Actual Rotary Displacement Request

[Barry Bordes]

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

[FerociousP]

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

[j9fd3s]

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

[Barry Bordes]

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

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.

[Barry Bordes]

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

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

[Libor]

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.

[Monkman33]

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

[RETed]

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

[t_g_farrell]

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.

[Barry Bordes]

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

[NoDOHC]

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.

[RICE RACING]

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

[RICE RACING]

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)

[My5ABaby]

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

[PhillipM]

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.


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

[t_g_farrell]

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.

[My5ABaby]

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

[Barry Bordes]

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/ll129/bbordes/Trochoidshapes.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/ll129/bbordes/Taurus_627-KLM_357MAG_009.jpg

[RICE RACING]

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:

[RICE RACING]

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) :)

[Libor]

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

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?

[RICE RACING]

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.

[Barry Bordes]

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:

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/ll129/bbordes/Trochoidshapes.jpg

[RICE RACING]

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

Nothing else :)

[scotty305]

Interesting question, Barry.

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 ECUs and stock ECUs are of this type 1080 deg true Wankel cycle, it is the only way to truly look at the engine.

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

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.

[RICE RACING]

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.


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/RotDev/pix/WankelCyclesLi.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.

[Monkman33]

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.

[RICE RACING]

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!!!!!!!!!!!

[j9fd3s]

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

http://www.rotarydevelopment.net/RotDev/pix/WankelCyclesLi.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.

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

[RICE RACING]

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!

[RICE RACING]

Wow! That does suck then.

Rice, that was a great demonstration with the pics.

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:

[Barry Bordes]

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!!!!!!!!!!!





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



http://i287.photobucket.com/albums/ll129/bbordes/Taurus_627-KLM_357MAG_009.jpg

[NoDOHC]

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.

[Barry Bordes]

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

[Libor]

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:

[RICE RACING]

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.

[RICE RACING]

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:

[RotorDad]

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:

.

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.

[Monkman33]

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!!!!!!!!!!!


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.

[Monkman33]

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.

[RotorDad]

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

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

[vex]

Fuel to the fire:

Engine displacement is the volume swept by all the pistons inside the cylinders of an internal combustion engine in a single movement from top dead centre (TDC) to bottom dead centre (BDC). It is commonly specified in cubic centimeters (cc), litres (l), or (mainly in North America) cubic inches (CID). Engine displacement does not include the total volume of the combustion chamber.

http://auto.howstuffworks.com/question685.htm
When the piston moves from top to bottom, it sucks in a certain amount of air. How much air it can suck in depends on how big around the piston is, and how far it moves when it goes from top to bottom.

Let's say that the piston in your car is 4 inches (10.16 centimeters) in diameter (also known as the bore), and it moves 4 inches from top to bottom (also known as the stroke). That means that one piston in your engine can suck in:

radius ^ 2 * pi * height = volume of a cylinder
5.08 cm (bore/2) ^ 2 * 3.14 * 10.16 cm (stroke) = 823.3 cubic centimeters

If your car has 4 cylinders, then it has a has a total displacement of:

4 * 823.3 cubic centimeters = 3,292.1 cubic centimeters, or 3.292 liters
A car manufacturer would round that up and say that your car has a 3.3-liter engine. This means that the displacement of this particular engine is 3.3 liters. If you were to turn the crankshaft of this engine through two complete revolutions, the four pistons would inhale a total of 3.3 liters of air.

http://www.wisegeek.com/what-is-engine-displacement.htm
Engine displacement refers to the volume swept by all the pistons in an engine. It does not include the volume of air above the piston where the initial spark fires. Engine displacement can be calculated with the following formula:

Engine displacement = π/4 * (cylinder diameter)2 * stroke * number of cylinders

Engine displacement is the volume swept by all the pistons inside the cylinders of an internal combustion engine in a single movement from top dead centre (TDC) to bottom dead centre (BDC). It is commonly specified in cubic centimeters (cc), litres (l), or (mainly in North America) cubic inches (CID). Engine displacement does not include the total volume of the combustion chamber.

Read more: http://www.answers.com/topic/engine-displacement#ixzz1ORxci7w9
http://www.answers.com/topic/engine-displacement

[Monkman33]

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.

[Monkman33]

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

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


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.

[Barry Bordes]

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

[RotorDad]

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.


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.

