MikeC

I know this topic has been beaten to death but this article provides a fresh pov. Everything I've read has always compared rotary and piston engines based on a "per crankshaft rpm" kinda basis. This doc shows why this is not valid.

http://mikeonline.cable.nu:1863/misc/rotor.doc

I'd be interested to hear peoples opinions on this.

DSMguywantsFD

so if it has 1.5 the gearing of a piston engine could you possibly alter the gears so it has lower gearing and more torque?

MikeC

If you changed the gear ratio that is inside the engine then the motor would not work. You could add another gear but there isn't much point because you can just compensate for it at the diff.

The rotary really has 1.5 times more torque than it is rated at because of this internal gearing and probably should be rated as such. This applies to most figures, the engine really revs 1.5x less than it's rated at, inducts 1.5x more etc. Power is the same because it's not affected by gearing.

This is the same as a piston motor with a 1.5x gearing up at the output shaft, it would produce 1.5x more revs, 1.5x less torque, have a 1.5x flatter torque curve and need a 1.5x lower diff ration to compensate. But really it's the same engine producing the same revs and torque.

spoolin93r1

so what's all that mumbo jumbo mean to a simple person?

maxcooper

MikeC,

I understand your perspective and I think you did a great job in presenting it. Before I go on, I want to stop here and recommend that anyone interested in this topic read and seek to understand MikeC's explanation. I think the great displacement debates are 99% people not understanding how the engines work and 1% people having different points of view on which parts of comparing displacement are most important. MikeC's document lays out a very reasonable argument that is technically sound. I just disagree on the other 1%.

MikeC's explanation may in fact be the best unqualified answer for the total displacement of a rotary, but it turns out to not be that useful when measured against the current state of the art for each type of engine. The redlines of performance-oriented engines of both types are on par with each other these days, so I think the 2.6L rating for the 13B is a reasonable compromise. Though I am perhaps just a rotary apologist -- with performance-oriented 6-cylinders revving to 8-9000 RPMs from the factory these days, I might just be making excuses for the production 13B having a redline of "only" 9000 RPM when it should be 13,500 RPMs. I believe the 13B is closer in weight to a 2.6L than a 3.9L piston engine as well, which also favors 2.6L as a useful displacement rating for the 13B if comparisons to piston engines is the goal.

I think the "compare by how much air is ingested per rotation of the main shaft (2.6L)" and MikeC's total displacement ratings (all chambers = 3.9L) are the most defensible. I am not a big fan of the 1.3L two-stroke explanation since I think it misleads people to think that the rotary combustion cycle is more like the common 2-stroke cycle than the full four cycle combustion process that it does indeed use. Though I do recognize that each rotor fires every two strokes, I still think it glosses over the issue of the real combustion cycle. The "Mazda says it's 1.3L" argument is weak but perhaps valid if you really only care about governmental regulations or want to stick it to domestic guys when you beat their 5.0L cars with just 1.3L. That's not my angle and the argument is usually presented in ignorance of how the engines actually work, so I don't like that one much. However, each explanation makes some sense in a particular context. So the debates continue...

Oh yeah, is it safe to use synthetic in my rotary? (PLEASE DON'T ANSWER THIS!)

-Max

maxcooper

The fact that the engines are connected to engine dynos directly by the output shaft (no gear up/down) may be another reasonable argument for the 2.6L rating if comparing to piston engines is a goal.

-Max

maxcooper

Well, I think it means there is no simple explanation. In what context do you wish to use the displacement rating?

-Max

MikeC

Thanks, I appeciate you taking the time to read it.

I think the 2.6 litre rating is probably fairer for racing etc because I think the rotary is an inefficient 3.9 litre engine and should be rated lower. But I'm really talking about the capacity of the engine mathematically.

Different engines of the same capacity have different redlines. There are many production performance sixes that have a redline of 6000rpm, it seems the rotary is tuned to run at lower rpm than it could because 13,500 would be too much.

BTW, do these 6s that redline at 9k have a capacity of 3.9 litres? My guess would be that they are smaller engines which tend to rev higher.

The 1.3L argument is not even worth a mention :-)

Gearing is irrelevant because you run the engine into a gearbox, then a diff and then change the ratio again with tyre sizes. You should be able to alter the gearing of any motor (by adding a gear at the crank) and still get the same capacity.

