spasso

This is probably a newb question but I was just wondering how the displacement is calculated on a rotary. I know this has probably all been discussed before but it just hit me that displacement would be calculated differently, as well as meaning completely different things between pistons and rotors.

I know for a piston engine it is the displacement of each piston using bore & stroke times the number of cylinders. Also, for the 4 stroke engine, each piston is only useful for every other revolution of the crank. I guess my thought is that if you add up the displacement, you would have say 2.0 Liters that is useful for every other revolution, the "wasted" revolution being the exhaust then intake strokes. So to complete all 4 cycles of internal combustion, each piston has to complete 2 revolutions.

A single combustion chamber of the rotor, on the other hand, completes all 4 cylces with three spins of the eccentric shaft.

On a 13B engine I would assume each rotor displacement is 654 cc. Now is that 654 cc for each combustion chamber created or is it 218 cc for each side of the rotor X 3 sides to get 654 cc for each rotor housing?

I hope this makes sense, my guess would have to be that each combustion chamber created is 654 cc meaning that each turn of the eccentric shaft causes 1308cc to be displaced.... or does it? I am too confused i guess.

Wankelguy

If you take both rotors and drop them in a full vessel of water, then measure the volume of water which has been DISPLACED, that's the displacement.

(Archimedes' theory) -WG

spasso

OK

MIKE-P-28

iTrader: (1)

I think its quite a bit more complicated than that, I wouldnt have wanted the task of figuring it out

REVHED

Spot on.

As you said before, a 4-stroke piston engine only uses half it's rated displacement per crank-shaft revolution. This means that a 13B displaces the same amount as a 2.6L 4 cylinder engine.

In fact, many people refer to rotaries as a 4-cycle 2-stroke engine. It has the 4 distinct phases of the otto-cycle yet it does the same amount of work as a two stroke.

mar3

iTrader: (8)

umm...serious or not?

error402

If you took that route, wouldn't you have to take a rotor housing seal off any holes, put on the ground and fill it with water. (Basically turning it into a bowl.) Then place a rotor in the rotor housing filled with water. Mesaure the amount of water left when the rotor is placed in.

-Error402

Wankelguy

Actually, a little of both.

FJ

You're right about Archimedes' theory of course, but what does the displacement of the rotor itself matter? Maybe I'm missing something. I did notice in your last post that it was half in jest, but what is the realistic half?

At any rate, the relevant question is the difference in capacity between BDC and TDC for one rotor face, per revolution, as compared to the bore and stroke of a piston engine and that is 654cc for the 13B. We don't need an extra revolution for the exhaust/intake stroke so the rotary gets compared to a four cylinder of twice the size, as mentioned, as we move twice the air/fuel per revolution.

Question: (semi-related to error402's post) If the rotor is placed in the housing with one face at BDC, and then is rotated to TDC, that would be 654cc displacement for that face. The total displacement at any given rotation, including the other chambers, would be what? Is there a constant displacement for the three chambers combined, or it varies?? I have a feeling it adds up to 2.6L always, but damned if I can prove it...

I've used the same description as Revhead: 4-cycle / 2-stroke and have had nothing but blank stares as a response. Takes a little explanation, and even then it's not always accepted/understood.

Bottom line though, rotaries just don't compare well to piston engines.
Also, the apparent "efficiency" of 135 HP per "1.3" liter (or more in the later generations) is really not particularly efficient.
Not that I lose sleep over it.

-John.

mar3

iTrader: (8)

He was joking...we just don't know him well enough yet to see when he's serious or not...it would be the same as saying the displacement of a boinger could be calculated by dropping all 16 pistons in a bucket of water and measuring what spilled...

Suparslinc

Uh Oh don't let Felix Wankel see this...
There was a huge and mostly pointless debate going on in the lounge about whether the 13B was 1.3 litre or 2.6

mar3

iTrader: (8)

Dear sir, you are virtually obligated to make the link now and pour some gasoline onto the barbeque fire!!

Suparslinc

FJ,
no, rotaries are not "efficient" and no one ever said they were. They are "powerful". There is a difference.
The surface area of the combustion chamber is too great for proper heat disappation or some ****.
Just a minor detail, but that whole efficiency thing is one of my pet peeves

FJ

That was my thought, and I didn't mean to sound serious in my response. But you're right...I sounded harsh.
I was just looking for a comeback on the "realistic half", in the same light as his other posts. Re-reading my post, it does sound harsh.
Not what I meant at all. I had a grin at his first post and just wanted to continue that train of thought (being silly) while adding my comments for spasso.

I certainly did not mean any disrespect towards Wankelguy.
My "humor" was lost ...but when I re-read it, I see what you mean... thanks for letting me know.

-John.

mar3

iTrader: (8)

The fuel mileage itself speaks volumes regarding efficiency...but ya gotta admit, for a half-keg, it sure can kick out a lot of ponies...

FJ

I believe that was my point... :
I agree they are powerfull...

-John.

Suparslinc

My mistake...
It was actually a choice between 2.6 or 3.9L

https://www.rx7club.com/forum/showth...ghlight=3.9%2A

The replies are so long-winded you will strain your eyeballs...but admittedly some guys put a whole lotta thought into it.

edit: alright FJ I misread your post... excuuuuuuuse me!

Pedestrian X

wouldent the displacement just be like what sumbody else said above, take a rotor houseing full of water plop a rotor in it and the amoun of water left in X2 (2 rotors) would be the displacement?

FJ

No problem.

-John.

FJ

Nope.

That would be the total displacement at a given position for the three rotor faces, times two for a two rotor housing, or times three for some lucky bastards.

Bottom line, only the one rotor face doing work counts.

On a rotary, it's one rotor face at a time, as mentioned above.
If it's a piston, only every second stroke counts, as one stroke up compresses, one down produces power. Then one up again to exhaust the burnt gas, and one more down to inhale the fresh air/gas. Repeat. Four strokes. But only one power (useful) stroke.

By the same token, only one rotor face per housing is doing work, not all three, so they don't all count on the displacement. The difference is, a rotary does not need to wait. While one rotor face is firing (on the expansion "stroke") the exhaust is already happening for the previous face. And the intake "stroke" is already happening for the third face.
No waiting for the exhaust and intake stroke to happen.

Saves a couple of strokes, which is why the rotary is considered a cross between a 2-stroke and a 4-cycle. Although only one face is doing work, it does not have to wait to exhaust on another "stroke" and then an intake "stroke"; as soon as it's finished the exhaust, it starts inhaling another intake charge, while the rotor face ahead of it is already firing.

But only one rotor face is doing work at a time, as only one piston is (on every second stroke), so only that displacement (one rotor face, BDC to TDC, times the number of rotors) gets counted in the displacement of the engine.

Hope it makes sense.

Corrections anybody?

-John.

Wankelguy

Pedestrian X,
NO, I was only joking! Sorry to mislead you. And John, no disrespect taken, mon ami.
My sense of humor rears it's hideous head from time to time. -Mike