[Monkman33]

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. :-)

[My5ABaby]

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

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:

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.
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:

[Libor]

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:

[cewrx7r1]

Forums!

Difference between me and Barry is I am right he is WRONG!
I get paid to teach people for a living!



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.

[My5ABaby]

Hey, nothing wrong with being a bit Narcissistic:o13:
A bit, no, nothing wrong.

A lot, yes, something wrong.

:ugh2:

[Monkman33]

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:


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.

[diabolical1]

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.
^ 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.

[Monkman33]

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?

[diabolical1]

^ 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.

[vex]

^ 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/airplane/Images/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.

[Monkman33]

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.

[vex]

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).

[diabolical1]

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.

[Monkman33]

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.

[vex]

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 :),Otto cycle is just the technical term of suck, squish, bang, blow 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.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.

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. 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.

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.
And that's fine to do--Just be aware that it is not equivalent to the standard practice of engine identification.

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.
Like many bright people have said before; it's all relative.
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.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.
If you want to compare the two engine types, you have to have a constant. That constant is TDC and BDC for all otto cycle engines. What is more constant than the point that transfers the power from the point of generation to the drivetrain?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.

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.
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.

[Monkman33]

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.

[NoDOHC]

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.

[vex]

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?

[Libor]

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

[Monkman33]

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?

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

[Barry Bordes]

http://www.grc.nasa.gov/WWW/k-12/airplane/Images/otto.gif
http://www.grc.nasa.gov/WWW/k-12/airplane/otto.html



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/ll129/bbordes/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/ll129/bbordes/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

[vex]

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/ll129/bbordes/450deglr0.jpg
[/COLOR]

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.

[My5ABaby]

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

[vex]

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 ;)

[Libor]

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

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

[Libor]

You mean like TDC to BDC ;)

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°.

[vex]

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?

[diabolical1]

Otto cycle is just the technical term of suck, squish, bang, blow.
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 ;))
The only thing I do not understand is where the 2.6L value comes from without using a multiplier. Anyone know?
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.

[Barry Bordes]

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.

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/ll129/bbordes/volumechangepistonvsrotary.jpg

[My5ABaby]

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'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.

You mean like TDC to BDC ;)
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...

The rotary engine as rated by Mazda is 1.3 liters because each individual rotor, following one face of one rotor through the complete cycle, has a swept displacement of 654cc or .65 liters. Multiply this times 2 rotors to achieve 1.3. Since this only accounts for 2 of the total of 6 rotor faces, we multiply our answer by 3 to get an actual displacement of 3.9 liters. However since the rotary engine is a 6 stroke engine and not a 4 stroke engine since it takes 3 complete eccentric shaft revolutions to fire all faces instead of the typical engine's 2, it only does 66% the work of a 4 stroke 3.9 liter engine. Calculating for this we divide 3.9 by 1.5 to get a total of 2.6 liters equivalent work to a 4 stroke piston engine. All of these, from a 1.3 liter in physical size package.

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

[vex]

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/ll129/bbordes/volumechangepistonvsrotary.jpg

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.

[Barry Bordes]

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

[vex]

Of consequence; this comes from a thread I had going for awhile.
http://www.rotarycarclub.com/rotary_forum/showpost.php?p=129975&postcount=35

[RICE RACING]

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:

[RICE RACING]

The problem with most of you is you are forgetting the Wankel shares a common combustion chamber, intake and exhaust port per rotor for 3 rotor faces.

What is not in dispute is that it takes 1080deg for the Wankel cycle to complete, it is in every proper text you will find on the topic. It does not take rocket science to then see it will equal 3 times the rated partial operating cycle some like to quote as the "capacity".

Sure its 1.3lt in one rev
And its 2.6lt in two revs
But I say again and again it takes three revs to complete the Wankel cycle, nothing more and nothing less. And it will displace 3.9lt for a 13B in that time. ;)

[RICE RACING]

FOr some who love comparisons or measuring to others (why would you I dont know but.........)

The 13B is like the following

Power density like a 1.3lt 2 stroke!
Fuel consumption of like a bad 2.6lt 4 stroke!
Racing durability similar to a equal powered 3.9lt!

Some old boys will get what I am saying with the above.
The Wankel has allot of good attributes with its very large areas and slow speeds and this shows up in its durability as a racing engine (on equivalence basis for time related "displacement" @ equal BHP levels) compared to reciprocating alternatives. When you understand how the engine works and why this is you quickly see its not magic but its an attribute of the Wankel Cycle, its slow speed, and big area to time relationships. Sure its inefficient on ANY equivalence measure (power density, peak speed, etc) but BANG for BUCK and DURABILITY wise its very very very hard to beat.

