MikeC

Kevin is right, they are 3 independant chambers around the rotor. While one is on the intake cycle the other could be on combustion and the other on exhaust. They are all doing something. Just like a piston motor, while one is on intake another is on combustion etc.

KevinK2

but as u know, the swept volume method for boingers is any swept cylinder volume, regardless of what part of the cycle it should be at. could be any of 4 strokes. just bore area times stroke.

At that housing position, 3 different faces will cause 3 unique swept volumes as defined by the different faces, and each will be fired in 3 separate events.

Evil Aviator

Why are you guys so obsessed with the rotor?

There is only one point in the housing that functions as an area of swept volume. At no time do any other portions of the housing function in the intake cycle. Yes, the next face of the rotor will cycle through this swept volume area, but it is still the SAME swept volume area, not a different swept volume area.

OK, the piston example didn't work because I think that you guys don't understand piston engines, either. Lets try a racetrack analogy. There is one racetrack with three cars racing around the track. How many race tracks are there?

KevinK2

I personally built a fully prepared scca d-production engine, and tuned the tripple dcoe's. I know piston engines.

the 3 rotor faces combine with that part of the housing for 3, distinct intake cycles. our crude unofficial definition just says stroke all pistons in all bores, or all rotor faces thru a stroke in the housing (intake if u wish). The rotor faces are analogis to the piston faces.

until you can bust free and recognize that your 'cylinders' are dynamic and are rotating around the housing, you will be forever in the dark on this.

-----------------------------

Do a 'special' radial piston engine, 4 cyl'rs 90 deg apart, in plane. the 'head' is a fixed ring (like housing), around and sealed at the cylinder ends. head has plugs at 3 o'clock, so tdc at 3 and 9 (exh and intake ports), and bdc at 12 and 6 o'clock. a planetary gearset allows the housings to rotate (like rotor) 90 degrees for 270 degre rotation of the crank. would this 4 cyl engine be rated at just 1 hole, where intake occurs?

although it takes 3 crank revs for all cylinders to be stroked up and down, displacement shold be 4 x bore area x stroke.

MikeC

Sure they don't function in the intake at that time but in your diagram the other 2 chambers are in the combustion and exhaust cycles. Three chambers working independant of each other just like three pistons. They are certainly not doing nothing. In a 3 cylinder piston motor the three pistons would be doing the exact same thing, one would be at the end of inlet, the other in the middle of combustion and another in the middle of exhaust.

There is no need to be rude.

This isn't really the best analogy. Are the cars the chambers, the apex seals or the housing? If they are the chambers then yes there are three. If they are the apex seals then there are three empty areas of track between them.

Did you read this this:

http://mikeonline.cable.nu:1863/misc/rotor.doc

MikeC

This is exactly what I was thinking of except with 2 banks of 3 cylinders. I didn't think of the having the spark plugs fixed in a single location, I had them fixed in the head of each cylinder. This leads to another idea, this engine could easily use ports instead of valves.

Hmmm, I could produce this engine and claim it is a 1.3 litre, kick everyone's butt in racing and throw the world into another 40 years of confusion.

Evil Aviator

... yet I am the only person participating in this thread who understands the displacement rating for a rotary engine, while everybody else's reasoning is that there is some conspiracy theory. I find this amusing. When your displacement calculation for a 13B comes up something other than 1308cc, doesn't that give you a clue that the calculation method is incorrect?

I wasn't being rude, I was making a statement that I believe to be true.

Since you know all about piston engines, let's try that example again...

I'm going to try to use simple numbers. Given a 4-stroke 1-piston engine, and the cylinder has 1.0 L of swept volume.

Intake Cycle: Piston travels down through 1.0 L
Compression Cycle: Piston travels up through 1.0 L
Power Cycle: Piston travels down through 1.0 L
Exhaust Cycle: Piston travels up through 1.0 L

Therefore, the engine's displacement is 4 x 1.0 L = 4.0 L

Now please tell me why this is incorrect.

KevinK2

1) I think your just defending the original NSU? rating method, which is fine. Just don't say it's a true, total swept volume method.

2) My equation for face displacement is 3xex(face area). It's not exact, 1% error. It is derived from known displacement per face, and known face area which is a function of R and w.

3) your 1 piston engine example is incorrect 'cause it violates your universal swept volume basis. Displacement is 1.0L, per swept volume logic ... bore area x stroke x number of holes. or, swept volume or 'displacement ' of ea hole/chamber x number of holes/chambers.

4) now comment on my example?

