http://www.drivingsports.com/site/2008/ ... uivalency/1 version
In a rotary engine, displacement is a bit trickier. Each rotor has three faces, or working chambers, and there are two rotors in a 13B engine. All six chambers have a displacement of 0.654 liters, but it takes three revolutions of the output shaft to bring all those chambers through a full cycle. Were it to be rated by the same method as a piston engine (chamber displacement times number of chambers), it would be called a 3.9 liter, which is hardly logical. The accepted approach is to treat each rotor housing as a cylinder and measure the displacement of one chamber per rotation. Interestingly, this happens to be precisely equal to capacity. In other words, on every revolution of the output shaft, the rotary engine moves a volume equal to its rated total displacement.
the real version
http://www.rotarywiki.com/index.php?tit ... splacementHere's a little info on how to properly figure out displacement on a rotary engine. Everyone argues that it is really a 1.3 liter while others argue that it is really a 2.6 liter engine. They are both wrong! If we look at how a piston engines volume is calculated we arrive at a displacement based on total swept volume of every piston added together. It is not based on rpm. On a rotary, displacement is figured using one rotor face in one complete revolution then multiplied by 2. This only leaves the total for 2 combustion chambers though and the rotary has 6! Since the volume of a 13b rotary is rated at 1.3 liters (only 2 combustion chambers) it really adds up to 3.9 liters!!! I can hear it now, "...but we only have 2 rotors!" So what! Like I said it makes no difference if there are 2 rotors with 6 faces or 6 rotors with one face each. the total is always 6 and the base numbers are only based on 2 chambers. The rotary merely does 3 times the work in a package 1/3 the size. It's just a 3.9 liter engine crammed into a 1.3 liter body. Just so none of you start a fight over this, I will explain this later so don't chastise me yet!!!
In case anyone is curious I did some math to determine what the 13B rotary would be sized at if it were a piston engine. The results are pretty neat. First of all the rotary would be a 3.9 liter, 6 cylinder engine. It would be a 6 stroke. Each cylinder would be 6.54" across (damn big piston!) but the stroke length would only be 1.18" in length peak to peak. Not much there. Interesting isn't it. Now just imagine a way to make all this work with only 2 intake runners!
In all fairness to the terms I have used, the word "stroke" can be interchanged with the word "cycle" since both technically have the same definition. The terms "periods", "quarters", or "phases" can also be used correctly. I merely wrote it the way I did to get a certain mental picture going.
I have already dealt with why the rotary engine is really a 6 stroke engine and why displacement is really 3.9 liters and not 1.3 liters. Now I need to explain why the rotary engine doesn't have the torque or horsepower of a good 3.9 liter engine or why it doesn't get the gas mileage of a 1.3 liter engine. The world has always wondered so here's why.
Remember that I stated that the true displacement of the rotary engine, if figured out according to the way piston engine volumes are calculated, is according to the total number of rotor faces and not the number of rotors, nor does it have anything to do with rpm. This added up to 3.9 liters for a 2 rotor 13B engine and not the published spec of 1.3 liters. They just crammed all 3.9 liters into a 1.3 liter body. If the engine is really a 3.9 liter engine then why doesn't it have the low end torque of a 3.9 liter engine? This has a very simple answer. Lack of leverage. OK, what the hell does that mean?
1.3litres yes
2.6litres yes
3.9litres yes
it took a long time coming
