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Re: Highreeving engine - at what rpm? - 09-18-2007, 04:41 PM

Quote:
Originally Posted by martinbo View Post
Ooo vedjo01

I don't think that by increasing piston speeds you will necessarily lose torque; torque is a product of displacement and not piston speed. Increasing piston speed will produce more power since this is a product of either:

- Stroke length
- RPM
Torque by itself is not RPM dependent, however engine torque is because of the cycle speed.


Engine Torque=P*A*R*Z
P= effective piston Pressure
A= Piston cross sectional area
Z= Number of pistons/Cylinders
R= Radius arm of the piston.

Because we are dealing with a dynamic state, the effective piston pressure is engine speed dependent.
The effective pressure is dependent on the volume of fuel+air mixture:heavier fuel will give more pressure, so will more air in the cylinder
The effective piston pressure is maximum at zero speed and exponentially decay with increasing engine speed. This decay is due to not having enough time for air to fill combustion chamber(turbo engines addresses this problem)

So then, why does torque increase at then decrease? The reason for this are the following:.

1. At zero speed, only half of the cylinders are generating torque, and since this is a cycle, you have to average for the entire cycle.
2. As you increase the speed, more cylinders/unit time contributes to the engine torque, however the torque generated per cylinder decreases exponentially
3. There reaches a speed where the torque generated per cylinder/stroke is so low that the overall engine torque starts to decrease.
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The Following 6 Users Say Thank You to chonkoa For This Useful Post:
Imhotep Evil (09-20-2007), martinbo (09-19-2007), siko (09-19-2007), Sunny (09-18-2007), Tycoon (09-20-2007), vedjo01 (09-21-2007)