# Thread: torque, RPM, friction, top speed and acceleration

1. this is what i know.

wheel RPM is on a ratio with engine RPM.
if the engine spins at 1000rpm and the wheel was directly linked to the engine, the wheel would also spin at 1000rpm.
torque does not increase with engine RPM, but basically creates an arc,
which varies hugely from engine to engine.
torque decides acceleration, depending on the weight of the vehicle,
and friction between wheel and vehicle.

i also know that if the torque exceeds the friction, the wheels will spin out.
what i dont know, is the numerical relationships here.

anyone got a reliable formula?

say if i put 150 Nm to a tyre of a diameter of 65cm, and the vehicle weights
3600kg, the friction of the ground is about 0,4 and the friction area is about 100 square cm, how fast is the vehicle going to accelerate?  2.

3. Here is a free program that looks like it will calculate most of what you want.
http://locost7.info/gearcalc.php

If the coefficient of friction is 0.4, the highest possible acceleration is 0.4 g, or 3.92 meters per second per second. That could only be obtained if you have all wheel drive or if practically all the weight is over the driving wheels as in a dragster. This comes from Newton's law F=ma and the formula for frictional force F_friction = N* coeff of friction where N is the normal force and is equal to the weight on the driving wheels, m*g.

What if the weight is distributed half on the rear, driving, wheels and half on the front, which are not driven? Then the maximum accelerating force is N*(coefficient of friction) = 0.5m*g*0.4 and the acceleration is F/m which is equal to 0.5m*g*0.4/m=0.2 g or 1.96 meter per second per second.

The area of the tire contact does not come into play in this calculation. The wide tires used on drag racers spread the heat out over more surface so the tire does not burn up too fast. It doesn't really increase the frictional force.

As far as the torque is concerned, if the wheels turn one revolution for each turn of the engine, and if there are two driving wheels, the torque will split between the two wheels and will be half the engine torque. Then the force on each tire tread will be equal to the torque on the wheel axle divided by the radius of the tire, and the combined force of the two wheels will be the accelerating force, if less than the maximum permitted by the coefficient of friction.  4. i just discovered that friction force is measured in newtons, so that helped me out a lot.
can i replace mass with ground pressure, when measuring the friction force?  5. Originally Posted by dejawolf
i just discovered that friction force is measured in newtons, so that helped me out a lot.
can i replace mass with ground pressure, when measuring the friction force?
No. Pressure is a measure of force per unit area and has the units newtons per square meter. You need the force in newtons. Mass has units of kilograms, so that's not a force either. To find the force exerted by a mass you need to multiply it by an acceleration to get your answer in kilogram meters per second per second, which is newtons.

The force exerted by the car on the road (its weight) is the mass multiplied by the acceleration due to gravity, g, which is 9.8 meters per second per second. Something that has a mass of 1 kilogram weighs 9.8 newtons.

The frictional force when the tire is sliding or on the verge of sliding, is the weight multiplied by the coefficient of friction.

Actually that's only true on a level surface. On a sloped surface you have to multiply the weight by the cosine of the angle of the slope to get the normal force, and the frictional force is equal to the normal force multiplied by the coefficient of friction.  6. hmm, so how do i factor in the PSI on the ground?
do i divide the force on the area?

i'm trying to figure out the traction for a tracked vehicle.  7. Yes, the ground pressure is the force divided by the area. But, as I mentioned earlier, the frictional force you calculate will not be dependent on the area.

A tracked vehicle with large track area and low ground pressure does have an advantage because it is less likely to sink in and bog down, but that won't show up if you are just calculating the frictional force using the coefficient of friction.  8. ok. well my thesis is that you can put more torque to a wheel with a larger ground contact area.
dont know how true that is.  Bookmarks
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