jeeperxj91

If a truck had a peak horsepower of 200 @ 5500 RPM and peak torque at 250 @ 4500 RPM, and you decide to apply even more RPMs this actually decreases the amount of hp and/or torque? How does this happen, is the combustion just converted all into useless heat due to inefficiency of engine when the engine putting out RPMs above the peak amount?

mike37

it is more of the ability to input air and full at the hi rpm and timing
cam duration

ulendo

I had a really decent mechanics teacher way back, who came up with a great example for this. picture yourself riding a bicycle at X rpm. if you double x, you can do more work, double it again, and do 4 times the work, BUT theres a limit to how fast your legs can pump, and thats the limit to how much work you can do. engines are dependent on any number of variables, including cam duration, lift, intake length, port shape, chamber shape, ignition timing, plug gap, spark intensity, intake pressure, and - a HUGE item on modern engines - emissions restrictions.

HP, btw is a derived number. HP=(torque*rpm)/5250 ( usually) torque you can measure directly in Ft/lbs, KW, etc, thus the low(er) HP numbers found on most diesels that never rev to the 'break even' rpm of 5250. Having worked with a dyno on a daily basis, I also found a footnote in the operating instructions to the effect that while an absolute 100% clean engine dyno will almost always show HP & torque crossing over at +/-5250 rpm, a chassis dyno, showing power at the wheels will not do so, as it reflects driveline & gearing losses.

hope that helps

Rob92XJ

The drop is due to being past the operating limits of the intake, cam , cylinder head etc...