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

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

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

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

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

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

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

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

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

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

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

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

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

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

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



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

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

For the 2 liter Toyota….
500cc

For the 6 liter LS motor…
750 cc

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

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

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

Barry

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

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

You mean like TDC to BDC ;)

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

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

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

I would look for advantages of wankel rotary elsewhere

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

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

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

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

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

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

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

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

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

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

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