wakeech

yeah, it's true... but this discussion IS going no where: those who're wrong are flat wrong. that's fine to make mistakes, but this isn't really something that's open to interpretation, it's pretty straightforward... but yes, maybe closing is a little harsh.

heh heh, i don't deserve so much courtesy but thanks very much 3R.

RedRotaryRocket

Dazz,

The Wankel IS a 4-stroke, 4-cycle* engine. It is not a two-stroke. There are two things I'd like to mention, which are constraints and definition

Constraints

What if I were to describe for you a four-stroke piston engine with two cylinders? This engine would also make one power stroke per output shaft revolution...just like the rotary. If I were to make this example, I'm sure you would give me hell about using more than one piston....and you'd be right. It is not meaningful to look at more than one piston. But the reason it is not meaningful has nothing to do with the number of pistons, but rather the number of combustion chambers. We are placing a constraint on the piston engine that we must only use one combustion chamber for comparison. To make an accurate comparison, we must apply the same constraints to both engines. This means that we can only look at what happens in one combustion chamber of the rotary. If we look at one combustion chamber, we see that the rotary does not provide one power stroke per revolution....in fact it is only providing one power stroke per three revolutions.

Definition

You have tried to define a two-stroke engine as an engine having one power stroke per output shaft revolution. This isn't quite right. The definition works if you talk about piston engines, but that is only because the geometry of the engine requires two-strokes per rev.

The real question is "What is a stroke?" A stoke is the combustion chamber going from minimum to maximum volume, or vise versa. It really has nothing to do with how many times the output shaft rotates. The number of output shaft revolutions is really just a consequence of the specific engine geometry, which is obviously different between rotaries and pistons. The output shaft rotates 180 degrees per stroke in a piston engine, but it rotates 270 degrees per stroke in a rotary.

So when somebody says an engine is a two-stroke or a four-stroke, what do they mean? They are really asking the question "How many strokes (expansions and compressions) does it take to complete one full combustion cycle*?" In a two-stroke engine, the combustion chamber goes from max volume to min volume (or vise versa) twice per full combustion cycle* A four-stroke engine requires four of these strokes to complete the job. In one rotary combustion chamber, the chamber must expand or contract four times to complete a cycle*. This means that the rotary is a four-stroke engine.

*Now that stroke is cleared up, I'll mention one more thing about definition. We should be careful about how we use the word cycle, because it is used in many different ways, and thus can be confusing. Cycle seems to be (correctly) used in three different ways:

1) To describe the complete combustion process, e.g. the Otto cycle. In this context, one cycle consists of intake+compression+power+exhaust.

2) To describe one of the components of the complete combustion cycle, e.g. the intake cycle, the compression cycle, etc. This is the context in which you have used it.

3) Used with the meaing of "stroke". This appears to be an accepted usage, although it makes things quite confusing. You can refer to a two-stroke engine as being a two-cycle engine. This is confusing because it's not implied whether you mean and engine which completes two strokes per combustion process, or an engine who's combustion process has only two components. This is an unfortunate use of the word, but who said the English language was perfect?

lefuton

here's a fairly recent discussion going on in the rx7 forum from Mr. Wankel. i think he's being more than fair in saying you can believe what you want and it really does matter how you measure things, ie, combustions per crank rotation, air pushed per crank rotation, rotor rotation, etc etc.

disclaimer: moderately technical

http://www.rx7club.com/forum/showthr...&highlight=2.6

zoom44

from how stuff works


"The combustion chamber is the area where compression and combustion take place. As the piston moves up and down, you can see that the size of the combustion chamber changes. It has some maximum volume as well as a minimum volume. The difference between the maximum and minimum is called the displacement and is measured in liters or CCs (Cubic Centimeters, where 1,000 cubic centimeters equals a liter). So if you have a 4-cylinder engine and each cylinder displaces half a liter, then the entire engine is a "2.0 liter engine." If each cylinder displaces half a liter and there are six cylinders arranged in a V configuration, you have a "3.0 liter V-6." Generally, the displacement tells you something about how much power an engine has. A cylinder that displaces half a liter can hold twice as much fuel/air mixture as a cylinder that displaces a quarter of a liter, and therefore you would expect about twice as much power from the larger cylinder (if everything else is equal). So a 2.0 liter engine is roughly half as powerful as a 4.0 liter engine. You can get more displacement either by increasing the number of cylinders or by making the combustion chambers of all the cylinders bigger (or both). "

