skc

I have heard that when measuring engine capacities on rotary engines you double the actual engine size.

This is aparently due to the way the chambers work and in one spin there are two lots of combustion.

Can anyone confirm this.

skc

playa78

i guess u can expect similar performance to 2-3 times it's capacity but it is still a 1.3L and not 2.6 =)

check this site out

http://www.rotaryengineillustrated.com

Hymee

iTrader: (1)

Nice question skc.

Someone (RotaryGod I think) put up a thread about this in the "tech garage". He put forward arguments that it was a 1.3 and a 2.6 and a 3.9. I don't think I completely understood the 3.9 litre argument though...

The 1.3L argument:

The displacement of an engine is normally described as the total swept volume of each combustion chamber. Example - an 8 cylinder engine with a 90mm (9 cm)bore and a 90mm (9cm) stroke - swept volume = pi * (4.5^2) * 9 * 8 = 4580 cubic cm = 4.6 litres. The exact same answer would be obtained for a single cylinder engine of 4.6L displacement.

A "13B" 2-rotor rotary engine: each combustion chamber = 654 cc. There are 2 combustion chambers. Therefore the swept volume of all combustion chambers = 654 * 2 = 1308cc = 1.3 litres.

The 2.6 litre argument:

The "4.6" litre 8 cylinder above - for one revolution of the crankshaft (output shaft), only 4 of the cylinders have combusted (since it is a "4 stroke"), so we effectively get 2.3 litres of "combustion". So 2 revolutions of the crank = 4.6 litres.

The 13B rotary engine... Each combustion chamber = 654cc. Each time the crankshaft revolves once, each rotor has gone around 1/3 of it's cycle (which produces 645cc of "combustion" on each rotor). Therfore in 1 revolution of the crank, we have 1.3 litres of "combustion". If we say 2 revolutions, to compare to the piston engine, we have effectively 2.6 litres of combustion.

I can't recall the 3.9 litre argument. edit: Yes I do. It regarded each rotor as 3 combustion chambers, so 645cc * 3 * 2 = 3924cc = 3.9 litres "swept volume". But I don't accept that argument for 2 reaons. 1) because not all of those 6 "combustion chambers" are fired in one (or 2 for that matter) revolution. and 2) not all 6 faces have "swept their volume" in one (or 2) revolution. It takes 3 revolutions of the rotary output shaft (eccentric shaft) to take place for each face of each rotor to fire once. So IMHO the 3.9litre argument is invalid.

But... What about 2-stroke engines? aka Motor Cycles/wipper snippers etc... (thinking about a single cylinder) a "250cc" 2-stroke engine still has a swept volume of 250cc, even though for 2 revolutions of the crankshaft it has had 2 firings (500cc of combustion). So the question is - why are 2-strokes not "doubled" when exporting their capacity?

The other (confusing?) thing is a rotary is not a "2-stroke", or a 2-cycle engine, but is a true 4-cycle engine.

Note: I have used the term "combustion" above losely, in that I really mean "combustible mixture that is ignited to produce power".

I hope this helps explain.

Cheers,
Hymee (going for post of the day again )

IKnowNot'ing

A shorter, however technically incorrect reply :

Its engine capacity is 1.3 l ;

It performs as well as a good 2.6 l ;

It has the fuel economy of a 3.9 l.

Hymee

iTrader: (1)

Or I could put this argument comparing it to my previous car...

It performs nearly as well as a 5.7 litre...

It has better fuel economy than a good 5.7 litre...



Cheers,
Hymee.

IKnowNot'ing

I was referring to European engines, you know, the ones with cams and valves on top of it...

(this message to be taken as seriously as my previous one please, thx)

rpm_pwr

Hymee, I'm a beleiver in the 2.6 argument myself, but the 3.9 argument is probably the second best and is actually fairly technically sound:
http://mikeonline.cable.nu:1863/misc/rotor.doc
With the obvious caveat that the vast majority of people don't have an understanding of translating torque through gearing so it's best not to have themthinking of 3.9 litre 6 bangers when they see a 13b.

Realistically it's best to argue for the 2.6l engine as then you can disregard gearing advantages.

-pete

BVD

That's not the reason, but here’s another way of trying to explain it all.

In one complete spin of each rotor there are actually 3 lots of combustion, not 2, because each rotor has 3 faces, each of which does its own full cycle of “Suck, Squeeze, Bang, Blow” (well, it sounded more innocent when it was first used… :D Intake, Compression, Combustion and Exhaust if you prefer…). The tricky issue here is that “capacity” is just one way of measuring engine “size”. And it doesn’t even precisely relate to the actually physical size, weight or shape of the engine, let alone torque, power, mechanical efficiency, etc.

The point is that capacity or “swept volume” just refers to the volume covered in one stroke of each piston on a piston engine. Generally, that would be of interest to give a rough indication of how much fuel/air mixture an engine can intake. But it came into widespread use at a time when it was mostly used to compare 4-stroke piston engines with other 4-stroke piston engines. It’s not so useful when comparing with rotaries or two strokes, and don’t even bother trying to equate it with jet or rocket engines.

Unfortunately, it says nothing about how efficient that engine is at actually filling that volume, or under what circumstances. Each engine (2-stroke, 4-stroke, and rotary) has a mechanically different way of doing the job, has a different power and torque curve, and is designed to run at different rev ranges of the output shaft (crankshaft). So comparing capacity, even relating it to the same number of output shaft revs, is a fairly crude kind of measurement.

A rotary engine can be seen as the rough equivalent of a 4-stroke engine with 2 combustion chambers, but 6 “pistons”. So if you only look at the combustion chambers it’s a 1.3 litre, but if you look at the complete volume swept by each combustion face then it’s a 3.9 litre engine. Both perfectly valid ways of looking at it, if you’re looking at other rotaries, but neither very useful for direct comparisons with piston engines.

However, if you look for one second of time at a 4-stroke piston engine doing 6,000 rpm and a rotary also doing 6,000 rpm (i.e. 100 revs a second of the crankshaft for both) you’ll see that the volume swept by inlet strokes on the rotary is twice that of the piston engine. This is the reason for saying that our 1.3 litre engines are the rough equivalent of a 2.6 litre.

Really, there are a lot more interesting and useful ways to compare what engines are actually capable of, and what you might call their “character”, than just the nominal capacity.

skc

Thanks for the great feed back guys.

We are dealing with a fairly ingenious and unigue engine that is difficult to pigeon hole....which is great and adds to the mystique of the vehicle.

I was not aware of the 3.9l arguement however, I can somewhat see their point.

skc

Hymee

iTrader: (1)

Bugger, I just submitted a beautiful argument for considering the engine to be a "1.3 litre rotary engine" and somehthing on my network went down and I lost the post

Consider yoursleves saved from the priveledge

Cheers,
Hymee

NT Rotor Head

From a registration point of view, the MVR in NT charges registration of rotary engined cars at the equivalent of a 2.6L or 2.4L piston engine. I tried to get around it as the rules actually state that registration is based on engine capacity. They had another piece of paper somewhere that said that rotary engines are to be listed as 2.6 and 2.4L for 13B and 12A respectively. My crusade was short lived!