Crusader_9x

In your theory does combustion happen in the other four chambers in the rotary like it does in a piston?

wwilliam54

only 1 can combust at a time in a rotor since they all share a compression component, where as the pistons can operate independately

stevenoz

iTrader: (27)

Displacement is the volume of the working chamber at BDC to TDC per revolution of the crank/eccentric shaft. Now for those of you who don't understand Bottom Dead Center, that is when the intake is at its biggest possible volume. TDC is chamber volume and is volume at its smallest. Gearing, cams, etc have ABSOLUTELY no effect on this, its the volume at BDC to TDC that matters. Now we know that this is roughly 40ci and since we have 2 chambers of volume per rev. that is where we get the 80ci or 1.3L.

The problem is that most guys with boingers are stuck on "No replacement for displacement" theory and want to figure their engines on full displacement. For a boinger this would have to consist of 2 revs of the crank. Each cylinder on an average 350 will consist of 43.73ci. If we use the 1 rev. method we would come up with 175ci ( 43.73 x 4 ) or 2 revs we get the full 350ci ( 43.73 x 8 ) that they want. Take your pick on what to use. I was always under the assumption that its one revs.

Either way if you want to use the 2 revs then you would get 4 working chambers withing the rotary and 160ci or 2.6 liters. There is no way to come up with 3.9. If you want to call it a 3.9 then you need to start adding another rev. to your pistons motors ( getting to 3 revs ), which would then just move your 350 to a 525ci 8.6 liter!!! I have no problem with people calling it a 2.6 if they are trying to compare it to a piston engine. I just don't get where this 3.9 comes from.

I have messed with rotaries for 15 years now, all of my driving life, and I have always used the 1 rev standard and will continue calling it a 80ci 1.3 liter. The thing is, I am not trying to compare this to anything else. Its unique.

I have argued this too many times and it always seems to be related to racing. Bottom line is call it what you want: 1.3, 2.6, 10.0. If you get outrun by a 2-rotor then you just got your *** handed to you by a pathetic, little *** motor that is no bigger than a milk crate. One that I can pick up and carry around with me. If it makes you feel better to give it more displacement, then go ahead.

In the rotary community it will always be considered a 80ci 1.3 liter motor because that is how we measure it. Most of us never try to compare it to a piston motor, its always the other way around.

Crusader_9x

thanxs those are my feeelings exactly.

KevinK2

1) we use the same area .. your HxAx(sqrt3) is plane area between apex seal contact points. Best to use std rotary terms:

H=w width
A=R radius to apex seal tip
C=e crank offset

looking from BDC to TDC in the compression stroke, rotor center starts at e below crank centerline, and ends at e to right of crank centerline. net displacement of face, normal to itself, is 2xe, relative to crank centerline, not 3xe per your calc's. rest of displacement is from the 'collapsing' outer housing surface, going from a bulging arc to one that nearly conforms to the rotor face.

2) gearing ... the 3 cases you note are irrelevant since they don't exist and will never be done. The .67 overdrive (greater than one implies speed reduction, ie 3.89 rears) of the 13B does exist by necessity. When compared to a 2.6L boinger, it simply means it's 2/3 thru it's engine cycle in 2 revs. Dyno curves are directly comparable since each engine has same 100%VE breathing at each rpm.

Defining capacity/displacement based on one completed engine cycle, as you suggest, would be fine as long as the revs per cycle were a part of the specification. FD would be 3.9L/3, VW 4 would be 1.8/2, and 250cc 2-stroke would be .25/1

KevinK2

from evil ..

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
quote:
--------------------------------------------------------------------------------
Originally posted by KevinK2
13B has 6 unique chambers (defined by unique rotor faces) that move air in and out of ports.
--------------------------------------------------------------------------------

... 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

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

"this gas was already counted during the two previous intake cycles"

when, how, and by whom were these two intake cycles accounted for?

you have just looked at one chamber for 3/4 crank rev.

same exact situation for any boinger .... make a similar pic for a 4 cyl with 2 cyl's up, then 2 cyl's down. looking at 1/2 crank rev, is displacement for engine just those 2 cyl'n, since others are in transient?

Simply put, NRU based 13B 1.3L rating is not based on the swept volume of all cylinders/chambers method. No big deal. It does wind up being the air injested (@100%ve) in one crank rev.

