rotary engine displacement ??
#1
rotary engine displacement ??
they are 1.3 liters right ? how are they measured ?
my brother insists that it is 650cc for each combustion chamber (each face of the rotor).. Im not sure so I don't coment.
anyone knows for sure?
my brother insists that it is 650cc for each combustion chamber (each face of the rotor).. Im not sure so I don't coment.
anyone knows for sure?
#5
Originally Posted by Parastie
It's 1.3L Volume. 650cc per rotor, not per rotor face. If you powered water into the engine it would take up 1.3L.
e=offset of the eccentric
i=length of rotor face
r=width of housing
don't quote me on this, though (might also be e+r+b)
#6
Originally Posted by Falcoms
I do belive you are wrong, sir. That is 650cc per face, not per rotor, being that it is measured by how much displacement occours when the working chamber is used once i.e. one face is measured. Also, the equation for calculating it out is sq. root of 3^3(e+i+r)
e=offset of the eccentric
i=length of rotor face
r=width of housing
don't quote me on this, though (might also be e+r+b)
e=offset of the eccentric
i=length of rotor face
r=width of housing
don't quote me on this, though (might also be e+r+b)
Mazda messured the engine to be 1.3L based on volume displacement.
#7
What fun it is!
You both are kinda right. Each rotor displaces 654 cc per crank revolution. And as you know, the rotor spins at 1/3 the crank speed. So it is 654 cc per face also. Then you get 1308 cc (1.3 L) for the whole engine.
The problem most people have is when they compare to 4 stroke piston engines. The 4 stroke "wastes" 1/2 of its displacement on two of the strokes. So if you want to compare a rotary to a boinger, consider the rotary as being roughly equivalent to a 2 stroke. Usually in normal cfm or mass flow equations, you will need to multiply the rotary (or two stroke) displacement by 2 to get the numbers to work out right. The "muliply by 2 part" is cause most of the equations are derived for 4-stroke boingers.
HTH,
Scott
The problem most people have is when they compare to 4 stroke piston engines. The 4 stroke "wastes" 1/2 of its displacement on two of the strokes. So if you want to compare a rotary to a boinger, consider the rotary as being roughly equivalent to a 2 stroke. Usually in normal cfm or mass flow equations, you will need to multiply the rotary (or two stroke) displacement by 2 to get the numbers to work out right. The "muliply by 2 part" is cause most of the equations are derived for 4-stroke boingers.
HTH,
Scott
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#8
Originally Posted by Parastie
If you want to get technical about this, but that messurement the engine is 2.6L because it does twice the combustions per rotation has a normal piston engine.
Mazda messured the engine to be 1.3L based on volume displacement.
Mazda messured the engine to be 1.3L based on volume displacement.
#10
Originally Posted by casio
2 is better than 4? that could be debated. besides, we're still 4 stroke. just different.
1308cc pwns joo. 1962cc pwns 1308cc.
1308cc pwns joo. 1962cc pwns 1308cc.
#11
yea, i've asked about 2-stroke before and was told why they're inferior to 4-strokes, but i have no recollection of the answer. i remember emissions output is a bit worse. i wonder what the biggest 2-stroke engine is.
2616cc; now we're just getting greedy. i wonder what the front of an rx7 would look like if it were made to fit a rotary "4-banger." its a big engine.
2616cc; now we're just getting greedy. i wonder what the front of an rx7 would look like if it were made to fit a rotary "4-banger." its a big engine.
#12
A rotary's displacement or swept volume is calculated just like a piston engine's, being the difference between maximum chamber volume and minimum chamber volume multiplied by the number of rotors. You don't multiply it by the number of chambers because they only one chamber is fired per engine revolution. This makes a 13B 1308cc.
In the automotive media it became common to multiply this by two because the rotary inhales this volume twice as often as a 4-stroke piston engine. Since the amount of air an engine breathes is almost directly proportial to it's power output, this gives a number that makes rotary-powered cars easier to compare to a piston-engine ones (and comparing cars is what the media does). This makes a 13B 2616cc.
In motorsport the rotary's swept capacity is usually multiplied by 1.8 as this gives a more accurate way to evenly class rotary-powered cars against piston-powered ones. This makes a 13B 2354cc.
