Rotary capacity, yet again ....
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Originally Posted by KevinK2
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.
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.
All you're really saying is that the way it's done because that's the way it's done. I've posted several reasons why it shouldn't be done that way yet no one has responded to those reasons. Plenty of piston motors don't use the same method of taking power from the pistons. Some don't have cranks at all, other's complete a full otto cycle in only one rev using a cam method. Other piston motor's complete a full otto cycle in 3 revs just like the rotary. But none of them are rated according to the 2 revs method. The manufacturers don't even try. Many engines have even included an internal gearing to gear up or down the output shaft but are still rated according to swept vol by no of pistons.
The funny thing is the rotary even includes a gearing inside of a factor of 3:2 :-)
Originally Posted by MikeC
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 still want to know what importance Karl Ludvigsen has to the rotary word? A well know automotive journalist is not an important item on a resume when it comes to rotaries. There are many automotive writers that don't know crap about rotaries but write about them like they do. I'm getting the sense that he's probably wrong in what he wrote if you are defending him. Karl Ludvigsen is a nobody. Sorry.
Last edited by rotarygod; Nov 8, 2006 at 10:54 AM.
Originally Posted by rotarygod
I still want to know what importance Karl Ludvigsen has to the rotary word?
He's far more accomplished than anyone on this site (that I know of).
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Originally Posted by MikeC
............ Plenty of piston motors don't use the same method of taking power from the pistons. Some don't have cranks at all, other's complete a full otto cycle in only one rev using a cam method. Other piston motor's complete a full otto cycle in 3 revs just like the rotary. But none of them are rated according to the 2 revs method. The manufacturers don't even try. Many engines have even included an internal gearing to gear up or down the output shaft but are still rated according to swept vol by no of pistons.
..........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.
..........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.
I would like to see details of these exceptions, if you have links or references, esp the 3 rev otto cycle.
The software is excellent, but the illustrations do not verify how close the volume change is in the case we discussed.
----------
to say "2 rev method" as I do is an oversimplification.
As I posted way back when I still had teeth (joke), piston pumps preceeded piston engines, and, I believe, are the basis of common total swept volume ratings, but the pumps had a one rev cycle at the time. The adopted math for enines is holes times area, but reflects what would be pumped by a similar pump with poppet valves in one rev. If you look at a 13B that way, you have to envision check valve'd housing ports, with 1 added pair of ports per housing near the spark plug location, to make it a pump. The "old school" pump-analogy displacement would be what it would pump in one rev, if converted to a pump:
1/3 cycle x 2 x 3.9L = 2.6L.
I think this is also the basis of the 2 stroke underating, as it was considered a "holes times area" for old fashion pump like displacement rating.
Last edited by KevinK2; Nov 8, 2006 at 11:40 AM.
Originally Posted by My5ABaby
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I’ve been following this debate with amusement for a while. Clearly, as a number of the folks on this thread have demonstrated, calculating “the displacement” of a rotary engine can be accomplished any number of different ways. Worse, these calculations can result in different answers, depending upon how you define the term displacement, whether you’re trying to make an apples-to-oranges comparison to piston engines, etc.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
-Mark in Tucson
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
-Mark in Tucson
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
Last edited by mhwarner; Nov 8, 2006 at 12:32 PM.
Originally Posted by rotarygod
He's written tons of books. Lots of people have. How is he a rotary expert again? I don't see a rotary book there. He's never been mentioned with the rotary in any way important in history. Chances are he wrote an article for a magazine about the rotary because he was asked to and he wrote what he learned about it in the short time he did his research on the topic. I'm sorry. When it comes to rotaries, he's nobody important. 65 more books on other cars won't change that.
Originally Posted by mhwarner
I’ve been following this debate with amusement for a while. Clearly, as a number of the folks on this thread have demonstrated, calculating “the displacement” of a rotary engine can be accomplished any number of different ways. Worse, these calculations can result in different answers, depending upon how you define the term displacement, whether you’re trying to make an apples-to-oranges comparison to piston engines, etc.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
-Mark in Tucson
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
-Mark in Tucson
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
Originally Posted by My5ABaby
I never said he was a rotary expert. But, outside of Mazda, who is? You?
Originally Posted by rotarygod
And it's been what I've been saying the whole time. He's correct.
Rotarygod, I wonder, using the mass airflow test, what would be the effective displacement of a 3.9 L 6-cyl, with a .67 OD gear before the output shaft as measured by the meter? How much effective displacement would you tune your ECU for?
I wouldn't be surprised if a turbo sized for a 2.6 L piston engine would be just as effective for a 3.9 L piston engine with a 0.67 OD gear attached before the output shaft.
Originally Posted by mhwarner
I’ve been following this debate with amusement for a while. Clearly, as a number of the folks on this thread have demonstrated, calculating “the displacement” of a rotary engine can be accomplished any number of different ways. Worse, these calculations can result in different answers, depending upon how you define the term displacement, whether you’re trying to make an apples-to-oranges comparison to piston engines, etc.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
-Mark in Tucson
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
-Mark in Tucson
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
Last edited by Roen; Nov 8, 2006 at 05:09 PM.
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Originally Posted by rotarygod
You still don't get it. If you changed to piston motor gearing to the output shaft, you've changed the way the engine works. It's not the same motor anymore! You absolutely, positively, must factor in crank rotation. You keep saying otherwise but you quite frankly are as dead wrong as they come. Your program is written based on a false assumption. You are cheating to get the results you want. You're just trying to justify it. I can't believe you are defending this position. It's wrong to the point of embarassment, especially as an ME student. I'm going to forward this on to my friend Rick (Engman) in Atlanta.
Originally Posted by rotarygod
I still want to know what importance Karl Ludvigsen has to the rotary word? A well know automotive journalist is not an important item on a resume when it comes to rotaries. There are many automotive writers that don't know crap about rotaries but write about them like they do. I'm getting the sense that he's probably wrong in what he wrote if you are defending him. Karl Ludvigsen is a nobody. Sorry.
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Originally Posted by rotarygod
He's written tons of books. Lots of people have. How is he a rotary expert again? I don't see a rotary book there. He's never been mentioned with the rotary in any way important in history. Chances are he wrote an article for a magazine about the rotary because he was asked to and he wrote what he learned about it in the short time he did his research on the topic. I'm sorry. When it comes to rotaries, he's nobody important. 65 more books on other cars won't change that.
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Originally Posted by mhwarner
I’ve been following this debate with amusement for a while. Clearly, as a number of the folks on this thread have demonstrated, calculating “the displacement” of a rotary engine can be accomplished any number of different ways. Worse, these calculations can result in different answers, depending upon how you define the term displacement, whether you’re trying to make an apples-to-oranges comparison to piston engines, etc.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
With that said, let me add some fuel to the fire. Putting aside the question of “displacement” for a moment, let’s just look at the amount of air that moves through a typical 13b engine at a given RPM (and, no, I’m not going to artificially add 1.5x gear multipliers, drive off of camshafts, or any other nonsense to try to compare the results to a piston engine; I’m just going to look at pure volumetric air flow.)
Assuming 100% VE, one face of a 3-sided 13b-sized rotor will ingest ~654cc during its intake phase. There are six of these faces ingesting air as the engine runs at a given eccentric output shaft RPM. If you work through the math at, say, 1000 RPM, a 13b engine will ingest 1308 liters of air per minute, or ~46 cubic feet per minute (CFM).
This volumetric flow rate is really the main thing anyone should care about. For example, if you were sizing a turbocharger for the 13b, you wouldn’t veer off on the tangent of trying to figure out its piston-engine-equivalent displacement before sizing the turbo, nor would you artificially add gear box multipliers or any other silliness. Instead, you would just want to know directly what the total CFM flow rate was for any given engine operating RPM. Period.
