I remember reading this sometime ago & some of you may have read it also.

http://rotarygod.com/index.php?title=RE:_Displacement

Fuel to the fire:

http://auto.howstuffworks.com/question685.htm
http://www.wisegeek.com/what-is-engine-displacement.htm
http://www.answers.com/topic/engine-displacement

All of those support what I am saying, even though worded oddly in one of them.

The example from howstuffworks.com is assuming a 4 stroke engine. However, if you were to look at the same engine without ahead of valvetrain on it, it would displace the same total in one crankshaft revolution. Their example speaks of "ingesting" total displacement over 2 revolutions" which is true, because the term "ingesting" really has nothing to do with measuring the displacement, that has to do with engine type, 2 or 4 stroke).

So, with the added support of your posts, I will stand even more firmly on the definition of displacement bing the total volume displaced through 1 revolution of the crankshaft/eshaft/centershaft regardless of engine type. The air that the valves allow to be ingested has nothing to do with the engine's actual displacement. That just determines th engine type.

I honestly don't know why "rotarygod" feels he has merit to claim anyone else is wrong. That's pretty much saying that the definition of displacement in relation to internal combustion engines is wrong.

Its always been in direct relation to 1 revolution of the crank, regardless of stroke (2, 4 or '6')

All these crazy arguments of multipliers are just relative arguments but are all lacking the fact that 2 and 4 stroke engines calculate displacement the exact same way. You don't need valves or a head to calculate displacement. just the bore are x the stroke x the number of pistons, which is going to net you exactly what I have been saying all along. 1 revolution worth of displacement. Starting position of the piston or rotor face DOESN'T matter as long as you go through EXACTLY one revolution of the crankshaft/eshaft.

So here are the key notes that need to be seen here:
Displacement is:
-based on 1 revolution of the crankshaft/eshaft/centershaft
-2/4/'6' stroke is irrelevant
-physical volume displaced per revolution regardless of stroke style
-multipliers are "rules or regulations" in place to even out the playing field due to the different power levels, and efficiencies of the different stroke types. (of course different methods of making power will yield different results)
-Nothing mystical or hidden.

I am not trying to make anyone look bad, only showing that this thread is turning into a magical goose chase of some mystical formula that must be used to calculate true rotary engine displacement.... The problem is, so many people are not willing to admit that the rotary CAN be rated in the exact same method as piston engines. And that is precisely what Mazda did.

Peter, please don’t take this so personal.

We had a saying in industry that all of us should apply on the forums.
And that is, ” It is OK to disagree. It is not OK to be disagreeable".

We are sharing ideas to further the Rotary's development.
You make your points well, especially the picture presentations.
We need to just supply facts to support our point of view and then let it go.

We just disagree on this point.
Barry

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.

Otto cycle is just the technical term of suck, squish, bang, blow Again, the issue isn't so much of what we're comparing, it's how the displacement is literally calculated. Note that all the links I posted concerning piston displacement use TDC to BDC of each individual piston. We can do the exact same procedure and get a displacement at the end of it, but whether or not it is what individual consider accurate is, as you have said, up for discussion.
And I agree. It's completely dependent upon how you read into the equivalences. You could take Rice's approach and claim each rotor face is a piston equivalent, which would net you a the 4L displacement. If however you consider the standard practice of TDC to BDC than you will always only end up with 1.3L (for a 13b). Another way I look at it that makes me lean towards the 1.3L displacement more than anything else is that the actual physical displacement is maintained in the same physical space.
And that's fine to do--Just be aware that it is not equivalent to the standard practice of engine identification.
Like many bright people have said before; it's all relative.
Actually you can. Take a snap shot of a piston in TDC. Then take another one at BDC. The resultant difference in volume is the displacement. You do the exact same thing for the rotor. TDC snap shot. BDC snap shot. Resultant displacement is the difference in volume between the two. There is no need to worry about crank shaft rotation.
That constant is TDC and BDC for all otto cycle engines.How about TDC and BDC since those are constant across all motor types. How would you measure the displacement when you have 3 eccentric shafts operating independently or at different rpm on the same engine? Using your definition you'd have to normalize the entire assembly rather than just look at the discrete values of TDC and BDC (think of it as the limit of integration: Vol_tot=integral(dv,vol_tdc,vol_bdc). Eccentric shaft revolution doesn't play any part of displacement.
And I agree that if you wish to calculate the combined displacement of all rotor faces that is the number you'll get. If however you wish to remain in standard practice you need only concern yourself with TDC and BDC of each rotor.

