In theory yes but if the output shaft was doing, say, 600rpm then the real 5 litre engine would be doing 10 oscillations per second where the fake 5 litre engine would be doing 100. So the volumetric efficiency would be much lower for the fake 5 litre engine and hence would pump quite a significant amount less.

Assuming you didn't know what was inside the engine or how it was geared you could get an approximation on it's displacement by increasing the rpm until it couldn't pump any faster. This is a much better indication of it's true displacement because it takes into account the loss of volumetric efficiency as the speed of the internal parts increase. It also takes into account issues such as friction and pumping losses. If you did this for the rotary it would peak out with a similar pumping capacity to that of a 3.9 litre six cylinder motor.

 

The question is why? An example would be helpful as well.

 

Ok, start with a 2 litre engine that revs to say 6000 rpm. We know it sucks 1 litre per rev and does 6000 revs per minute, so will need a turbo that can flow 6000 litres per minute (this is a simplified example of course).

If we take that engine and hide a gearing up of 2:1 inside the engine then it is still a 2 litre engine but from the outside things look a little odd. From the outside the gearing throws out every measurement we could take, eg it now inducts 500cc per rev but revs to 12000rpm. It has half the torque (with the same power) and takes 360 degrees for a piston to go from TDC to BDC.

BUT, the turbo that we should use for it will still be same. This is because 500cc per rev x 12000 rpm is the same figure for flow as 1000cc per rev and 6000 rpm. Applying a gear at the output affects many parameters but not turbo size.

 

I guess that answers my question about why a turbo sized for 2.6L piston engines work well for rotaries as well.

 

The incorrect rating of those few should not be the basis of essentially considering gearbox output shaft (and gear selected) in determining displacement.

The internal gearing of the wankel (3 revs per engine cycle, vs 2 for common boinger) is ***essential to the function of the engine***, and not a simple non-essential internal reducer or increaser.

The fact that a 3/2 overdrive slapped on a 3.9L V6 (esp one with 1.8" stroke) results in identical-to-wankel stroke cycle durations and displacements for a rotation of the false output ( who thought of that 1st, u or me? ) is not reason enough to consider the 13B a 3.9L engine. You still need to rev the 13B's output 50% higher than the 3.9 boinger's true output shaft to match 100%VE cfm.

False internal gearing is easy to spot. The "similar pump" method for displacement rating (the basis for 2 rev ingestion rating for boingers) is sufficient. I'll send a request for an SAE standard, after a I finish ripping apart someones paper that ASME asked me to review.

 

That internal gearing of the rotary is NOT essential. It could have been designed a different way and still functioned identically. Granted it would be less than ideal but it would still be possible. Piston engines do exist that have an internal gearing that is more than just a gear. There is a radial aircraft engine that uses a sliding cam mechanism which produces all 4 strokes in 1 rev. Should that engines capacity double?

Not entirely. Both output shafts are doing say 9000rpm (the 13B and the geared up 3.9s) but *internally* both engines are going along at 3000 otto cycles per second for each chamber. Looking at it this way both engines have this gearing up, for one engine it is very obvious because you can look and see the gear. For the other engine it's a little more hidden so we let them get away with it. If someone geared up a 3.9L six we would laugh if they suggested it was a 2.6 but not for the rotary. Although it's not really that hidden, there is a big **** 2:3 cog inside the engine :-)

BTW, we both thought of the geared up 3.9 if I remember correctly. I first came up with the idea many years before going on this forum and when I first spoke to you about it here you knew about it already.

I thought a previous post of mine addressed this issue pretty well, it might have got lost as a new page was started. Assuming we have 2 pumps, one with a 500cc piston with the output shaft geared down 10:1 and another with a 5 litre piston and standard output shaft. Measuring both engines from the outside they pump 5 litres per rev so if we don't know what's inside them we would assume both are 5 litre. But if we use them to pump air and slowly incease their revs the 500cc pump will hit its limit much sooner. As the internal speed of this engine is 10x higher its VE will drop dramatically. If we measure the max flow rate of both engines we'll get an indication of the true capacity as the 5 litre engine will flow approx 10 times the 500cc engine. (naturally it won't be exact but it will give an indication).

