Actual Rotary Displacement Request
 

Request to use Actual Rotary Displacement

If someone started talking about a 2.6 liter Rotary would they be referring to a 4 rotor or are we dealing with someone who doubles displacement of a two rotor?

In the interest of clarity I believe that we should describe the displacement of our rotary engines by its actual scientific size.

Some very experienced individuals have doubled and sometimes even tripled its size. This has become confusing to new impressionable members trying to communicate ideas correctly.

One would think that a physical measurement like displacement would be rudimentary but on a rotary it is more complicated than π X radius² X stroke X number of cylinders. Finding max volume from trochoid and peritrochoid shapes is a lot tougher.

I won’t bore you with formulas but all of the manufacturers signing licensing agreements to develop, and those producing Wankel engines including Alfa Romeo, American Motors, Citroen, Ford, General Motors, Mercedes-Benz, Nissan, Porsche, Rolls-Royce, Suzuki, Toyota and of course NSU and Mazda…and motorcycle manufactorers Sachs, DKW/Hercules, Norton and Suzuki… add John Deere, Artic Cat, Curtiss-Wright, also miscellaneous outboard and unmanned arcraft manufactures all agree on way actual displacement is determined and refere to them acordingly.

And yes, some sanctioning bodies use a multiplier which penalizes the Rotary as they do the two-stroke to help even out the competition. Displacement, however is a scientific measurement not up for opinions.

What size would Felix Wankel or Kenichi Yamamoto say that it is?

My Thoughts,
Barry

So.... which side are you agreeing with?

i have thought about this, and there are a few ways to look at it, none of which is that great IMO. my thoughts started with howard colemans assertion that 255hp in an 159CID engine will have a lot of cylinder pressure. which might be true in a PISTON engine, but the rotary is different enough that we can cast some doubt.

first off, the Mazda rotary is a 4 stroke engine. there are 4 distinct strokes, and there are 2 TDC's and 2 BDC's, just like a piston engine. unlike a piston engine these 4 strokes happen in a different physical location from each other.

also unlike a piston engine, the mazda rotary takes THREE rotations of the eccentric shaft to have a power event on all the working chambers in the engine. a piston engine needs two rotations. this is why the displacement game is apples and oranges.

mazda rates a 13B with 360 degrees of eshaft rotation, 2x654cc (13B, 12A is 573), if you rate a piston engine on 1 turn of its crank, a 350 chevy becomes a 2.85 liter engine....

so its easy to rate a rotary on 2 turns of the eccentric shaft, which gives 4 working chambers worth, 4x654cc or 2.6. in the real world this seems to be close.

however what about those other 2 chambers? we count all 8 cylinders on a v8, but ignore a one third of the engine when we do the 255/159CID calculation. 3 rotations of the shaft is 654cc x 6 = 3924cc.

apples to oranges... a rotary piston engine IS different than an uppy downy motor, and this is one of the fundamental ways.

mike

13B = 1300cc or 1.3 liter
20B = 2000cc or 2.0 liter
R26B= 2600cc or 2.6 liter

Simply put, it is what the factory engineers say that it is.

On the contrary, if a 4 stroke doubles its process to order to enhance its charge (read as slight supercharge) into its cylinder should its displacement not then be doubled?

And yes, Howard is one of the knowledgeable people gone astray, probably through dealing with the SCCA as they slowly handicapped the Rotary out of competition with Porsche.

Barry

What is point of this? It doesn´t matter if someone calls 13B 1.3L or that its comparable to 2.6L 4 stroke engine or when whole engine is used it equates to 3.9L. All three variants are correct, but only last is fully understandable.

First variant is good for manufactures to simply describe engines configuration.

