Outkast187

LOL right....It was a 2 second search, and was direct and to the point. I trust them more than I trust a keyboard mechanics theories.

Don't dig yourselves in any deeper.

rotary.enthusiast

Yeah, I know. I've read that article a long time ago--howstuffworks is generally a good site, but they certainly aren't infallible. I don't suppose simple concepts like the definition of how the displacement of an engine is measured matter to you huh? You can postulate performance gains all you want, changing the size of the recess in the rotor faces does not change the displacement of the engine.

maxxdamigz

iTrader: (1)

Well, this keyboard mechanic is about to prove you entirely wrong.

The volume of the combustion chamber at any given time is not synomous with any representation of the displacement of the engine. The displacement of a single cylinder is determined entirely by pi²*bore*stroke. In simplistic terms, it is piston face area times distance traveled by piston. If you have a pipet with 2 cc's in it already and you draw the plunger back 8 more CCs, you have caused a diplacement of 8 ccs and not 10. Thus, the displacement is determined by the change in volume and not the total peak volume. In the rotary, the definition is no different. At peak volume (bottom dead center for a piston), the dish in the rotor has a volume of X. The volume of the chamber not in the dish is Y making total volume X + Y. At top dead center (for a piston) or minimum volume, the dish in the rotor has a volume of X. The chamber has a volume of Z making total volume X + Z. Displacement is defined by maximum volume minus minimum volume. In this case, it would be (X + Y) - (X + Z) = displacement. That reduces to X + Y - X - Z or Y - Z = displacement. Where is the rotor dish volume in the final equation? It's not there because it doesn't matter. It never changes and thus is not part of the displacement equation. Furthermore, because of the increased dish volume, the expulsion of exhaust gases will be weaker, the exhaust hold over between cycles will be higher, and the adiabatically compressed charge temperature at top dead center will be lower. In short, the engine will make less power. The only gain is that with lower peak charge temps, you can increase boost pressure until the charge temps at top dead center near the knock/detonation point.

If you want, I can probably dig up a "howstuffworks" reference. Actually, it appears in http://auto.howstuffworks.com/engine7.htm that they don't even acknowledge stroking an engine as a means of increasing it's displacement.

http://en.wikipedia.org/wiki/Engine_displacement
There - you will see that combustion chamber volume and compression ratio are not part of the displacement equation. It also mentions that displacement is defined as "the volume that is swept as the pistons are moved from top dead center to bottom dead center."

rotary.enthusiast

Outkast187

Great more piston math.

So, let me get this straight....
You think the only way to increase displacement on a rotary(2-rotors) is increase the width of the rotor..(I assume you agree on that, but no surprized if thats wrong also).

maxxdamigz

iTrader: (1)

You can increase the width of the rotor or the central gearing ratio scale. The gear on the inside of the rotor is mated to the gear on the eccentric shaft in a 3:1 ratio. This must be maintained. So, you wind up with a bigger eccentric shaft, bigger rotors, and a larger case. It is my contention that were the rotor reduced to 3 imaginary dimensionless planes that extended from the center of the rotor out to the apex points, that the displacement of the engine would still not change.

edit - my discription of the geometry change is sloppy at best. I have reference books on this at home I can probably dig up. All this stuff was described at length in the Herges History of the Wankel Engine.

rotary.enthusiast

You just really aren't very bright are you? You can increase the width of the rotor, or change the dimensions of the rotor housing and rotors. The former would roughly correlate to bore, the latter to stroke. I know, "more piston jargon." Sorry, just trying to make it analogous to something you might actually understand.

Perhaps you should do some research on the 16X engine and how Mazda increased the displacement there.

maxxdamigz

iTrader: (1)

Throughout Mazda's history, there have only been 3 different eccentricity/rotor diameters used.

First in the list (but not chronologically), there was a limited application 13A which had a rotor diameter of 120mm and an eccentricity of 17.5 mm. The idea behind the higher eccentricity was stronger low end torque.

Next and most recent is the 16x whose dimensions I don't know but I believe is also a higher eccentricity/rotor diameter setup.

All other Mazda rotary engines from the 10A to the 13B use a rotor diameter of 105mm and an eccentricity of 15 mm. The 10A was something like 60 mm wide. The 12A was 70. The 13B family is 80 mm wide.

chickenwafer

iTrader: (9)

And you think you know everything, huh?

You've already been successfully pwned, but I'll still clean up.

If changing the compression ratio alters the displacement, then how do T2 FC's and FD's have the SAME displacement (654cc per rotor) yet different compression ratios (9.0:1 and 9.5:1, respectively)? I guess Mazda fucked that up, huh?

And how is when you put a metal head gasket on a piston engine, the displacement doesn't increase, but the compression ratio decreases?

Um, der.

Outkast187

Exactly my point, now maybe you should research the 16x a bit more....

And yet another site
http://www.encompassholdings.com/RETI.htm

Looks like I found 2 places that support me....and you guys have only gave me a couple examples of piston engines, lol. Lets see some proof, its odd that no site actually backs up your opinions.................
So, till you find actual proof, you can blow smoke up each others asses for all I care. Maybe a ***** displacement comparison is next?

chickenwafer

iTrader: (9)

I already have proof. It's called the FC, FD, and the RX-8. Different compression ratio, different sized pockets on the rotor faces, and yet the same displacement.

It's funny because that website you linked to and the quote your provided proves our point more- it says the pocket alters compression ratio.

All this arguing doesn't matter because you were already dis proven by MATH.


And nice comment about the ******, you always know someone is defeated when they resort to lewd comments and shoot themselves in the foot.

paulmasoner

iTrader: (1)

FAIL

FAIL

FAIL

FAIL
FAIL


your VERY quickly on the road to being the biggest ignorant idiot i've ever seen post here....

chickenwafer

iTrader: (9)

Damn, that's 5 FAILS in one thread.

Isn't that a record or something?

paulmasoner

iTrader: (1)

no he's done worse in the housings thread and in his REW RX-8 thread, as well as other OP's threads....

chickenwafer

iTrader: (9)

Oh snap!

He has a bad reputation that lingers like a wet fart.

paulmasoner

iTrader: (1)

ewwww, i just did that too... and i've been eating wings and hot everythng for a few days. kinda made my eyes water...

rotary.enthusiast

I love the provided "evidence." If it's on the internet it must be true right? How about the fact that these are relatively simple concepts, and anybody with a rudimentary understanding of an internal combustion engine and the terms typically associated with it could figure it out on their own? No? OK, Il'l play along.

http://members.tripod.com/~HARDEBECK/engine.htm


Not that some crap on somebody's website actually strengthens the argument, but in fact the information there is correct. By increasing the pockets on the rotor faces, you're not increasing the amount of volume being displaced, you're merely increasing the amount of space it is being compressed into. The maximum volume is increased, but so is the minimum, resulting in a net gain of 0 displacement. Why does the difference between the minimum and maximum volume matter you ask? Because that is how much air/fuel will be pulled into the engine by the rotor on the intake cycle.

It's so incredibly simple. But I'm sure you still don't believe it... that's fine. Please go machine your own "high performance" rotors and let us know how it turns out. I love being proved wrong.

WingleBeast

displacement is just that, the amount of air displaced. since the pockets are in the rotar at all times, during compressions and not, that air volume is constant, it is not displaced.

your talking about engine volume, which is a very different thing


so maybe we tuners know something about engines after all, afterall we do spend most our time working on getting max HP from them

chickenwafer

iTrader: (9)

I think this thread must be archived as one of the most swift and incredible Pwnages in the history of teh internetz.

WingleBeast

i can scan my automotive engineering degree and post it too if he wants.

im glad howstuffworks is backing him up, when i see at least 10 or so sucsessful car tuners throwing out all sorts of math, but hey what the hell is math. most likely dosent matter just like pistons dont matter. its too bad that internal combustion engines physics dosnt change

maxxdamigz

iTrader: (1)

I had posted this on another forum because I wanted to know specifically how to calculate displacement for a rotary engine. Sadly, I can't do it with pure calculus as I don't want to look up the integration rules for trig functions. Instead, I used a simple numerical method in excel which is turning into a strong personal preference.



The Mazda rotary engine (all rotaries from 10A to 13B except for the 13A and 16X) have an eccentricity (e) of 15 mm and a rotor diameter (R) of 105 mm. The width of a 13B rotor is 80 mm. Rotor width does vary from 10A to 12A to 13B.

If one were to plot the data points of all values of alpha from 0 to 360, the epitrochoid (rotor housing shape) for a mazda rotary engine has the following equations:

X = e*cos(3*alpha) + R*cos(alpha)
Y = e*sin(3*alpha) + R*sin(alpha)

If you would like to know the location of the 3 rotor tips at any given alpha, the equations are:

X = e*cos(3*alpha) + R*cos(alpha + 2*n*pi/3)
Y = e*sin(3*alpha) + R*sin(alpha + 2*n*pi/3)
where n is the number 0, 1, or 2 depending on which rotor tip you want to solve for.

If you take the rectangular summation of the Y equation for the epitrochoid from alpha = -60 to alpha = 60 and subtract out the integral of the Y equation minus -67.5 from alpha = 210 to alpha = 330, you will have a representation of the swept area of the rotor. Then, multiply by the engine's rotor width (converting from mm³ to cm³), you will have the full displacement of the rotor which is 654 cm³ for a 13B family engine.


*A little note - the equations for the tips of the rotors are important because you can use them to define lines for a theoretically equilateral triangle rotor. You can then subtract the area occupied by this triangle in your numerical methods. This explains the 67.5 in the min volume equation. In the max volume equation, the side of the rotor is vertical and does not intrude into the boundary of the numerically summed area.

**If you want info on rectangular summation, it's a rather boring and simple concept for brute force integrals of known equations.

StealthTL

iTrader: (1)

So let me recap - the inside of the housing is PLATED?

S

rotary.enthusiast

05rex8

iTrader: (7)

this thread is fantastic.....or craptastic...

meh...same difference

swoope

iTrader: (5)

damn you stealth.

you got the whole monty, rum and diet coke out the nose.

and to all, no meat to be removed from the rotors to lower compression..

a rumor existed about 3 years ago that someone respected did some low compression rotors for our engine. seems it was just a rumor..

beers