Why Do Rotaries Suck?

[mad082]

i see what you did there, LOL. sure, base it of a 360 degree rotation. that means you halve the displacement of a 4 stroke and the rotaries get to stay at 1.3L, LOL

[GT-R32]

well why is that the two strokes problem to have to suffer a capacity disability?(or are we basing things on 720degrees now)And wheres my extra capacity as ive lost some from having an exhaust port!!!!!

Remember to pull excess apex seals out if you are going off 360 degrees.

[Sydneykid]

That is indeed the fair displacement comparison

Most of the world has recognised this and in motorsport categories where rotaries are allowed to compete...bar some other variables that influence it...the displacement of a rotary is approximately doubled to put it into fair displacement categories with 4 stroke engines.

But they don't put them into fair displacement categories. A 13B should be included with 7.8 litre 4 stroke engines, that's 3.9 litres of pumping capacity times 2, because it's not a 4 stroke.

This is also where the turbo sizing question becomes interesting, and supports the 7.8 litre capacity equivalence. If it wasn't 7.8 litres how could it spin a T04Z using only 3,000 rpm? What the alternative, 1.3 litres? We all know that there is no way a 1.3 litre engine is going to spin a T04Z at any rpm, let alone at 3,000 rpm.

It's quite simple, an engine is an air pump and we all know how much air a rotary pumps and it sure as hell ain't 1.3 litres.

Perhaps the most important thing to remember in all this displacement arguing...is you can call a 13B X displacement if you want...its actual displacement is so irrelevant in the real world it's not funny. The end power/torque outputs and the fuel economy of the engine are what its customers actually give a shit about. Who buys an RX7 and argues with the salesmen about the displacement of the engine? You want to know how fast it goes and how much it's going to cost you. I fail to see how Mazda can successfully LIE about 1.3 litres to make the engine seem attractive, when the thing has the fuel economy of a V6/V8? What's the point of lying about its displacement if fuel economy and power output are going to shut you down? There's no advantage to listing it as 1.3 over 2.6 or 3.9. The moment someone asks Mazda how many litres per 100km the thing uses, the lie is all over? Rotaries are their own engines, the displacement is whatever the inventor wants to attach to it. For all other areas where displacement IS a relevant factor, e.g. motorsport...we have formulas for working out the displacement advantage/disadvantage. End of story!

And that's all we need to know...the rest can be agreed to disagree.

My experience is exactly the opposite, ask the average Joe how big the engine in his car is and there is a good chance he will tell you straight away. Some will look at the badge on their car. In comparison ask the average Joe the power and torque outputs of his engine and he look at you blankly. Capacity is relevant to rego, insurance, road tax, for expense claims purposes the Tax Office even quotes costs to run a car by its engine capacity. Nobody outside the enthusiast uses horsepower and torque, it's all about capacity. Ask a pushrod lover what's different about an LS1 compared to an LS2, you can bet the first response you get is 5.7 versus 6.0 litres. I don't know about you but I haven't ever seen torque output badges on cars.

I've been playing with rotary engines for a lot of years and I know full well that Mazda’s claim (for a 13B) of 1.3 litres, 4 stroke and 9,000 rpm is rubbish. The real world results don't support it, then when we get down and dirty with the facts we find they don't support it either.

1. does it ingest air like a 1.3 litre? No it doesn't

2. does it use fuel like a 1.3 litre? No it doesn't

3. does it spin up large turbos like a 1.3 litre? No it doesn't.

That’s where I started in my quest for the truth many years ago. The real world logic just didn’t stack up, not even close, so I went searching for the facts. I completely ignored Mazdas words, actions, drawings, diagrams, visual representations etc, because they would obviously be slanted towards supporting their version. I sought out independent views, unbiased opinions, reasoned measurement and unconstrained opinion. That’s why I found Phil Irving’s published engineering papers so significant. Even more so the unbelievable lengths Mazda went to to lampoon someone who was an internal combustion engine luminary. Having done that research and my own measurements, I’m left with the conclusion that Mazda has at the very least distorted the facts for marketing reasons or at the very worst lied through their teeth.

Cheers

Gary

[spx25t]

I will start this off with I am not wading into the argument but something I have noticed when reading the first 13 pages is that the main argument has been centralised around the 13b being a 3.9L motor for one complete revolution.(not disagreeing with that) but due to one complete revolution taking 1080 degrees to do so.

wish to confirm something with a piston motor the entire process happens over 360 degrees yes/ no

if yes see below if no ignore the rest

so going a different route instead of comparing a rotary to a piston engine if you were to compare a piston engine to a rotary would you then have to work on a piston engine completing three complete revolutions (360 degrees x 3 ) to equal one Rotary revolution.

if so then would a 2.6l be considered a 7.8 litre in engine displacement.

this is just me thinking out aloud trying to see both sides.

Rob

[Smity42]

a 4 stroke takes 2 revolutions so 720 degrees

Hence why we have been saying the 3.9L rotory 13B is roughly equivalent to a 2.6L 4 stroke. 720/1080 = 2/3.

What logic sydneykid is using to now double that 3.9L and get 7.8L (or 6.8L ) is totally beyond me...

Edited September 24, 2009 by Smity42

[GT-R32]

You honestly believe a 13B is like a 6.8L V8? (even then his maths is astoudingly bad) You better call CAMS and FIA and let them know you have discovered something amazing that they don't know! What a conspiracy!

How much air does a 13B ingest per 360 degree revolution? Look at the e-shaft mate. Then think. Then think again. And again if you need to.

720 degrees?

1080 degrees?

It's not rocket science and this is starting to get embarrassing for you Gary.

[ylwgtr2]

i see what you did there, LOL. sure, base it of a 360 degree rotation. that means you halve the displacement of a 4 stroke and the rotaries get to stay at 1.3L, LOL

well i must admit there was a "slight"prodding for the rotary there....however i was more going for the two strokes there!(possibly rotary )

[Sydneykid]

You honestly believe a 13B is like a 6.8L V8? (even then his maths is astoudingly bad) You better call CAMS and FIA and let them know you have discovered something amazing that they don't know! What a conspiracy!

How much air does a 13B ingest per 360 degree revolution? Look at the e-shaft mate. Then think. Then think again. And again if you need to.

720 degrees?

1080 degrees?

It's not rocket science and this is starting to get embarrassing for you Gary.

Oops, nobody's perfect, not even me, 7.8 litres it is.

Pardon me if I blow it back at you.

How much air does a 1.3 litre 2 stroke ingest per 360 degree revolution? Look at the crankshaft mate. Then think. Then think again. And again if you need to.

720 degrees?

1080 degrees?

How much air does a 1.3 litre 4 stroke ingest per 360 degree revolution? Look at the crankshaft mate. Then think. Then think again. And again if you need to.

720 degrees?

1080 degrees?

You argument relies totaly on the eccentric shaft revolutions having some effect on the actual measurement of a rotary engine's true capacity. When in no other internal combustion engine is that the case.

As for why did I double the 3.9 litres,? Well you started the "doube the 2 stroke's actual capacity to get the equivalent for a 4 stroke" scenario.

For instance, if you raced 2 stroke 250cc bikes again 4 strokes, what would the 4 stroke engine size be?

I was just following your lead. So you agree with me then, that doubling a 2 strokes capacity to get some equivalency is not practical?

Cheers

Gary

[FineLine]

The one leg is both the same

[DRD-00F]

You honestly believe a 13B is like a 6.8L V8? (even then his maths is astoudingly bad) You better call CAMS and FIA and let them know you have discovered something amazing that they don't know! What a conspiracy!

How much air does a 13B ingest per 360 degree revolution? Look at the e-shaft mate. Then think. Then think again. And again if you need to.

How much air does an RB26 ingest per 360 degree revolution?

You better make a call yourself!

My point here is that just cause something has done 360 degrees, it doesn't always mean it has completed 100% of its combustion cycle.

[Smity42]

sydneykid, that logic is flawed, as a rotory is not actually a 2 stroke.

Allow me to introduce the following table of engine characteristics:

this is just a quick table i did up of things off the top of my head. You could add more.

You seem determined to call a rotory a 2 stroke because it shares points 8 and 9 with a 2 stroke, completely ignoring that it shares 10 with a 4 stroke, and the rest it has nothing at all in common.

I will give you that it is MORE like a 2 stroke than a 4 stroke based on that table, sharing 2 characteristics rather than 1, but it has more not in common that it does in common.

Now ask yourself, what is the DEFINING characteristic of a 2/4 stroke motor? Point 6 of course! Unsurprisingly, a rotory doesnt fit either of them. You can introduce point 7 and call it a 1-revolution motor, or you could just call it a wankel.

Right, now, why do we double a 2 stroke displacement when comparing to 4 stroke? This is where point 5 comes into play. Because the two stroke only turns the crank half as far with one cycle, in order to compare it with a 4 stroke fairly we must turn it twice, hence, we double it. By the same logic, a rotor has turned 1/3rd as much again with one cycle, so to fairly compare it with a 4 stroke we must take 1/3 off it.

So, once more: it is a 3.9L Wankel engine. It can be roughly compared to a 2.6L 4 stroke or a 1.3L 2 stroke.

I will agree with you it is most definitely not the 1.3L 4 stroke mazda claim it is

Edited September 24, 2009 by Smity42

[DRD-00F]

It seems like everyone is muddying up the waters unnecessarily.

Here's how I see it:

We say a 3.9L 2 stroke engine is 3.9 Litres

We say a 3.9L 4 stroke engine is 3.9 Litres

Yet...

We say a 3.9L Wankel cycle engine is 1.3 Litres?

Both piston engines are judged and titled on their literage by how much air one chamber holds, timed by the number of chambers, making the number of rotations the crank shaft does irrelevant. This is a FAIR comparison. So why is it fair to say a 13B is 1.3 Litres when it hasn't even completed it's own wankel cycle? Every other engine has to complete it's combustion cycle to get the literage, so why is a wankel so special?

In reality, it is a 3.9L wankel motor. Everyone keeps saying it's a different engine so it gets different rules. Well they're right. This rule says that it does 1080 degrees of rotation to complete it's combustion cycle. In comparison only, it's a completely fair thing to say.

In motorsport however, I would pit it up against 2.6L 4 strokes, which is what some governing bodies have done.

[Sydneykid]

sydneykid, that logic is flawed, as a rotory is not actually a 2 stroke.

Allow me to introduce the following table of engine characteristics:

this is just a quick table i did up of things off the top of my head. You could add more.

You seem determined to call a rotory a 2 stroke because it shares points 8 and 9 with a 2 stroke, completely ignoring that it shares 10 with a 4 stroke, and the rest it has nothing at all in common.

I will give you that it is MORE like a 2 stroke than a 4 stroke based on that table, sharing 2 characteristics rather than 1, but it has more not in common that it does in common.

Now ask yourself, what is the DEFINING characteristic of a 2/4 stroke motor? Point 6 of course! Unsurprisingly, a rotory doesnt fit either of them. You can introduce point 7 and call it a 1-revolution motor, or you could just call it a wankel.

Right, now, why do we double a 2 stroke displacement when comparing to 4 stroke? This is where point 5 comes into play. Because the two stroke only turns the crank half as far with one cycle, in order to compare it with a 4 stroke fairly we must turn it twice, hence, we double it. By the same logic, a rotor has turned 1/3rd as much again with one cycle, so to fairly compare it with a 4 stroke we must take 1/3 off it.

So, once more: it is a 3.9L Wankel engine. It can be roughly compared to a 2.6L 4 stroke or a 1.3L 2 stroke.

I will agree with you it is most definitely not the 1.3L 4 stroke mazda claim it is

There are 10 tests, a rotary is like a 2 stroke in 3 of those tests

3

8

9

But it's only like a 4 stroke in 2 tests.

3

10

To me that means it is most like a 2 stroke.

But as usual the devil is in the detail. The main problem I have with the table is #6, you have called it "piston strokes". Obviously a rotary is going to fail that test, it doesn't have pistons. So you separate #6 (pistons) and #7 (rotors). Yet in #4 you use the equivalency "numer of faces per piston/rotor". Same again in #5 with "crank/eccentric revolutions", so being fair you should also use the same equivalency in #6 and combine it with #7.

My suggestion for equivalency in #6 would be "number of combustion face cycles to complete combustion". In which case a rotary engine is 1, a piston 4 stroke is 2 and a piston 2 stroke is 1. Hence a rotary is the same as a 2 stroke piston engine in test #6.

Cheers

Gary

Edited for my bad maths.

[scathing]

In motorsport I'd run them in their own class. As people have said, they are two very different engine types and so comparisons are pretty hard to draw.

[Sydneykid]

It seems like everyone is muddying up the waters unnecessarily.

Here's how I see it:

We say a 3.9L 2 stroke engine is 3.9 Litres

We say a 3.9L 4 stroke engine is 3.9 Litres

Yet...

We say a 3.9L Wankel cycle engine is 1.3 Litres?

Both piston engines are judged and titled on their literage by how much air one chamber holds, timed by the number of chambers, making the number of rotations the crank shaft does irrelevant. This is a FAIR comparison. So why is it fair to say a 13B is 1.3 Litres when it hasn't even completed it's own wankel cycle? Every other engine has to complete it's combustion cycle to get the literage, so why is a wankel so special?

In reality, it is a 3.9L wankel motor. Everyone keeps saying it's a different engine so it gets different rules. Well they're right. This rule says that it does 1080 degrees of rotation to complete it's combustion cycle. In comparison only, it's a completely fair thing to say.

Well whatya know, there's another one who understands.

In motorsport however, I would pit it up against 2.6L 4 strokes, which is what some governing bodies have done.

Unfortunately they don't, it's 2.3 litres.

Cheers

Gary

[GT-R32]

Oops, nobody's perfect, not even me, 7.8 litres it is.

Pardon me if I blow it back at you.

How much air does a 1.3 litre 4 stroke ingest per 360 degree revolution? Look at the crankshaft mate. Then think. Then think again. And again if you need to.

720 degrees?

1080 degrees?

You argument relies totaly on the eccentric shaft revolutions having some effect on the actual measurement of a rotary engine's true capacity. When in no other internal combustion engine is that the case.

As for why did I double the 3.9 litres,? Well you started the "doube the 2 stroke's actual capacity to get the equivalent for a 4 stroke" scenario.

I was just following your lead. So you agree with me then, that doubling a 2 strokes capacity to get some equivalency is not practical?

Cheers

Gary

A rotary is not a two stroke. Okay, I see where you are getting mistaken here. Every single engine in the world's displacement is measured using crank degrees. It's just that they are all mostly 4 strokes so you don't need to have that understanding and hence your comment above. The rotary is not excluded from this. It is this ensures it is accurate and further that it's the relative measure.

A two stroke completes its Otto cycle in two distinct 'strokes' and does this in 360 degrees.

A four stroke completes its Otto cycle in four distinct 'strokes' and does this in 720 degrees.

A Wankel completes its Otto cycle for all rotor faces (which also have 4 distinct 'stokes') in 1080 degrees.

For every 360 revolution a 13B Wankel inhales 654cc x 2 = 1308cc. You then apply this volume for whatever 'equivalent' you require above.

That's what was trying to show you with regard to the 2/4 stroke comparison (250cc 2 stroke = 500cc 4 stroke). It's all about volume of air per shaft revolution for relative comparisons. I encourage you to do some reading on this.

How much air does an RB26 ingest per 360 degree revolution?

You better make a call yourself!

My point here is that just cause something has done 360 degrees, it doesn't always mean it has completed 100% of its combustion cycle.

Errrr... That's been my point all along. I'm just trying to ensure that crank degrees are used as the measure as that is the only accurate way to measure displacement. You can then run equivalents off this.

[Smity42]

your main problem is I called it piston strokes?

2-stroke

4-stroke

get a dictionary, or better yet, google the definition of 2 stroke and 4 stroke.

You can't 'suggest your own equivalency' when that's not how the engine is defined.

Flat out, its not a 2 stroke.

You can't even rename it to 'cycle' and call it 2-cycle (Which you have been so intent on doing), as there is no '2' involved in the rotary's method of combustion anywhere.

I did already agree with you it is MORE like a 2 stroke than a 4 stroke. I don't think there is any doubt about that, and in fact I've always held that opinion.

P.S. It has point 1 in common with both of them, which is why i left it out in my comparison. You conveniently left it out of just 4 stroke to make your argument look better.

I'm out, you can't save some people from themselves. It's been fun guys, a good discussion for the most part

[Sydneykid]

your main problem is I called it piston strokes?

2-stroke

4-stroke

get a dictionary, or better yet, google the definition of 2 stroke and 4 stroke.

You can't 'suggest your own equivalency' when that's not how the engine is defined.

Flat out, its not a 2 stroke.

You can't even rename it to 'cycle' and call it 2-cycle (Which you have been so intent on doing), as there is no '2' involved in the rotary's method of combustion anywhere.

I did already agree with you it is MORE like a 2 stroke than a 4 stroke. I don't think there is any doubt about that, and in fact I've always held that opinion.

P.S. It has point 1 in common with both of them, which is why i left it out in my comparison. You conveniently left it out of just 4 stroke to make your argument look better.

I'm out, you can't save some people from themselves. It's been fun guys, a good discussion for the most part

Hang on, after all it's the Otto Cycle we are talking about here. If we are really trying to compare rotaries with piston engines, then using the term "strokes" just makes the comparison impossible. I don't think anyone should fall for the cop out, "hohoho a rotary engine doesn't stroke, so it can't be a 2 stroke hohoho". If you aren't truly trying to compare, then there is your get out of jail card, no strokes in rotary. I just get a bit sick of that same card being played all the time. If you are truly trying to compare, then at least substitute something that is applicable. I choose "cycles", sure it's not perfect, but it is applicable, it is logical and it can be understood by most people. If you don't like my choice for equivalency, that's fine, just pick one of your own.

Anyway, I think you missed the point I was making, you found a method of equivalency in others, but not #6 and #7. So why not have a go at it?

Cheers

Gary

[bigmikespec]

I am enjoying this... it is like Gary is giving a bunch of apes a hammer and watching them slowly figure out what it does.

[Sydneykid]

Every single engine in the world's displacement is measured using crank degrees.

No they are not.

If they were, then a 2 stroke/cycle 1.3 litre would be a 2.6 litre

Or a 2.6 litre 4 stroke would be a 1.3 litre.

Nobody measures an engines capacity dividing or multiplying by the number of degrees the cranskaft rotates.

Actually that's not true, Mazda does, but only for rotaries, they don't do it for their 4 stroke engines.

It's just that they are all mostly 4 strokes so you don't need to have that understanding and hence your comment above. The rotary is not excluded from this. It is this ensures it is accurate and further that it's the relative measure.

Accurate? What's more accurate than how much the pump (the rotor) actually pumps? The eccentric shaft doesn't pump, the rotor does.

A two stroke completes its Otto cycle in two distinct 'strokes' and does this in 360 degrees.

A four stroke completes its Otto cycle in four distinct 'strokes' and does this in 720 degrees.

A Wankel completes its Otto cycle for all rotor faces (which also have 4 distinct 'stokes') in 1080 degrees.

For every 360 revolution a 13B Wankel inhales 654cc x 2 = 1308cc. You then apply this volume for whatever 'equivalent' you require above.

That's what was trying to show you with regard to the 2/4 stroke comparison (250cc 2 stroke = 500cc 4 stroke). It's all about volume of air per shaft revolution for relative comparisons. I encourage you to do some reading on this.

Let me help you out here with some numbers,

A 3.9 litre two stroke completes its Otto cycle in two distinct 'strokes' and does this in 360 degrees. Eveyrone calls it a 3.9 litre engine.

A 3.9 litre four stroke completes its Otto cycle in four distinct 'strokes' and does this in 720 degrees. Everyone calls it a 3.9 litre engine.

A 3.9 litre Wankel completes its Otto cycle for all rotor faces in 1080 degrees. Mazda call it a 1.3 litre engine.

Now explain that to me again, because it looks to me like one rule for Mazda and another rule for everyone else.

Errrr... That's been my point all along. I'm just trying to ensure that crank degrees are used as the measure as that is the only accurate way to measure displacement. You can then run equivalents off this.

That's been my point all along. I'm just trying to ensure that eccentric shaft degrees are not used as the measure as they are irrelevant to everyone else except Mazda.

Cheers

Gary