Try monitoring mass airflow on a running engine sometime and compare to it a comparably sized piston engine. It matches up with a 2.6. Any calculations that you try to do match it up with a 2.6. That's what it directly compares to no matter what you consider it.

 

Although I would love to consider my engine 1.3L (and I tell people it is), I'd say it's really 2.6L. Everything points to that being what the displacement really should be classified as.

 

Both of these statements are not correct because they both rely on the output shaft as a reference points (air inducted per 2 rev of shaft). The article you sent My5ABaby stated why you can't use the output shaft as a reference point, because it is a reference point that can change. The simplest example is if you took the drive off the camshaft of a piston motor, the capacity of the engine wouldn't magically double. Similarly you could take the drive off the rotation of the rotor in the rotary and get wildly different figures (the 13b would measure as 7.8 litres of 2 revs!)

The real reference point is the completion of 1 otto cycle. For this piston motor capacity stays the same no matter what output reference you use and the same for the rotary. Using the 1 otto cycle for the 13b gives 3.9 litres.

For your statement rotarygod that no matter how you measure it it always comes to 2.6 litres, that is not true. There is only one method to get to 2.6 litres and that is to measure air inducted over 2 revs, which can fairly easily be shown to be flawed. On the other hand there are many method that lead to 3.9 litres:
1) 654ccs x 6 chambers is 3.9 litres
2) Air inducted over 1 otto cycle is 3.9 litres.
3) Total volume of chambers at any point in time doubled will give 3.9 litres (for a piston motor the volume above the piston at any point in time is always half the displacement).
4) Air inducted at max rpm (assuming 100% ve) will give an approximation of 3.9 litres. This is a bit hit and miss as 2 different 2 litre engines could give different figures but for the rotary it will indicate 3.9 more than 2.6.
5) Air inducted over 2 revs corrected for the 270 degree stroke gives 3.9 litres. For any engine that an otto cycle does not complete in 2 revs a correction to the measured capacity is needed. For example, taking the drive off the camshaft of a 2 litre motor will make the air inducted over 2 revs increase to 4 litres, but the effective stroke will be only 90 degrees. So 4 litres / 180 x 90 gives the true capacity of 2 litres. If we do the same thing with the rotary we get 2.6 litres / 180 x 270 which is 3.9 litres.

 

Nice thread and very interesting arguments. For my .02$, The rest of the driveline is built around crankshaft RPM. With the rotory competitors not being that much higher than the built piston engines they are racing against.

Practical application has it's place too!

 

That's true, the rotary drivetrain has a lower diff ratio to compensate for the 270 degree stroke (higher rpm). But that is only because of the method mazda chose to take the drive off the motor. If they'd used a different method then the drivetrain would be different.

 

Trust me, no matter how I look at it, it doesn't always come out to 2.6 liters. I just understand from what frame of reference certain comparisons need to be made from. There is no one correct answer. There are actually 3 correct answers, all different. If you look at it from an airflow over time standpoint, it does. That's what you set standalone ecu's to. Mass airflow doesn't lie and I and many others have seen the proof of this. From an airflow standpoint, it's a 2.6 L.

Mazda rates the engine as 1 complete revolution of the eccentric shaft. The entire engine displaced 1.3 liters in 1 revolution. This is how Mazda figures it out. In 2 complete revolutions it displaces, 2.6 liters. However it takes 3 complete revolutions of the eccentric shaft to use every face of every rotor. That would be 3.9 liters.

If we are measuring engine displacement the same way piston engine's do, we'd measure the swept volume of every rotor face. That takes 3 complete revolutions of the eccentric shaft to make happen and that is 3.9 liters. However a piston engine displaces all of it's cylinders over only 2 complete revolutions of the crankshaft. From a direct comparison of how to tuen an ecu and how to measure fuel going into the engine, it is a 2.6 liter. From a functionality standpoint that's what it is. Depending on what frame of reference you are talking about, it can be a 1.3, a 2.6 or a 3.9 liter. All answers are correct!

