jame88

So "1 complete cycle of the pumping media (the piston/s)" for a 4-stroke is the piston going up, down, up, down?

Ok so its: 2-stroke = amount of air for 2 rotations of output shaft 4-stroke = amount of air for 2 rotations of output shaft Wankel = amount of air for 3 rotations of output shaft or: 2-stroke = amount of air for 1 rotations of output shaft 4-stroke = amount of air for 2 rotations of output shaft Wankel = amount of air for 1 rotations of output shaft Which is what everyone used before it was suggested otherwise. (Also 2.6L for a 13b is a good way to compare to a 4-stroke, but the engine itself has little to do with 2.6L) Guess it just depends on if you believe the evil insane LIES invented by the greedy heads of the evil multinational corporations in their secret plan for world domination.

Sure is debatable, refer to my post above and have a close look at this 2-stroke animation and see if you can see what i mean. 2-stroke

I agree. Sorry to repeat but the only reason i wouldn't say 3.9L is because in that method of measuring capacity, a 1000cc 2-stroke would be a 2000cc. I think this thread is slowly coming most to an agreeance.

Perfectly right there, but the "how much it pumps per cycle" seems to work for the current ratings of 4-stroke and 2-stroke, and it can relate to a Wankel, so i thought it is a good way of rating it. Perhaps the Engineers at Mazda thought similar.

This seems like it will go on forever, but ill reply to a few things. Pretty much right there just had to say if you turn the output shaft of a 4-stroke piston once, it will only have pumped air in, the other will have pumped that much air in and out. The rotor has only rotated 120 degrees, but as far as the chambers are concerned here, its in the same state. Kind of the equivalent of shoving different piston in a bore each cycle of a 4-stroke or something. Right, except a more accurate way of saying this would be you measure the capacity by rotations (not orbits) of the rotors, I measure the capacity by orbits of the rotors. Seeing as the orbits are what is translated into power in a Wankel i see no reason to use rotations. Just to clarify i'm using rotations and orbits as different motions here, not interchangibly. Agreed

It is getting repetitive, thats why im trying to only say things that haven't been said before. I raised the point that there is no gear ratio between the eccentric shaft and rotor, noone argued, so thats over now. Back to the capacity thing. A 4-stroke engine's cycle is 2 rotations of the output shaft long A 2-stroke engine's cycle is 1 rotations of the output shaft long Do we all agree on that? In this cycle they both pump their capacities worth of air. Your argument is that a Wankel take 3 rotations of the output shaft for it to pump out the same air it pumped in, right? Something noone here has seemed to notice is that a 2-stroke engine actually take 2 rotations of the output shaft to pump out the same air it pumped in. You could argue that a 1000cc 2-stroke engine should be rated at 2000cc. But noone does. Of course the 3.9L Wankel, 2.6L 4-stroke, 1.3L 2-stroke way of comparing works, i'm trying to argue what a Wankel should actually be rated at.

Has the "cycle" of an engine got to do with the combustion cycle or where things are in the engine? If it was just about where things are in the engine, a 4-stroke piston cycle would only be one rotation of the shaft. Wouldn't work very well though. I do see your point, but i don't think the fact that it takes 3 combustion cycles for the rotor to do a complete rotation is somewhat irrelevant.

That page demonstrates the Wankel cycle perfectly. Everyone should look at that and observe that the power is coming from the "orbital" motion rather than the "rotational" motion, before they again say a Wankel only does 3000rpm. Ok now that we have all learned that part, onto the capacity debate. I would define "capacity" as the amount of air an engine would pump in one "cycle" of its operation. Would anyone disagree? Now lets define "cycle". I would call a cycle when something is in a initial state, does something, and returns to its initial state. Wikipedia defines a "cycle" as: "...one complete occurrence of the event which repeats" Nobody here is arguing that a 4-stroke engine takes 2 complete rotations of the output shaft to complete its cycle. Nobody here is arguing that a 2-stroke engine takes 1 complete rotations of the output shaft to complete its cycle. At the end of each cycle, both engines are doing exactly what they were doing at the start of their respective cycles. At the end of a 4-stroke cycle of a 2L engine, the engine has pumped in 2L and pumped out 2L. At the end of a 2-stroke cycle of a 1L engine, the engine has pumped in 1L and pumped out 1L. No arguments there? Now lets play spot the difference (courtesy of RICE RACING) In both of these pictures, the top left chamber has almost finished inletting, the bottom left chamber is about to exhaust and the right chamber has compressed the air/fuel, waiting for a spark (or two). In terms of combustion, both of those pictures are in the same state, however they are one output shaft ratation away from each other. After a 13b Wankel has completed 1 full rotation of the output shaft, the engine has pumped in 1.3L and pumped out 1.3L. In 3 full rotations of the output shaft, it will have done this cycle 3 times. Hopefully this has explained my point of view. I'll be interested to see how people will debate this. Please back your arguments up with evidence.

