# Formula for finding the optimal weight for a given torque and vice versa

Whenever i see videos of two trucks pitting with each other to see who has a stronger, i always see comments that sounds like:

If you load the other truck more it would beat the other one

That truck is an all wheel drive and the other is 4 wheel drive

It got me thinking how to you calculate the optimum weight for a given torque of a four wheeled vehicle.

To make everything simple lets assume that out truck is a Tesla truck where each motor can generate torque equally on all the wheels and the wheels are evenly spaced. If the the motor on each wheel can lets say generate a 1000kg-cm of torque how would i find the optimal weight so that the truck can push to its maximum potential.

My follow up question would then be. What happens if the other truck has double the torque at half the weight which one will be able to push better? What would then be the formula for the pushing power?

• Can a light breakdown truck rescue a heavier truck? Answer is yes... Jan 5, 2020 at 20:39

It is more question of the friction than torque.

The truck with the greater torque will be pushed back the moment its tires start to skid.

Torque on a tire will be delivered to the pavement very effectively as long as static friction conditions are involved (no slipping). But in the case of slipping the kinetic friction is the active force and most of the work of the torque goes to heat and burning the rubber.

In most of the tires on a paved dry road, the kinetic friction coefficient is around 0.3 to o.4 and disastrously lower on wet roads.

This implies a truck with a smaller torque if loaded say two times as much on each tire as the truck with higher torque will win. $$F_{friction}= weight*\mu$$

• So weight is the determining factor for these kind of bouts, which means that whichever is heaver would win, so long as their engine/motor can give enough torque to spin the wheel? If i undstand the equation correctly whoever has more F would win. Is it safe to assume that whichever truck with the lower F would most likely/ definitly skid?
– Jack
Jan 5, 2020 at 17:47