Airflow through a rotary is a very different thing. The rotary is not as efficient as a piston engine. The average street piston engine makes about 10 hp or so per every pound of air ingested. An S2000 engine is over 11 hp per lb of air. A pound of air at seal level on a dry 60*F day is about 14 cu ft worth of air. The average 13B rotary (nonturbo engines only for comparison sake) makes about 7.5 hp per pound of air used. The Renesis is more efficient and makes about 8.5 hp per pound of air used. Keep in mind these are numbers that have been verified from engine dyno testing and come from both mass air flow measuuring but also with manomoeter measurements and conversions. We can see right off the bat that we need to flow more air to make the same power as an equivalent sized piston engine. Unfortunately our lack of efficiency is also going to hurt our mileage compared to other engines.

Let's plug in some numbers to test this so we can have a little fun. It's math time!

Here's a formula to know: ((Displacement in cu.in.) X (RPM)) / 3456

This is going to tell us how much cfm we could flow at 100% volumetric efficiency. Let's start with saying it's a 1.3L engine. That's 80 cu in. Lets also say we have a max of 8000 rpm. This gives us (80 X 8000) / 3456 = 185.185 cfm. No one is going to believe our engine flows that little amount of air. Let's try the 2.6 L number

(160 X 8000) / 3456 = 370.37 cfm. That's actually pretty close. For comparison sake I should have used 6500 as that's the power peak on an n/a 13B. That changes things to 300.93. Divide that by 14 (how many cu ft of air equals 1 lb) and we get 21.49. The last 13B was rated at 160 hp so we divide 160 by 21.49 to equal 7.45 hp per pound of air used. Yep that's pretty close to reality.

Now let's move on to the 3.9 liter size. (240 X 8000) / 3456 = 555.56 cfm. That may sound right to some but it actually isn't. Your engine can hit this efficiency level but it must be ported big. Even the math is wrong on that! Why? What does the 3456 stand for? 1728 is the number of cu in in 1 cubic foot. Remember we already know how many cu ft of air equals 1 pound of air. It's 14. 1728 X 2 = 3456. The 2 stands for how many times around the crankshaft rotates to get all of the engine displaced. For the 1.3L number I should have used 1728. For the 3.9L number, I should have used 5184. Let's try that again.

(240 X 8000) / 5184 = 370.37 cfm. Wait a second, that's what I got when I figured it out as a 2.6 liter! Engine rotation is absolutely important to know! Let's have some more fun. Let's try to get the Renesis number. It is rated at 238 from the factory. I know some claim it isn't doing it but it is EASILY capable of it! It's peak power is at 8500 rpm. I'll even do tha math according to the 3.9 number but I must use 1728 X 3 to compensate so I may as well just use 3456 and 2 instead. It's the same outcome.

(240 X 8500) / 5184 = 393.52. Divide this by 14 to get 28.11. Rated power is 238 so we divide it by 28.11 to get 8.47 hp per lb of air used. That's pretty much dead on to the test results. Hot damn math does work! You can find out how much air any engine uses according to this simple formula.

From an airflow standpoint the rotary is a 2.6 liter engine. Of course I've set up ecu's according to this number and they all work so what do I know?

I am arrogant enough to say I completely understand rotary airflow and size comparisons but then again when you are absolutely correct, you can afford to be a little arrogant!

 

Rotarygod, i've read many many many similar posts to yours. Everything in your post is based on the output shaft. Have a read over your post, the terms "per rev" get used a lot and there are lots of comparisons with rpm etc. I stated in my previous post, and it was also stated in the article from Karl Ludvigsen that this is not correct. You CANNOT use the output shaft as a means to compare rotary to piston motors. This is because it is possible to change what the output shaft does while still keeping the basic engine the same. Again, the simplest example of this is taking the drive off the camshaft of a piston motor, it is going to throw out all your figures. Likewise, taking the drive off the rotation of the rotor is going to change all your figures (making the 13b 7.8 litres I believe). Because of the 270 degree expansion as opposed to 180 what you measure at the shaft is out of wack. If you're going to use the output shaft you need to adjust for the 270 degree expansion.

There is only 1 correct answer and that is 3.9 litres for the 13b.

Trusting mazda's rating is like trusting a cigarette company that smoking doesn't cause cancer.

The 2.6 litres for an ECU works only by pure co-incidence. This is because the real capacity of 3.9litres * 180 / 270 = 2.6.

Your figures don't quite make sense there. HP is a rate of producing energy where pound of air is not a rate. We can't compare a quantity of air used to a rate. This is like saying it takes $1000 to run a company, which begs the obvious question for how long? Obviously the $1000 needs to be per some time frame. eg $1000 per day to run a company.

However, I do agree that the rotary is likely to be less efficient volumetrically and with mileage.

this is all a fancy way of looking at the amount of air injested per rev of the output shaft. We know that's going to be 2.6 litres per 2 revs. This is a fact and not up for dispute. What is up for dispute is whether this relates to the engine being 2.6 litres or not. I've already covered why it's not, you CANNOT use the output shaft as a method of comparison.

Setting up an ecu is not going to give you an understanding of why the engine is a certain capacity. In fact it's going to mislead you. Yes, the 2.6 litre figure works for the rotary but a 4.0 litre figure would also work for an s2000 with drive taken off the camshaft (try plugging the figures into all your formulae above and you'll see what I mean). That does not mean the s2000 is a 4 litre engine, it just means for the purpose of the ecu we could consider it as such.

I don't think you do. I have studied this very topic extensively as part of my mech eng degree and have studied it over the last 14 years on and off. I've done a fair bit of research into this topic and spent many hours searching for various articles in dusty library basements. I also wrote this software, which required a very intimiate knowledge of the maths involved. (http://mikesdriveway.com/engineapp). The problem is every man and his dog on newsgroups thinks he understand the issue absolutely. :-) While you have a much greater knowledge than most of the people on these groups you do not understand this issue completely.

 

If you (or anyone else) wants the article I was referring to by Karl Ludvigsen i'm happy to email it. I'd post it on a website but the article specifically states that it can be freely emailed but not made available via a link. Not sure why but that's what it says. My email is rx7club at mikesdriveway.com. If anyone is reading this in a years time or so and that email address is no longer valid (due to spam) check mikesdriveway.com for my new address or check the bounce notification.

Cheers,
Mike

 

while I dismissed rotarygod early on with his declared "6 stroke wankel" engine abusing the stoke term, i now find his argument is more reasonable. my take is that rotors and rods both drive a fundamental output shaft (not a cam shaft) that in it's most basic form will define the revs needed to complete one complete otto cycle for the total engine .... all piston or rotor faces. That shaft revs 3 turns for a wankel, 2 revs for boinger, a one rev for a 2-cycle engine.

Now, with no sae, astm , of jis standard to define displacement, you don't ignore the world domination boinger history. you leave simple swept volume (one way swept area pistons x #cylinders) for 2 and 4 stroke boingers, keeping 2 stoke underated. For the rotary, suck it in and get the 3.9L total cycle rating, and take 2/3 to match the boingers aspiration potential in 2 revs. Power edge goes to smokers, followed by boingers, trailed by wankels.

Now to sharped the pencil and match burn duration and firings per rev, add on a 1.5 speed increaser to a mega oversquare 3.9L v6, and you match the wankle otto cycle times at the altered output shaft on the V6..... lowered torque but same hp at 50% higher revs.

I was 1st drawn to the 1st gen rx7 as light weight coupe, 550/50 weight,and rwd with funky watts linkage. My 3rd gen was eventual follow up to seeing those gtp's finishing first at the 24Hr LeMans race.


Mike C, I'll pm you for the article.

 

I think the focus on size here is on the fact that an engine's total displacement is figured based on the total swept volume of all of it's chambers. A 2 stroke and 4 stroke engine are figured out this way and their rotation has nothign to do with it. I think that's what he's getting at. For a rotary that is in fact an answer of 3.9 liters. That's all fine and dandy and call it what you will, it doesn't flow the air of a 3.9 liter piston engine. At least not over the same amount of time. We can't have work, without having a time frame of reference. This is why torque is not a measure of work.

I understand what is being said in regards to crankshaft not having any effect on displacement. As far as is being concerned, it's not movement but rather how many faces combined with their total swept volume that is the only thing being considered. That's fine. On paper. Putting it into use is a different story altogether though. We need a time frame of reference to make a comparison. We need to displace a certain amount of air over a certain amount of time to base a comparison against other engines. To do this we need a common standard. Our amount of time is measured in revolutions per minute. Engine RPM is based off of eccentric or crankshaft revolution only. If we have 2 engines running, in order to compare them in terms of size, they must be at the same crank rpm. Remember this is a comparison of them from a functionality standpoint over time. How many chambers each one has is irrelevant. When you do a comparison in this manner, a rotary engine displaces the same amount of air per time as a 2.6 liter piston engine. You can absolutely compare them this way. This is not a coincidence. The rotary just doesn't use it as efficiently. From this standpoint, the engine is a 2.6 liter. This is verified with ecu tuning, carb sizing, etc... This is not a coincidence.

Mazda calls their engine a 1.3 liter and who are we to argue with Mazda?! Therefore it's a 1.3. Total swept displacement is a 3.9 liter. That answer is correct too!!! 3 correct answers.

In order to get an accurate answer to the displacement question, we need a frame of reference as there is always something that will not directly compare in the same manner. This is no different than comparing a 2 stroke to a 4 stroke engine. Sure we might say that engine A is a 2 liter 4 stroke and engine B is a 2 liter 2 stroke, but from a functionality standpoint over time, which one displaces more air? Perspective is needed. They obviously do a different amount of work per time. I know this is the comparison that is being made with the rotary and the piston engine. That's fine but it doesn't tell us everything and when it comes to tuning or figuring out intake or exhaust design, it doesn't tell us much. The nice thing about the formula I provided is that it takes time into account and is self adjusting. Just enter in the correct amount of time (rpm's to get all cylinders/ rotor faces fired).

FWIW: I went to school for ME as well.

 

A 4 stroke engine's displacement is measured on the completion of 1 otto cycle (all 4 strokes), which is 2 revs. If we use 2 revs for the rotary we are using only 2/3rds of the otto cycle.

This is exactly the same as measuring air inducted over 2 revs multiplied up by a factor. If the 13b inducts 2.6 litres over 2 revs then it will induct 260 litres at 200 rpm over a minute. Just the same figures multiplied up. Again, it's basing figures off the output shaft that could easily be geared up or down depending on the method used to take power from the engine. The output shaft is not a valid frame of reference. It's just like using the output of the gearbox as a frame of reference with the gearing fixed at the factory.

It not really a relevant comparison. We wouldn't compare a truck engine at 1800rpm to a 2 litre piston motor at 1800 rpm. We would compare at the max power point or redline. The rotary revs higher than pretty much every engine on the market by a significant margin and if we measured the max airflow we'd get a figure higher than all 2.6 litre engines on the market.

This is a co-incidence and why it is a co-incidence was covered in Karl Ludvigsen's article. It's basically because of the 270 degree stroke reducing the effectiveness of the 3.9 litres down to 2.6 litres over 2 revs. Email me for the article.

Mazda have a HUGE amount to gain from rating their engine lower than what it actually is. ANYTHING mazda say on the topic should be at least viewed with skepticism.

You're going back to using the output shaft without providing any reasons why using the output shaft is valid. I've provided reasons why using the output shaft is not a valid comparison but you have not responded to any of those reasons. I know full well that the 13b inducts 2.6 litres over 2 revs, and I know full well that it inducts the same as a 2.6 litre engine if both are at 5000 rpm. These are facts that cannot be disputed. What can be disputed is whether this relates to displacement. I've given reasons why these facts DON'T relate to displacement but you have not responded to those reasons.

You completed a bachelor of mechanical engineering?

 

I'm a Psychology major if that counts.

Anyways... why do we keep using piston engine measurements on our rotary engines? Is that necessary? Is there another way to do it?

 

rotarygod - lots of people argue with mazda.

Need I remind you what dynamometer tests and overly optimistic horsepower ratings have lead to?

 

The engines are actually very similar in many ways. I often say that the rotary and 4 stroke piston engine have more in common than a 4 stroke piston engine and a 2 stroke piston engine. This might sound a bit mad at first but if you ignore the shape of the chambers and look at both engines mathematically many similarities appear that the 2 stroke doesn't share.

The piston engine measurement of capacity actually applies very well to rotaries and does give an exact answer and only 1 answer as long as you don't use the output shaft as a means of comparison.

It's quite interesting to compare a 13B rotary to a 3.9 litre 6 cylinder piston motor that has a gearing up of 1.5 times applied at the output shaft. Because the piston motor has this gearing it appears to take 3 revs instead of 2 to complete an otto cycle, output torque is lower but output rpm is higher etc and pretty much every last detail you can measure is the same, eg:

both have 6 chambers/cylinders
both take 3 revs of the output shaft to complete an otto cycle
both fire twice per rev
both induct 2.6 litres of air per 2 revs
if you summed up the volumes of all 6 chambers/cylinders at any point in time they would both have a volume of half of 3.9 litres no matter what angle the crank was at.
if you graphed the individual volumes of each chamber/cylinder as the crank rotated both would draw the same graph.
both would have a stroke of 270 degrees.
both would appear to rev higher than most other engines on the market
both would have lower torque but feel like they have more than they are rated at.
both would have a lower otto cycle rate than output shaft rpm by a factor of 1.5. eg for both engines running at 9000 rpm they would cycle through 3000 otto cycles per minute per chamber.

On the other hand if you compare the rotary to a 2.6 litre piston motor then things are all over the place:
1 has 4 chambers, the other 6
1 takes 2 revs to complete an otto cycle, the other 3
both fire twice per rev
both induct 2.6 litre over 2 revs
the sum of volumes would be 1.3 for 1 and 1.85 for the other
the graphs of the individual volumes of each chamber by rpm would look nothing like each other.
1 has a stroke of 180 degrees and the other 270
1 revs much higher than the other
both might have similar torque.
Otto cycle rates would be the same, although not for each chamber.

(Edit)
I forgot to mention, with the comparison with the 2.6 litre piston motor the only similarities are the ones that were chosen to be similar anyway. For example, both induct 2.6 litres per 2 revs but we chose to compare with a 2.6 litre engine. They both fire twice per rev but we chose a 4 cylinder. The similar torque and otto cylces also relate to the figures chosen. Basically what I'm saying is that it couldn't be a worse comparison, only those figures that we set to match do match.

The other interesting thing to look at is the torque fluctuation graph. This is a graph of torque over 1 otto cycle. For example with a single piston engine there is a big spike in torque as that 1 piston fires. Over the next 3 cylcles there is negative torque as the piston uses energy to go through exhaust, induction and compression. With a 2 cylinder motor there are 2 torque peaks per otto cycle. For a 4 cylinder motor there are 4 peaks etc. Guess which engine the 13B's torque graph looks like? It's unmistakely a 3.9L 6 cylinder motor with a gearing up of 1.5 times. Again it looks nothing like a 2.6L 4 cylinder motor.