wow. very well thought out response. I will definitely take much of this into consideration now. (yes, I am admitting that my perspective on this topic has shifted a bit)

My only argument would be that at this point, each rotor face could be considered the equivalent of a piston as to tdc and bdc. Or perhaps it is tdc/bdc of the crankshaft rotation since each piston or rotor face would be in a different position. These are just merely perspective questions to further define baseline definition and procedure and not to discredit or debase.

As I stated before, it is merely a difference in frame of reference. Each rotor housing undergoes one full cycle during each revolution, but each rotor face will require 3 rotations for a complete cycle.

Because both the housing and the rotor are required to displace air, both frames of reference are valid.

I think that Peter, Barry and Vex have done a very good job of illustrating the concepts that we all need to take away from this discussion. This is a good thread.

No it's not, and you can't prove I've done a good jorb!

Yeah, I don't understand why people were getting all hot and bothered by this. The only thing I do not understand is where the 2.6L value comes from without using a multiplier. Anyone know?

Food for thought
http://www.rotaryeng.net/Ansdale-displacement.pdf

I´m inclined to 3.9 definition:suspect: Simply saying this is wankel engine and full engine is utilized only after 3 revolutions:p

I would agree completely. I cant think of any method that would make the 2.6l number viable.

What is the value of Line 1-2 for a 13B Rotary?

(Using a VE of 100)it would be 654cc...

and it would look like this, but the VE would actually be lower because of the still opened intake port and the trailing spark-plug hole spit-back.

http://i287.photobucket.com/albums/l.../450deglr0.jpg



For a KX 500 2-stroke...
500cc... and it would look like blue bore depicted. It's actual VE would be much lower also because both intake and exhaust are open for part of the stroke (Consider what the effective compression ratio while starting).

http://i287.photobucket.com/albums/l...2-stroke-1.jpg

For the 2 liter Toyota….
500cc

For the 6 liter LS motor…
750 cc

These pictures are mostly for the new guys that are a little afraid to raise their hands and get into the discussion... yet.

I opened Rotarygod's web site. It is well done and he makes his case for 2X and 3X displacement, but you will notice that all references to the different engine sizes are exactly what the manufacturer calls them.

This was the point of my initial post… if we don’t use a standard way of describing the Rotary’s displacement in our discussions we will be confusing ourselves, especially neophytes.

Barry

That's an idealized otto cycle Barry. For all engines there's going to be descrepancies and variations from that. I don't think there's a p-v diagram out ther for the 13B at least not to my knowledge. I'll have to check the SAE papers.

I still vote to compare apples to apples we use power/engine weight and/or size (physical, not displacement...).

You mean like TDC to BDC ;)

Rotary maybe had slight edge in this, some 40-50 years ago:07:

Bare engine may appear compact and light, but whole package is what counts. Engine, intake and exhaust manifolds, cooling system, muffler system, it all counts.

We can´t just say that certain engine is compact and light, when all needed accessories through their bulk and added weight doesn´t make it such viable powerplant.

I would look for advantages of wankel rotary elsewhere

I understand your reasoning but doesn´t such approach ignores that rotor housing has two TDCs and BDCs?

