Actual Rotary Displacement Request

[vex]

Originally Posted by Monkman33

the reason the displacement is usually calculated on 1 rotation is because that is the best way to compare on a consistent basis.

While I agree that you bring up very good points, I also maintain my stance. However, as usual, every darned thing is always a bit different when it comes to these engines.

I doubt we'd ever see a thread like this on any other type of forum.... aside from jet turbines. :-)

By the way, that diagram is great!

As far as the argument that the engine is technically a 3.9l is extremely strong because of that. However, I think that figure would be divided by 3 to make it equivalent to its piston engine counterparts. But I am no expert.

When it comes down to it, the only argument that I see as truly weak is the argument that it could be a 2.6l. 1.3 for1 crank rotation. 3.9 for total of all faces regardless of number of rotations.

I am too the point where I am willing to accept either of those.

I don't think you're getting it. Ignore crank and eccentric shaft rotation. Take the rotor from TDC to BDC. The same process is done for pistons. TDC to BDC. Don't worry about the faces of the rotor but only consider tdc and bdc of the rotor and the volume that is 'ingested'. Do this for every rotor in the engine and you'll get the Mazda displacement as well as the equivalent displacement of the piston engine. Again the procedure is held constant across all engines (not just rotaries, not just 4-strokes, not just 2-strokes, and not just 6-strokes).