DMBfan

Pretty sure lots of people have seen this one but didnt see it here so check this one out:
http://www.rx7.com/techarticles_displacement.html

Enjoy

Still Waiting

Thanks for the Links, I'm sure that will help clear up a lot of confusion which is exactly what I was trying to convey.

Zoom 44 - I wasn't trying to compare litteraly strokes vs rotary movement in the basic principle operation you're probably thinking. I was saying since it opperates in an entirely different way you can at least assume that:

The number of combustions per complete turn of the crankshaft from a rotary is the same as a 4 cyclinder. You will get 2 combustion phases per turn of the crankshaft for both type engines.

And that each engine combusts a certain volume of air during this combustion phase.

One is more efficient than the other when the number of crankshaft revolutions (displacement of air) is set at a standard number of revolutions versus power generated. It just happens that Mazda rates their rotary displacement at 1.3L which is one crankshaft revolution vs how a 4 stroke engine is always rated at 2 crankshaft revolutions (2 complete revolutions of the crank to always fire off all pistons). See the difference?? Don't blame me why it's 2 revs, I didn't create the standard. Just saying that people should be aware of how these displacements were obtained and what has to be recalculated to make a direct comparison from the same reference point.

In my opinion I would just think it would be more correct to compare efficieny by the air consumed per turn of the crank. Just wanted you rotary enthusiasts to look at it from a different technical perspective and not be confused why 1.3L cannot be direcly compared to the way pistons displacement is measured. I think my approach of 2.6L (for NA 13B's) I'm trying convey is the more accurate.

Neit Jnf and RG understands what I'm trying to say. RG shows the loss of efficiency by the rotary requiring more air (and consequently more fuel) to create similar power.

As for behaving like a 3.6 liter? Let me think about that.

missinmahseven

I've given up trying to compare the wankel to a pistonmotor... I compare it to a jet.. all about the air it can move. :D

Still Waiting

RotaryGod - are you saying that your rotary requires a fuel trim equivlent to that of a 3.6L? I would think that if you compared it in terms of 2 revolutions for 2.6L of natural aspirated air and for a simple comparison versus an S2000 motor displacement output (2.6L versus 2.0), just for a baseline wouldn't you calculate that you would need an extra 0.6L more equivlent volume of fuel in the correct stochimetric ratio versus what a 2.0 4 cyclinder requires? That's just for a baseline off the start and it can be trimmed from there. But 3.6L equivlent of fuel seems way off? Does the rotary design require that much more fuel or is the ecu's standard for a piston requirement slightly off calibration? At any point air is air and fuel is fuel and it should be somewhere around 13/14:1 ratio.

The only way I can think of you needing extra fuel is either you are running a turbo (you drive an 13B-REW correct?) and are compensating for extra CFM and/or on the rich side. If you are driving an NA rotary, I don't know why the engine would have to be run this rich. 3.6L equivlent fuel trim on an NA rotary, I can't even imagine how that could even fire up and make power? Is that where the real efficiency loss is at? I hope you're not refering 3.6L of fuel to an NA setup.

beachdog

All advanced philosophy aside, displacement of a piston engine does not take cycles or crank revolutions in consideration. It is piston swept area multiplied by the number of cylinders. Period. Same calculation for a two stroke piston engine or a 4 stroke piston engine.

Still Waiting

You have to take the number of revolutions into consideration. Are you going to calculate a 2 stroke with 1 revolution or 2? I would think one. I don't recall anybody letting a 2 stroke run through 2 revolutions and calling it's displacement at that double volume swept measurement.

A four stroke can hypotethically sweep all air into cyclinders in 1 rev... it would just need 2 revolutions to combust all that air. See what I'm saying.

The problem is the number of crankshaft turns as seen here also by 2 stroke isn't standardized for a direct comparison as evident also by Mazda's 1.3L rating towards a 4-stroke design.

It's all a marketing thing I suppose....

Anyway I think I'm confusing more people than I'm trying to debate on. Just go out and enjoy your cars! :p

beachdog

Just to poke a hole (not maliciously) at the logic of using crank revolutions to calculate displacement...

Some old motorcycle engines were 360 degree twins. Both pistons rose and fell together, intake, combust, exhaust together. Did they have twice the displacement of 180 degree twins on alternate revolutions?

wedge357

I told you I smell a TROLL!

Still Waiting

What does troll have to do with anything I posted or discussed in this thread? It was all technical and factual data... no BS. I just wanted to make sure people understood the idea of "standardization" when comparing one thing to another.

What's wrong with debating on how one views the point of reference of "displacement" achieved? It just seems that it's harder to explain the concept to a lot of people out there so I just gave up already because I and some few others out there know what's going on down to the itty bitty technical details and I think it's gotten too far for some of our readers.