Ok i appreciate you saying that. I have however read through this whole thread, thats why i decided to post. For those of you who still doubt what i'm trying to say, have a look at this youtube vid: This video explains it well, and is much easier than trying to explain with words in a post. Does anyone still think there is a step up gear ratio between the rotors and eccentric shaft? The stationary gear pinion and the ring gear on the rotor keep the rotor where it should be for the Wankel cycle, much like the cylinder bore keeps a piston where it should be. A Wankel needs this 3:1 (no other) ratio to operate properly just as a piston engine needs a bore.

Im glad someone is actually trying to understand. I know Sydneykid will never back down after this long but hopefully i'm getting through to someone. To explain the kind of motion im talking about here think about a amusement park ride. Im no artist but hopefully this will get the point across. People are in a cart which is rotating about point B on the end of a large boom, which is in turn rotating at point A in its center, held in place by a frame. The cart is rotating around point B at rpm 1 and is also rotating around point A at rpm 2. Hopefully this has cleared this up. Getting there. Anyway i have a life to live away from my computer for the rest of this day. Ill make an effort to explain a few other things in the near future.

*facepalm The cycle of a Wankel is a much more complex motion than the simple sliding crank of a piston engine. The rotors are rotating about two different points, think of the earth rotating around its axis while orbitting the sun. Do you at least now understand the pinion gear in a rotary is fixed and there is no gear ratio between the rotor and eccentric shaft? I don't mean to offend but how do you manage to reply in this thread so much? You write a very impressive amount in this thread, i don't know how you find the time.

Alright ill reply to a few more things, but i seriously can't reply to every point you have raised or else each post in just going to get longer and longer till it gets ridiculous and noone will even bother read it (probably already has) maybe just tell me what you disagree with the most. Sorry if i have confused anything by using "Angular speed", it just means how fast something is rotating. Rpm is a measure of angular speed. I just don't like using rpm to refer to how fast something is rotating, its like saying "What's the degrees today?" or "What's the meters from here to over there?" and so on. Sorry if thats being a bit pedantic. I agree noone cares about piston speed vs apex seal speed. Yes, if you were to take the speed before your imaginary step up ratio, the power will not change no matter what the ratio is, if the speed is decreased, the torque is increased, pretty simple stuff there. Power = Torque x Speed This is of course irrelevant, as there is no gear ratio between the rotor and eccentric shaft. Ill try to have a read of that as soon as i can. The viscosity of Mazda's marketting department is irrelevant. I like a good joust too, and im willing to try to explain my argument as long as people are actually going to listen. Please try to keep this as a technical discussion, not a flame war, as attempting to personally offend me doesn't make you correct.

OK well i'll start with the speed argument. Heres the animation posted earlier in this thread, also on wikipedia I can see where many of you may how got confused here. There is indeed a ring/pinion gearset which has a 3:1 ratio in the Wankel engine. The ring gear is attached to the inside of the rotor, but where you seem to have misunderstood is the pinion gear is FIXED to the rotor housing. NO POWER is transmitted through this gearset. This gearset's only purpose is to force the rotor to move in a Wankel cycle. All power is transmitted from the rotor to the eccentric shaft via bearings on the lobes of the eccentric shaft (the white circle in the center of the rotor in the animation). You are correct in that the rotor is rotating about its center (point A rotating about point B) at 3000rpm when the engine speed is 9000rpm. This is however about as relevant as the speed in rpm of the camshaft, timing chain, water pump etc in a piston engine, as no power is produced at the output shaft of the engine from this rotation. Power is produced in a Wankel from the rotation of the center of the rotor (point B) about the centre of the output shaft. This rotation is at 9000rpm and should be where the speed of the rotors in taken from. This is an eccentric shaft for a 13b engine, notice the offset lobes which the rotor rotates on, and the lack of gears of the shaft. Hopefully i have explained this better and those of you who care to learn will benefit. As for the other debates (combustion cycle, capacity etc) i will explain these further when i have a chance, as there is only so many hours in a day.