If we do the same thing with a 13B and a 2.6 litre 4 cylinder engine we'll find the 2.6 will peak out sooner than the 13B. While the 2.6 might reach it's peak at 6000rpm the rotary will reach it's peak at 9000rpm (all things being equal such as intake efficiency etc). The 13B will pump 50% more than the 2.6 and hence pump the same as a 3.9 litre 6. We could increase the 2.6 litre up to 9000rpm but it's VE would have dropped significantly compared to the 13B so will never reach the pumping capacity of the 13B.

 

Hi Kevin,

Have a look at this animation. The engine in the middle apparently exists although I couldn't find any links. It has 2 banks of 3 radial pistons of 654cc each (the other bank is not shown). The centre of the engine rotates at 1/3rd crank speed similar to the way a rotor rotates. The intake and exhaust are in the same locations and it has a 270 degree stroke. Piston surface area is the same as that of a rotor's face although the crank is 45mm diameter instead of 30 for the rotary. Given that this engine has 6 x 654cc pistons for 3.9 litres but only inducts 2.6 litres per 2 revs, what is the displacement of this engine?



http://mikesdriveway.com/misc/radial...comparison.gif

This animation is from my software which runs much more smoothly and can be controlled with the mouse. For anyone who has already downloaded it you'll need to download it again to get this animation (clicking check for updates in the app will download the new version also but not the new help file).

http://www.mikesdriveway.com/engineapp

 

Mike C:
That internal gearing of the rotary is NOT essential ...K> it is essential for the certifed wankel ... It could have been designed a different way and still functioned identically. Granted it would be less than ideal but it would still be possible....K>.. musy be practical, not just possible, but I'd like to see details it ...

Piston engines do exist that have an internal gearing that is more than just a gear. There is a radial aircraft engine that uses a sliding cam mechanism which produces all 4 strokes in 1 rev. Should that engines capacity double? ...K> sounds essential internals for that design ...this handfull of engines would be rated like a 2 stroke boinger, and actually have a 2x advantage over a 5.0, but I suspect big props and low revs., and not competitive in max hp vs same rev displ rated boinger.


Not entirely.
Yes entirely, add the increaser for cycle comparison at same rpm, but it's not essential, and 13B must spin 50% higher than the pure 3.9 to match cfm. case closed......
....Both output shafts are doing say 9000rpm (the 13B and the geared up 3.9s) but *internally* both engines are going along at 3000 otto cycles per second for each chamber. Looking at it this way both engines have this gearing up, for one engine it is very obvious because you can look and see the gear. For the other engine it's a little more hidden so we let them get away with it. If someone geared up a 3.9L six we would laugh if they suggested it was a 2.6 but not for the rotary. Although it's not really that hidden, there is a big **** 2:3 cog inside the engine :-)

BTW, we both thought of the geared up 3.9 if I remember correctly. I first came up with the idea many years before going on this forum and when I first spoke to you about it here you knew about it already. K> yup



I thought a previous post of mine addressed this issue pretty well, it might have got lost as a new page was started. Assuming we have 2 pumps, one with a 500cc piston with the output shaft geared down 10:1 ..k>.. STOP RIGHT THERE ...I have made it clear that non-esssential reducers/increasers are bogus. I deal with agitated vessels or extruderes or crushers, all start with a high speed motor, go through 40-100:1 reducer then slowly drive an agitator, extruder scre or crusher at high torque and low speed. You don't consider the motor muscle any differently based on the mega torque after the moster gearbox.


If we do the same thing with a 13B and a 2.6 litre 4 cylinder engine we'll find the 2.6 will peak out sooner than the 13B. While the 2.6 might reach it's peak at 6000rpm the rotary will reach it's peak at 9000rpm (all things being equal such as intake efficiency etc). The 13B will pump 50% more than the 2.6 and hence pump the same as a 3.9 litre 6. We could increase the 2.6 litre up to 9000rpm but it's VE would have dropped significantly compared to the 13B so will never reach the pumping capacity of the 13B.[/QUOTE]

You don't want to compare the best production 2.2L engine with the (bigger) best 2.6L rotary (rx8).

You are trying to tie basic displacement rating to max hp potential ... not logical when within a breed, my old 2L triumph maded 1/3 the hp of a modern s2000, at about 1/2 the rpm.

You start with a common, basic displacemt rating, and see how well or differently the various designs deliver, hp, tq, reliability, and mpgs. Don'y try to wrap up all those ouputs with a basic displacement rating, plaese?

 

MikeC: Hi Kevin,

Have a look at this animation. The engine in the middle apparently exists although I couldn't find any links. It has 2 banks of 3 radial pistons of 654cc each (the other bank is not shown). The centre of the engine rotates at 1/3rd crank speed similar to the way a rotor rotates. The intake and exhaust are in the same locations and it has a 270 degree stroke. Piston surface area is the same as that of a rotor's face although the crank is 45mm diameter instead of 30 for the rotary. Given that this engine has 6 x 654cc pistons for 3.9 litres but only inducts 2.6 litres per 2 revs, what is the displacement of this engine?

K> First, this is very much like the case I gave to evil aviator (defends displ based on 1 spark per housing per rev, ie .65L per housing x 2 housings =1.3L ... 2 revs 2 sparks) So per him, it's a 1.3L. Per my historic pump method, you get the same 2.6L as the I got with the 13B.

 

are there any actual conventions for rating the displacment of rotary engines, besides the one use by mazda?

There must be some other existing convention, as the renisis has won 2 awards for international engine of the year in the 2.5-3 liter class. http://media.ford.com/newsroom/featu...?release=17973

BTW in order to get around a rule preventing cars under some displacmemt figure from producing to much HP in taiwan. Taiwan Rx8 imports are officialy rated by mazda at 2616 CCs.

http://www.mazda.com.tw/ (click on rx8, the displacment is in the first page)

 

Kevin, you kind of confused me with those posts. Not to be an ***... but [q uote] and [/q uote] (without the spaces of course) are your friend. I found them pretty hard to read...

So Mazda changed the displacement due to monetary concerns. Same thing that NSU did 40+ years ago. Mazda can't really be looked to for a true source of rotary displacement. Somebody in Mazda that leaves then writes about it... sure... But for now, why would you rate an engine you sell bigger? Simple, you wouldn't, whether it be the true displacement or not.

P.S. Nice find on the Taiwan thing...

 

I was running out of time and took a shortcut ... All was cut from MikeC' s animated post, and ....K> bla bla bla were my responses. But it was abit messy, and will avoid it in the future.

 

I figured it out, just a bit rough.

 

Ok, got the Wankel article in English. This quote I thought might be of interest.

"Therefore, as regards their volumetric displacement and power impulses, two-chamber planetary-rotation engines are equivalent to either a three or twin-cylinder reciprocating-piston four-stroke engine."

Eh, .

 

I like the animation. Alot! It leads me to believe in what Mike C. hit on in his original post.

Lets think about the visual. I get:

1) Half the cylinders hit per crankshaft revolution in the piston engine.

2) The three-piston radial gets three hits per crankshaft revolution.

3)The two rotor Wankel engine gets two hits per crankshaft rev.


The piston engines displacement has been wrongly given for so many years! Just think of the shock wave when my I start calling my 2.2 liter Honda a 1.1!

By the way. That makes the 2.6 liter V-6 a 1.3. Imagine that.

 

I don't think it needs to be practicle. For any theory it needs to stand up to all the what-ifs. Saying that it wouldn't be practicle or does not currently exist isn't valid IMO.

These engines would function just like a normal piston motor and produce the same power as other engines of similar displacement (using sweptvol x no of pistons). However, by your thinking they would have twice the displacement.

Ok, so what if this gearing was essential to the engine? I can easily design an engine with any effective gearing I like.

hmmmm. You don't consider the motor muscle any different because of the gearing. Interesting :-) Why then do you consider the rotary differently after it's 1.5x gearing? :-)

 

The assumption with this animation is that the engine on the right has a gearing up of 1.5x applied to it. The rotary and the radial don't. The other assumption is that only half of each engine is shown and that all engines have 654cc swept volume per working chamber. In this case all engines are very similar, eg:
- all have 654cc swept vol
- all have 6 working chambers
- all have 270 degree stroke
- all take 3 revs to complete an otto cycle
- all fire once per rev
- all of the engines are running slower internally than externally. eg at 9000 rpm all of the engines will be doing 3000 otto cycles per second.