I don't know if there are many or any on this forum who get WTF I just said but hopefully it will ring true to someone who has more than a shitty web page with fictional customer base and delusions of self praised status.

Wankels are a cool motor its only the people into them in the majority that are weird!

[Barry Bordes]

FOr some who love comparisons or measuring to others (why would you I dont know but.........)

The 13B is like the following

Power density like a 1.3lt 2 stroke!
Fuel consumption of like a bad 2.6lt 4 stroke!
Racing durability similar to a equal powered 3.9lt!

Some old boys will get what I am saying with the above.
The Wankel has allot of good attributes with its very large areas and slow speeds and this shows up in its durability as a racing engine (on equivalence basis for time related "displacement" @ equal BHP levels) compared to reciprocating alternatives. When you understand how the engine works and why this is you quickly see its not magic but its an attribute of the Wankel Cycle, its slow speed, and big area to time relationships. Sure its inefficient on ANY equivalence measure (power density, peak speed, etc) but BANG for BUCK and DURABILITY wise its very very very hard to beat.

Wankels are a cool motor its only the people into them in the majority that are weird!

It is interesting that you should describe the Rotary that way.

While watching the 24hrs of Le Mans I had a similar thought. In the parallel universe where the wankel displacement is measured after one full rotation the rotor, it could be argued that Rotor RPM would also be used and not the PTO shaft (eccentric shaft) RPM.

This would necessarily change our perspective of the Mazda win in 1991. Of course Mazda should have beaten Jaguar, Mercedes, and Porsche.

Its engine was much larger at 7848 cc, and it loafed along at 2300 RPM with a red line of 3000RPM.

It won with its superior fuel economy of this slow turning large displacement rotary.

Barry


http://i287.photobucket.com/albums/ll129/bbordes/Mazda787B.jpg

[FerociousP]

Wankels are a cool motor its only the people into them in the majority that are weird!

I think everyone in this thread shows that to be true! lol

[RETed]

WTH...
Why did that post reply delete got undone?

This is just a reminder that this is a TECH section.
Anything NOT related to tech should NOT be in here.
Personal beefs should go to PM - don't air your dirty laundry in here.
I'm tired of editing replies, so I'm just deleting them en masse at this point.
I'd hate to get rid of this thread just because a few people can't follow rules.
There's good information in this thread, and it's getting muddied by a few immature kids.

I'D HIGHLY SUGGEST YOU USE THE IGNORE MEMBER FUNCTION IF YOU JUST CAN'T STOMACH READING THEIR REPLIES.
What you can't read can't hurt you.


-Ted

[My5ABaby]

Love you............................

:suspect:

[vex]

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 ... :)Why? I don't like counting grains

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. I'm still not seeing it. With the 3.9 and 1.3 I understand it as the references are stated, but I can only achieve 2.6 with a multiplier. My contention is that if a multiplier must be used then it is less accurate in that it inherently makes assumptions about some process that doesn't come in to play. (I'll have to read over Peters post in more detail, but I was not convinced by the other links attempting to explain it)

[RICE RACING]

Barry your PTO shaft idea is a little off base.

We are not talking about "gearing down" the actual cycle itself, that is a non negotiable part of any engine cycle be it 2 cycle, 4 cycle or Wankel.

I can see what you and everyone else including Mazda is saying (1.3lt for 13B chamber capacity), that is a given. But its an odd engine by nature that it has common combustion chamber, intake and exhaust ports ;). I only look at all motors in their complete "cycles". See my earlier posts.

Equivalence (just based on the physical displacement to time across all three established engine types allows them to race).

The Mazda 787B only had "good economy" cause it had Group B weight breaks on its side V's the Sauber and Porsche and Jaguar cars, nothing more, it was a smaller and lighter car, and had allot less power (250+bhp less than the Sauber in race trim!)... It was very light weight with moderate power, but with EXCELLENT ROTARY ENGINE RACE DURABILITY. Mazda's have always expolited the rules to get a favorable advantage in racing, the one biggest factor they have had over others is astounding reliability and durability in race trim, rather than any outright speed. :biggthumpup:

[Libor]

The Mazda 787B only had "good economy" cause it had Group B weight breaks on its side V's the Sauber and Porsche and Jaguar cars, nothing more, it was a smaller and lighter car
This is correct. Mazda was allowed to weight 830kg compared to 950kg of Porsche and 1000kg of Sauber and Jaguar. Huge difference:suspect:

and had allot less power (250+bhp less than the Sauber in race trim!)

