Originally Posted by RICE RACING (Post 154033)

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!

Originally Posted by diabolical1 (Post 152902)

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 ... :)

Originally Posted by RICE RACING (Post 154157)

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

Originally Posted by RICE RACING (Post 154157)

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

Originally Posted by scotty305 (Post 154218)

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/mendama...57623971081040

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.

Originally Posted by Barry Bordes (Post 150200)

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

Originally Posted by RICE RACING (Post 154537)

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.

Originally Posted by Barry Bordes (Post 154560)

Originally Posted by RICE RACING (Post 154723)

"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:

Originally Posted by RICE RACING (Post 155041)

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:

Originally Posted by Libor (Post 155090)

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.