Actual Rotary Displacement Request

[My5ABaby]

Originally Posted by Libor

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

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.

Originally Posted by vex

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...

Originally Posted by RotaryGod

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]

Originally Posted by Barry Bordes

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)

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]

Originally Posted by vex

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_...5&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

[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]

Originally Posted by 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.

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

[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............................

[vex]

Originally Posted by diabolical1

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.

[Libor]

Originally Posted by RICE RACING

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

Originally Posted by RICE RACING

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!

[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/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.