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Old 01-26-2017, 12:55 PM   #1275
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Member#: 395882
Join Date: Jul 2014
Chapter/Region: MWSOC
Location: GR, Michigan
2016 WRX


Originally Posted by wadeh View Post
Yes. I get this.

But would those higher cylinder pressures not result in an increase of the calculated torque (work) that is reported by the dyno?

In other words, isn't this effect accounted for in the dyno graph? When we say that an engine produces X amount of torque at 2000rpm and the same amount of torque at 5000rpm, that it's seeing the same cylinder pressures accounting for all of this?
No, because that assumes peak pressure is directly related to torque, which is untrue. Think about a single cylinder engine. The shaft doesn't produce a constant torque because it only has firing pressure 1/4 of the time. On top of that, the peak torque might not occur at peak cylinder pressure because of the small ATDC crank angle. Anything near TDC will have a really small moment arm to translate the force into torque. Work is Force * Distance, sort of like how torque is Force X Distance (scalar vs vector multiplication). If you graph the force (pressure) vs the crank angle (distance), its the area under the curve that will tell you how much work (torque) the engine is producing at a given RPM. As the piston comes down, the pressure is reduced, but the length of the moment arm increases, like so:

More concisely, BMEP (brake mean effective pressure) is a calculated value based on torque output and engine geometry. Actual cylinder pressure is all over the place and some calculus turns it into constant power.
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