The truth is one can't accurately compare a rotary's swept volume with a piston's. There are too many other differences to use just this one number. In my exerience people who actually argue about whether or not a 13B really is 1.3L usually don't fully understand what they're talking about.
In the automotive media it became common to multiply this by two because the rotary inhales this volume twice as often as a 4-stroke piston engine. Since the amount of air an engine breathes is almost directly proportial to it's power output, this gives a number that makes rotary-powered cars easier to compare to a piston-engine ones (and comparing cars is what the media does). This makes a 13B 2616cc.
In motorsport the rotary's swept capacity is usually multiplied by 1.8 as this gives a more accurate way to evenly class rotary-powered cars against piston-powered ones. This makes a 13B 2354cc.
The truth is one can't accurately compare a rotary's swept volume with a piston's. There are too many other differences to use just this one number. In my exerience people who actually argue about whether or not a 13B really is 1.3L usually don't fully understand what they're talking about.
#13
all these different ways of calculating rotary displacement is just nonsense. when someone asks me the displacement of a rotary, i tell them:
12A>1146cc>1.1L
13B>1308cc>1.3L
look at the plates in the engine compartment.
12A>1146cc>1.1L
13B>1308cc>1.3L
look at the plates in the engine compartment.
#14
my definition from rx8club.com:
a motor is defined as having completed an entire combustion cycle when all combustion chambers have had a combustion event (fired their spark plugs). in a 2 stroke motor, that means once up, once down (one rotation of the crank shaft). in a 4 stroke piston motor, that means twice up and twice down (two complete rotations of the crank shaft). in a '4 stroke' (otto cycle) wankel motor, that means 120 degrees of rotation for each rotor (one full rotation of the eccentric shaft).
being that each rotor in a 13B mazda wankel engine displaces at theoretical maximum 654cc's per 120 degrees of rotation, and it has 2 rotors, it is a 1.308 L engine. case closed, and one more reason no one should listen to idiots who write in car magazines who think they know what they're talking about (thinking of the SCC article).
when it comes to calculating rate of mass flow, you need is a rate (which is, yes dependant on displacement, but not solely), so all this bullshit about the 'real' displacement is pretty pointless, especially since it really is 1.3L.
a motor is defined as having completed an entire combustion cycle when all combustion chambers have had a combustion event (fired their spark plugs). in a 2 stroke motor, that means once up, once down (one rotation of the crank shaft). in a 4 stroke piston motor, that means twice up and twice down (two complete rotations of the crank shaft). in a '4 stroke' (otto cycle) wankel motor, that means 120 degrees of rotation for each rotor (one full rotation of the eccentric shaft).
being that each rotor in a 13B mazda wankel engine displaces at theoretical maximum 654cc's per 120 degrees of rotation, and it has 2 rotors, it is a 1.308 L engine. case closed, and one more reason no one should listen to idiots who write in car magazines who think they know what they're talking about (thinking of the SCC article).
when it comes to calculating rate of mass flow, you need is a rate (which is, yes dependant on displacement, but not solely), so all this bullshit about the 'real' displacement is pretty pointless, especially since it really is 1.3L.
#16
Senior Member
Joined: Jun 2003
Posts: 494
Likes: 0
From: Minden, NV
^^^
I always just tell people that if you are going to compare apples to apple, and measure piston engine displacement the same way you measure rotary displacement then the displacement is 1.3L. But it functions so much differently displacement doesn't matter much.
OMG, did I just say it? A replacement for displacement??!!??? bwahahahahahahahahaha
I always just tell people that if you are going to compare apples to apple, and measure piston engine displacement the same way you measure rotary displacement then the displacement is 1.3L. But it functions so much differently displacement doesn't matter much.