Mikelikes2drive

^^^ ppl here hate trolls... i ono why
i guess they love the exclusiveness , but i love it when ppl from hondatech come over or other car forums and share their knowledge. i'm pretty sure most of us frequent other car forums.
so dont worry about it :D
i found ur write up intruiging, but I dont have enuff knowledge to fully understand it all, but i get it somewhat i think...

rotarygod

I am saying exactly what you think I am. The nonturbo 2 rotor rotary engine requires the same amount of fuel as a 3.6 liter piston engine. I know this since I am running a standalone ecu on my RX-7. In it you have to program what size engine you have. The most obvious thing to do is to compare how much air it theoretically displaces as compared to a piston engine in the same amount of revolutions. It's the same amount of time we are concerned with. Since the rotary displaces 1.3 liters per revolution and so does a 2.6 liter piston engine, we need to tell it that we have a 2.6 liter engine. If you do this though you are WAY underfueled. You will add nearly 40% more fuel across the entire map structure just to get the car running decent. This makes plenty of sense though when you realize the amount of air needed to make 1 hp and the differences between the piston engine and the rotary. As I said in the other post a rotary needs around 10 lbs of air to make 1 hp where a piston engine needs around 7 lbs of air to make 1 hp. From this knowledge alone we can see that a rotary needs about 40% more air to make the same power levels. If I am entering in 2.6 liters for an engine size and have to add 40% more fuel everywhere to make it run right, the numbers fit. So rather than adding fuel all day and guessing at it, we just tell the ecu that we have 40% larger than 2.6 liters which is 3.6 liters. suddenly the car runs easily on the base map setup it gives you since it is now fueling an engine that size. There are many ways to compare a rotary to a piston engine. Is it a 1.3 liter, 2.6 liter, 3.9 liter? In truth the comparison that really matters is it's output level. What does it really compare to? From it's firing per number of times per revolution it compares to a 4 cylinder. The standalone ecu agrees as it fires perfectly when you enter in a 4 cylinder. From a fueling standpoint it is a 3.6 liter piston engine in size as that is the amount of fuel that it needs to run good. What does the Renesis compare to? A 4 cylinder, 3.6 liter piston engine. They both fire at the same intervals per revolution and they both require approx. The same amount of air and fuel. do they make the same amount of power? OK now we've got to go back and do different calculations again. You'll find that a rotary and a piston can never directly compare to each other in every aspect so you will never get one right answer. It just depends which aspect you want compared.

Still Waiting

Very interesting rotarygod. Thank you for the informative reply. Can you help me further understand my new found confusion?

So if I'm reading your post correct assume both naturally aspirated 13B and a 2.6L 4 cyclinder are spinning at the same exact rpms, theoretically both engines consuming the same amount of air, CFM correct? I would think so.

Now if we want to compare when both engines are making the same power (different from same rpm) the rotary will need to be at a higher rpm (consuming more air and consequently more fuel) to make the extra necessary matching HP versus the more efficient piston engine at a lower rpm.

Now I can understand the above mentioned scenarios. I'm confused on now how you say we are underfueled by approximately ~40% throughout the whole rotary rpm band. Have you correlate this fuel difference to the real time operation of a 2.6L piston throughout all points in an base line rpmband? Is it pretty consistant in it's additional Gallons Per Minute rate?

If this is so, is it truth to deduce that therefore the rotary's design and operation is more volumetric efficient than a piston design? The fact that you are ~40 leaner throughout the whole rpm band (and I'm assuming this would compared to a piston's 2.6L fuel mapping) would indicate that you are getting extra air per revolution than the piston engine, hence the reason you are too lean? Am I correct or is there something else going on?

RPM is RPM and CFM should be constant. So within 4 rotar faces (2.6L) we are somehow packing in there an equivlent 3.6L of air and appropriate fuel mix? Boost condition?? How is this possible? Please explain if you know?

Still Waiting

Rotarygod you there?? Do you have an answer to help clear up my confustion?

Can anybody offer any rational explanation?

BlueEyes

I don't know that he is a troll. If he is, he is going about it awfully politely and asking reasonable questions.

Ike

^noob

Ike

Bah, that was meant for Abbid!

QBallz

This is like compairing an circle to a square. Hey they are both shapes!
The liter rating only refer to max area displaced by the combustion chamber, nothing more nothing less.

If you want to compair it to computers then its like the MHZ. I mean intel slams out chips that are 3+GHZ, while AMD's are still in the 2GHZ range, but hey AMD's for the most part kick Intel's ***.

zoom44

to make the same power the rotary has to ingest 10% to 15% more air. so at the same revs and the same displacment the rotary makes less power because of its inherrenlty less thermal efficient design (shape of the combustion chamber) and its inherent torque disadvantage (the smaller amoount of leverage supplied to the output shaft because of the smaller distance it is being cranked from).

so if we keep the displacement the same the rotary has to have more rpms( to ingest more air) to get the same power as the piston. which is just fine because revving is the rotary's forte. and since it weighs less and is smaller (in general) it doesnt need the same amount of power to move itself about.

rotary crazy

I prefer to compere the engines fisical size how big it is an how much it weihts and what power it makes, I think is more fair this way.

In my experience the renesis is about the same size and waight as a 1600cc to a 1800cc honda engine, older 13b are more heavy.

Still Waiting

Zoom44 I know the rotary has to work more (higher revs) to create the same amount of power compared to a piston engine. That is what I stated earlier, that in doing so I can understand the increase in fuel consumption becuase you're consuming more air through increased revs to make up for the inefficiency of the rotary's natural design just to match the piston.

