I just posted the link showing the views of another, I know nothing of his credentials or why he believes what he does. Just figured that it was related to the discussion.

:-) Wasn't making a stab at you. the linked website was labeled 'rotarygod.' Was most definitely not making any reference to you. :-)

Food for thought... http://www.ncbi.nlm.nih.gov/pubmedhealth/PMH0001930/

Although I feel that you have some knowledge Rice, your approach and interpersonal skills are terrible. It's quite unfortunate. :uhh:

Hey, nothing wrong with being a bit Narcissistic:o13:

This topic went to much from its track and I already really don´t know what is being argued.

Rice is right in everything he said. Some people may take it to personally, that he suggesting that 13B has capacity of 3.924 L, when everyone and his mother knows, that 13B is 1.308 L.

Byt why arguing?

This keyboard war doesn´t change fact, that 13B is fully and without any artificial factor comparable to:
1.3 L two stroke
2.6 L four stroke

Should I post calculations of engine power, torque, BMEP etc.? Which clearly states that number of power pulses per time or revolution is essential?:beatdeadhorse5:

Remember the old saying: "Those who can't do it, teach it."
There are many bad teachers in the world who think they are good.
Getting paid to do a job is not proof of being good at it. Just look at most politicians.

It does not matter how you determine the displacement of an engine.
To compare engines of different designs; a fair method is to measure how much air is pumped through it in "N" revolutions. And this was mentioned.
Now you can not apply this to jet or rocket engines, or electric motors. course.

A bit, no, nothing wrong.

A lot, yes, something wrong.

:ugh2:

You are talking "equivalences and similarities" in relation to different styles of internal combustion engines... not Physical Displacement per revolution of the crankshaft.


The reason a rotary is comparable to a 1.3l 2 stroke is that it fires its full displacement each revolution.

The reason it is comparable to a 2.6l 4 stroke is because it separates the strokes and ignites the "equivalent" amount of displacement as a 2.6l 4 stroke each revolution.

But using your logic, a 2.6l 4 stroke is comparable to a 1.3l 2 stroke, and a 5l 2 stroke is comparable to a 10l 4 stroke. The fact of the matter is they are different types of ICE's, which means they will have differences and similarities. Doesn't change how displacement is calculated.

If 2 stroke versus 4 stroke changed how displacement was measured, then you better let the engine industry know they've been doing it wrong. lol. a kx500 is in fact a 500cc 2 stroke engine, but it you changed the ignition and head to a 4 stroke system... would that automatically make it something else? No. the bore area and stoke didn't. it still displaces 500cc in one crankshaft revolution. Tis is how it is done.

I honestly don't see the argument here.

Even in your examples: "engine power, torque, BMEP" is not used at all in calculating displacement. Guess what it is used for? Taxation and racing classes, to even the playing field.

And "power pulses per revolution" again has nothing to do with displacement. If it did, then a 500cc 2 stroke engine would only be a 250cc 4 stoke engine. But, this is inherently incorrect, as again, the stroke (distance of piston vertical travel) and bore area did not change.


Please don't take this post as an attack, but as you said, a 13b is just that. a 1.3l. while it may share similarities with different sized engines due to its type, it is still just a 1.3l. Under no standardized definition of internal combustion engine displacement will a 13 be anything other than 1.3l. Similarities do not change how physical displacement is measured.

Remember, displacement is derived from a constant that all of the mentioned engines share: The crankshaft/centershaft/eshaft rotation is used to transfer power from the engine to the rest of the drivetrain. This is the reason displacement is measured by one rotation/revolution of this crankshaft/centershaft/eshaft.

^ this!

i'm not an engineer. i have mechanical experience, but no paper credentials. i just want that out of the way before i offer my thoughts.

i've read through this thread a couple times now (and i read the rotarygod link last night) and it seems to me that no one of these assertions is wrong because they clearly state the context for which they claim.

the 1308 chamber derivation is irrefutable. the 2616 argument is just as valid in the context of the 2 rotor derivation. finally, the 3924 is also valid in terms of absolute displacement because it's the only conclusion that takes all 6 chambers into account regardless of their phase during a crank rotation.

i think the only true argument left is which, if any, is MORE right. i'm not qualified to make that determination. hell, i'm still trying to find out how 80 and 240 yield 1308. however, while i don't dismiss any of the others, i will say that i'm tending to lean toward the 3.9 assertion now. i think a part of the lingering ambiguity is simply us not knowing when and where to draw the line with the reciprocating engine comparisons. rotaries are different, there's no getting around that.

the points made for 3.9 are compelling (to me) simply because none of the piston calculations leave any cylinder uncounted, while the 1.3 and 2.6 assertions for the 13B do. that said, none of the piston calculations require 3 revolutions - though, in all fairness, revolutions are not relevant. when you plug bore and stroke numbers into the volume of a cylinder formula, then multiply by the number of cylinders, none of that takes revolutions of the engine into consideration. it's just the number for the engine's absolute capacity. crank lobe angles/phasing have no bearing. my MR2 Turbo was 1998 cc, not 500. my Audi engine is 2671 cc, not 445.

