Horsepower/torque help must figure out thrust?

Question

First, if i have an engine with 6.5 horsepower at 3600 rpm, the torque is measured from the outside of the crankshaft of diameter 3/4", right? so if i put a sprocket on to drive a chain i have to divide my torque of 9.48 ft-lbs by the ratio between 3/4 and the diameter of the sprocket? does this affect horsepower AND torque or does the hp stay the same due to the increased linear speed?

Also, if i am supplying a certain amount of torque and hp to the rear drive sprocket, how do i figure out how much "thrust" i am supplying to the ground, say if its 40 ft-lbs on the sprocket and the wheel is 2 times bigger, do i divide 40 ft-lbs by 2 and get 40 pounds force, or what?

Ive been trying to figure this out for awhile so the more detailed the better, thanks.

Answer 1

Horsepower and Torque output by the engine does not change, the only thing that changes by changing the diameter of sprockets and wheels is the force developed by the torque at the sprocket and at the wheels.

If your engine has a torque rating of 9.5 foot lbs, and you put a1 foot diameter sprocket on the shaft then the force on the chain on the sprocket would be 9.5 foot lbs/0.5 foot sprocket radius = 19 pounds.

If your engine has a torque rating of 9.5 foot lbs, and you have two rear wheels then the torque delivered to each wheel is 9.5 foot lbs/2 = 4.75 foot lbs each, and if each wheel is 2 feet in diameter, then the horizontal drive thrust against the road of each wheel is 4.75 foot lbs/1.0 foot tire radius = 4.75 pounds.


The torque delivered to the sprocket is the same torque that is delivered to the wheels, or 9.5 foot lbs.