I though that these outputs were used only in qualification?
I´m sure you know how it was!:bigear:

[scotty305]

At one of the Sevenstock tech talks (two or three years ago?), someone from the Mazdaspeed team mentioned that both Mercedes cars were consistently faster than the 787B, but they had engine problems and did not finish the race.

I came across a nice set of photos from that race:
http://www.flickr.com/photos/mendaman/4974579060/in/set-72157623971081040

Every serious race team tries to exploit their class rules as much as possible in order to win. In my opinion, it would be a great honor to have a new rule written because your team had found and exploited a good loophole in a previous set of rules.

[GorillaRE]

At one of the Sevenstock tech talks (two or three years ago?), someone from the Mazdaspeed team mentioned that both Mercedes cars were consistently faster than the 787B, but they had engine problems and did not finish the race.

I came across a nice set of photos from that race:
http://www.flickr.com/photos/mendaman/4974579060/in/set-72157623971081040

Every serious race team tries to exploit their class rules as much as possible in order to win. In my opinion, it would be a great honor to have a new rule written because your team had found and exploited a good loophole in a previous set of rules.
This may be true but I think it only applies due to Mazda keeping a 9000rpm limit during the race, for longevity reasons. The 787B was capible of 150-200 hp more with the tested 10500rpm limit!

-J

[Barry Bordes]

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

This would probably be a good time to reiterate my initial request of having us use the Industry Standard in referring to rotary sizing when corresponding with one another.

Interesting side note… I checked the websites of both Rotarygod and Rice.

It was reassuring to note that they both use the requested Industry Standards for clarity. There were no 26B or 39B 2-rotors noted.
Barry

[RICE RACING]

There is no question at all that its a 13B.

There is also no question at all that it displaces 3.9lt when you count the whole engine.

They are two different things, its only when you are talking equivalence to other types of engines you need to know how much it displaces in time (without repetition)... like for your formulas (all based around 2 stroke and 4 stroke) then you just use 1.3 or 2.6 respectively.

It's the common shared chambers that allow all the 3 rotor faces *per rotor* to do their work (Wankel cycle) which confuses many people & to be honest its not really worth debating as we are all really talking about the same thing.

As you well know when you do a health check on the Wankel (two rotor) you will get in most cases 6 different readings as each chamber is different most times (seal wear, tolerances and rotor cavity disparity) ALL of these go to providing work and a health engine overall and ALL need to be counted you will agree to validate that assessment. Otherwise we would only ever need to measure two faces? *stirring pot*...... You and everyone get what I mean here who has anything ever to do with Wankels.

This is why I count all faces and rate the true capacity of the engine as per its cycle, no matter what type it is. And that is the basis of my points on the topic. So its a 13B that displaces 3.9lt over its Wankel cycle.

[FC Zach]

This is why I count all faces and rate the true capacity of the engine as per its cycle, no matter what type it is. And that is the basis of my points on the topic. So its a 13B that displaces 3.9lt over its Wankel cycle.

Well said! short, simple, and to the point. That's easy enough for me to comprehend.

[Barry Bordes]

From Webster’s New Collegiate Dictionary… displacement\ c: the volume displaced by a piston (as in a pump or engine) in a single stroke; also: the total displaced by all the pistons in an internal combustion engine.

The key phrase being “ in a single stroke”.

Notice that using the dictionary definition of displacement with your “personal convention of a full cycle” to determine displacement of a 4-stroke... the 720º required for the full cycle would incorrectly doubled its displacement.

Barry

[RICE RACING]

From Webster’s New Collegiate Dictionary… displacement\ c: the volume displaced by a piston (as in a pump or engine) in a single stroke; also: the total displaced by all the pistons in an internal combustion engine.

The key phrase being “ in a single stroke”.

Notice that using the dictionary definition of displacement with your “personal convention of a full cycle” to determine displacement of a 4-stroke... the 720º required for the full cycle would incorrectly doubled its displacement.

Barry

"all pistons" = "all faces" :117: apply that one :Chevy_anim: you then will see that all faces and or otherwise "displace" a volume.

