Porting, VE, Intake, & a little Math...
02-15-08, 07:44 AM
#1
Nikki-Modder Rex-Rodder
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Porting, VE, Intake, & a little Math...
Will someone do us all a favor, and list all the various porting configurations for both the 12a & 13b engines, and give the corresponding changes to any variables they believe to be changed within the engine that make it more powerful.
I'm talking about a genuine measurable variable used in an engine to ultimately calculate it's output.
What changes? I'm just curious as to what everyone thinks, before I continue...
I'm talking about a genuine measurable variable used in an engine to ultimately calculate it's output.
What changes? I'm just curious as to what everyone thinks, before I continue...
Last edited by Sterling; 02-15-08 at 07:56 AM.
02-15-08, 10:41 AM
#2
Sterling,
Not exactly sure what we're going for here, but this should at least be a start on the basics.
http://www.turborx7.com/portingpictures.htm
I have no doubt that the more knowledgeable members will show up and fill in the gaps.
Jamie
Not exactly sure what we're going for here, but this should at least be a start on the basics.
http://www.turborx7.com/portingpictures.htm
I have no doubt that the more knowledgeable members will show up and fill in the gaps.
Jamie
02-15-08, 04:24 PM
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I'll bite.
The goal with any of the intake porting styles is to improve VE which is directly related to torque. This is done by extending the opening duration of the port and by improving the flow characteristics of the port and runner. This allows more air to flow into the engine and increases VE.
There is another benefit. As the ports get larger and open earlier, more overlap is created between the intake and exhaust port. This improves chamber filling at mid to high rpm as the vacuum generated by the exhaust pulse helps pull the intake charge into the chamber which increases VE even further.
The goal with any of the intake porting styles is to improve VE which is directly related to torque. This is done by extending the opening duration of the port and by improving the flow characteristics of the port and runner. This allows more air to flow into the engine and increases VE.
There is another benefit. As the ports get larger and open earlier, more overlap is created between the intake and exhaust port. This improves chamber filling at mid to high rpm as the vacuum generated by the exhaust pulse helps pull the intake charge into the chamber which increases VE even further.
02-15-08, 06:33 PM
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I've seen 60% for a stock port up to over 100% for a peripheral.
To do it properly you would need to measure the actual airflow with a mass airflow meter.
VE = measured airflow / theoretical airflow x 100
You can also calculate it if you have access to dyno charts and some brake specific fuel consumption figures.
To do it properly you would need to measure the actual airflow with a mass airflow meter.
VE = measured airflow / theoretical airflow x 100
You can also calculate it if you have access to dyno charts and some brake specific fuel consumption figures.
02-15-08, 06:46 PM
#6
Nikki-Modder Rex-Rodder
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-Any hard reference material to substantiate this, REVHED?
What I'm really looking for is a comprehensive list of the VEs of various porting configurations for the 12a & 13b.
I can't seem to find a collective database that has this info.
You say 60% VE for a stockport while Craig's Page says 70%.
I think that for the purpose of where I'm going with this, most of us can agree that even the best super-ports probably wouldn't get over 110%.
And where I'm going with this is right down there in my new sig; to calculate the actual air/fuel mixture requirements of any rotary engine at a given RPM and VE. The equasion should illustrate that no matter how generously you might inflate those values (without getting downright impossible), most people will be quite surprised at the actual size of the carburetor they really need to fully satisfy the requirements of their engine.
What I'm really looking for is a comprehensive list of the VEs of various porting configurations for the 12a & 13b.
I can't seem to find a collective database that has this info.
You say 60% VE for a stockport while Craig's Page says 70%.
Originally Posted by Graig's Page
Rotarys make high power for their size because of their volumetric efficiency, which is a measure of the volume of air going into the engine divided by the capacity of the engine. (for example a 1.0 liter engine at 100% volumetric efficiency intakes 1.0 liters of air per revolution). Obviously, more air going in means more fuel and thus more power.
Most stock piston engines only have a VE of 50-60%. A stock rotary is about 70%. Peripheral ported rotarys can go over 100% (which means that it is more air than the engine's volume). This phenomemon does not occur in normally aspirated piston engines (except in highly modified race engines). It does occur in turbocharged engines fairly often though.
Most stock piston engines only have a VE of 50-60%. A stock rotary is about 70%. Peripheral ported rotarys can go over 100% (which means that it is more air than the engine's volume). This phenomemon does not occur in normally aspirated piston engines (except in highly modified race engines). It does occur in turbocharged engines fairly often though.