-----------

the commonly used swept volume method is independent of crank revs needed to sweep air into all distinct chambers. it does not count the stroke for ea piston/rotor_face twice. 13B has 6 unique chambers (defined by unique rotor faces) that move air in and out of ports. it takes 3 revs to complete 4 cycles in each of all 6 chambers. swept volume for ea face is .65L which relects the volume change in that chamber during an expansion or a compression stroke. total swept volume is 6 x .65 = 3.9L.

swept volume method displacement rating is consistent with the 100% VE breating capacity in one complete engine cycle. this takes 3 revs for 13B, 2 revs for 5.0 stang, and one rev for a 2 cycle weed wacker.

SPOautos

Actually I know someone that has taken a internally stock 13b with no mods other than a bridge port (no oil mods, ect) up to 15K rpms a handfull times, in addition it would see 12K every weekend and acording to him that engine lasted for well over a year.

I was doing pretty good with the article till I got down to the end where all the math is haha

Anyway, To me it seems like the actual physical size of the engine is 1.3 liters. That fact that it can consume more air per revolution than a standard 1.3L pistons engine is just due good design.

Its really like trying to compare apples to oranges and the problem with comparing pistons engines and rotary engines is that racing sanctions tend to use the formula that bests suits thier needs.

I personally feel like it should go by physical chamber size. You figure the size of a chamber multiply by the number of chamber (in our case 6) and wola, you have the size of the engine. If it takes in more air than its piston partner than so be it, they need to redesign thier engine

STEPHEN

Crusader_9x

Ohk one of two things that evil has said so far is true. Now which one is it to you? What you believe in is wrong and the fact that you dont understand this yet is getting old im my opinion. either this is true:

Which basicaly says that you dont count the number of sides that a rotor has you only count the number of rotor housing and then multiply the area of one of these housings(cylinder) by two.

Or if you dont wanna believe that then this would be true(which it aint so that means.....):


So either youstart believing that the rotary is a 1.3l motor and leave it alone. other wise your telling us that evrey onther 5.0l motor out there is really a 20.0l motor? get a life.

KevinK2

how about none of the above.

the logic of .65L per 'housing' would equate to considering 5.0 V8's as '2 bank' engines. since one piston in each of the 2 banks displaces/sweeps .625L, then a 5.0 is really 2 x .625 = 1.25L. Stang's have heck of a lotta power and torque for just a 1.3L engine!

swept volume method has no direct ties to just the intake cycle. sorry.

Crusader_9x

double.

Crusader_9x

no you dont understand what i said. if you wanna count the rotary cc 6 times instead of 2 then u need to count each cylinder 4 times instead of 1. not one time for each set of 4 cylinders.

Or you can just belive what it really is, 1.3l. you get that by only counting each cc once like your suppose to. now if you dont understand that then keep your mouth shut and go argue on ur piston driven motor forums.

one cylinder does not equate to one bank. i have no earthly idea how you came up with that. one housing equals one cylinder, not rotor.

wwilliam54

A radial is couted by the # of cylinders becuase they each have a the parts needed to run
a rotor face shares the combustion side with all 3 sides, and only one side can combust at a time, so 1 sides displacement per rotor

KevinK2

" .... one housing equals one cylinder"

easy to come up with it ... you just did.

MikeC

No, it's you have misunderstood kevin completely. What he is saying is the a single rotor/housing is 3 independant chambers just like one side of a v6 motor is 3 independant cylinders. Calling a rotor/housing one single chamber is like calling one half of a v6 a single piston.

This is totally uncalled for and just shows your ignorance. If you are really that smart then download this doc and explain the calculation of swept volume.

http://mikeonline.cable.nu:1863/misc/rotor.doc

That is obvious. I don't mean this rudely but you don't really have a grasp of the concept. Read the above doc thoroughly and then get back to me. Have a look at the comparison to the geared up 6 cylinder motor. Don't you at least think it's unusual that every single figure matches?

MikeC

This is not true, each chamber is always doing something, they never sit idle. They are always either on inlet, compression, combustion or exhaust. Just like in a piston motor.

Same as a 3 cylinder motor, only one piston can combust at any one time, while the other 2 would be on inlet, compression or exhaust. Exactly the same as the rotary.

MikeC

Just to remind everyone, this thread is about this document. If you want to argue, it should be about why you think it's wrong. If you want to argue in other directions then a new thread should be started.

http://mikeonline.cable.nu:1863/misc/rotor.doc

KevinK2

Mike,

Hope you can resolve my 2e vs 3e question for the d value in your calc's.

Capacity confusion is likely due to a lack of an offical standard for determining it. I have not found any SAE, ASTM, JIS, or DIN standards for this.

I think the best method to rate displacement (suggested by felix m on the big list) is air consumed at 100% ve in one output shaft rev. This would make the 13B a 1.3L, but such a std will never be agreed to. Air flow in 2 revs is next best thing, as this is consistent with all 4 strokes, past and present. Btw, your paper is about 4 strokes, so mabe a 2 stroke disclaimer is needed.