The troubling thing in a Wankel that people don't get (even if they dont understand the cycle) is that is has a common housing that is shared across three faces per rotor :cheers2:

Therein lies your path to understanding, rather than looking for an obscure definition in a dictionary :ack2:

[RICE RACING]

Baz,

Would you only measure 2 pistons in a 6cly 3.9lt for a customer and give it a clean bill of health?

so follows

Would you only measure 2 rotor faces in a 13B and give it a rubber stamp pass BDC style?

You would measure all pistons and report on each.
Mostly everyone inc me and you would measure all rotor faces and report on each.
Why would we do this? if its only using 2 faces and is only 1.3lt in capacity? Surly by Mazda convention and everyone else's then we are mad and we could have saved money by leaving out one apex seal per rotor (potential saving there!) just as we could leave out 4 pistons in the 6cyl and save weight too :)

PROBLEM: Engines function and displace volume over the operating cycle and total sum of working chambers. This is how all internal combustion engines work. 2 stroke, 4 stroke and low and behold the Wankel too :smash:

[calculon]

Out of curiosity, what does everyone think that the displacement of an LS2 is?

[Barry Bordes]

"all pistons" = "all faces" :117: apply that one :Chevy_anim: you then will see that all faces and or otherwise "displace" a volume.

The troubling thing in a Wankel that people don't get (even if they dont understand the cycle) is that is has a common housing that is shared across three faces per rotor :cheers2:

Therein lies your path to understanding, rather than looking for an obscure definition in a dictionary :ack2:

Peter,
Maybe it would help if I told you that I agree that a full cycle for the rotor is its complete revolution or three revolutions of the eccentric shaft.
My point is… what does that matter?

In the example of the .357 revolver both views are expressed.
The criminal has two main concerns as he runs from the crime scene.
1- What is the size and velocity of the slug coming at me?
2- How many shots does this guy have before a reload is required?
Both valid concerns… Now let’s say that we add an ammunition belt to supply bullets…. Then the only concern is the size and velocity of the slug. Our engines have continuous belts of ammunition between fuel stops.


Our attempt to make power is done by igniting a specific volume (654cc) which in turn works on a lever. We want to have this happen as often as possible. The best we can do in our case is once each revolution.

Counting in three’s won’t affect anything.

Barry

[RICE RACING]

It has an effect if one chamber (or face) is buggered, Image a S@W 357 mag with a stuffed chamber one will always not work properly, shoot a bad group :dunno: it's why we count all the parts that make up the 6 shots, just like in our donkey engines. :)

All the chambers count to make a whole rig that will work well, faces, pistons does not matter, we dont just look at one alone, we look at all parts of the system :)

We are talking the same things I think :ack2:

[My5ABaby]

It has an effect if one chamber (or face) is buggered, Image a S@W 357 mag with a stuffed chamber one will always not work properly, shoot a bad group :dunno: it's why we count all the parts that make up the 6 shots, just like in our donkey engines. :)

All the chambers count to make a whole rig that will work well, faces, pistons does not matter, we dont just look at one alone, we look at all parts of the system :)

We are talking the same things I think :ack2:
The totality of the gun (i.e. whether it works, the total chambers, how well it works, if all the chambers are working, etc.) has nothing to do with it's displacement (i.e. caliber). Unless I'm not understanding it, your argument seems to be referencing the engine as a whole in terms of whether it works or not and how you determine that versus how to determine displacement. When measuring the displacement of a piston engine, do we care if one of the cylinders is screwed up or if a valve isn't working?

[Libor]

I would be interested how we should approach "displacement" of unnusual engines. Like Ilmor "5-stroke" where three cilinders are creating working unit - two outer are regular and middle one is bigger and extracts addition power from still expanding exhaust gasses.

By definition of displacement from above, such engine would be described as sum of displacements of individual cilinders. But only two cilinders have Intake and are doing pumping work of working fluid.

Displacement on its own means nothing without considering working cycle.

[RICE RACING]

I would be interested how we should approach "displacement" of unnusual engines. Like Ilmor "5-stroke" where three cilinders are creating working unit - two outer are regular and middle one is bigger and extracts addition power from still expanding exhaust gasses.

By definition of displacement from above, such engine would be described as sum of displacements of individual cilinders. But only two cilinders have Intake and are doing pumping work of working fluid.

Displacement on its own means nothing without considering working cycle.

:iamwithstupid: