the force at the driven wheels) times the speed of the vehicle. Some power will be lost in the transmission due to friction (Even if your motors are connected directly to the wheels, there is the wheel bearing's friction to consider). This transmission loss is usually estimated as a percentage of the input power.

The resistance forces are the following: Rolling resistance, aerodynamic drag force and the inertia (mass X acceleration. Not present when speed is constant).

The maximum constant power output (assuming an adequate torque output vs speed), will also dictate an appropriate time to reach your 2 ft/s (i.e. acceleration desired).

Using this simulator, assuming an 8000 lb vehicle with a 0.75 friction coefficient and a 0.004 rolling resistance, you get a 0-2 km/h (≈ 2 ft/s) in about 5 s with 0.25 hp at the wheels and a top speed of 4 km/h (At this speed, drag and weight distribution is pretty much irrelevant, unless something is really out of the ordinary).

If you want to do it within 1 s, you need 1 hp (which could also give you a 17 km/h potential top speed if power is delivered appropriately).