Why Do Rotaries Suck?

[Sydneykid]

With every combustion the output shaft completes one cycle. SK has been trippling his numbers because it takes 3.9l of air for the first side of the rotor to return to its origional position (the other 2 sides following afterwards). If we measure the cycles by full turns of the e/shaft then 1.3l is correct. [/url]

If we measure the cycles by full turns of the cranskshaft then a 2 litre 4 stroke is really a 1 litre.

Cheers

Gary

[Sydneykid]

Has the "cycle" of an engine got to do with the combustion cycle or where things are in the engine?

If it was just about where things are in the engine, a 4-stroke piston cycle would only be one rotation of the shaft.

Wouldn't work very well though.

I thought you had it there for a minute, but then, dissapointment....

I do see your point, but i don't think the fact that it takes 3 combustion cycles for the rotor to do a complete rotation is somewhat irrelevant.

How does one side of the rotor do a complete combustion cycle without the other 2 sides also doing complete combustion cycles?

Cheers

Gary

[Jez13]

What if the rotor was a hexagon? Who cares. Measure the cycle of the e/shaft and the crankshaft returning to is original position.

Funny though because the rotor has three equal sides so when one combustion completes it becomes in its exact original postition only its not with the original rotor face. I think thats enough to fill the definition "returns to its original state". Just because its not the same rotor face doesnt mean it hasn't returned to its original position.

[Jez13]

How does one side of the rotor do a complete combustion cycle without the other 2 sides also doing complete combustion cycles?

Cheers

Gary

I know how. Just dont focus a full combustion cycle on one rotor face. Refer to my last comment.

[Sydneykid]

What if the rotor was a hexagon? Who cares. Measure the cycle of the e/shaft and the crankshaft returning to is original position.

OK, so on that basis a 4 stroke 2 litre is really a 1 litre as the cranskshaft has returned to its original position.

Funny though because the rotor has three equal sides so when one combustion completes it becomes in its exact original postition only its not with the original rotor face. I think thats enough to fill the definition "returns to its original state". Just because its not the same rotor face doesnt mean it hasn't returned to its original position.

Tell me how the same rotor face can return to it's original state without the other 2 sides also returning to their original states?

Cheers

Gary

[Sydneykid]

LOL

Get your hand off it mate. RICE Racing is miles ahead of you. I posted the link from 2008 where he cut up a rotor to educate people on its cycle. He's been saying all this stuff since the early 1990s. The thing is he understand relative time scale, considering the engiune as a whole unit, which you do not. He gave up arguing as you merely assume you are correct.

Actually no, he's miles behind me, I have known exaclty how a rotary works since I rebuilt my first 12A in 1972. Then, as now, I know exactly what is going on and I fail to see any relevance in time when it comes to capacity. It's very much like saying a 4,000 rpm engine has half the capacity of an 8,000 rpm engine because it takes twice as long. Time is completely irrelevant in capacity measurement for that very reason.

Cheers

Gary

[Jez13]

OK, so on that basis a 4 stroke 2 litre is really a 1 litre as the cranskshaft has returned to its original position.

Tell me how the same rotor face can return to it's original state without the other 2 sides also returning to there original states?

Cheers

Gary

Why does it have to be the same rotor face if after one combustion its in exactly the same position? It doesnt matter that its a different rotor face because it ends up in the same state.

Edited September 30, 2009 by Jez13

[Smity42]

OK, so on that basis a 4 stroke 2 litre is really a 1 litre as the cranskshaft has returned to its original position.

Yes, we CALL it a 2L but when we COMPARE it to a 2-stroke we DO halve it and say it is the equivalent of a 1L 2 stroke.

in the same way we can CALL a 13B a 3.9L, but when we COMPARE it to a 4 stroke we multiply by 2/3 (2.6L), or when we COMPARE it to a 2 stroke we multiply by 1/3 (1.3L).

The opposite argument would be to CALL a 13B a 1.3L and when we COMPARE it to a 4 stroke we multiply by 2, or when we COMPARE it to a 2 stroke we do nothing. However I agree with you that it should be called a 3.9L.

The actual measurement of capacity is a little inconsistant even between 2 and 4 strokes. But what is relevant is having a way to compare them, and we do this using relative timescale, to acheive the ratios above.

Going back to the pump analogy someone used a while ago (it might even have been you sydneykid), if I wanted to compare two pumps what would i look at? First i would look at how much each pumped with one revolution of the input shaft. Its not fair to spin one twice and one once. Then i would look at what is the maximum speed i can spin each one (rpm). If they both pump the same amount in one turn of the shaft, but one is capabable of spinning twice as fast, it has the potential to pump more. Then you would get into how much energy is required to turn the shaft of each one (this is where you get into efficiency, etc). You could then start to look at the physical size of the pumps, etc.