In my opinion, the rotary is equivelent to a 6 cylinder 3.9 litre motor that has a gearing up added at the output shaft. If I took a 6 cylinder piston motor and geared the output shaft up by 1.5 times it would produce 1.5 times less torque and 1.5 times more revs, but would you really consider that the motor is revving 1.5x higher and producing 1.5 times less torque?

felix_is_alive

i believe even mazda doesn`t know what it`s got
everybody will know details about a specific engine (piston) beeing sixes eights or fours , but when it comes to rotaries you always hear "oh yeah ...THAT THING"
i guess you are what you are , in this case .....different

MikeLMR

where did you get the 1.5 gearing from ? 1 revolution of the output shaft = 1/3 of a rev for the rotor and visaversa 3 revs of the output shaft = 1 rev of the rotor

wwilliam54

they rate 2 cycle engines the same as 4 cycles
and rotaies have the same advantages of a 2 cycle

keep it the same

Kenku

Couldn't get to the word doc, first off. So my statements are going to be less specific than they otherwise would be.

The reason Mazda calls it a 1.3L is pretty self-explanatory, really. The maximum amount of water you can fit in one working volume is 750ccs or thereabouts, and this is multiplied by the number of working volumes. By the exact same method of measurement used by piston engines, it's 1.3 liters, period.

Directly comparing it to piston engines gets silly... hell, comparing anything based on displacement is silly. You'll note that most sanctioning bodies didn't use the 2x formula or anything else so simple, they adjusted it to whatever it took to keep everything competitive. For example, in the "Race" classes of Midwestern Council's autocross series, they're divided by displacement. Rotaries are rated at 1.8x.

I posit the following question. What difference does it make what displacement it is? Racing bodies already have equivalency formulas that have no direct basis in displacement, but instead based on performance. Mazda rates them at what their displacement is by the same reasoning used to determine piston engine displacement. Exactly what use is displacement? Oh yeah, marketing.



On kind of a seperate question, I'm wondering what engines you're thinking of, Max. Highest revving factory 6 I've heard of is the M3's motor, and that peaked at 7900 though I think they've detuned it slightly because of reliability problems. As for weight, I have to strongly disagree. I've done quite a bit of research about engine weights at various points, and I can say with a pretty good degree of certainty that the NA 13B is right about on par with some light weight 4-cylinders, for example Honda's B-series engines.

Shinobi-X

Outstanding document MikeC. I've read it, and I plan on reading it again to (re)form some of my own theories and opinions. While I can't say I agree with all that has been written, valid points have been raised in favor of all displacement arguments thus far.

MikeC

The rotary and piston engine are actually quite similar. They both work by having chambers expand and contract in a sine wave and both are 4 stroke.

The rotation of the rotor itself isn't that important. What is important is the rate at which the chambers expand and contract in relation to the output shaft. This is 1.5 times different to a piston motor. The rotary chamber expands over 270 degrees while the piston motor expands over 180. So 270/180=1.5. The other way to look at it is if the output shaft of the rotary is doing 9000rpm then the chambers are expanding and contracting 6000 times a minute, which is 1.5:1. If a piston motor is doing 6000rpm the the chambers are doing 6000, which is 1:1.

This is if you think of each rotor/housing as a single unit, but really it is 3 independant chamber that don't interact with each other. The fact that they share the same rotor has little meaning, they are just as independant as 3 pistons.

Try http://mikesdriveway.com/misc/rotor.doc

There are 6 working volumes of 654ccs in a 13B. A rotor/housing contains 3 independent chambers.

Using a factor of 1.8x or whatever is fair enough but the true displacement can be calculated. Why? Why not :-)

Kenku

Yeah, that URL did it.

A couple points you bring up I have a problem with.

You say that using the "air consumed per 2-revs of output shaft" methodology is incorrect because it fails to take into account the internal gearing. The thing is, it isn't. For that methodology, there's two things that matter; air flow and output shaft speed. Period. You point out that if you were to look at the same measurement off of the camshaft of a piston engine, it would be equivalent to something twice the displacement. That's perfectly true... looking in terms of airflow at RPMs though, it works out the same either way. Thing is, looking at the rotary engine in terms of airflow per output shaft revolution allows direct comparison to other high performance engines; we have the same fuel requirements, airflow requirements and so on as a 2.6L engine tuned to a similar level of performance.