tribal azn2

the fact of the matter is that the renesis has a 1.3 liter displacement. u cant just go changing it just to compare it with a piston engine. its absolutly ridulclious. its like saying a meter isnt longer then a foot because a meter consister of 3 feet(approx). but the fact of the matter is that a meter IS longer then a foot regardless. get it? comparing rotaries to piston engines is like comparing piston engines to rocket engines, its absolutly pointless

Sputnik

I'd be closing alot of threads if that were the case.

If anyone out there is getting tired of the thread, then don't bother responding, and let it drop.

---jps

Dazz

There is no simpler way to explain it than this. Or maybe there is, but with peoples lack of understanding I don't think it would matter how simple it was they still would not understand!

Lets try to be very very simple about this, for the slow ones.

Lets have a single rotor rotary engine, a single piston two stroke and a single piston four stroke engine.

Lets say all three have a capacity of 1,000cc (that being the maximum chamber volume) and all have 100% volumetric efficiency to make it very simple.

So, assuming all three engine are about to begin their intake stroke, we rotate the eccentric/crank shafts through 360 degrees.

The rotary will inhale it's 1,000cc, the 2-stroke will inhale 1,000cc, and the 4-stroke will inhale 1,000cc.

On the next rotation of the eccentric/crank shafts through 360 degrees, the rotary will inhale it's 1,000cc again, the 2-stroke will inhale it's 1,000cc again, and the 4-stroke will exhaust it's 1,000cc.

On the next rotation of the eccentric/crank shafts through 360 degrees, the rotary will inhale it's 1,000cc again, the 2-stroke will inhale it's 1,000cc again, and the 4-stroke will inhale it's 1,000cc.

On the next rotation of the eccentric/crank shafts through 360 degrees, the rotary will inhale it's 1,000cc again, the 2-stroke will inhale it's 1,000cc again, and the 4-stroke will exhaust it's 1,000cc.

So you can see quite plainly, that both the rotary and the two-stroke will inhale their full volume every revolution, where as the four-stroke will inhale it's full volume only every second rotation, so for the same rpm, the rotary and two stroke will inhale double the capacity of the four stroke engine of the same physical capacity.

This is why you can compare a rotary to a two-stroke as far as capacity is concerned in comparison to a four-stroke engine. With "compare" being the crucial word.

You could in all honesty argue that a normal four-stroke engine should only be classed as being half of it's calculated displacement as it only inhales it's displacement every second rotation of the cankshaft.

But seeing as how the humble four-stroke has been around longer in automotive applications, it is the constant by which other engines are compared to.

All I am trying to get across, is to explain that the rotary engine, capacity wise, is the equivalent of a normal four stroke engine of twice it's size, as far as capacity is concerned when the engine is running.

Yes the engine is 1308cc in swept capacity, so it is a 1.3 litre engine, but opperationally it will consume the same volume as a four-stroke twice it's size at the same rpm. It's quite simple.

I know this gets boring and frustrating, but as someone who has been working on these engines for about 15 years now, and who learned from a person who has been working on rotaries since the early 70's, I get annoyed at the misinformation that it out there.

It makes explaining to people who are new to these engines difficult becasue they have heard all of the rumours and they then become confused as to what is right and what is wrong.