Although I like to used the 3.9L-6 boinger with .67 OD before the output shaft as a theoretical equal to the 13B, it does not exist. Each engine must be evaluated at it's actual output shaft. I don't see that any 'gearing' correction is needed, even for a new engine design that may take 4 revs to fire all chambers. Gearing corrections would not change hp produced, only torque. Bottom line is engine hp produced, per L of air injested (in 2 revs@100%ve). And 2.6L is right for 13B at 2 revs. 3L is also right for a 900hp 18000 rpm F1 V10 ..... a very impressive 3L (263 ft-lbs at 18K !).

your graphs are informative, but the 6 cyl should only be 6 bumps in 720 degrees, and the 13B 6 bumps in 1080 degrees.

Crusader_9x

look im sorry if my responses came off inmature but i was just responding to a childish subject in a childish manner. i have only been into the whole rotary thing for anbout 4 years now which is no where near as long as alot. in the time that i have been here this subject is by far the most annoying and beat to death. if you were right then the people and companys that designed and built this thing would aggreee with you. but since they dont then why should anyone else? they were the smart guys that invented this motor and made everything up by changing the rule book which means i believe they are right. once again you can not compare the rotary to a piston with piston formulas.

one rotor and housing and not 3 indepent cylinders its one.

Shamrock.James

Ok i finally had time to read it it's quite intereting and i do agree with some of the points made.

I find it interesting with the 1.5x factor especially with regards to the torque and MAX rpm not that it matters as the gearbox will convert it anyway...

But apart from that what does all this **** tell us... absolutley nothing! you can argue about it all day but weather you say your 13b is 1.3l or 3.9l it isn't going to affect it's output now is it??

i think the whole capacity of an engine thing is the bigest waste of space in the world...

It's how an engine performs that maters, it's torque, power output, max RPM, throttle response, torque curve,
fuel effiency and reliability is what makes an engine.

NOT how many CC/cubic inches/litres capacity it has.

KevinK2

but the 'wise ones' from mazda do agree, Mazda recently accepted an award for the rx8 engine in the 2-3 liter 4-stroke class. tell you something? search for thread. The original NSU rating of 1.3 was loosly based on firing 2 chambers in 1 rev, and had nothing to do with total swept volumes.

Evil Aviator

I am just using the intake cycle as a reference point. Any of the cycles could be used to define the swept volume, but I think the intake cycle is easiest to understand because it happens first, and it would have the most congruence with different engine types.

Which cycle do you prefer to define the swept volume?

Assuming that I understand the model correctly, and each cylinder experiences an intake cycle at some point during the engine operation, the rated displacement would in fact be 4 x bore area x stroke.

OK, this is what I was looking for. My question was not a test, it was designed to narrow down your sticking point. It appears that the number of holes is where we have a problem.

... and this is the problem. While each housing does have three chambers which are divided by the three apex points of a rotor, only one of these chambers is valid for the displacement value, as the other two are simply transient. You cannot count the gas in the other two chambers because this gas was already counted during the two previous intake cycles.

MikeC

I think it is an incredibly interesting subject.

Come on Crusader, they have a *huge* vested interest in rating this engine lower than it really is. The *last* people you would believe is them! I've said this before but I'm not sure if it was in this thread, but taking mazda's word for it is like believing a health report from a tobacco company.

Yes you can, they are very, very similar. Did you have a look at the comparison in the document, how do you explain that every figure matched the piston engine?

This is what I used to think but I realised why it was wrong.

That's correct. The max torque and rpm is just a measure at one point of the drivetrain, gear ratios change both for free. The 1.5x gearing is removed in the diff ratio.

Sure, but it's an interesting topic.

See my reply above.

Evil Aviator, this is *exactly* what a piston motor does. While 1 chamber is at the end of the inlet stroke (as in your pic), another is part way through combustion and another is part way through exhaust. Every chamber is doing something at all times, they are not idle.

MikeC

I'm not sure I follow you here. Can you explain it in more steps. I think the difference in formulae is because I'm using the distance from the centre of the rotor to apex seal tip to calculate cross sectional area, where you would be using distance between 2 apex seals. It would probably be easier to email me at rx7club@mikesdriveway.com

There are a few problems with this:
This would turn the 3.9l 6 with a .67 OD into a 2.6l engine.
If you took the drive off the camshaft of a piston engine then its capacity would double.
Your rotating radial engine would have a different capacity depending on how fast you rotated it.
If they took the drive off the rotation of a rotor (assuming this is possible) then capacity would change again.

A method that does not change when the output shaft ratio changes is needed OR you need to even up this ratio by removing it from the engine. For the 13B this can be done by calculating the air inducted for a complete cycle (3 revs) or by adding a 1.5x gearing down to the rotary.