A bit farther on, you say "Once the speed of an engine becomes too large the piston or rotor face has moved away from the fuel before the fuel has had enough time to push on the face. This is the main reason power of an engine drops off after a certain rpm. So the speed at which fuel can burn determines the number of times a chamber can expand and contract per minute." I'm sorry to say, but you're completely wrong. The speed at which the powerband drops off is because of breathing restrictions, on gasoline engines anyway. I attended a technical seminar a few years back on the effects of combustion chamber swirl on flame front propogation in a high speed research engine; IE a single cylinder motor approximating a Formula 1 engine. At 19,000 RPM, the flame front had reached the entire mixture by something like 20 degrees after the plug lit.

Yes, 19k RPM. Current F1 engines are making a bit over 900hp out of 3 liters, NA. That can't be done on rotaries (well, Mazda rotaries) because of breathing limitations.

Oh yeah and just FYI, I can't think of a series offhand where you'd be restricted to a certain final drive ratio except for things like showroom stock where Mazda would have done that for you anyway.


Well, here's my thoughts overall, having read it. The theory you have come up with is technicially correct; yes, moreso than existing equivalence factors. You did a very good job bringing up your points and supporting them. It's ultimately not very useful though. The 2x displacement is used because it's handy; rotaries spin the output shaft about as fast as a built piston engine, so we have an easy thing to compare to to find parts. The standard displacement was defined to be the way Mazda rates it back in the early days of development of the things. Hell, I'm willing to bet there's some written SAE standard about such things around somewhere.

My personal view is that I don't really care to bother to equate it to any displacement. It would confuse the people at Mazda to ask for parts for a 3.9L rotary, the bits for a 2.6L one would cost more than I can afford, so I'm going to call it a 1.3. Or, they want to call it a 2.34L engine for autocross, so that works too, whatever. I'm rather more interested in power output than displacement.

MikeC

This sort of comparison is fairly flexible. For example, it would be possible to determine that a F1 engine that runs to 19,000 rpm is twice the capacity of an engine that revs to 9500 but actually has the same capacity.

You need to compare the rotary to other engines with a swept volume of 654ccs. This is a reasonably large piston and I doubt there would be any production engines that rev to 9000rpm. Most will be more in the range of 6k to 7k.

Great point, this is the sort of info I'm after to try to make my article more accurate. However, the same thing still applies. If the rotary is doing 9000rpm then the fuel has the same time to enter the chamber as it has with a piston engine doing 6000rpm. This is the reason the rotary revs higher.

Yeah, I should probably remove that bit.

I accept that the rotary can be rated at the 2x displacement (although I would call this 2 thirds ) for racing but the actual capacity really is 3.9 litres.

Kenku

Well, true. Again, see my general disdain for displacement as a measure of these things.

Quote:

You need to compare the rotary to other engines with a swept volume of 654ccs. This is a reasonably large piston and I doubt there would be any production engines that rev to 9000rpm. Most will be more in the range of 6k to 7k.

Or less, true. Of course, horsepower peak is below 9k too on all stockport motors but the Renesis. The new version of the S2000 and the BMW M3 both have pistons with swept volumes of 550ccs. Ooh! Porsche 911 GT3 has a swept volume of 630ish per cylinder, and a horespower peak at 7900!

Quote:

Great point, this is the sort of info I'm after to try to make my article more accurate. However, the same thing still applies. If the rotary is doing 9000rpm then the fuel has the same time to enter the chamber as it has with a piston engine doing 6000rpm. This is the reason the rotary revs higher.

Well, I'm just kind of saying that I don't see rotaries revving all *THAT* much higher. The Renesis, sure, but even that's only incrementally higher than some production boingers. You could argue that that's just street cars and that race tuned engines are turning higher RPMs, but... well, the factory race cars had about the same RPM ceiling as the Renesis.

Quote:

I accept that the rotary can be rated at the 2x displacement (although I would call this 2 thirds ) for racing but the actual capacity really is 3.9 litres.

Well, I say *an* actual capacity really is 3.9 liters. Wankel rated the things at the way Mazda does, so it could be argued that he gets to say. I'm not disagreeing with you, merely saying that precedence is against you, no matter the fact you're right.

Shamrock.James

Sorry guys i will read it later from what i saw:

The only reason a 13b is more than 1.3 litres is the same reason the R26b is more than 2.6....

everyone else couldn't hack losing LEMANS

MikeC

True, displacement is a very poor measure of power. Someone once explained this to me better than I could have:

"The latest top of the line Detroit Diesel MBE 4000, which the company touts as "Big Power in a Small Package", with the 450 performance package (including turbo) makes 450 hp from a 12.8L inline six. It weighs "just 2,040 pounds". Big power in a small package. Good for transplant into Vettes."