And this I believe is one of the biggest hurdles that the rotary engine faces. I'm sure that a lot of people have been turned off owning a rotary powered vehicle becasue of the amount of rumours and mistruths that are around, and that is a great shame.

mrWankel

actually, because Felix Wankel is the inventor of the rotary engine, A LOT of people would want to use his name......it's common sense. And my name on rx7club.com is rx7turboracerx, btw. i sold my 3rd gen because it was getting too expenssive, and i would rather have an RX-8.

lefuton

one revolution on a 1000cc 4stroke ohv piston motor will only inhale 500cc of air given 100% volumetric efficiency... that and 100% volumetric efficiency is hard to visualize on a 2 stroke =)

Dazz

How do you figure that? I'm keen to see your technical reason!

maxcooper

There seem to be two viable definitions for displacement:

1. The amount of air ingested relative to rotations of the main shaft.

2. The geometric displacement of the engine in one turn of the main shaft, ignoring how much air it ingests.

1. For the first defintion, there is still some variability to contend with, since there is no number of main shaft rotations given.

1a. One answer is to rotate the shaft as many times as you need to to have all chambers complete their cycle. With this defintion, you get these answers:

A 2.6L 2-stroke piston engine ingests 2.6L (at 100% VE, good luck ) in ONE crankshaft rotation, completing the combustion phase for all of its chambers.

A 2.6L 4-stoke piston engine ingests 2.6L (at 100% VE) in TWO crankshaft rotations, completing the combustion phase for all of its chambers.

A 13B rotary ingests 3.9L (at 100% VE) in THREE eccentric shaft rotations, completing the combustion phase for all of its chambers.

This is technically correct, but with all the different numbers of main shaft rotations, it makes comparisons kind of difficult. Rating the rotary at 3.9L in three turns doesn't lend itself to comparisons with other engines that take less rotations.

1b. Another way to go is to compare the engines in just one or two or three rotations.

ONE ROTATION: A 2.6L 2-stroke piston engine ingests 2.6L (at 100% VE, good luck ) in ONE crankshaft rotation, completing the combustion phase for all of its chambers.
TWO ROTATIONS: Using two rotations would be kind of silly because all the chambers would fire twice, but it would ingest 5.2L (at 100% VE) in two rotations.
THREE ROTATIONS: Using three rotations would even more silly because all the chambers would fire three times, but it would ingest 7.8L (at 100% VE) in three rotations.

ONE ROTATION: A 2.6L 4-stroke piston engine ingests 1.3L (at 100% VE) in one rotation of the crankshaft, but you are only using half the displacement.
TWO ROTATIONS: A 2.6L 4-stoke piston engine ingests 2.6L (at 100% VE) in TWO crankshaft rotations, completing the combustion phase for all of its chambers.
THREE ROTATIONS: A 2.6L 4-stoke piston engine ingests 3.9L (at 100% VE) in THREE crankshaft rotations, but that would be kind of silly since half of them will fire twice in the process.

ONE ROTATION: A 13B rotary ingests 1.3L (at 100% VE) in one eccentric shaft rotation, completing the combustion phase for only 1/3 of its chambers.
TWO ROTATIONS: A 13B rotary ingests 2.6L (at 100% VE) in two eccentric shaft rotations, completing the combustion phase for only 2/3 of its chambers.
THREE ROTATIONS: A 13B rotary ingests 3.9L (at 100% VE) in three eccentric shaft rotations, completing the combustion phase for all of its chambers.

This can be useful for comparing engines because it gives a good indication of power output since the amount of air ingested is normalized to some fixed number of main shaft rotations. But it takes a different number of rotations for each kind of engine, so you end up using some chambers more than once or some none at all depending on how many rotations you choose for the comparison and which engine designs you are comparing.

2. Using the geometric displacement in one main shaft rotation is nice because the only caveat for estimating power output is whether the engine uses a 2-phase or 4-phase combustion cycle.

A 2.6L 2-stroke piston engine uses its full 2.6L geometric displacement at 100% VE in one crankshaft rotation.

A 2.6L 4-stoke piston engine uses only half of its 2.6L geometric displacement in one crankshaft rotation. The rating is 2.6L, but it only ingests 1.3L in one turn of the shaft.