OMG, did I just say it? A replacement for displacement??!!??? bwahahahahahahahahaha
#17
Originally Posted by wakeech
my definition from rx8club.com:
a motor is defined as having completed an entire combustion cycle when all combustion chambers have had a combustion event (fired their spark plugs). in a 2 stroke motor, that means once up, once down (one rotation of the crank shaft). in a 4 stroke piston motor, that means twice up and twice down (two complete rotations of the crank shaft). in a '4 stroke' (otto cycle) wankel motor, that means 120 degrees of rotation for each rotor (one full rotation of the eccentric shaft).
being that each rotor in a 13B mazda wankel engine displaces at theoretical maximum 654cc's per 120 degrees of rotation, and it has 2 rotors, it is a 1.308 L engine. case closed, and one more reason no one should listen to idiots who write in car magazines who think they know what they're talking about (thinking of the SCC article).
when it comes to calculating rate of mass flow, you need is a rate (which is, yes dependant on displacement, but not solely), so all this bullshit about the 'real' displacement is pretty pointless, especially since it really is 1.3L.
a motor is defined as having completed an entire combustion cycle when all combustion chambers have had a combustion event (fired their spark plugs). in a 2 stroke motor, that means once up, once down (one rotation of the crank shaft). in a 4 stroke piston motor, that means twice up and twice down (two complete rotations of the crank shaft). in a '4 stroke' (otto cycle) wankel motor, that means 120 degrees of rotation for each rotor (one full rotation of the eccentric shaft).
being that each rotor in a 13B mazda wankel engine displaces at theoretical maximum 654cc's per 120 degrees of rotation, and it has 2 rotors, it is a 1.308 L engine. case closed, and one more reason no one should listen to idiots who write in car magazines who think they know what they're talking about (thinking of the SCC article).
when it comes to calculating rate of mass flow, you need is a rate (which is, yes dependant on displacement, but not solely), so all this bullshit about the 'real' displacement is pretty pointless, especially since it really is 1.3L.
Thats the way I feel too however, when comprared to a 4 stroke piston engine, a 1.3L rotary will breath like 2.6L piston engine due to how much air the rotary will move in 2 rotations of the e-shaft. This is one of the reasons why these rotarys will spool a large single fairly decent.
#19
Originally Posted by Parastie
If you want to get technical about this, but that messurement the engine is 2.6L because it does twice the combustions per rotation has a normal piston engine.
Mazda messured the engine to be 1.3L based on volume displacement.
Mazda messured the engine to be 1.3L based on volume displacement.
just because it takes in it's full displacement every rotation doesn't mean it now has double the displacement because the max amount of volume stays the same regardless of how many revolutions it takes.
another thing if anything though with your though of logic a rotary should be 1.3L while a piston motor would be rated at half :p
#20
Yes the 13B is 1.3L I also agree. It is the rotary haters who like to argue this. They cannot stand that the rotary seems to out do 99+% of the beloved V8s when it comes to HP per cube. Every official book/manual by Mazda says its a 1.3L so I will listen to them before some idiot running off at the mouth on here. I will agree the power is comparable to a 2.6 or so liter piston motor but I will never agree with those who say it is 2.6L by measure of displacment. You cannot rate dispalcement on output of power and I agree just because the rotary can make full use of its displacment also does not change displacment. So to all you rotary hating traders you can SUCK IT!
#21
I'm not a rotary hater, and I think there are some very good arguments for the 2.6L displacement of the 13B. There are so many long threads about this topic, however, that I will leave you to read them and decide for yourself which displacement rating is appropriate in which contexts.
-Max
-Max
#23
It's not hard to figure out how so much power can be produced by so few cubes:
First, the power "event" (not "stroke") lasts a full 270 degrees of e-shaft rotation vs. only 180 degrees of rotation for the boinger's power stroke. (for the sake of simplicity I'm ignoring valve and port timing factors).
Second, there is no valve train to drive in the rotary so no hp gets wasted on that particular task. Read it and weep, boinger fans!
Oh, and it's 1308cc, not 2616.
First, the power "event" (not "stroke") lasts a full 270 degrees of e-shaft rotation vs. only 180 degrees of rotation for the boinger's power stroke. (for the sake of simplicity I'm ignoring valve and port timing factors).
Second, there is no valve train to drive in the rotary so no hp gets wasted on that particular task. Read it and weep, boinger fans!
Oh, and it's 1308cc, not 2616.
#25
^ Yeah, 1.3L converts to 80.1something CI. Also, aviator, I was talking about the 4 rotor engine, which uses 4 13b housings and rotors, coming out to a grand total of 2616cc. just FYI.