There are other factors coming into play at high rpm such as apex seal speed but if you can get a piston engine with swept volumes of 654ccs to rev to 10k then it should be possible to get a rotary to rev to 15. This, of course assumes that the thing doesn't explode before it gets there. Most likely it will, which is why the rotary doesn't always do 1.5 times the revs.

Letting wankel or mazda set the standard is a bit like cigarette companies setting tabacco laws. You've gotta at least be sceptical

Steel

Well.. now i'm confused... is this a question of POWER, or a question if DISPLACEMENT?

KevinK2

Mike,

I fully agree with the comparison with a 3.9L 6 cyl boinger, with integral .67 OD before the output shaft. I include my note from this thread last year (08-25-03):

https://www.rx7club.com/showthread.p...hreadid=217838

" one rotor equals 3 pistons, but on a 'special' crankshaft with overdrive.

for the 13B, the best piston engine analogy is a very short stroke 3.9L 6 cyl, with a .67 overdrive gear built into the output end of the crank, before the flywheel.

at the flywheel, this gives the correct one combustion cycle per rev ( edit: per rotor, ie 2 per rev for each engine), and very smooth 270 deg power cycles for each piston, and 1.3L charge air displacement per rev.
my $.02 "

Not sure on your swept volume calculation though. Mabe I read it wrong. Point P displacement, normal to the rotor face, absolute or relative to the crank center, is simply 2xe over 270 degrees of e shaft rotation. This is intuitive from observing illustrations of TDC and BDC positions. The rest of the expansion is due to apparent motion of the housing surface. The effective displacement of the rotor face (projected) area is 3xe, but that includes the effect of the dynamic combustion chamber ... housing surface.

Evil Aviator

I think I can explain displacement in simple terms so that you guys can stop beating this dead horse. Do you understand the difference between a piston and a cylinder? Well, displacement is based on the cylinder, not on the piston.

For example, a 4-Stroke V8 may be rated at 5.0L based on the displacement of the swept volume of all its cylinders, no matter how many times the pistons produce a combustion cycle per crankshaft revolution. Now make that same engine into a 2-Stroke, and it is still rated at 5.0L, regardless of the fact that the pistons now have twice the combustion cycles as before.

OK, if you are with me so far, now look at the Wankel engine in the same manner. Its displacement is based on the swept volume of the rotor housing (cylinder), and it doesn't matter how many times the rotor (piston) produces a combustion cycle. The fact that the rotor has 3 faces has absolutely no bearing on the displacement of the rotor housing. In a similar manner, if there were such thing as a piston engine with one cylinder and three pistons that alternately traveled into the cylinder, its displacement would be based on the swept volume of that one cylinder, not on the 3 pistons.

Does this make sense now?

KevinK2

Got it. Forget combustion events per rev and total revs needed to complete a full engine cycle. Total swept volume ... bore area x stroke x number of holes. Problem is bore area is projected rotor face area (same as defined by mike), effective stroke is 3xe, and number of dynamic holes is 6. That gets you 3.9 L. In this case it takes 3 revs to actually do it, with 2 revs for the 5.0 4 stroke, and 1 rev for the 5.0 2 stroke. At least that makes all 3 ratings different from each other, regarding the 100% VE beathing rating per rev.

If you think a 9 lb cast iron 3 faced rotor wobbling aroung the e-shaft at 1/3 the e-shaft rpm is a single piston ... oh well to each his own. I visualize 6, clever, dynamic combustion chambers.

MikeC

FINALLY!!!!! SOMEONE AGREES!!!!! I'M NOT CRAZY!! I'M NOT CRAZY!!!! I'M NOT CRAZY!!!!!

Once you understand this concept it is so obvious it is correct. Every other theory always has something that doesn't quite fit but this explains everything. It's one of those things that just clicks.

BTW, how did you arrive at this conclusion?

I'll have to have a look at it again tommorrow.

Evil Aviator

Yes

The displacement rating is based on static conditions, not dynamic conditions, otherwise you would need to also include other factors like compression ratio and volumetric efficiency.

No, you are getting back to the rotor issue, and that is causing the confusion. Forget the rotor. There is only one swept volume are per housing. The other two areas of volume NEVER function in the intake cycle.