A 13B rotary, which uses the 4-phase combustion cycle, actually has a geometric displacement of 2.6L in one turn of the eccentric shaft. There is some contribution from each of the chambers, but the total geometric displacement in one turn equals just 2/3 of the total chamber volume for for the two rotors. It uses half of its geometric displacement for power production, just like the 4-stroke above that shares the same 4-cycle combustion cycle. That means it ingests just 1.3L of air for that single turn.

This rating method is useful because you need only know what combustion cycle the engine uses (2-cycle or 4-cycle) to estimate the engine performance.

Mr. Wankel posted about this on the RX-7 Forum, and I was frankly kind of skeptical that the rotary actually had a 2.6L geometric displacement in one turn of the shaft. It is hard to picture, so it is kind of hard to believe. However, careful analysis reveals that it is indeed true. Mr. Wankel posted some pictures in this thread:
http://www.rx7club.com/forum/showthr...hreadid=176121

And I came up with a written proof for the 2.6L of geometric displacement on the 13B in the other thread that goes like this:
1: Rotary is 4-cycle
2: A 4-cycle has four phases of the combustion cycle
3: To complete the cycle, the chamber goes from min volume to max volume (intake), max volume to min volume (compression), min volume to max volume (power), and max volume to min volume (exhaust)
4: The chamber geometrically displaces its full volume twice to complete the full cycle (compression and exhaust phases)
5: A 13B has six 654cc chambers, so the total geometric displacement that occurs in course of firing all of the chambers is 6 chambers x 0.654L per chamber x 2 geometric displacement to complete 4-cycle combustion = 7.848L of geometric displacement for all the chambers to complete the full cycle
6: It takes 3 full rotations of the eccentric shaft for all the chambers to complete the combustion cycle
Therefore: 7.848L total geometric displacement / 3 rotatations of the main shaft = 2.616L of geometric displacement per rotation of the eccentric shaft


ALL of the displacements arrived at by these defintions are technically correct. But obviously each defintion gives different answers, and that seems to lead to arguments about which is the best defintion that sometimes end up looking like arguments about how the engines operate. How the engines operate is totally independent of how we choose to define the word displacement. Understanding each of these defintions and how they relate the actual operation of the engines is the most important thing to understand. Choosing a defintion that you like better is a separate issue, and I think any of these choices is a defensible position. I hope we don't get into the same arguments here that we did on the RX-7 Forum about what is the best definition.

It can be confusing about how these different defintions lead to the ratings I gave above. I am darn sure that the ratings I gave are correct for each defintion, but I realize that it may not be 100% clear how the engines operate or how that operation yields the given ratings. Please indicate which defintion of displacement you are interested in if you want to discuss why an engine was rated a certain way. I'll be glad to respond to posts like that. But I won't argue about which is the best defintion, because I don't really care about that!

Sorry for the long post, but this is more or less a complete summary of what took many pages on the RX-7 Forum, minus the "what is the best defintion for displacement" arguments.

-Max

ZoomZoom

Excellent post Max!

wakeech

yup, super-keen. thanks.

lefuton

maxcooper explained it much clearer than i ever would have =)

Dazz

That was all very entertaining, and basically proves what I said.

And lefuton, I feel that you are still mssing the point with the 4-stroke engine capacity per revolution.

It only injests it's full volume every second revolution, as on the alternate revolution it is exhausting it's volume.

This DOES NOT mean it injests half of it's displacement every revolution. If that was the case every revolution there would be an intake and exhaust cycle! Doesn't work like that.

Maxcooper went to an aweful lot of trouble and effort, to say the same thing I did in a few simple sentances that I hoped you would all understand. But he got one thing wrong.

"A 2.6L 4-stoke piston engine uses only half of its 2.6L geometric displacement in one crankshaft rotation. The rating is 2.6L, but it only ingests 1.3L in one turn of the shaft"

That is wrong. During it's intake cycle or "1st stroke" it will injest it's full 2.6L as the piston goes from TDC to BDC. During it's Compression cycle or "2nd stroke" it will compress it's 2.6L mixture as the piston goes from BDC to TDC, during it's power cycle or "3rd stroke" it will ignite the mixture and the piston will go from TDC to BDC, then during it's exaust cycle or "4th stroke" it will exhaust the mixture as the piston goes from BDC to TDC.

Just becasue it only inhales it's full mixture every second revolution, DOES NOT mean that it inhales half of that every revolution. Just thought that needed clearing up as we don't want people getting more confused. :D

lefuton

oh dear, lets try this again.

2.6L 4 cylinder 4 stroke engine with overhead valves. that means if you measure the volume from tdc to bdc of one piston you will get .65 litres of volume. are you with me? .65 litres per piston x4 pistons gives you a 2.6L engine.

lets say the crank is at tdc. 2 pistons are at tdc and 2 pistons are at bdc. lets take the motor in the 2nd gen eclipse for example since i'm familiar with those and have taken it apart many a time much to our dismay ><. piston 1 is bottomed out and just finished a combustion stroke, piston 2 is topped out and just finished an exhaust stroke, piston 3 is topped out and just finished a compression stroke, piston 4 is bottomed out and just finished an intake stroke.

turn the crank a half revolution, 180 degrees. now piston 1 just did an exhaust stroke (lost .65litres of air), piston 2 just finishes an intake stroke (inhales .65litres of air), piston 3 just finishes a combustion stroke (all valves closed, no air comes or goes), piston 4 just finishes a compression stroke (no air in or out). so far half a crank revolution and only .65litres of air in and .65 litres of air out.

turn the crank back to tdc. piston 1 is bottomed out again and just finished an intake stroke (.65 litres of air in), piston 2 is topped out and just finishes its compression stroke (no air in or out), piston 3 is topped out and just finishes its exhaust stroke (.65 litres of air out), piston 4 is bottomed out and just finishes a combustion stroke (no air in or out). another half crank and .65 litres of air in and .65 litres of air out.

one full revolution of a crank on a 4 stroke 2.6L 4 piston motor and 1.3L of air comes in, and 1.3L of air goes out. it takes *two* revolutions of the crank to inhale 2.6L of air.

technically you *could* time the valves so that the motor would inhale its full displacement within one crank revolution but that would be...dumb, as you would have a very lumpy motor that's hard to start and have a high idle speed.

or, if you would so please, explain to me how a 2 stroke and 4 stroke motor of the same displacement both inhale their full displacement in 1 crank revolution. how does 1 crank revolution account for all *four* strokes in any given one piston.

Dazz

If you look at my previous post, we were talking about single rotor and single piston engines, not a 4-cylinder engine vs two rotor etc.

If a piston engine has a 2.6 litre single cylinder capacity, then on it's intake cycle it will inhale 2.6 litres volume assuming 100% VE, right!

Nowhere did I mention in the example that it was a multiple piston engine so you are simply trying to make up the rules as you go along.

Read what I wrote below one more time, carefully, and then come back to me.

"Lets try to be very very simple about this, for the slow ones.

Lets have a single rotor rotary engine, a single piston two stroke and a single piston four stroke engine.

Lets say all three have a capacity of 1,000cc (that being the maximum chamber volume) and all have 100% volumetric efficiency to make it very simple.

So, assuming all three engine are about to begin their intake stroke, we rotate the eccentric/crank shafts through 360 degrees.

The rotary will inhale it's 1,000cc, the 2-stroke will inhale 1,000cc, and the 4-stroke will inhale 1,000cc.

On the next rotation of the eccentric/crank shafts through 360 degrees, the rotary will inhale it's 1,000cc again, the 2-stroke will inhale it's 1,000cc again, and the 4-stroke will exhaust it's 1,000cc.

On the next rotation of the eccentric/crank shafts through 360 degrees, the rotary will inhale it's 1,000cc again, the 2-stroke will inhale it's 1,000cc again, and the 4-stroke will inhale it's 1,000cc.