No one would argue that power equates work per time. Discussed engines are positive displacement pump. So what will happen when you rotate 13B and any 4 stroke 2.6L for same number of revolutions? It pumps same amount of air - granted with same VE%. So they are fully comparable - no handicap. I can´t recall any sanction body which would use this true equalizer - SCCA IIRC classed rotaries with 1.7 multiplier - probably lower thermal efficiency was taken into account... Or BK motorsports effort in ALMS - Courage C65 with 20B - class allowed for 3.5 atmo engines and intake restrictor. They were allowed although 20B equates to 3.9 and also were permited to run larger restrictor. So there isn´t any handicap, exactly opposite - sanction bodies are kind enough, to account for wankel engines disadvantages.

But only last one is really true, it takes 3 revolutions but all parts of engine went through all cycles and everything adds up.

I thought someone had matched up the exhaust pulse timing with a 2.6liter inline 6?

Need to stop equating the rotary engine with a piston engine...
It's just not going to work.

If we take the definition of "displacement" for a piston engine, it's the volume displacement of combustion / intake / exhaust / etc. cycle given for two crankshaft revolutions.

If you try and apply this to the rotary engine, one side rotor is in the middle of one of the cycles.
Do you count this as a "half"?
That really doesn't make sense...


-Ted

I'm going to offer a non-techincal reason for why mazda and various other
makers of rotary products quote displacement the way they do. For the Mazda
case it makes all the 2 rotor engines less that 2 liters in displacment. From a
marketing perspective this is a big win especially in Europe and Japan where
heavy taxes get levied for anything over 2 liters.

I think trying to compare a rotary by displacement to a piston is a mistake
because they are so different in how they behave as they develop power.
I think the only way to compare them is by power and torque. A 13B is
very similiar in power to a 6 cylinder except that the torque is low at low rpms.

I'm talking a stock NA sort of 13B. I know if you slap a turbo on it and stuff
its a v8 killer but thats a whole different animal.

I am saying that the Rotary’s displacement is a physical and measurable size.

A poor example would be a 22 cal. rifle. If it fires twice is it a 44 cal? No its physical size stays the same. And that size is measurable.

Terms like: comparable, relative to, or sort of like, are great for helping others understand, but have no place in a scientific discussion .

For example, if you “cc” the Rotary’s combustion chamber at TDC and multiply that times the compression ratio. What would you expect the resultant displacement to be?

Or another example more to the point… I am doing some in-chamber testing… trying to optimize burn rate and pressure… we need the dimensions of the working parts…. the surface area of the rotor, the offset of the eccentric shaft, and the displacement for each degree of rotation. The changes of displacement will be compared to the actual pressure read by the pressure sensor. Then formulas convert this force to torque and in turn HP, VE, burn rate, location of peak pressure, etc.
What displacement do I use?

The three internet suggestions of the rotary displacement would yield: 500HP for 1300cc, or 1000HP for 2600cc, or 1500HP for the 3900cc.

Which should we use?
Barry

I agree Barry, the engine displaces 1.308 liters per revolution. If I tell a piston engine guy that I am making 200+ WHp out of a Naturally aspirated 1308 cc engine, they about do backflips. So I have to explain that it displaces the same amount of air as a 2.6 L piston engine.

I teach this subject professionally at an educational facility for a salary, would you like to know my thoughts ?

I decided to put this together for the new players who struggle with understanding what a wankel cycle is about and also the true capacity of the engine, a picture tells a thousands words, so I cut up a rotor and a shaft and marked them taking a photo at every 90 degree's of main shaft rotation, following a chamber from firing to firing or one full Wankel Combustion Cycle.

0 degree's TDC No1 chamber firing
http://img401.imageshack.us/img401/839/0degso8.jpg

90 degree's
http://img229.imageshack.us/img229/1204/90degxr0.jpg

180 degree's
http://img229.imageshack.us/img229/4618/180degaw0.jpg

270 degree's
http://img140.imageshack.us/img140/2852/270degym6.jpg

360 degree's (one revolution of crank)
http://img440.imageshack.us/img440/2682/360deghi2.jpg

450 degree's
http://img208.imageshack.us/img208/6769/450deglr0.jpg

540 degree's
http://img212.imageshack.us/img212/5224/540degjo3.jpg

630 degree's
http://img221.imageshack.us/img221/2944/630degtv4.jpg

720 degree's (two revolutions of crank)
http://img380.imageshack.us/img380/2440/720degvr8.jpg

810 degree's
http://img401.imageshack.us/img401/3505/810degho8.jpg

900 degree's
http://img401.imageshack.us/img401/1644/900degxj4.jpg

990 degree's
http://img87.imageshack.us/img87/6586/990degcj4.jpg

1080 degree's Wankel Cycle is complete ! (after 3 full revolutions of the crank shaft) No1 chamber firing again
http://img87.imageshack.us/img87/3117/1080degbg5.jpg

From the above you can see each individual separate chamber (3 per rotor) only fires after 1080 crank shaft degree's has elapsed,, this is why the Wankel is so different to ANY other type of engine, 2 strokes fire each individual chamber once every 360 degree's and 4 strokes fire every individual chamber every 720 degree's.

If you look at a 13B with its 654cc per Individual chamber capacity (thus 1308cc) you can see it aspirates this ONCE every single revolution thus you can compare the 13B to a 2 stroke if you must do so on an equivalence basis (but remember you are not counting the other 2/3rd's of the combustion faces!

Now if you compare it to the much more common 4 stroke engine you can see that 2 faces ONLY are being counted in the engine and thus it has aspirated a total of 2616cc over 720 degree's of crank shaft rotation....... nice little bit of info there but it still misses a whole 1/3rd of the engine!

Finally the ONLY TRUE way to look at a Wankel Rotary is to view it in its own cycle! (and not comparing it to something that it is NOT!) this is only over 1080 degree's of crank shaft rotation, where ALL of the working faces can be accounted for (just as when you do a compression test to see if the poor little donk is healthy or not) :) For it is only when the entire engine has complete one full cycle of work can it thus be rated, be that as functional or in its true capacity sense. You will then see that the humble 13B is indeed 654cc x 3 working faces x number of rotors ! = 3924cc.

Equivalence capacity to time scale (revolutions) for 13B engine, has one power pulse per 360 degree's per rotor
1308cc 360degree's (2 stroke)
2616cc 720 degree's (4 stroke)
3924cc 1080 degree's (Wankel Rotary)

Would apples to apples be hp/lb (of engine weight)? :)

It's not that because for taxation reasons it gets classed as a 2.6L anyway...

Wow! That does suck then.

Rice, that was a great demonstration with the pics.

Of course Mazda says it's a 1.3L. It would be stupid for them to call it anything else from a marketing standpoint, regardless of taxation.

I don't want to buy a 3.9L or 2.6L engine that makes 146hp. I want to buy a 1.3L engine that does.

Yes Peter I agree that it would take three revolutions of the eccentric shaft to have all faces of the rotor see fire.

Following that logic the other trochoid shapes would need four or five revolutions to show all faces and hence we would have to multiply them by 4 or 5.

I am saying that in all of these shapes that the displacement is the volume of the largest cavity created by the two corresponding surfaces (actually 4 counting end plates).

This would hold true for a two-stroke and a four-stroke also.

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

As an example the caliber of this revolver doesn't change if it holds different numbers of slugs. It is still a 357.

http://i287.photobucket.com/albums/l...357MAG_009.jpg

Barry, ALL of the total sum faces fire once in 1080 degree's, nothing more nothing less.

Like I tell my students, you need to take time out and see the pictures, if you don't get it at first do not worry you are not the first and not the last to get confused on how internal combustion engines work.

2 stroke
4 stroke
Wankel

Its pretty simple really. :conehead:

For EVERYONE remember this.

In 360 deg
ALL CYLINDERS in a 2 stoke fire once! no matter if its single or a V16!

