Actual Rotary Displacement Request
 

Request to use Actual Rotary Displacement

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

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

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

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

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

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

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

My Thoughts,
Barry

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

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

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

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

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

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

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

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

mike

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

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

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

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

Barry

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

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

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

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

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

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

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

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


-Ted

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

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

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

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

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

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

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

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

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

Which should we use?
Barry

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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