 

And rightly so. Mazda have a huge amount to gain from rating their engines at 1.3 instead of 2.6 or 3.9. I don't mean to pick on anyone but to believe mazda on this issue is pretty gullible.

I'm not sure what you mean there.

 

Mazda has over-rated its cars performance... at least in the states.

Also, all 4 stroke piston engines have to turn over TWICE to complete the otto cycle, not three... just fyi.

 

I've provided reason and proof for everything I've talked about. And who the hell is Karl Ludvigsen and how is he important to the rotary world? Is he just some guy that wrote a paper on the rotary? I've done that.

Yes, I went to school for ME at the University of Houston.

 

MikeC:
"You're going back to using the output shaft without providing any reasons why using the output shaft is valid. I've provided reasons why using the output shaft is not a valid comparison but you have not responded to any of those reasons"

I have provided sufficient definition, consistent with current engine designs, to use the true output shaft in displacement discussions. No slipery slope there. It allows simple ingested air per rev comparisons. But, logic would suggest the 4 stroke wankel be rated like .......... all other 4 stroke engines that dominate the earth, diesel or gas, regarding injested air in 2 revs (or total swept volume for boingers) .... 2.6L.

"if you graphed the individual volumes of each chamber/cylinder as the crank rotated both would draw the same graph." ref 3.9L 6 with 1.5 OD.

I would be surprised if this is true, you are saying the slider-crank kinematics give same dynamic rotary volume history, relative to point in the cycle.

 

It wouldn't.

 

Not really, all you've proved is the rotary inducts 2.6 litres of air per 2 revs. We knew that already.

He is a well published automotive author. Check amazon for the 200+ results.

Can you post a link?

You didn't answer my question, did you complete a bachelor of mechanical engineering degree?

 

I wasn't referring to you but .... why should we use the 2 revs method? That was chosen because it's the time taken to complete all 4 strokes for each piston. The rotary does not complete all 4 strokes in 2 revs. Some piston motors use a different method to take power and the engine completes all 4 strokes in 1 rev of the output shaft. Should that engine be rated at twice the capacity of what it really is?

For the rotary the volume in the chamber follows a pure sine wave. The piston motor is slightly off a pure signwave but only by a small amount. Besides this minor difference the 2 engines follow the same volumes if the output shaft of the piston motor is also geared up by 1.5 times. Download my software, you can see it happening in real time. The volumes in the chambers are always equal at every point in time between a piston motor and a rotary. The is for a 3.9 litre 6 cylinder motor. You can also see the comparison between the 2.6 litre 4 cylinder motor and notice that nothing similar is happening at all.

http://mikesdriveway.com/engineapp

 

Download my software and see for yourself in real time. A 3.9 litre 6 cylinder piston motor and 13B follow each other exactly over time. This is assuming the piston motor has a 1.5x gearing applied at the output shaft.

 

I think you missed my point. There are many 4 cyl. with 4.10 final diffs. Factory. And, with simaler trans gearing.

The "practical" application is very simalar to low displacement six cyl. or high displacement 4 cyl.

The formula seems to be working.

 

The 4.1 diff is common in 4s due to the lack of power. I doubt you get any cars on the market with 250hp factory and a 4.1 diff. I doubt you get any that are even close. Except maybe the s2000 which seems to be the 1 anomoly.

 

As I said before, it's because one way or another most gas or diesel 4-strokesthat dominate the roads use this method, and the rotary is a 4 stroke.

If it's just an acedemic exercise, you can go with ar injested in one rev, or total swept volume in a complete engine cycle (unofficial piston engine definitins), and correct later for differences in total engine cycle.

Dealing with the real histoic ratings of 2 cycle and 4 cycle motor, the 2.6 13B is same terminology as the 5.0 stang, and has same disadvantage to a 5.0 or 2.6L two cycle smoker. I'm not in searce of a new system that confuses historyic data and records, but one for the 13B that belnds in nicely.