Common sense would tell that we examine only intake part, but who knows:suspect:

I think that wankel engine should be treated as wankel - whole termodynamic cycle is completed only after 3 revolutions. And of course 3 such cycles will be completed, just shifted by 360°.

That's all displacement is measuring. The amount of air injested by the engine when each piston (whether you want to consider a piston the rotor itself or the faces there of is inconsequential) goes from TDC to BDC. We don't count the TDC to BDC of the Compression/Expansion of the piston engines, why would we for the rotary?

Why do we care again about the thermodynamic cycle or how many revolutions it's completed in; when we're worried about displacement?

no, i was familiar with the term. i had just never seen the graph you posted and i figured i would spend hours trying to get, but after looking at it that night, i had it in less than an hour and confirmed my understanding of it via Wikipedia (for what that's worth - wink, wink ;))
get your grain of salt ... :)

as i understand it, they are using a multiplier - based on the 720 degree theory. by itself, it does seem arbitrary though.

it's funny, when i first got into rotaries (back in the mid 80s), some people used to say two rotor engines were equal to 2.4 liter, 4 cylinder engines (most people i knew primarily messed with 12As at the time), but extending that way of thinking to a 13B, you'd get 2.6L. the thinking was that rotaries were more akin to 2-strokes in nature, so they multiplied by 2. i don't know where the 4 cylinder thing came from.

if you are inclined to think of each rotor face as cylinders (which i know you don't), then with a 2.6L 4-banger, you have exactly two-thirds of a 3.9L 6. so in that context, i guess it makes sense ... sort of. come to think of it, i think i just found the 2616 theory less valid.

Vex, I think the 2-3 line and 4-5 would change only slightly.
The piston engine's rod ratio will affect its shape vs. the Rotary's sine wave movement. (Dotted line-rotary, from Yamamoto's book)

http://i287.photobucket.com/albums/l...onvsrotary.jpg

I'm not saying the rotary engine is compact and light. I'm saying that what is needed to make the engine run as intended (i.e. manifolds and engine) is what should be measured. If, as it sits in the car, a 13B weighs 350lb (or whatever), that's what we measure. If, as it sits in the car, an LS6 weighs 450lb (or whatever), that's what we measure.

Like the weight of the engine including manifolds and such (excluding other drivetrian - tranny etc.) / the horsepower and torque.

Perhaps all this is taking it too far off subject though...

This makes sense to me. Perhaps it's because it's put in non-engineer speak, but whatever.

I can't tell if we're missing each other on this. I'm not understanding what you're asking.

What I posted was a P-V diagram of the Otto Cycle (4-stroke cycle). The amount of pressure contained within the piston and the amount of volume it sweeps will be the only determining factor so long as combustion and heat rejection are kept the same.

What you provided is only one part of the P-V. We'd still need the pressure it sees prior to combustion, after combustion, and the adiabatic expansion and heat rejection.

Since no engine is ideal in the Otto Cycle the steps from compression up to combustion would be isentropic (it's a fair assumption; not great, but not as bad as adiabatic). After combustion it would be isentropic to the point of exhaustion--then depending on if there's a turbo in the way we'd have different pressures going on there. Intake would be dependent on stuff outside of the engine.

I meant only that the Rotary's diagram would look similar.
Barry

Of consequence; this comes from a thread I had going for awhile.
http://www.rotarycarclub.com/rotary_...5&postcount=35

Anyone of you nobodies could just as easily spend some of your time instead of trying to pass analysis on me spend some time on trying to figure out how a rotary works!

Now there is a novel concept!@

Equivalence does not equal displacement, and the Wankel is the ONLY engine where its cycle is not acknowledged nor are ALL of its working elements............... this is not that hard really to comprehend.

for those of you that are stuck or are uneducated in the formal prerequisites to pass qualified commentary then you can stick to your partial false analysis and use the well worn out equivalence factors I mentioned in my second post.

Is any of them Wankel? Is any of them complete? NO! Practice and learn and you may get it one day :icon_tup: