t-von

iTrader: (8)

Well here goes nothing! LOL

IMO shouldn't displacement be calculated based on a "SINGLE" combustion/spark event per cylinder/chamber? In a 1.3L rotary, we all know that the spark plugs in both cambers can only fire 1.3L at any given time. Think about this example. Lets take a 8.0L v8 piston engine. Regardless of engine revs, we know that if all 8 spark plugs fire "ONCE", 8.0L's of displacement would have been used. Same thing with the 1.3L rotary. Regardless of revs if both leading spark plugs fire "ONCE" the displacement used would only be 1.3L. Therefore any "ADDITIONAL" sparks in that particular chamber/cylinder shouldn't be counted as "ADDITIONAL" displacement. To classifly a rotary as a 2.6L would only be possible with an "ADDITIONAL" spark of both cambers. If you spark the individual 8 cylinders of the piston twice it now becomes a 16.0L engine. This is the reason why I agree that the rotary is a 1.3L engine.

MikeC

So what would happen if you devised a method of using only 2 spark plugs for the 8 cylinder motor, say 4 cylinders with the top of each cylinder joining in the one location.

t-von

iTrader: (8)

Ok I can visualize what your talking about. OK assume we continue to use the 8 piston design. Lets say that you now have 2 main joining cylinders (the actual place of combustion)that has 1L of displacment each and 4 pistons each that "eventually" join that one cylinder. Both cylinders regardless of how many pistons enter the main cumbustion cylinder will only fire off 1L of displacement per cylinder. That makes it a very inefficiant 2.0L engine because of the "shared" cumbustion area and addition rotating mass.

Here's one for ya. Visualize a regular v-twin motor cycle engine. Lets say that the engine is a 2.0L. 1L per 2 cylinders. If it was possible to divert one of the pistons to a "single" cylinder. The cumbustion area in that cylinder would only be 1.0L's. Even if you were able to shove 10 pistons into that one cylinder, it still would be a very inefficiant 1.0L engine.

MikeC

I didn't mean that the pistons share the same displacement area, just the same spark plug. If you took the v8 and mounted a single spark plug per bank on a sliding mechanism, it would be possible to slide the spark plug over to the piston when it needed firing. This would be a total waste of time and effort but would be possible. Would this engine go from being an 8 litre down to 2?

This is bascially what the rotary does except that it bring to chamber to the plug instead of the plug to the chamber.

SPiN Racing

Not the Chamber.. the Rotor Face... Tvon is right I believe.. You are once again confusing Cylinders... The V8 Model above had 8 Individual cylinders.. The rotary still has only two..

Spark Plugs is fine and all and is a good method... But.. Mike will still argue it.. no matter how you put it across..

He he OK GOing back to the garage to finish porting my 13B 2 Rotor.. or as Mike may say.. 26B.. or maybe 39B.. depending on what theory he is trying to promote at the minute..

Hmm sooo Then Evil's 3 Rotor in my garage would be a 4 Liter? Or 6 Liter?? ooooh Nasty..... Twin turbo 6 Liter making 280PS.. Pretty crappily running Big Block.. even with Twin Turbos..

MikeC

I'm not confusing anything. The rotary has six chambers. If you read any technical document from the experts they always talk about individual chambers.

Well, there are some obvious flaws in the idea of calculating capacity based on a single ignition event. Should I not reply?

Yes, the 3 rotor is 5.85 litres 9 chamber engine, that's why they make so much power. Maybe it's a bit low as a stock engine but they have the potential for big horsepower. I don't see too many stock 2 litre engines producing 280 hp with 600 being fairly easy to obtain.

KevinK2

Ok, but then, as I said it should be qualified as 3.9L in 3 revs.

KevinK2

Since no official displacement rating exists, you could ignore the other 4 rotor faces that are actively involved in the 4-stroke process, and just count the two that fire, in sequence, from each housing. That would give you 1.3L from 2 of the 6 active chambers, based on letting each spark plug fire once.

Firing actually corresponds directly to displacement. 13B fires twice per rev, so displacement is 1.3L PER REV. A 4 cyl boinger, with .65L per cyl, would be 2.6L, displaced in 2 revs, all chambers fired. A 2-cycle version of the boinger would be 4 firings per rev, for 2.6L in one rev.

But, this method, like the current swept volume method, does not provide consistent displacement per rev, as 4-stroke boinger is rated for 2 revs, and 13B for 1 rev.

And it's easy to imagine a .65L rating for a special 2.6L boinger, using your logic. Go back to the old valve in body design, like a B&S lawn mower engine. Now for inline 4, a single long slide plate, with one spark plug, forms the top of all cylinders. As each cyl is ready to fire, the plate is slid to put the spark plug over the firing cylinder, without loosing pressure or vacuum in the other 3 cylinders. 1 spark plug on the single 'block', .65L per combusion event, therfore .65L rating. As this 'spark' method ignores 4 of the 6 13B chambers, in this case 3 of the 4 active cylinders is ignored, and displacement is based on 1/2 rev.

A geat 'spark' method would be sparks per rev x chamber volume being sparked. Ths puts the 13B at 1.3L, and a 5.0 stang at 2.5L, and the 250 hp 2L S2000 at 1.0L. A 250cc 2-stroke bike would be 500cc.

Evil Aviator

Even more important than displacement, I have just determined that the 2Gen RX-7 is actually 16 feet long, as opposed to the 14-foot figure as reported by Mazda! Since there is no standard for measurement of a car's length, I took off my shoes and walked the distance myself, and sure enough, 16 feet! This obviously means that there is a conspiracy in which Mazda is trying to make the car sound smaller than it really is, much like the engine displacement conspiracy. I'm going to check the weight tomorrow to see if Mazda is lying about that, too.

KevinK2

Don't be surprised to find you weigh less than expected, since you are full of gas which you freely flatulate around here as fact.

It is interesting to see that when confronted by logical, factual information to clarify this subject, some respond in the Spinaltap movie tradition .... " but this amp has a dial that goes up to 11, so it's louder than all other amps that just go to 10." (sorry for ref if u didn't see the movie)

Evil Aviator

The sustain...listen to it...

I was actually thinking the same thing for quite some time now, but still you guys don't get it.

KevinK2

great flic

what I and a few others do get is the very creative concept of 6 dynamic chambers that act just like 6 cylinders for an otto cycle engine, but share some common surfaces.

live well and prosper in the false notion that a cylinder is eqivalent to a rotor housing.

btw, why did mazda accept the engine award in the 2-3L class? answer: they know their rating method is not consistent with ratings for all other 4 stroke engines.

FDNoi2egard

Ok, when I started reading this, I had a good theory. Somehow in the five pages, my brain pretty much shut down on the subject. The following is a desperate attempt to grasp the concept that was floating through my head earlier. Here goes. These are all idiot definitions btw because math makes my head hurt badly.


Displacement in my mind is the amount of air contained inside an engine at any given time. Therefore, if, at ANY given time you were to stop ANY engine, during ANY part of its ignition cycle, it would contain the same amount of air. Regardless of how many cylinders, rotors, or whatever, the engine only has a capacity to contain a certain amount of air. This is all assuming the density of the air is the same and all other factors are equal.

Thinking of it this way leads me to believe that the math should be extremely simple for anyone knowing the proper formulas. Output speed should not matter at all. The engine can still only displace a certain amount of air at any given time, regardless of how high its RPM's are. So, if you visualize an engine, piston or rotary, you stop it somewhere in its duty cycle and figure the volume of each individual chamber is capable of holding; you have just successfully deduced how much the displacement of the engine is.

What all this leads me to believe is that, since at any given time in a 13b there are six chambers containing relatively the same amount of air(remember: compressed air is still the same amount of air); then the displacement of a 13b is the volume of one of the chambers, multiplied by six. This obviously results in a final displacement of 3.9L.

I know this has been said a lot of times before, but I didnt notice any that put it as simply as this. I am not a physicist or an experienced engine builder, so I am sorry if this information is majorly misleading. This just seems the most common sense way to figure the displacement of an engine. In my mind, displacement equal volume, period.

Guerillah

Just throwing this in as it has always bugged me, if the rotary really is effectively a 3.9L then why doesn't it spool a turbo like it is? Seems more along the lines of a 2.6L when it comes to spooling turbo's.

KevinK2

Ok, freeze frame displacement concept. 1st must ignore the non-swept 'combustion chamber' volumes. gases can be exh or intake.

piston and rotor faces all vary in position from tdc to bdc, so on average, they are 1/2 way through stroke. consider 13B or 4 banger with .65L per hole. 13B = .5 x 6 x .65 = 1.95L !!! It is a novel rating. Banger is easy to visualize, with 2 pistons down and 2 pistons up, so instead of usual 2.6L rating it's 1.3L, as would be a 2 stroke version.

method fails to account for the 50% longer time it takes to fill the chambers, vs a boinger, so derate the 13B by 1/1.5 and you get the old 1.3L back.

I think t-von had basic concept right, but should just look at ( number of firings per rev ) x ( displacement per chamber/hole ). This is an ideal (100%VE) displacement in 1 rev rating. Double this to get ideal disp in 2 revs, which is effectively how all 4-stroke boingers are rated.

KevinK2

spools like a 2.6 boinger 'case it breaths just like one. to suck it's full 3.9L through all 6 chambers, it takes 3 crank revs. the 2.6 boinger does 2.6L in 2 revs.

rex u.k

Well i see it like Evil does and like MikeC does, it can be 1.3 or 2.6 or 3.9 but it can also be 8.7, i mean i can't really see a 1.3 or 3.9 giving 6MPG so i believe it's 8.7
Is this a good theory?

Sorry just trying to introduce a bit of light heartedness.

BTW, i like the way people plug theirselves on this forum,
Degrees,building race engines for 20 billion years, or you would know such and such a thing if you had been building formula1 race engines *LIKE ME!* or yes i am a scientist thermoassholeerectum scientest,blah ,blah,blah.

Sounds like a conspiracy.

MikeC

That's true but the number of revs is not relevent. It would be possible (but not practicle) to build a piston motor that did 100 revs at the output shaft for each expansion of a chamber. The real speed of the motor would still be the speed of the pistons, not the speed of the output shaft. I know the example sounds rediculous but this is exactly what is happening with the rotary, only it is a factor of 1.5, not 100. The real speed of the rotary is the speed of the chambers, not the output shaft. When the output shaft is doing 9000 rpm the chambers are doing 6000.

Another intelligent post from evil......

Maybe you should re-read some of your posts and mine.

Exactly.

This is an interesting concept. If you add up all the air volume in a rotary at any point in time you will get less than 3.9 litres. I don't know what it is but say its 2.2 litres. If you did the same thing with a 6 cylinder 3.9 litre piston motor you would also get 2.2 litres.

If you compare them at the same revs then they will be the same, but a rotary revs higher.

If you want proof:
http://mikeonline.cable.nu:1863/misc/certificate.jpg

MikeC

I look at the rotary the other way around, the shaft does not drive the chambers, the chambers drive the shaft and the chambers are essentially controlled by the intake. If you supplied a rotary and a 3.9l piston motor with the exact same intake conditions and supplied both motors with an equivelant horsepower load then the chambers would settle on a certain speed. Say it was 3000 expansion/contractions per minute, the output shaft of the piston motor will be doing 3000rpm but the rotary will be doing 4500. So the volume intake per second is the same.

Another way to look at it is not that the rotary has 50% longer chamber expansion time for the same revs, but that at the same chamber speed the output shaft is doing 50% higher speed. I consider this to be a very important distintion.

KevinK2

me: Ok, but then, as I said it should be qualified as 3.9L in 3 revs.

no engine has non-essential internal gearing, and your examples would easily be filtered out in a true L/rev based system.

the "speed of the chambers" sounds alot like rotor speed, which is 1/3 crank speed. I think your concept of real speed of a rotary muddies the water.

KevinK2

"So the volume intake per second is the same."

This gets to heart of the matter. Your simple 3.9L 13B rating, with no qualification as to 3 revs to displace that air, is really based on it's assumed ability to outrev the piston engine by 50%. In your paper, all factors except stroke cycle duration were ignored in assuming that is the case. IMHO, any 'displacement' method should be independent of size and number of rotors or pistons, for simplicity and sanity. I could easily construct a 3.9L V12 that revs to 9K that would pass emissions and put out a bunch more than 240 hp. If you argue that the 3.9L for comparrison must be a 6 cyl, then that engine's power capacity is limited by mechanical dynamic loads, and not combustion cycle duration as your paper assumed.

You shouldn't make different displacement rules for different piston or rotor sizes, although it's true per my example that piston size matters, as does number of valves, regarding maximum breathing capacity. May be that's where we differ. I look at the 4-stroke displacement rating as a fundamental common basis for engine comparison, not an exact predictor of maximum hp capacity.

The time factor aspect was relative to my reply to t-von's concept.

///////////////

I think your paper concludes that the 13B should be rated only on the total swept volume as 3.9L. And you imply that for any specific car speed, it must rev 50% higher than any 3.9L 4-stroke boinger to really match it's power. True?

MikeC

I didn't mean the speed of the rotor, I was talking about the speed at which the chambers expand and contract. If the output shaft is doing 9000rpm then the chambers are expanding/contracting 6000 times a minute. IMHO the true speed of the motor is 6000. If you looked at the 6 cylinder motor with overdrive it's the same situation, the output shaft could be doing 9000rpm but the pistons are only doing 6000. In this case it is very clear that the motor is really doing 6000 and the output shaft is doing 9000, it's not so clear with the rotary but true just the same.

It's based on the assumption that the motor can theoretically do 50% higher rpm. It probably doesn't due to mechanical limitations. The fact that it doesn't achieve this in the real world doesn't have any bearing on it's actual capacity, if 2 3.9 litre 6 cylinder motors had 2 different top rpm they would still be rated the same.

True but for the purpose of calculating capacity all other factors are irrelevant. It might be reasonable to scale the capacity for racing because the rotary can't achieve it's true potential rpm limit. Say a 3.9L six can redline at 7500rpm it will induct 7500x3.9/2=14625 litres per minute at 100%VE, a rotary running at 9000 redline will induct 6000x3.9/2=11700 (notice the chamber speed needs to be used). So the equivelant capacity for the rotary would be 3.12L. This would probably be a fair rating for racing but I think the true capacity is 3.9L.

100% .

Evil Aviator

Which award are you referencing?

You can debate comparison methods of different types of engines forever, and never get anywhere. My point is that there is a set standard for a displacement rating. A "class" rating is NOT necessarily a "displacement" rating. The displacement rating of the 13B is 1.308L, and that is that. A "class" rating can be anything that you want it to be.

If Mazda, a business, received a large award that yielded them good press, do you really think that they would not accept the award due to semantics?

KevinK2

https://www.rx7club.com/showthread.p...d&pagenumber=1

2.5-3L displacement

Evil Aviator

Oh geez, you guys are not referencing this website, are you?
http://www.ukintpress.com/engineoftheyear/