I guess this is a 'black box' approach, where the internal workings of the pump are irrelevant, if we want to compare them we need to look at how they acheive what they are designed to do.

So going back to our engines with this logic (black box approach):

Lets take a 13B, a random 2.6L 4 stroke, and a random 1.3L 2 stroke. Lets pretend we don't know how these work, we just want to compare them. What is an engine designed to do? turn an output shaft.

If we turn the output shaft once, the 13B will have pumped 1.3L of air, so will the 2 stroke, so will the 4 stroke. So they are roughly the equivalent capacity, even though they are labelled differently.

Next, you would compare the max RPM of that output shaft. For the 13B this IS 9000 RPM. For the piston engines, it would depend but on average it would be a little less than that. What the internals of the engine are doing here doesn't matter remember, black box approach. The output shaft is what we are comparing.

Now to compare the engines, you would have to look at the power/torque graph over the whole RPM scale, as well as fuel economies, size and weight of the engine, etc. But you have a fair comparison to do this as the capacities are EQUIVALENT.

So, what if we were to put a 1:3 ratio on the output of the 4 stroke piston engine? Suddenly, its only pumping 1/3 as much air with each rotation of the crankshaft, but is capabable of spinning 3 times as fast. So you would have to take that into account when comparing to other engines, it's power curve would be the same, but spread over 18000 rpm instead of 6000 (say). It would make less torque, but spin faster. Gear ratios do not give an engine an outright advantage, they just change the rpm range it operates in.

Likewise if we imagine for a second a wankel that doesnt have this inherit 1:3 ratio. Suddenly when I turn the output shaft once, its going to have pumped a full 3.9L of air. Nice. BUT, its now only capable of doing 3000 rpm, so I have to take this into account when comparing it to another engine.

Basically I guess what I am saying is the workings of the engines are so different between piston and rotary engines, the only way we can compare them is with this 'black box' method. The engines still do the same job, that is to turn the output shaft, so we compare how they do this. If one has an internal gear ratio it does not matter, it will give it an equivalent capacity advantage but limit its RPM, or vice versa.

Edited September 30, 2009 by Smity42

[Birds]

No offence intended towards you guys, but all you are doing is reiterating the same points that we (I) started making about 20 pages ago. Don't get me wrong, it's nice to see there are other people who understand the concepts too...but it's pointless reposting it again and again if 20 pages later this guy still isn't getting it. Do you think after 31 pages worth of thread he's going to suddenly change his mind? Not going to happen my friends. You're arguing with a brick wall. Let's just leave it be and let the learned people who read this thread make their own conclusions based on what is already written in it.

[Jez13]

Yes, we CALL it a 2L but when we COMPARE it to a 2-stroke we DO halve it and say it is the equivalent of a 1L 2 stroke.

in the same way we can CALL a 13B a 3.9L, but when we COMPARE it to a 4 stroke we multiply by 2/3 (2.6L), or when we COMPARE it to a 2 stroke we multiply by 1/3 (1.3L).

The opposite argument would be to CALL a 13B a 1.3L and when we COMPARE it to a 4 stroke we multiply by 2, or when we COMPARE it to a 2 stroke we do nothing. However I agree with you that it should be called a 3.9L.

The actual measurement of capacity is a little inconsistant even between 2 and 4 strokes. But what is relevant is having a way to compare them, and we do this using relative timescale, to acheive the ratios above.

Going back to the pump analogy someone used a while ago (it might even have been you sydneykid), if I wanted to compare two pumps what would i look at? First i would look at how much each pumped with one revolution of the input shaft. Its not fair to spin one twice and one once. Then i would look at what is the maximum speed i can spin each one (rpm). If they both pump the same amount in one turn of the shaft, but one is capabable of spinning twice as fast, it has the potential to pump more. Then you would get into how much energy is required to turn the shaft of each one (this is where you get into efficiency, etc). You could then start to look at the physical size of the pumps, etc.

I guess this is a 'black box' approach, where the internal workings of the pump are irrelevant, if we want to compare them we need to look at how they acheive what they are designed to do.

So going back to our engines with this logic (black box approach):

Lets take a 13B, a random 2.6L 4 stroke, and a random 1.3L 2 stroke. Lets pretend we don't know how these work, we just want to compare them. What is an engine designed to do? turn an output shaft.

If we turn the output shaft once, the 13B will have pumped 1.3L of air, so will the 2 stroke, so will the 4 stroke. So they are roughly the equivalent capacity, even though they are labelled differently.