BUT... what I was trying to point out that rotarygod stated he needed an extra ~40% more fuel throughout his entire rpmband!? That means rev for rev in a rotary vs piston, the rotary is somehow needing more fuel because it's recieving more air in 2 revolutions (0.6L x 4 rotar faces = 2.6L) than a rated 2.6L piston engine??? I don't get where this extra air is coming from (requiring his extra needed fuel triming) when referencing from the same rpm point or any other rpm point. If the rpms are held constant, the CFM's should be as well theoretically be the same. Remember he said he needed more fuel throughout the entire rpm band when inputing 2.6L value in his ECU program.

Assuming that NA engines run around 13:1 air/fuel ratio. Ideally perfect would be ~14.6:1. Now if he is saying he needs an extra ~40% more fuel, his ratio (if my math is correct) is now ~9.2:1!! I would think this is a lil rich for anybody's NA tastes.

My only logical explaination is maybe the ecu is starting point is offsetted negatively because you entered 2.6L as your displacement and it still isn't supplying enough fuel... but needs another 40%? That's a big shift for changes. Maybe injector profiles aren't assumed correctly in the ecu mod? Smaller injectors at lower psi will need to be setup to do more cycle duty work. That would mean the rotary renesis engine is really injesting the proper 2.6L of air per 2 crank revolutions to begin with and the fuel programing was improperly setup/offset to begin with.

I don't know... that's why I want to hear from rotarygod. With your ECU programing of a value entered at 2.6L, what is the air/fuel ratio? What is it after you trim it with an additional 40% fuel on top of the default 2.6L value?

zoom44

because the ecu is using a map programmed for a PISTON engine so it has to be adjusted to get teh power out of a rotary.
dont assume- measure. the RX8 runs at 14.5 or very close in closed loop mode. not 13. your math is incorrect because you have misunderstood what he wrote. the a/f ration need not change- just the amount of air and there for fuel that is required at the particular rpm.

Still Waiting

I ask you this then, does the ideal stoichiometric ratio of 14.6:1 apply to the actual chemical combustion process regardless of engine design? Why then would a piston's map need to be that dramatically different than a rotary's if the fundamental of the ratio is still 14.6:1? Both engines should be consuming 1.3L of air per one revolution. That combine with the 14.6:1 should equal to the same amounts of fuel per ratio. Either he's getting more air than actually is (the reason he needs more fuel) or the ecu is inherently saying that the piston is less volumetrically efficient at scavenging (getting less air and not needing as much fuel for the piston).

I based it off 13:1 to be on the safe side. I thought anything at/near 15:1 would be risking detionation, especially more so since your apex seals are exposed to this shock.

The a/f ratio does change... he said he needed MORE fuel.. approx 40% throughout the whole rpmband (from the default ecu mod input of 2.6L) so he can get his intended (don't know if it's proper) a/f correct to make his needed power.

Only rotarygod and tell us what his a/f ratios were before fuel trimming and after and determine whether his ecu wasn't calibrated with this fuel system setup. This would be based off the 14.6:1 air/fuel principle.

zoom44

because

A. as has been stated before the rotary 1.3 liter is not the same as the piston 1.3 liter.
B. just because it is rated at 1.3 liter that doesn't mean it can actually inhale that much air each time. although there are times that the Renesis is doing better than 100% because of the charging effect of the tuned intake.

c if you had 1.3 liters of of air fuel mixture and the ratio of air to fuel inthat mixture was 14.6:1, you ignited it and completely combusted the entire 1.3 liters- it would make the same power every time. however in an engine it is hardly ever 100%volumetric efficient and the inherrent difference in the shape of the combustion chambers makes for diferences in the the flame front propogates and the percentage of combustion completion.

what RG said in a nut shell is this- according to the way the maps are made for a 2.6 liter piston engine in his megasquirt hios 13b engine wont run well. in order for it to run well on that megasquirt he has to use maps that are written for a 3.6 liter piston engine.

the only thing this shows is that the piston engine rated displacement cannot be compared directly to the rotary engine rated displacement.

rotarygod

I understand what is being asked and it is indeed perplexing. Logically if we have an engine that is truly comparable to a 2.6 liter engine in terms of displacement but less efficient in terms of how much air it uses vs how much fuel it uses, we would think that you'd need less fuel to keep it running since you can't have more air. That would seem to make sense. For some strange reason it doesn't work this way though. When I entered in a 2.6 liter engine in my ecu, and yes I assure you it isn't a calibration problem within the ecu code, the car ran extremely lean. I could not keep it running. It had just enough that I could force it to stay running by modulating the throttle heavily. The cat got so hot it was glowing red and I even had the radiant heat from the cat catch a rubber hose on fire. Way way too lean. Enter in as a 3.6 liter engine and the car runs very nice. I'm not going to try to question it. It works and that is how it compares. Because I can't directly explain it or because it sounds backwards to me doesn't change the fact that it works this way. The bottom line is this, a rotary engine is not directly comparable to a piston engine. It can be made to in certain aspects but you can never narrow it down to 1 standard.