SIPSOIL

Does anybody here know why Mazda didn't choose an epitrochoid with positive curvature? There are two areas where the apex seals are accelerated towards the center and at high rpm the springs will not be able to counteract this effect.

I lack the motivation to figure out the exact value but when we plug in the values a:200 b:100 h:38 here

Epitrochoid

we get positive curvature for all possible rotor positions. The difference in chamber shape and volume at TDC can be compensated by reshaping the surface of the rotor.

Of course, for the same displacement the diameter of the engine will be larger, but I can't believe that this is the only reason.

Mazda is considering increasing and the stroke the eccentricity for the 16X and stroke, worsening this effect.

MAZDA:16X | The Rotary Engine

rickeo

iTrader: (1)

I have no idea what you just said but i'm subscribing anyway.

Rootski

iTrader: (1)

iunno. build it, see what happens, maybe get rich. it worked well enough for some **** in the '50s.

SIPSOIL

Hm...

I'm in MENSA so in my head I just built one. Single rotor, 24cm long (3x current rotor), 2000cc displacement, three peripheral intake and two peripheral exhaust ports, 7 plugs. Fuel is injected after the exhaust port is closed. For lean burn fuel is only injected into the center port.

A689 spring steel sheets, bars and pipes, outside wrapped in fiberglass, with lots of bolts to adjust the shape of housing and rotors.

ECU: Beaglebone

So far so good! Now let me make one for real...

[hammer, hammer, grind, grind, solder, solder...]

SIPSOIL

Fast-forward ten years...

The most awesome Wankel engine to end all Wankel engines is complete! Here's a pic:

madbouncy

Are your sure you can get enough compression? Usually when your K value, in this case a/h gets lower you loose max compression. If you find the maximum arc and ignore the rotor recess for the rotor shape it'll give you your maximum theoretical compression. Also, all the rotory books I've seen do talk about it but they are more worried about the angle the apex seal gets to which gets worse as K gets larger.

SIPSOIL

Isn't it the other way around? The larger a/h is, the less compression and the less variation in apex seal angle. In the extreme case of h=0, i.e. a/h=inf we have a circle, zero compression and a constant apex seal angle.

Additionally, rotor face do not have to be circular arcs. At the cost of displacement we can always get infinite compression by making the face a segment of the housing!

The only thing I can come up with is that the initial rate of change of chamber size might be significantly lower for higher eccentricities. The idea being that the longer chamber stays 'small' after ignition, the more mechanical energy can be extracted, because the MEP is higher.

madbouncy

From any time I've run the formulas it shows that compression goes down as K goes down, (i.e. eccentricity goes up) However as compression goes down the apex seal angle does get better, so I agree with you there.

h=0 would be a singularity and not an actual case. I'd have to run it as h-->0 and a>>h to see what it's doing before it gets to that point, I assume it's going to look very much like a vertical line before it goes to the singularity.

It doesn't have to be an arc but an arc will give you the maximum compression. It should be an arc because the point that sticks down in the middle of the two lobes is going to define the maximum size of your rotor as it goes past it. I'm not really sure what you mean by making the face a segment of the housing, if you could easily get more compression you wouldn't have had companies going to such extremes trying to make a diesel rotary, especially ones that basically had one engine feeding the other to get the necessary compression.

Sorry I can't run anything through actual formulas right now, I'm out of town and everything is on my laptop at home. Working off memory right now so I won't guarantee anything, but you can always plot out compression ratio vs K and see how the actual graph looks.

SIPSOIL

"The maximum compression will be obtained if the flanks of the rotor are shaped exactly according to the inner envelope."

Source:
http://www.rotaryeng.net/NSU-rotating-combustion.pdf p.3, "Compression Limits"

BTW, when the curvature is positive there is no "point that sticks down" on the minor axis, which was the point of the original question.