Next, you would compare the max RPM of that output shaft. For the 13B this IS 9000 RPM. For the piston engines, it would depend but on average it would be a little less than that. What the internals of the engine are doing here doesn't matter remember, black box approach. The output shaft is what we are comparing.

Now to compare the engines, you would have to look at the power/torque graph over the whole RPM scale, as well as fuel economies, size and weight of the engine, etc. But you have a fair comparison to do this as the capacities are EQUIVALENT.

So, what if we were to put a 1:3 ratio on the output of the 4 stroke piston engine? Suddenly, its only pumping 1/3 as much air with each rotation of the crankshaft, but is capabable of spinning 3 times as fast. So you would have to take that into account when comparing to other engines, it's power curve would be the same, but spread over 18000 rpm instead of 6000 (say). It would make less torque, but spin faster. Gear ratios do not give an engine an outright advantage, they just change the rpm range it operates in.

Likewise if we imagine for a second a wankel that doesnt have this inherit 1:3 ratio. Suddenly when I turn the output shaft once, its going to have pumped a full 3.9L of air. Nice. BUT, its now only capable of doing 3000 rpm, so I have to take this into account when comparing it to another engine.

Basically I guess what I am saying is the workings of the engines are so different between piston and rotary engines, the only way we can compare them is with this 'black box' method. The engines still do the same job, that is to turn the output shaft, so we compare how they do this. If one has an internal gear ratio it does not matter, it will give it an equivalent capacity advantage but limit its RPM, or vice versa.

Its not a 3.9L. Yes it can take 3.9L of air but it only combusts 1.3l per cycle. Thats if I convince the public that a cycle is made per combustion and not once the same rotor face has completed a full revolution.

[910trx]

No offence intended towards you guys, but all you are doing is reiterating the same points that we (I) started making about 20 pages ago.

It is getting a bit repetitive... so are we going in circles, orbits or strokes now?

Ive learnt so much by reading all of this, and its given me enough ammo so next time my mate raves on about how great his rotor is i can slam him, or, if someone tries to tell me a rotor sucks i can slam them back.

I just hope to god they have not been reading this as intently as I have over the past week

i still like rotors - for what they are

[Birds]

Yes, well if nothing else, arguments serve to expose higher truths and provide education to those uninformed

[Smity42]

Its not a 3.9L. Yes it can take 3.9L of air but it only combusts 1.3l per cycle. Thats if I convince the public that a cycle is made per combustion and not once the same rotor face has completed a full revolution.

The definition of capicity seems to be a little ambigous, especial when dealing with something not piston based.

But call it what you like, the equations to relate it to its piston brothers are the same. It compares to a 1.3L 2 stroke or a 2.6L 4 stroke

[jame88]

It is getting repetitive, thats why im trying to only say things that haven't been said before.

I raised the point that there is no gear ratio between the eccentric shaft and rotor, noone argued, so thats over now.

Back to the capacity thing.

A 4-stroke engine's cycle is 2 rotations of the output shaft long

A 2-stroke engine's cycle is 1 rotations of the output shaft long

Do we all agree on that?

In this cycle they both pump their capacities worth of air.

Your argument is that a Wankel take 3 rotations of the output shaft for it to pump out the same air it pumped in, right?

Something noone here has seemed to notice is that a 2-stroke engine actually take 2 rotations of the output shaft to pump out the same air it pumped in.

You could argue that a 1000cc 2-stroke engine should be rated at 2000cc.

But noone does.

Of course the 3.9L Wankel, 2.6L 4-stroke, 1.3L 2-stroke way of comparing works, i'm trying to argue what a Wankel should actually be rated at.

[jame88]

This seems like it will go on forever, but ill reply to a few things.

If we turn the output shaft once, the 13B will have pumped 1.3L of air, so will the 2 stroke, so will the 4 stroke. So they are roughly the equivalent capacity, even though they are labelled differently.

Pretty much right there just had to say if you turn the output shaft of a 4-stroke piston once, it will only have pumped air in, the other will have pumped that much air in and out.

Tell me how the same rotor face can return to it's original state without the other 2 sides also returning to their original states?

The rotor has only rotated 120 degrees, but as far as the chambers are concerned here, its in the same state. Kind of the equivalent of shoving different piston in a bore each cycle of a 4-stroke or something.

It's actually quite simple, you measure capacity by eccentric shaft revolutions. I measure capacity by revolutions (oribits if you prefer) of the pumping medium, ie; the rotors. In one revolution (orbit if you prefer) of the 2 rotors a 13B pumps 3.9 litres.

