Actual Rotary Displacement Request

[diabolical1]

^ understood.

but what i'm saying is the crank rotations are not relevant - at least, if they are, i don't see it.

for example, my manipulation of the 86 mm x 86 mm MR2 engine, simply yield roughly 499.5... cc per cylinder, which i then multiply by 4 and get 1998 cc. engine revolutions never figured into it. if this engine block were used with a 2-stroke or 4-stroke head, it's still 1998 total capacity.

the rotary does require 3 complete revolutions to fire all 6 faces, but i don't see how that affects total capacity, which is in fact 6 chambers, and it is the volume of those 6 chambers that yield it's total capacity/displacement.

it's just the way i see.

[vex]

Originally Posted by diabolical1

^ understood.

but what i'm saying is the crank rotations are not relevant - at least, if they are, i don't see it.

for example, my manipulation of the 86 mm x 86 mm MR2 engine, simply yield roughly 499.5... cc per cylinder, which i then multiply by 4 and get 1998 cc. engine revolutions never figured into it. if this engine block were used with a 2-stroke or 4-stroke head, it's still 1998 total capacity.

the rotary does require 3 complete revolutions to fire all 6 faces, but i don't see how that affects total capacity, which is in fact 6 chambers, and it is the volume of those 6 chambers that yield it's total capacity/displacement.

it's just the way i see.

The confusion with everyone is in how to view the rotors and what they are displacing in comparable fashion to a piston engine. Point-in-fact, you can not. As Rice has said it's a different cycle (though technically speaking, it's the same cycle as any other 4-stroke engine just with a different configuration--just look at the p-v diagrams).


http://www.grc.nasa.gov/WWW/k-12/airplane/otto.html

Each face sees that otto cycle after 3 revolutions of the crank shaft. There's no two ways about it. That's what each face sees. Thus the total displacement for all 6 faces comes out to the 4L or so that Rice posted up.

Now here's where I think everyone is getting messed up: As I posted previously each piston engine calculates displacement by going from tdc to bdc of each individual piston. Now lets compare this to the 13B. How many times do we have to turn the crank shaft to go from TDC to BDC for the front rotor housing (note I didn't say each individual face)? 1 right? 1 turn takes the front rotor from TDC to BDC, while the rear rotor is 180* out of phase (thus displacing the same amount). From which we have the standard nomenclature of 1.3L of displacement.

Does this make sense to everyone?

This method is scalable as well. To think of this clearly do not consider the rotor faces as pistons, but rather the rotor as a whole as equivalent to a piston (or think of the rotor housing as the piston equivalent sleeve). If the 20B were to be calculated using the method described above you would simply take the TDC to BDC of the front rotor (aka 1 revolution). Since all other rotors move the same amount with that 1 revolution they displace exactly the same amount of air. If you do not like that procedure simply take TDC to BDC of each rotor (they only take one revolution each) and it will displace the same amount of air. This is where the Mazda Displacement rating came from. This is the number you use (at least on the Haltech) when you put in the engines displacement. There's no black magic here.

[Barry Bordes]

Originally Posted by vex

http://www.grc.nasa.gov/WWW/k-12/airplane/otto.html

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.





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).



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

[vex]

Originally Posted by Barry Bordes

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.


[/COLOR]

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.

[Barry Bordes]

Originally Posted by vex

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.

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)

[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