In 720 deg
ALL CYLINDERS in a 4 stroke fire once! no matter if its a single or a V16!

In 1080 deg
ALL faces of a rotary wankel fire once! no matter if its a single rotor or a 4+ rotor!

Which ever formula you decide to use in your PISTON applications you need to know which EQUIVALENCE displacement to use (2 stroke, 4 stroke or wankel) there are no aftermarket ECU's set up to run the true Wankel cycle or count the individual rotor faces based of its ACTUAL cycle (which is only 1080deg, nothing less)......... so you just need to know how a motor works and apply the other piston engine formulas to the wankel as appropriate (as I showed all of you) :)

Peter, I was curious about something. Would there be any benefit in engine control with ECU of such abilities? Like better fueling of individual chambers?

There are some of us who are using this kind of bespoke *aftermarket* true rotary ECU and have been doing so for a long time! ;)

Yes there are major benefits!
Mazda racing ECU's and stock ECU's are of this type 1080 deg true Wankel cycle, it is the only way to truely look at the engine.

Otherwise you do need to do alot of conversions to adapt all the other ECU's, not to hard to do, but most dont get it, especially when you are looking at things like control of injector opening and closing degree's etc.

Actually example (3) (shown below) would take 1440º and

example (4) would take 1800º to complete all faces of its trochoid shape.

Different rotor/stationary gear ratios would be used to accommodate the 4 in 3 and 5 in 4 trochoid mesh.

Barry


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

I am ONLY refering to the Wankel Rotary I pictured and as we all use, drive and modify ;)

Nothing else :)

Interesting question, Barry.

Mazda stock ECUs are only capable of measuring 360 degrees of eccentric shaft rotation; since the trigger wheel is mounted on the eccentric shaft it cannot discern the difference between one rotor face vs the other three. As far as the ECU is concerned, the rotary may as well be a two-cylinder two-stroke that displaces 1.3L per 360 degrees.


In a similar fashion, you could compare a 1.3L two stroke piston engine to a 2.6L four-stroke piston engine... both ingest the same amount of air in 720 degrees, but the 1.3L two-stroke occupies less space and will weigh less assuming similar materials are used to build both engines. Is this starting to sound familiar? If I'm not mistaken, the two-stroke should have less torque at low RPM greater fuel consumption than the four-stroke. I'm not an engine expert (rotary or piston) so please correct this if I'm wrong.

No mate they use 1080 in programming and can measure it with a 360 wheel on the crank ;)

http://www.rotarydevelopment.net/Rot...elCyclesLi.jpg

You may not know where each individual rotor face is but that is not an issue so far as knowing the actuall Wankel Cycle and when to trigger things.

Guys, displacement by definition is the volume displaced by one revolution of the crankshaft or e-shaft in this case. doesn't matter if it is 2 stroke, 4 stroke, or not.

You guys are all going off on this "when does it ignite, and when does it do this or that?" It's all moot.

total volume displaced over 1 crankshaft revolution = Displacement. It is consistent across the board. There is no Mazda conspiracy to hide its actual displacement. A 6.0 liter engine is a 6.0 liter engine, whether its 2 stroke or 4. You can't use the "Getting back to face 1" argument due to the trochoidal nature and the 3:1 ratio. If you do use this argument, then you have to divide the final result by 3 because of the 3:1 ratio.

Now, what is the one consistent thing in all the engines? output RPM. Displacement is based on 1 of these revolutions. So any other attempt to define definition by more than 1 revolution is incorrect. 1 revolution is the consistent factor across the board. Te reason racing bodies use a multiplier is to even the playing field just like they do for 2-stroke engines. The advantage a 2 stroke has is that is happens to fire its full displacement per revolution, just as a rotary does, where as a 4 stroke only fires half of its displacement per revolution.

Don't confuse revolutions of the rotor versus the crank/eshaft. You'll end up chasing ghosts.

It's not a argument, its FACT! ALL INTERNAL COMBUSTION ENGINES (except the rotary!) are rated on ONE CYCLE OF WORK FOR THE COMPLETE ENGINE.

Mazda conveniently choose to not rate the whole engine, if you or others can't get that then you need to move onto another area of interest I suggest, one you can understand ;)

The Wankel Rotary is a 1080deg cycle, nothing more nothing less!

You btw f*** your own argument cause a 4 stoke is NOT rated after only 360 degrees! cause it HAS NOT COMPLETED ITS CYCLE OF OPERATION! it is rated ONLY AFTER 720 degree's (suck sqeeze bang blow) it only SUCKS once in 720 degree's! and ALL piston faces are counted to rate the displacement of the whole engine! not 1/3rd of them, or 2/3rd's of them! but ALL OF THEM!......... The irony is only in the rotary world where people want to only count 1/3rd of the combustion faces and rate it as a 2 stroke engine, but its NOT! its a wankel and thus they only want to count one face (which misses out 2/3rds of the rest of the engine). if only you could do this on a BDC built half bridge!!!! then when it drops an apex seal on each rotor he can then tell you its only a 2 stroke engine like people in this thread and the other 2 apex seals and four combustion faces are not required LOL!!!!!!!!!!!

ah yes, good point! one of the rotary ADVANTAGES is that each stroke is 270 degrees instead of 180 like a piston engine.

you are wrong though, piston engines suck all the time :smilielol5:

mike

Just remember

Each engine is only *honestly* rated for displacement AFTER (ALL OF ITS ELEMENTS) have completed one cycle of work.

Elements = ALL individual combustion faces
Cycle = defined by power pulse (or simply spark firing) *remember suck sqeeze bang blow*

It's not a hard concept to fathom honestly ......... some always struggle cause they don't know how the 2 storke or 4 stroke or Wankel actually work though!

Define the cycle!
Count the faces!
There is your answer for displacement!

Equivalence (like I mentioned in my second post) can be used to compare the 13B to other non wankel cycles (as defined by time to displacement), IE displaces 1308cc in 360deg or 2616cc in 720deg, but it only does its wankel cycle in 1080deg which my friends = 3924cc in 13B form, nothing less!

no worries :001_302:

and yes taxation is based on equivalence (to std common/majority 4 stroke reciprocating engines) and thus in 720deg the 13B does inhale (suck!) or partially displace 2616cc as shown............... but as we all know and can see it has NOT completed its full Wankel Cycle in that time, this only happens in 3 full revolutions of 1080deg.

:party:

OK, so my guess is that this is a 2499 magnum!



http://i287.photobucket.com/albums/l...357MAG_009.jpg

Peter and Barry, I have too much respect for both of you to stand idly by while you dispute so vehemently.

It's Ok to disagree, but let's not turn this into an personal argument. This is a case where the only important factor is that we understand how the engine works. It is accurate to say that a rotary engine completes an intake, compression, power and exhaust stroke on each rotor during a single rotation of the engine. However it requires 3 revolutions of the engine for all faces of each rotor to see all four strokes.

The difference between you is that Barry is looking at a single rotor housing as the displacement-providing chamber, while Peter is looking at the rotor as the displacement-providing chamber. Neither of you is right or wrong, it is a difference of perception.

How each of us slices this up depends on personal preference and nothing more. There is no right answer here.

I appreciate the information and clarification provided in this thread, but I don't want two knowledgeable and intelligent members of this community bogged down in this senseless argument.

Let's leave it with the cycle explanation and keep this thread informative.

Hey DOHC, who are you calling knowledgeable? :lol:

Just one more thought to keep in mind.

We have all driven different size engines with standard transmissions. When you let the clutch out on a rotary what size engine does it feel like?

A- 3900cc engine.
B- 2600cc engine
C- 1300 cc engine

This is not valid Barry and you know it.

Reasons why rotary engines feel soft, and they really are at low RPMs have roots in inherent engine configuration and its design flaws.