And where I'm going with this is right down there in my new sig; to calculate the actual air/fuel mixture requirements of any rotary engine at a given RPM and VE. The equasion should illustrate that no matter how generously you might inflate those values (without getting downright impossible), most people will be quite surprised at the actual size of the carburetor they really need to fully satisfy the requirements of their engine.
02-15-08, 07:01 PM
#7
Absolute Power is Awesome
VE isn't static, it changes with RPM, what you end up with is a curve and it will depend on the entire system, intake inlet to exhaust pipe tip. There's some info in Yamamoto that might be helpful.
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02-15-08, 07:26 PM
#8
Nikki-Modder Rex-Rodder
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I had a chart of a 13b FI that had the VE curve on it, but I can't seem to locate it anywhere.
02-15-08, 07:40 PM
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http://not2fast.com/turbo/glossary/turbo_calc.shtml
i'm coming up with 80% ish numbers for a stock 12a, 110% ish for a 12a P port.
the two important things are that the stock port makes peak power at 6000rpms, and the p port makes power @9k. so its not only 30%, but 30% and 3000rpms HIGHER in the rev band.
also the porting style affects the BSFC, worst ever is like an s5 13b, stock they run way rich, bsfc is like .8... P ports are like .5, very good.
so thats my guesstimates, ive looked at a lot of dyno sheets, good non chassis dyno ones, will have the bsfc numbers
i'm coming up with 80% ish numbers for a stock 12a, 110% ish for a 12a P port.
the two important things are that the stock port makes peak power at 6000rpms, and the p port makes power @9k. so its not only 30%, but 30% and 3000rpms HIGHER in the rev band.
also the porting style affects the BSFC, worst ever is like an s5 13b, stock they run way rich, bsfc is like .8... P ports are like .5, very good.
so thats my guesstimates, ive looked at a lot of dyno sheets, good non chassis dyno ones, will have the bsfc numbers
02-15-08, 07:41 PM
#10
Absolute Power is Awesome
You can back calculate it based on a torque curve from a dyno. It's not exact, but it might help. These can start you off.
http://ezinearticles.com/?Volumetric...ained&id=47639
http://www.installuniversity.com/ins...n_9.012000.htm
It's a fairly easy job to make up a spreadsheet that converts a torque curve to VE curve.
http://ezinearticles.com/?Volumetric...ained&id=47639
http://www.installuniversity.com/ins...n_9.012000.htm
It's a fairly easy job to make up a spreadsheet that converts a torque curve to VE curve.
Last edited by purple82; 02-15-08 at 07:49 PM.
02-16-08, 12:23 AM
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what units for your formula?
Is this based on sae in/lb or metric?
Do you use the fictitious 70 cu in for a 12A or the real equivalent 140 cu in?
(1.15 liters or 2.3 liters) Is the result in CFM at 1.5 in vacuum?
Do you use the fictitious 70 cu in for a 12A or the real equivalent 140 cu in?
(1.15 liters or 2.3 liters) Is the result in CFM at 1.5 in vacuum?
Originally Posted by Sterling
Displacement x Maximum RPMs x VE + 15% = Carb Size
02-16-08, 12:42 AM
#12
love the braaaap
Since double the displacement is generally the accepted proper airflow displacement, I would think for a 12A you would use 140 cu in and for a 13B you would use 160 cu in. But even that may be off.
Using the above calulation and a 70% VE with 140cu in at 7000rpm, a 12A needs about 482 CFM, not far off what most people consider optimum. A 13B works out to 550cfm under the same 70% and 7000rpm.
Using the above calulation and a 70% VE with 140cu in at 7000rpm, a 12A needs about 482 CFM, not far off what most people consider optimum. A 13B works out to 550cfm under the same 70% and 7000rpm.
Last edited by 85rotarypower; 02-16-08 at 12:51 AM.
02-16-08, 01:28 AM
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so a 12a is 140cu in
02-16-08, 06:52 AM
#14
Nikki-Modder Rex-Rodder
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For the purposes of my equation, the "real" 70 cubic inches is used.
OMG- Is THIS why everyone insists on overcarbing their rotary?
02-16-08, 06:58 AM
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The simple equation I came up with applies directly to the rotary engine. Nothing has to be halved or multiplied by two. Mazda rated these engines in cc, and THAT"S WHAT THEIR DISPLACEMENT IS!
No need to bring the complications of the piston engine into the thread to muddy up a simple, straight forward equation meant for the rotary.