On the next rotation of the eccentric/crank shafts through 360 degrees, the rotary will inhale it's 1,000cc again, the 2-stroke will inhale it's 1,000cc again, and the 4-stroke will exhaust it's 1,000cc.

So you can see quite plainly, that both the rotary and the two-stroke will inhale their full volume every revolution, where as the four-stroke will inhale it's full volume only every second rotation, so for the same rpm, the rotary and two stroke will inhale double the capacity of the four stroke engine of the same physical capacity"


By the way, I think you need to learn about how a two-stroke piston engine works before making such daft statements like.....

"or, if you would so please, explain to me how a 2 stroke and 4 stroke motor of the same displacement both inhale their full displacement in 1 crank revolution. how does 1 crank revolution account for all *four* strokes in any given one piston."

When you figure it all out get back to me

maxcooper

Dazz,

I agree with your posts, and I was thinking of a 4-cylinder 4-stoke piston engine when I said that it ingests half its displacement in one rotation of the crank. I think it is most reasonable to consider the average amount of air ingested per rotation anyway, which works out to half the rated displacement for any number of cylinders.

-Max

Dazz

You are absolutely right in saying that the 4-stroke "on average" ingests half of it's volume per rotation. It's just some people on here read but don't understand exactly what they are reading and then come back with dumb and more importantly incorrect statements.

lefuton

mrrrr i guess i missed that crucial 1 cylinder part, as a matter of fact i do understand how 2-stroke engines work which is why i made the comment. look at it from my eyes, here i am scratching my head going how on earth does this guy think a 2stroke motor and a 4 stroke motor inhales its full displacement every revolution. alright alright so you were talking about 1 piston only.

btw, im not making up rules as i go along, was just a misunderstanding of what we're working with. now, just so that we're on the same page. agree that a 1.3l rotary effectively breathes as much as a 2.6l 4stroke? at the same rpm.

maxcooper

Yes, a 1.3L rotary breathes the same amount as a 2.6L four-stroke piston engine at the same RPM. The rotary has a 4-phase combustion cycle, just like the 4-stroke, which makes the 1.3L rating a bit strange when considered on the technical merits. The 1.3L rating is what it is, though, and isn't going to change no matter what we come up with, so it is important to understand what it means from the technical perspective.

-Max

Dazz

Glad it's finally all cleared up now

StealthTL

iTrader: (1)

It's as clear as an azure sky of deepest summer.
But I don't even buy the fact that it is a high revving, low torque engine!
Let me tell you how I see it - it's a 3.9 litre motor, with six power pulses per complete cycle. (Like a six cylinder.)
To get the engine thru all it's phases and stokes, back to the starting point, takes three revolutions of the 'output' shaft and six intake volumes of 600+cc. I dont buy the high revs at all, the rotors are only rotating at 3000 revs when the 'output' shaft is at 9000, because of the internal 3:1 planetary gearset in each rotor.
It is a similar situation as if a V6 motor had its 'output' shaft to the gearbox from its overhead cam; not a true indication of 'actual' engine revs.
The planetary gear triples the output shaft speed to 9k, dividing the torque-per-cycle by three, which then is rectified immediately by the gearbox, back to a realistic 450 ft/lbs, at 3000 actual rotor revolutions! Now tell me that I am not DEEPLY in denial !!!!

zoom44

actually i feel like i can ask this know that you have mentioned it without looking too stupid. in all of the discussion previously i think everyone talked about the rpm of the output shaft. shouldn't we actually measure it by the rpm of the rotors themselves. if i missed someone making this point earlier i apologize.

Dazz

Gee just when it was all cleared up and put to bed someone comes along with more whacky thoughts!

RPM is always measured at the crank, or eccentric shaft in this case.

We do not measure piston speed or distance, nor do we count camshaft rotations, and if you start to measure things other than the shaft be it crank or eccentric for rpm then nothing will make sense.

And StealthTL, yes you are deeply in denial and need some serious help!