So I define a cycle of a rotary engine as one complete revolution (orbit if you prefer) of the rotor. You on the other hand define a cycle of a rotary engine as one complete revolution of the eccentric shaft.

Right, except a more accurate way of saying this would be you measure the capacity by rotations (not orbits) of the rotors, I measure the capacity by orbits of the rotors. Seeing as the orbits are what is translated into power in a Wankel i see no reason to use rotations.

Just to clarify i'm using rotations and orbits as different motions here, not interchangibly.

Funny though because the rotor has three equal sides so when one combustion completes it becomes in its exact original postition only its not with the original rotor face. I think thats enough to fill the definition "returns to its original state". Just because its not the same rotor face doesnt mean it hasn't returned to its original position.

Agreed

[Smity42]

It is getting repetitive, thats why im trying to only say things that haven't been said before.

I raised the point that there is no gear ratio between the eccentric shaft and rotor, noone argued, so thats over now.

Back to the capacity thing.

A 4-stroke engine's cycle is 2 rotations of the output shaft long

A 2-stroke engine's cycle is 1 rotations of the output shaft long

Do we all agree on that?

In this cycle they both pump their capacities worth of air.

Your argument is that a Wankel take 3 rotations of the output shaft for it to pump out the same air it pumped in, right?

Something noone here has seemed to notice is that a 2-stroke engine actually take 2 rotations of the output shaft to pump out the same air it pumped in.

You could argue that a 1000cc 2-stroke engine should be rated at 2000cc.

But noone does.

Of course the 3.9L Wankel, 2.6L 4-stroke, 1.3L 2-stroke way of comparing works, i'm trying to argue what a Wankel should actually be rated at.

And therein lies the problem. I can find NO definition of capacity that doesn't relate to: piston bore * stroke * number of pistons.

This obviously does not work for rotarys.

People have differing opinions of what capacity actually means when you take pistons out of the equation. Depending on how you define it, you could get 1.3L, 3,9L, or even 2.6L if you try hard enough.

Edited September 30, 2009 by Smity42

[GT-R32]

And therein lies the problem. I can find NO definition of capacity that doesn't relate to: piston bore * stroke * number of pistons.

This obviously does not work for rotarys.

People have differing opinions of what capacity actually means when you take pistons out of the equation. Depending on how you define it, you could get 1.3L, 3,9L, or even 2.6L if you try hard enough.

How many power pulses or 'power strokes' are there for the respective figures?

3.9L = 6

2.6L = 4

1.3L = 2

[Smity42]

Pretty much right there just had to say if you turn the output shaft of a 4-stroke piston once, it will only have pumped air in, the other will have pumped that much air in and out.

True. But when you look at a motor as having more than one piston, half will have pumped air in and half out so it works out pretty much the same.

The rotor has only rotated 120 degrees, but as far as the chambers are concerned here, its in the same state. Kind of the equivalent of shoving different piston in a bore each cycle of a 4-stroke or something.

Right, except a more accurate way of saying this would be you measure the capacity by rotations (not orbits) of the rotors, I measure the capacity by orbits of the rotors. Seeing as the orbits are what is translated into power in a Wankel i see no reason to use rotations.

Just to clarify i'm using rotations and orbits as different motions here, not interchangibly.

Agreed

See my post above. Once again, it all comes down to what does capacity mean? Yes we are repeating ourselves here, and really, call it 1.3L or 3.9L, it doesnt really matter as long as you know how to relate that to how piston motors are rated you can make valid comparisons and assesments

[jame88]

And therein lies the problem. I can find NO definition of capacity that doesn't relate to: piston bore * stroke * number of pistons.

This obviously does not work for rotarys.

People have differing opinions of what capacity actually means when you take pistons out of the equation. Depending on how you define it, you could get 1.3L, 3,9L, or even 2.6L if you try hard enough.

Perfectly right there, but the "how much it pumps per cycle" seems to work for the current ratings of 4-stroke and 2-stroke, and it can relate to a Wankel, so i thought it is a good way of rating it. Perhaps the Engineers at Mazda thought similar.

[Smity42]

How many power pulses or 'power strokes' are there for the respective figures?

3.9L = 6

2.6L = 4

1.3L = 2

Good point, and if you want to go down sydneykid's path of calling it a 'stroke' motor, which i still disagree with as stroke is a piston movement, you could then say:

its a 1.3L 2 stroke

or a 2.6L 4 stroke

or a 3.9L 6 stroke

I'd prefer to see the word 'movement' or yeah 'power pulse' instead of 'stroke' however