You are completely correct there, i must have read incorrectly ill edit that. As for the other many points everyone made, To be continued...

To reply to the OP's question, Wankels don't suck, they are powerful, small and simple. The only reasons i can think of is that they once had sealing problems (pretty much fixed these days), and there arent as many around as piston engines.

Hey guys, Sorry to pop up here and argue in my first post on this forum, ill try to be as nice as i can. I've read through this whole thread, and it seems in the last 26 pages, not too many of you seem to know exactly whats going on with a Wankel and i thought id better clear up a few things, so people dont start believing incorrect information like what Sydneykid has said. Seeing as Sidneykid has basically said the same thing over and over again, ill just correct what was said in his first post. While it's hard to compare a Wankel engine's cycle to that of any sort of piston engine, i can't see too many similarities to a 2-stroke cycle, but i can see plenty of similarities to a 4-stroke engine. In its cycle, it intakes, compresses, ignites, and exhausts, thats four by my count. I can't really find any similarity to a 2-stroke engine, could you please point them out. Also most lawnmowers are 4-stroke engines. Now this one you seem to really not understand. In a Wankel revving at 9000rpm, the rotors are rotating around the eccentric shaft at 9000rpm and around their own center at 3000rpm. By your reasoning, the pistons in a piston engine are really doing 0rpm. Also i don't know where you dreamed up this 3:1 ratio between the eccentric shaft and rotors, the only gearsets in a Wankel are the ring gear on the rotor and the fixed gear on the housing. This is a 3:1 ratio and keeps the rotor rotating about its center at one third of the engine rpm, and is essential to make the rotor move in the Wankel cycle, but it does not step up the speed of the rotor to the eccentric shaft. A Wankel could not use any other ratio but 3:1, or else the rotor would not move in the Wankel cycle. The angular speed of the rotors around the eccentric shaft is the only speed that matters, The angular speed of the rotors around their center being a third of the engine speed is just a part of the Wankel cycle, just as the speed of a camshaft of a piston engine is half that of the crankshaft and should not be an indicator of the engine speed. Why? In one cycle of a 2000cc 4-stroke engine, the engine "pumps" 2L of air, in one cycle of a 1000cc 2-stroke engine, the engine "pumps" 1L of air, and in one cycle of a 13b Wankel, the engine "pumps" 1.3L of air. I don't see any problems with that. I can't really think of many 1.3L 4-stroke piston engine that put out the same power as a 13b. If you can, please tell me how much fuel they use, when they are driving a car of the same size/weight. I didn't know engine capacity was rated by what size turbo they take, perhaps it is due to the better design of a Wankel. I don't see how they have "OK" power output when they are up there with the rb26dett stock, can make 1000hp+, have been made to run 7s 1/4 miles etc. I would suggest they use 2 spark plugs for pollution reasons, and also because it is far easier to fit more spark plugs in the combustion chamber of a Wankel than in a piston engine head. I also don't understand how they have "lousy torque output". A stock 13B-REW has 314Nm (more than a rb26dett) and a 20B-REW has 402Nm, with an impressive 380Nm at just 1800rpm. Also, if we were to go by your incorrect way of measuring the angular speed of a Wankel, then a 20B would be putting out 1140Nm at 600rpm, which i wouldn't call lousy. They are either correctly a torquey, high revving engine or incorrectly a extremely torquey low revving engine. Modified Mazda Wankels often do lack low-down torque because they have enlargened ports, which will increase high-revving speed with the compromise of low-end power, just like a piston engine. This also increases fuel comsumption (just as in a modified piston engine) due to the overlap in intake and exhaust ports being open, which is where Wankels got their reputation for chewing through fuel. Most figures i've seen for the compression ratios for a Mazda Wankel are around 9-10, how is that low? They respond well to turbocharging due to their free-flowing design, no valves to get in the way. I don't think this needs much explaining, i doubt anyone really cares if Wankels fit well into front wheel drives, they are a performance engine, they go in performance (rear wheel drive) cars. Of course they save space, they are smaller than piston engines and they need big gearboxes because they are powerful. If a compact V6 doesn't need any bigger engine bay fit one to a r100. If Mazda has been lying to the world for the last 40 years then wouldn't someone intelligent person have picked it up and done something about it before you? Well hopefully thats cleared things up for everyone, Sydneykid, if you need me to explain anything further or you would like to point out where you have actually said anything correct, just tell me.