Ok assuming:
Any engine that produces 300 ft-lbf of torque at 3000 rpm has 171 hp at that speed. Regardless of it's stroke, bore, displacement, supercharger, etc. Also, any engine that makes 300 ft-lbf of torque at 5000 rpm will produce 286 hp.
This is due to the simple relationship between power and torque:
power = (torque * rpm) / 5250
where power is in horsepower, and torque is in ft-lbf. Also note - at 5250 rpm, hp is equal to torque numerically.
I know that my standard GT TDi Pd 150 produces max 150 at the flywheel and max torque of 245 ft Ib. Now the question is if we can work out at what RPM the engine produces its max BHP does that mean that its best to shift up gear rather than reving to make use of all the 245 ft Ib of torque again.
From the calculation above: power = (torque x rpm) / 5250
therefore 150 (Max BHP flywheel) = (245 x X) / 5250
I get 3200 RPM value for X ??? Does this mean theoretically that max power is at 3200RPM ??
Discuss.
Any engine that produces 300 ft-lbf of torque at 3000 rpm has 171 hp at that speed. Regardless of it's stroke, bore, displacement, supercharger, etc. Also, any engine that makes 300 ft-lbf of torque at 5000 rpm will produce 286 hp.
This is due to the simple relationship between power and torque:
power = (torque * rpm) / 5250
where power is in horsepower, and torque is in ft-lbf. Also note - at 5250 rpm, hp is equal to torque numerically.
I know that my standard GT TDi Pd 150 produces max 150 at the flywheel and max torque of 245 ft Ib. Now the question is if we can work out at what RPM the engine produces its max BHP does that mean that its best to shift up gear rather than reving to make use of all the 245 ft Ib of torque again.
From the calculation above: power = (torque x rpm) / 5250
therefore 150 (Max BHP flywheel) = (245 x X) / 5250
I get 3200 RPM value for X ??? Does this mean theoretically that max power is at 3200RPM ??
Discuss.