That's possibly true but it doesn't really matter, it's a convention. The question is how to apply that convention to the rotary :-)

 

I thought I answered that.

 

And. The piston engine always uses half of its rated displacement per rev regardless of # of cyl.

It all works fine as is except the piston displacement volumn. Should be half total.

 

The radial engine in the animation is just a fancy way to gear up a piston motor. This can be shown by:
1) Removing the ports and giving each piston valves (while still keeping everything spinning at 1/3 crank speed)
2) Placing all the pistons inline (again while keeping everything spinning at 1/3rd crank speed)

What we've now got is really just a standard inline 6 cylinder motor where the entire motor itself rotates at 1/3rd crank speed. The effect of rotating the entire motor in this way is the same as simply gearing the output shaft up by 1.5x. (the rotation of the body of the engine just has the effect of increasing the speed of the output shaft, hence being equivalent to a gearing). So what the animation is trying to show is that the rotary is equivelent to the radial motor and the radial motor is equivelant to a geared up 3.9 litre 6 cylinder motor. That you already knew.

But, if we take the radial motor and replace the ports with valves then the rotation of the body doesn't have to be 1/3rd crank speed and can be anything we like. According the air inducted over 2 revs method I can now set the displacement of the engine to be anything I like. By rotating the body very fast the displacement could be brought down to almost nothing and by rotating it backwards the displacement could be made very large. However, calculating the displacement over a complete otto cycle will always give the same value.

 

I'm sorry dude. I'm can't follow you into "The straight six at 1/3 RPM.

???!!!.

Hits per crank-rev. Done deal. Including partial displacements/odd cylinder in odd cylinder piston engines in a ratio of total displacement still equals half total.

Even that bastard the V-twin, with it's odd-ball, two-rev to get 720* of complete cycle still has to comply to the rules. It's displacement, because it is a piston engine, is the total of both cyl.

And, as I stated before, should be half.

Who cares how it's configured? It's all about times per rev.

Why do you think V-12's are attractive?

Nobody wants a real big one cylinder.

 

It's a exercise to look at things a different way. The working chambers in a rotary rotate around the centre of the crank at 1/3rd crank speed. From TDC to BDC in a rotary the chamber has moved 90 degrees. This extra 90 degrees gets added to the traditional 180 degrees and causes the rotary to have a 270 degree stroke.

In a piston motor the chambers don't rotate around the centre of the crank, they are fixed. But what would happen if we made them by rotating the entire motor at 1/3rd crank speed? I know that might sound stupid but suspend disbelief for a sec and consider it. From TDC to BDC the motor has now rotated 90 degrees, just like the rotary. And like the rotary that extra 90 degrees has caused the stroke to become 270 degrees. If we use a 3.9 litre 6 for the comparison we find that pretty much anything else we can measure is also equivelent to the rotary. Both engines now induct 2.6 litres per 2 revs, both fire twice per rev etc etc.

I can see what you're saying about the half displacement, quite possibly the convention is wrong. But it doesn't really matter, that is the convention we have and it's well established.

 

You really have to put the crack-pipe down. If anything, the crank turning at three times rotor speed is detrimental to torque.

The internal gearing really doesn't help! But, still two hits per rev.

It's a acts like a 2.6 four more than a 2.6 six.

 

It's funny you should say that, your posts make little sense and are quite scattered. :-)

That is correct, torque is 2/3rds of what it would otherwise be. Shaft speed is up 50% so power is the same.

Without the gearing it would fire 3 times per rev.

There is no question it acts more like a geared up 3.9 litre 6, in just about every conceivable way. It is nothing at all like a 2.6 litre 4. Download my software and have a look for yourself. There is a comparison with a 2.6 litre 4 and another with a 3.9 litre 6. You can see that the 13B and 3.9 litre 6 are functionaly equivelent.

http://mikesdriveway.com/engineapp.