fc3sdrifterz

Basically, it goes like this:

How do you figure the displacement of a "regular" engine? Its the displacement of each cylinder added together.

Or, the volume of each combustion chamber in the combustion cycle. In a regular 4 cycle piston/crank engine, that is the power strokes over 2 turns of the crankshaft, not one. If you only turned the crank over once, only 1/2 of your cylinders would have fired (and made power) as the other half would be going through an exhaust stroke.

If you turn the crank once on a rotary (say a 13b), you have used up 1.3l of displacement (firing 1 face on each rotor, being .65L each). But...thats not fair to say right, since we compare a piston engines displacement to 2 turns of the crank (you get a fire on 2 rotor faces). For 2 turns of the crank, you get a firing sequence on two rotor faces per rotor, so its really 2.6l of displacement, not 1.3l.

If you call a rotary a 1.3l and say "ooo, mazda's rotaries make 200hp out of a 1.3l" then you have to say a 5.7l ls1 is really a 2.85l, or a 1.8l honda making 180hp is really .9l.

CrackHeadMel

iTrader: (2)

Well this is the first time somebody has put it this way..

t24todd

iTrader: (7)

damn well that seams to make sense...i still like telling people i have a small sports car thats only 1.3l and gets shitty gas mpg... the confused looks are priceless...then you tell them its a rotary and they are still confused

dj55b

iTrader: (2)

some food for thought in the amateur FC area :P

https://www.rx7club.com/general-rotary-tech-support-11/some-info-you-guys-debate-215701/

dial8

iTrader: (4)

No your wrong. Read this
https://www.rx7club.com/rotary-car-performance-77/good-info-read-rotary-engine-displacement-699272/

*edit, beat me to it.

davemo

there's gotta be a hole somewhere in this argument and if i wasn't supposed to be writing an essay instead of cruising the forums in the library, i would endeavor to find it...

amhorach

or read this:

http://www.rx7.com/techarticles_displacement.html

Jager

iTrader: (10)

I got a book from 1986 and it tells how Mazda goes about figuring out their displacement.

I'll post it up when I go back to Bemidji.

Evil Aviator

Yes, it is the displacement of each cylinder added together. Note that this is regardless of how often each cylinder fires, so 2-stroke and 4-stroke piston engines are rated in the same manner despite the fact that the 2-stroke fires twice as much as 4-stroke of equal displacement. Also note that displacement is based on the cylinder, not on the piston. Therefore, if you had a 2-piston 4-cylinder engine, or a 4-cylinder 2-piston engine, you would multiply the swept volume times each cylinder, not each piston. This is what confuses most of the rotary engine displacement nitwits.

Another confusing part of displacement is the misuse of the value. Some countries use it for taxation, and many racing organizations use it for classifying output. Displacement doesn't work out well for these applications when the engines are radically different from each other, such as in cases where 2-stroke piston, 4-stroke piston, supercharged engines, and rotary engines are compared to each other. A better use for displacement is in the calculation of the flow rate, which is a much more useful value when comparing engines or choosing performance parts.

Of the links posted so far...

These links are wrong.
https://www.rx7club.com/showthread.php?t=215701
https://www.rx7club.com/showthread.php?t=699272

This link is a comparison as opposed to actual displacement.
http://www.rx7.com/techarticles_displacement.html

If you have doubts about this subject, any method of calculation for a 13B engine is wrong if it yields anything other than 1.3 L (1308cc).

This is used to calculate the flow rate, not the displacement.

V = Z * Vh

V = Total Displacement
Z = Number of Rotors
Vh = Volume of a Single Working Chamber

NZConvertible

To add to what Evil said about flow, if you compared a 13B (2 x 654cc = 1308cc) to a 1308cc 4-stroke piston engine and assumed both had 100% VE, the 13B would flow twice as much air at the same engine speed, because it has an intake stroke every engine rev instead of every second rev. But the VE of a rotary is generally lower than that of piston engine, so the few motorsport categories that allow rotaries to compete against piston engines use an equivalence factor of 1.8 rather than the expected 2.0. So a "1308cc" 13B is classed as 2354cc and a "1962cc" 20B is classed as 3532cc.

Delphince

I think you meant to say "stroke" in there somewhere rather than "piston" or "cylinder" because I'm having an interesting time imagining an engine with any ratio of cylinders vs pistons that is different from 1:1...

wrankin

And just to make it more confusing, the actual "power stroke" on a piston engine is less that 180-degrees of crankshaft output, whereas on a rotary it's significantly longer in duration (depending on the porting of the exhaust).

-b

rotary8

13b engine is 2.6 litres of displacement 20b 3.9 litres and 4 rotor 5.2 litres of displacement if any body tells you anything else they are on crack .

gio64

iTrader: (1)

This is completely wrong. The displacement of a 13B is 1.3l (or whatever the exact number is).

Using an equivalence to compare it to a piston engine can very well yeld the 2.6 value, but that is not the displacement of the 13b, it is only what you can consider it equivalent to when you compare it to a piston engine.
Just like somebody else has already said, there is more to this discussion than just combustion chamber size and number of power strokes over one crank/eccentric shaft rotation. As a matter of fact, if you are attempting to determine scientifically what the comparable size is, even the flow rate might not be the most accurate comparison. The "stroke" affects the comparison too and not only in terms of the actual offset from the centre of rotation, but also in relation to the duration and the angle at which the force is applied to the crank/eccentric shaft during the power stroke.


Somebody has used the 2 stroke engine for comparison. It would be possible to build 2 engines with identical number of cylinders, with same bore and stroke, but one could be 2 stroke and the other 4. In this situation, the displacement of either engine would be identical. However, the 2 stroke would fire twice as many times as the 4 stroke in any given amount of time/rotations of the crank. This doesn't mean that the 2 stroke is twice the size. As a matter of fact, 2 strokes cannot be equated to double the size 4 stroke counterparts. Because of the engine design, no valves, a very different flow in and out of the engine, the 2 stroke has much less sealing ability than a 4 stroke, therefore resulting in a less efficient engine, if comparing 1 cubic centimetre of displacement on the 2 stroke to 1 cubic centimetres of a 4 stroke engine.

If what you say were true, then the 13b would be comparable also to a 2.6 L 2 stroke piston engine, and I would imagine you very well know this is not true.

Similar considerations should be applied in the comparison between 4 stroke piston and a rotary engine.

I read somewhere that technically, you could read the 13b as a 3.9 L displacement engine. This is because there is another criteria you could consider in attempting to compare engines: the amount of revolutions of the crank that are necessary to complete the cycle; the article I read said that since each rotor face needs to complete 1 entire rotation of the rotor to complete one combustion cycle, this means that the number of eccentric shaft rotations necessary to a rotary to complete a cycle is 3. And this would mean that the 13b would have ingested 3.9 L of air/fuel mix by the time that would be over, which would make it a 3.9 L engine. Please forgive me if I quickly regurgitated this little piece of technicality, but, mistakes or not, the point is, there are several ways you could look at things.
The bottom line is that the displacement is the volume of the combustion chamber, so a 13B is 1.3 L displacement. All the comparisons and technical interpretations one could go through and argue in whatever way are irrelevant if we're talking about displacement.
If you want to start analyzing, then even 2 equal 4 stroke engines, one with 5K rpm redline and another with a 10K rpm redline have a different displacement, since one can fire twice as many times than the other in the same amount of time...
I hope you get the picture...

rotary8

The key for comparing the displacement between the 4-cycle engine and the rotary engine is in studying the degrees of rotation for a thermodynamic cycle to occur. For a 4-cycle engine to complete every thermodynamic cycle, the engine must rotate 720° or two complete revolutions of the crankshaft. The rotary engine is different. The engine rotor rotates at 1/3 the speed of the crankshaft. On two rotor engines, front and rear rotors are 180° offset from each other. Each rotation of the engine (360°) will bring two faces through the combustion cycle (the torque input to the eccentric shaft). This said, it takes 1080° or three complete revolutions of the crankshaft to complete the entire thermodynamic cycle. Obviously, we have a disparity. How can we get a relatable number to compare to a 4-stroke engine? The best way is to study 720° of rotation of the two-rotor engine. Every 360° of rotation, two faces of the engine complete a combustion cycle. 720° will have a total of four faces completing their cycle. 40ci(654cc) per face times four faces equals 160ci or 2.6L. That’s a well-reasoned number and now gives us something to be able to compare to other engines. In addition, since four faces passed by in the comparison, it’s like a four cylinder engine.

Now we know, the 13B compare well to a 2.6L 4-cylinder 4-cycle engine.

gio64

iTrader: (1)

Very well put. And I agree.
But it is for comparison reasons only, not to designate what the displacement is. What I was and am trying to say, is that while a realistic comparison is one like this one I quoted, at the same time, this has nothing to do with the actual engine displacement.

Evil Aviator

I wrote it correctly. Yes, you will probably have a difficult time imagining an engine that does not have the same number of pistons and cylinders. This is why a lot of the people on this forum do not understand the displacement of the rotary engine.

Here is an example... a 12-cylinder 6-piston engine:
http://www.yachtforums.com/forums/te...iami-show.html

Yes, I have been exposed to a lot of weird stuff that you guys probably didn't even know existed.

This has absolutely no bearing on the subject.

Please read my post again or take a college course on the subject. Or, if you want to really make it simple, just read the owner's manual, factory service manual, or read the number on the engine... and put down the crack pipe, lol.

DarrenS

someone tried to convince me that a rotary is really double of what they rate it and mazda lied.. maybe they were thinking that it puts out twice as much exhaust or produces twice as much combustion as a 1.3L piston engine..?

gio64

iTrader: (1)

Not quite.
They are simply saying this: for a 4 stroke engine, it takes 2 complete revolutions of the crank to ingest and spit out whatever displacement the engine is called for (let's say, 1.3 L, for the argument's sake). If you look at the 13B, it will be capable of ingesting and spitting out 2.6 litres within 2 revolutions of the eccentric shaft. Therefore, some people say it is equivalent to a 2.6 litres engine. But this is a very big simplification of a comparison, not of what the displacement is.
Like was said before, a 1.3 L 2 stroke could have identical number of pistons, identical bore and stroke as a 4 stroke 1.3 L engine. Its displacement is 1.3, in spite of the fact that it will ingest and spit out 2.6 L over 2 revolutions.
You see, one thing is the displacement of the engine, another is the way the engine uses that displacement while it functions. If we want to try to make a comparison based on volumes of mix used, sure, we can start reasoning in those terms, but this doesn't change the actual displacement of the engine.
If you were to compare, you should then factor in efficiency and overall engine design. The eccentric shafts tend to offer a smaller leverage in comparison to a piston engine, the power stroke durations are different, the way the flame travels in the rotary combustion chamber is different, there are so many factors in the equation that most of these are gross simplifications.

Evil Aviator

Yes, there is the conspiracy theory touted by those who don't know any better, as well as the misapplication of the displacement term as I stated earlier.

Delphince

A double-headed piston. Now that's just cheating. Ok, I'll give you that one, but it made your explanation a little hard to follow for anyone that hasn't seen what you're talking about before. Though in making the point that non-conventional motors like rotaries and your crazy contraption still follow the same basic displacement rules comes through loud and clear.

Edit: I retract the "crazy" adjective. Now that I think about it, it's not that different from the drive piston of a steam locomotive. But I've never seen such a thing applied as a combustion engine.

ronbros3

boy has this one been beat to hell, anyway who says how many rotations of the engien output shaft has to make to come up with a socalled displacment , or is it swept volume , or some other!

t-von

iTrader: (8)

People need to stop worrying soo much about total displacement. All you need to know is that a 13b moves 1.3L's of air every e-shaft rotation. Regardless of how many rotations it takes to fire an engines full displacement, what's important is the air moved per every single crank rotation. Both the piston and rotary's rpm's are measured by crank revolution. So what your looking at is the amount of air being moved per rotation per engine. This is the easiest way to compare the two. This is why the 1.3L 13b is comparable to a 2.6L piston engine. They both move the same amounts of air each e-shaft/crank shaft rotation.

CarbonFD3Assault

I think,.. that this is just a bit rotarded...

misterstyx69

iTrader: (142)

Pure crap..like Saying that it shouldn't be called a Rotary Because Actually it is Triangles in a Circular Motion.It should then be Triangular Volume Displacement instead of "cylinder Volume displacement".so Unless they can figure out how much Room the Rotor Takes up as a Triangle,then they Won't get a real Displacement of Volume for the Engine....See?I can make no sense Too!