I always thought engines were at there most efficient at or close to their peak torque rpm, which would put it right around 1,800-2,000 rpm on the Cummins.
Cam
Cam
Originally posted by jponder
I'll bet almost any truck will see a gain of 5MPG just by slowing from 70 to 50. I bet that truck that was running 80 would gain 7-8MPG by slowing to 50. [/B]
Originally posted by nwilliams98
Fd=Cd*Af*V^2
Power=Fd*V
Therefore the amount of horsepower required to go a certain speed increases with the cube of velocity. You need 8 times as much horsepower to go twice as fast. This of course neglects rolling friction but applies quite well to your planes.
BTW. I'm doing my Mech E thesis on light truck drag for the Army's national automotive center. It basically consists of a CFD optimization. So far I've reduced the drag on a Ram by 33%. I'll post the results, and methodology to the board when I finish. (If the Army does not mind).
Originally posted by nwilliams98
I'm not sure of the exact Cd but think it is around . 55.
I'm running simulations to calculate the viscous and pressure forces directly so the Cd will be calulated later once I settle on a shape.
At 70mph you are looking at about 1000 N of drag for a stock 4x4.
I've never thought the Fd equation (mentioned before) was a perfect fit using Cd as basic shape fudge factor anyway. There is probably dependence between Cd and V.