Now, if you absolutely won’t give up the fight and you have to know what size piston engine ingests this same amount of air at this same output shaft RPM, the calculation is trivially easy. The result, of course, is a 2616cc engine. Not surprisingly, turbochargers that work well for high-revving 2.6-liter piston engines work equally well for 13b rotaries.
Now, there is not one single 2.6 litre 4 cylinder engine on the market that revs anything like the rotary. Remember the rx8 goes to 9000rpm stock. The is a huge amount of revs for a factory engine, especially one with 654cc swept volume per chamber. If we used the turbo for a 2.6 litre standard revving 4 cylinder engine it would be way short of what a rotary needs. Because the rotary revs close to 1.5x higher surely a turbo suited to a standard 3.9 litre motor would be more what it required.
Originally Posted by mhwarner
-Mark in Tucson
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
PS. No, I'm not a rotary engine expert, but, yes, I'm an automotive author, a member of SAE, and a licensed mechanical engineer.
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Originally Posted by slo
Hey Mike C why don't you either post the article or post a link to the article. I'd like to read it.
PS, I get a lot of spam and sometimes accidentally delete genuine email, if you don't get a reply within 12 hours please email me again.
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Originally Posted by KevinK2
Q Plenty .... what % of total physical existing piston engines are "plenty", knowing there are millions of 4 stroke diesels and si engines on the road and in industry? What production engine that people buy takes power off a cam shaft, vs the crank shaft? That is no different than taking power off the waterpump, oilpump, etc.
"Many engines have been built with internal gearing (aviation engines and Alfa Romeo, Mercedes-Benz, Porsche and BRM Grand Prix engines) and no one has even suggested that their gearing should have the slightest effect on their rated displacement."
Originally Posted by KevinK2
I would like to see details of these exceptions, if you have links or references, esp the 3 rev otto cycle.
Originally Posted by KevinK2
The software is excellent, but the illustrations do not verify how close the volume change is in the case we discussed.
Originally Posted by KevinK2
to say "2 rev method" as I do is an oversimplification.
As I posted way back when I still had teeth (joke), piston pumps preceeded piston engines, and, I believe, are the basis of common total swept volume ratings, but the pumps had a one rev cycle at the time. The adopted math for enines is holes times area, but reflects what would be pumped by a similar pump with poppet valves in one rev. If you look at a 13B that way, you have to envision check valve'd housing ports, with 1 added pair of ports per housing near the spark plug location, to make it a pump. The "old school" pump-analogy displacement would be what it would pump in one rev, if converted to a pump:
1/3 cycle x 2 x 3.9L = 2.6L.
I think this is also the basis of the 2 stroke underating, as it was considered a "holes times area" for old fashion pump like displacement rating.
As I posted way back when I still had teeth (joke), piston pumps preceeded piston engines, and, I believe, are the basis of common total swept volume ratings, but the pumps had a one rev cycle at the time. The adopted math for enines is holes times area, but reflects what would be pumped by a similar pump with poppet valves in one rev. If you look at a 13B that way, you have to envision check valve'd housing ports, with 1 added pair of ports per housing near the spark plug location, to make it a pump. The "old school" pump-analogy displacement would be what it would pump in one rev, if converted to a pump:
1/3 cycle x 2 x 3.9L = 2.6L.
I think this is also the basis of the 2 stroke underating, as it was considered a "holes times area" for old fashion pump like displacement rating.
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Originally Posted by Roen
I'm lost as to why the it's irrelevant what the gearing is for the turbocharged application? Someone want to explain that part to me?
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Originally Posted by My5ABaby
I also have the article (trying to claim it MikeC? j/k).
Now... I'm off to go look up Wankel's article and see what he says!
Now... I'm off to go look up Wankel's article and see what he says!
Originally Posted by MikeC
Sorry :-) I have kind of run amock with the article you sent me :-) If you've got any other's I'd be very interested to see them. :-)
The problem with this thinking is again it doesn't take into account the gearing at the output shaft. We could gear the output shaft down by a factor of 10:1 and suddenly a 500cc piston would look like a 5000cc piston. But this isn't going to pump anything like a real 5litre piston.