as for the dust-up, by now Rice should know i respect him very much. i actually like his un-PC style, it's a breath of fresh air to me - not to mention sometimes it's just bloody hilarious (see my sig :)). being somewhat socially inept myself, i won't comment on his interpersonal skills and i certainly won't judge him, but i think he should just step back and reasses the discussion. i don't see where anyone attacked or offended him (i know its not my call, but it is my observation) so he should return to the discussion with no hard feelings.

if you count every face, then you are using 3 rotations of the shaft, which now means you are using a different method of calculating displacement than every other engine.

This is why there is a standard: What does an engine displace per revolution of the shaft?

^ understood. :)

but what i'm saying is the crank rotations are not relevant - at least, if they are, i don't see it.

for example, my manipulation of the 86 mm x 86 mm MR2 engine, simply yield roughly 499.5... cc per cylinder, which i then multiply by 4 and get 1998 cc. engine revolutions never figured into it. if this engine block were used with a 2-stroke or 4-stroke head, it's still 1998 total capacity.

the rotary does require 3 complete revolutions to fire all 6 faces, but i don't see how that affects total capacity, which is in fact 6 chambers, and it is the volume of those 6 chambers that yield it's total capacity/displacement.

it's just the way i see.

The confusion with everyone is in how to view the rotors and what they are displacing in comparable fashion to a piston engine. Point-in-fact, you can not. As Rice has said it's a different cycle (though technically speaking, it's the same cycle as any other 4-stroke engine just with a different configuration--just look at the p-v diagrams).

http://www.grc.nasa.gov/WWW/k-12/air...mages/otto.gif
http://www.grc.nasa.gov/WWW/k-12/airplane/otto.html

Each face sees that otto cycle after 3 revolutions of the crank shaft. There's no two ways about it. That's what each face sees. Thus the total displacement for all 6 faces comes out to the 4L or so that Rice posted up.

Now here's where I think everyone is getting messed up: As I posted previously each piston engine calculates displacement by going from tdc to bdc of each individual piston. Now lets compare this to the 13B. How many times do we have to turn the crank shaft to go from TDC to BDC for the front rotor housing (note I didn't say each individual face)? 1 right? 1 turn takes the front rotor from TDC to BDC, while the rear rotor is 180* out of phase (thus displacing the same amount). From which we have the standard nomenclature of 1.3L of displacement.

Does this make sense to everyone?

This method is scalable as well. To think of this clearly do not consider the rotor faces as pistons, but rather the rotor as a whole as equivalent to a piston (or think of the rotor housing as the piston equivalent sleeve). If the 20B were to be calculated using the method described above you would simply take the TDC to BDC of the front rotor (aka 1 revolution). Since all other rotors move the same amount with that 1 revolution they displace exactly the same amount of air. If you do not like that procedure simply take TDC to BDC of each rotor (they only take one revolution each) and it will displace the same amount of air. This is where the Mazda Displacement rating came from. This is the number you use (at least on the Haltech) when you put in the engines displacement. There's no black magic here.

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).

vex ~

again, in as much as i understand what you have said in your post (i'll still need some time to understand the Otto cycle graph you posted :), but i haven't read the accompanying link yet) i can't disagree with what you're saying. in fact, i agree wholeheartedly. i guess the problem with me is that i agree with all the points of view because the lines of comparison are blurred.

however, it still comes down to our individual perceptions on where the lines of comparison (rotors vs. pistons) are to be drawn. i suppose the best thing may be making no comparisons at all, but it's probably not going to happen.

i've never thought of the housing surface as one cylinder (sleeve), so that is yet another point of view to consider - and in that context it is quite consistent with 1308. it's new, at least to me it is. i don't see it as any more (or less) right as any of the other points of view.

that said, it makes sense with the TDC-to-BDC definition of displacement. the eccentric shaft only allows one TDC per rotor and that would also be the reason why you said the rotor should be treated like a piston. i get that. again, it's not that i reject any of the other assertions, i just find myself gravitating to the 3.9 more than the others.

for what it's worth, regardless of what i posted before (or in the future) about the 3.9 displacement theory, i do still consider our beloved 13B to officially be a 1.3L powderkeg of fury - just not in an absolute sense. as far as what to consider any of the rotary engines, i believe Mazda's final and absolute determination trumps mine any day and i'm good with that.

You cant "just ignore" the rotation factor. its the only way to create something consistent and precisely equivocal between the two styles of internal combustion engines. If you want to compare the two engine types, you have to have a constant. What is more constant than the point that transfers the power from the point of generation to the drivetrain?

If you don't want to compare the rotary to a piston engine, then i feel that 3.9l could very well be more accurate in the aspect that in 3 rotations it does displace 3.9l.