Even Kenichi Yamamoto states exact reasons why is it so. Gas leakage through numerous gaps of gas sealing coupled with 1.5 times slower working cycle - losses losses losses:beatdeadhorse5:

Forums!

Difference between me and Barry is I am right he is WRONG!

I get paid to teach people for a living!

He comes to forums cause no one will pay him to teach people :fawk:

You can bury your head in the sand as much as you like, if you cant accept equivalence or know what an internal combustion cycle is or how all elements of an engine are accounted for then there is not much hope left for you to learn. It is not a hard concept, for some though I agree its a mountain v's for others a mole hill.

Good luck to you, one day you will learn something, I suggest you go to school and spend less time on the internet.

I'll leave you with one simple point to ponder!

Why when you do a compression test do you measure ALL THREE FACES over 1080 degree's? per rotor?

Why is the engine classed as not healthy and due for rebuild if ANY ONE of these three faces per rotor is below another by any great margin?

Why wont the engine function normally with only two apex seals per rotor if its only 1308cc for a 13B?

Why
Why
Why
Why
Why??????

Cause Rice racing is RIGHT!@

You are WRONG!

:beatdeadhorse5:

:seeya: Have fun in your thread :seeya: when you want to learn about engineering feel free to E-Mail me :seeya:

Okay where I do agree with Rice on how he is coming to his conclusion, the fact that he is getting paid to teach something means absolutely nothing. To be honest there are more then a few people out that get paid to do a job & know nothing about what they are doing including teaching. So that comment literally could be seen as ego talking not knowledge.

I see nothing wrong with a healthy discussion & info provided with proof/facts, but I do agree with NoDOHC that insults do nothing for the topic.

Wow.... you are so off base its amusing.

ALL engines, 2 stroke, 4 stroke, rotary... are using only one rotation of the crank shaft to measure displacement.

If you use your logic, then the 6.0 liter ls2 engine would really be a 3 liter 2 stroke engine if you built a custom head for it and ignited fuel every revolution? ... I am sorry, but you are wrong. The bore and stroke never changed therefore, your logic is home to a MAJOR fallacy.

Wow.. you are arguing that somehow the method of measuring displacement has anything to do with which apex seals are required. To measure the displacement of a rotary engine, you would total the amount of volume displaced by one face of each rotor and add it up.

Just like in a piston engine, you would take the displaced volume of EACH cylinder (notice how 4 stroke or 2 stroke doesn't matter) and add them together.

Displacement calculation DOES NOT take into account how many revolutions it takes to actually fire each piston or rotor face. I have a lot of respect for you rice racing, but you are very far off base here. Displacement caluculations are simple:

How much volume is displaced by the engine (regardless of type) in 1 rpm of the crankshaft/eshaft IS your displacement. Its fact, non-arguable.
You can do displacement calculations without the heads on. it doesn't matter if it is 2 or 4 stroke. Hence why they do all 2 and 4 stroke piston engine calculations like so: stroke x bore x # of pistons = displacement. which would be the EXACT same as: displacement of each piston added together x 1 rpm. The 1 rpm factor must remain the same between all engines to have an effective and consistent method of calculating displacement.

The key thing here is that people are arguing different things. "taxation displacement" has nothing to do with the actual definitive displacement of an engine. And determining displacement by the argument of "every rotor face/cylinder needs to fire" is faulty at best. For this to be even remotely true, then you would have to calculate 2 stroke and 4 stroke displacements differently. The fact of the matter is: This just doesn't happen. So these imaginary figures based on taxation methods or requiring every cylinder or rotor face to fire are just that. Imaginary figures that do not take into account he constant of 1rpm. Without this constant, there is no mathematical consistency, and then the game becomes fault riddled.

Displacement is universal to all engines as the displaced volume from 1 crankshaft/eshaft/centershaft revolution. 2 stroke or 4 stroke or rotary does not change this.