MikeC

We've just used different methods to calculate the same thing. The distance I've calculated is not the same thing you've calculated. You are calculating the distance that the point P moves from the centre of the crank. I'm calculating the distance it moves perpendicular to the face in both x and y directions. It's difficult to describe but if you move the crank a very small amount then the face of the rotor will move in a certain direction by a certain amount. The face won't be moving in the same direction it is facing so you have to work out the component of movement in the direction it is facing. This will be the swept volume for that small degree of crank movement because your calculating how much volume that face swept through. If you add up all these small movements over 270 degrees you get the swept volume. It's just easier to add up all the little movements first and then multiply by face area.

I haven't checked but does the 3e or 2e give the correct result? If it's 2e then there must be an error in my maths somewhere (it's been 15 years since I originally wrote it.

Here's another way to look at it. Assuming all things being equal, such as intake efficiency, the rotary will rev to 1.5x higher than a comparable piston motor (assuming no valve bounce, friction etc). This will happen because the chambers on both engines are going at the same rate. At redline the rotary will be consuming as much air as a 3.9l piston motor.

BTW, I don't think that the fact that the engines don't exist makes them irrelevant. If I produced a 3.9l 6 with a .67 OD and tried to claim it was a 2.6 I'd be laughed out of town. But this is exactly what the rotary is, it's just a bit better hidden so no-one can see it.

KevinK2

Mike,

1) I understand your calc'n method now, and see how it can/should exceed 2e. You found a 3e 'stroke' of the rotor face that gives about the right chamber volume. I'm just not sure that the implied swept volume calc has a rational basis.

My problem is that the face motion is only part of the expansion/compression process. I did some heat transfer area calc's for the 13B, and determined chamber effective radius values for tdc and bdc to get the right compression ratio and displacement. R@tdc was 7.67", and R@bdc was 3.61". This change is housing shape is part of the chamber volume change, but not accounted for in your method, imho.

2) 13B rev advantage is there, but not an exact 1.5x. The advantage is offset some by thermal inefficiencies, and mabe port size limitations. In theory, the 1.5x higher matches 3.9L boinger hp capability, but not in reality. Stock engines: RX8 has 240 hp at 8500. M3 3.2-6 has 333 hp at 7900. Audi A6 4.2-8, scaled down to 3.9L, would be 316 hp at 7K.

Gearing corrections just muddy the water, and really are w/o merit.

3) again, the 13B has the 1.5 effect by necessity, it's no trick, as you know. If a new 2.6L/2-revs engine concept with 4 revs to complete it's full cycle, can rev to 10k, pass emmisions, and match M3 and A6 numbers, it should stay rated at 2.6L, and not be called a 5.2L due to implied rev capability based on 4 revs to complete 4 strokes.

MikeC

I can see where you are coming from. It's difficult to explain but try this. Start with the engine at BDC and draw a line across the rotor face. Move the crank one degree and draw a line across the rotor face again. Color in this area and calculate it's area (if you assume straight rotor faces this is much easier). Any parts of the page that have been covered by the rotor should be counted as positive area and any parts that have been uncovered up should count as negative. Do this 270 times and add up all the positive areas and subtract off all the negetive area. This will be the swept volume. You'll see that the outside of these areas is in the shape of the housing.

I'll leave this for now because we probably won't ever agree. Your the only person I've met who's managed to recognise the similarity to the .67 OD 3.9l engine, so I'm suprised you disagree with me on the capacity.

DP13B

[QUOTE]Originally posted by MikeC
[B]if the output shaft of the rotary is doing 9000rpm then the chambers are expanding and contracting 6000 times a minute, [quote]


If the the output shaft performs 3 RPM to the rotor's 1 RPM, then with the output shaft turning at 9000 RPM is is rotor not turning at 3000 RPM ?

It's a good thing Rice has not been involved yet

Here is his take on the question http://ausrotary.dntinternet.com/for...t=stroke+cycle

He supports the 3.9l theory, just puts in another way.

MikeC

It's not the speed of the rotor that's important but the speed of the chambers. For a piston motor doing 6000 rpm at the output shaft the chambers are expanding/contracting 6000 times a minute, so 1:1. A rotary needs to be doing 9000 rpm at the output shaft for the chambers to be doing 6000, so 1.5:1


hmmmm, 6 stroke, interesting ..........

KevinK2

I did a simple case of a horizontal line, rotated to vertical. First case pivot at end, 2nd case pivot at 1/4L and some negative area. d is the arc of point a mid lengh. Both cases worked for swept area, consistent with your method. You have define the curved 3e stroke that exists, which I have argued vs those that said rotary had no stroke.

I'm still thinking about the different housing shape at tdc and bdc, it's possible/likely there is no + or - adjustment to volume change needed, or mabe displacement is not exactly the swept area calculated .... like .67L vs .65L?

My 3.9/.67 thing came from simple cycle diagrams of 13B vs boingers, and what it would take to match the FD. I also came up with huge bore/stroke ratio, which is not a torquey way to go. I agreed to a qualified 3.9 rating, but still think air injested per rev should be the common theme for displacement ratings. Many race groups treat 4-valve engines diff than 2 valve engines ... a simple rating system just lets the better engines shine, like the s2000.

MikeC

You've got me stumped here, I know the calculation is correct I'm just not sure how to explain it. The swept volume is the volume that the rotor face pushes through over 270 degrees. It doesn't really matter about the housing.

maxcooper

NSXs (V6) have had 8000 RPM redlines for a while (since 1991?). The Nissan Skyline (I6) has an 8000 RPM redline. Um, you already mentioned the E46 M3 (I6) with a redline of 8000 (power peak is 7900). Okay, so maybe there aren't that many high-revving factory 6 cylinders. But then you compare the weight to 4 cylinder engines, that's not fair! (not a flame -- just kidding around).

-Max

KevinK2

piston engine ... Vmax is volume swept by piston + comb chamb volume. Vmin is just that same comb chamber volume. For rotory, there is no consitently shaped 'combustion chamber' volume. If that volume in the piston engine changed from tdc to bdc, it would contribute to net displacement.

I suspect your rotor face displacement accounts for full displacement, a byproduct of the rotary kinematics. Mabe there are some Vmax and Vmin calcs out there that would confirm this.

Jeff20B

iTrader: (13)

So RICERACING calls it a 3 rev 6 stroke? Heh, I've always called it a 3 cycle engine because you have three rotor faces with a 3:1 gearing to the shaft and three distinct areas inside the engine (intake port area, plug area, exhaust area). This of course does NOT count how many rotors are in the engine, which I think RICE is saying (I didn't spend very much time reading that forum, so meh).

I'm all for the liquid measurement method saying the 13B's rotor is 654cc.

heb09

Theres one very easy way to explain all of this. If it aint a rotor, it aint a motor! And my motor is 1.2L.

MikeC

If the head of a piston engine changed shape then I guess that wouldn't change the swept volume because it can still be calculated as the amount of volume that the piston face pushes through. If you assume a straight rotor face it should be easy to calculate the min and max volume to get the swept volume via a different method.

MikeC

Here you are comparing all the top revving 6s to the rotary. They probably vary from 5000 to 8000 redline for factory 6s. The problem is there is only one rotary so you don't know if it is one of the higher reving rotaries or not. A rotary should be able to rev 1.5 times higher than it's equivelant piston motor (one with 654cc pistons) but maybe it doesn't because it has been tuned to rev lower due to wear limitations.

KevinK2

point is pure swept volume would no longer equal the displacement of that cylinder.

Kenku

Hmm... R34 Skyline has the HP peak at 6800, 2004 NSX has the peak at 7100. Mazda called the S5 cars HP peak at 7000, and the RX-8 is... well, 8500. Good examples, though obviously all of these things are pretty high-level cars. Hell, the NSX comes from the factory with titanium conrods.

And crap, I had to go back and check a couple times to figure out what my point was. Oh yeah, right, you're right that there are some high-revving 6s pushing equivalent RPMs. But hey, if we used some of their technology... forged titanium rotors! That was a joke, as I do remember what happened with the experimental titanium seals.

And you're right, it's not fair... we should get motors that weigh the same as the various 6-cylinders and put out the same power per pound as existing rotaries... from the factory! IE, highly tuned 3 and 4 rotor cars! *grumblemutter*

MikeC

Looks like you are correct, it really depends on the definition of displacement, which I think is the difference between max and min volume. In the case of rotary you need to take into account the volume of the spark plug holes.

I think with the change in the housing shape between the 2 apex seals it is taken into account with the swept volume calculations.

MikeC

Looks like you are correct, it really depends on the definition of displacement, which I think is the difference between max and min volume. In the case of rotary you need to take into account the volume of the spark plug holes.

I think with the change in the housing shape between the 2 apex seals it is taken into account with the swept volume calculations.

MikeC

Looks like you are correct, it really depends on the definition of displacement, which I think is the difference between max and min volume. In the case of rotary you need to take into account the volume of the spark plug holes.

I think with the change in the housing shape between the 2 apex seals it is taken into account with the swept volume calculations.

MikeC

Looks like you are correct, it really depends on the definition of displacement, which I think is the difference between max and min volume. In the case of rotary you need to take into account the volume of the spark plug holes.

I think with the change in the housing shape between the 2 apex seals it is taken into account with the swept volume calculations.