02-16-08, 08:03 AM
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iam pretty sure "a piston engine fires its rated displacment in 2 rotations of its crankshaft"i belive its 4 isnt that why they call them 4 stroke motors. iam like 75% sure its 4 i rember a interview with the guy from banks .
i have heard this 1 rotation of a rotor makes the e shaft turns 3 times.
correct me if iam wrong.
i have heard this 1 rotation of a rotor makes the e shaft turns 3 times.
correct me if iam wrong.
02-16-08, 08:11 AM
#17
Nikki-Modder Rex-Rodder
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I updated the formula in my sig. I had forgotten the conversion from cubic inches to cubic feet.
The eccentric shaft spins three times for every one rotation of the rotor.
That means in one rotation of the eccentric shaft, the two rotors have each taken in one intake charge.
In a 12a engine, that's a theoretical intake charge of 2 x 35 cubic inches.
The piston engine formulas for carburetor sizing take into account that half of the strokes are not intake strokes.
They do this by calculating theoretical carb size the way I have, and then dividing that by 2.
Typically they do this within a simplified formula by multiplying the conversion factor from cubic inches to cubic feet by 2.
So where I use 1728, which is the number of cubic inches in a cubic foot (12 x 12 x 12), they typically use 3456, which is 2 cubic feet.
-They also are not adding a percentage at the end of the equation. I only do that to further bolster the fact that people are over-carbing their rotaries, because by doing the math and even adding an unnecessary, generous percentage, the engines requirements are shown not to meet the size of carbs people are installing.
The eccentric shaft spins three times for every one rotation of the rotor.
That means in one rotation of the eccentric shaft, the two rotors have each taken in one intake charge.
In a 12a engine, that's a theoretical intake charge of 2 x 35 cubic inches.
The piston engine formulas for carburetor sizing take into account that half of the strokes are not intake strokes.
They do this by calculating theoretical carb size the way I have, and then dividing that by 2.
Typically they do this within a simplified formula by multiplying the conversion factor from cubic inches to cubic feet by 2.
So where I use 1728, which is the number of cubic inches in a cubic foot (12 x 12 x 12), they typically use 3456, which is 2 cubic feet.
-They also are not adding a percentage at the end of the equation. I only do that to further bolster the fact that people are over-carbing their rotaries, because by doing the math and even adding an unnecessary, generous percentage, the engines requirements are shown not to meet the size of carbs people are installing.
Last edited by Sterling; 02-16-08 at 08:20 AM.
02-16-08, 09:18 AM
#18
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http://www.musclecarclub.com/library...rburetor.shtml
CFM (amount of air the engine needs) = Displacement (in cubic inches) X Maximum RPM / 3,456
http://www.mortec.com/carbtip1.htm
CFM = ( Cubic Inches x RPMs / 3456 ) x VE%
http://www.4secondsflat.com/Carb_CFM_Calculator.html
-a Javascript carb calculator that automatically doubles displacement to accomidate piston engines
http://golenengineservice.com/calc/calccarb.htm
-another of the same
http://www.rdrop.com/~/larry/download/formulas.pdf
Engine size (cid) x maximum RPM / 3456 = CMF
http://www.nightrider.com/biketech/c...mulas_page.htm
-for bikes; racing carb cfm = rpm x displacement / 3456 x 1.1
http://users.bigpond.net.au/webstuff...selection.html
(CID x RPM) / 3456 = max CFM
...need I go on?
Here, for those who still aren't quite on board...
http://web.archive.org/web/200005261...r/12ptdyn.html
This is the dyno for street ported 12a with a Yaw Super Street Nikki.
Note the peak VE, and the peak CFM, as well.
What all this illustrates is that a Holley 600 CFM is absolutely, inarguably, too damned big for a 12a or a 13b rotary engine.
G'ahead and do the math for a bridge-ported 13b with pinned rotors that spins to 10,000 RPM and a VE of 110%, and then the Holley 600 makes sense.
For more realistic rotary configurations, using an oversized Holley, where the power band cannot be tuned by the user, it may perform particularly well in a narrow band up high, but it will downright suck everywhere else. You may be able to learn how to "drive around" those short comings and exploit the narrow performance band a Holley gives by changing your style, but putting an oversized carb on an engine is a guaranty that there will be very little low end power.
My point here is not to sell carbs. Go get a Holley, I don't give a **** who you get your carb from. But at least get the right damned size, and stop telling newbies who are confused about rotary engine displacement to get carburetors sized for piston engines.
CFM (amount of air the engine needs) = Displacement (in cubic inches) X Maximum RPM / 3,456
http://www.mortec.com/carbtip1.htm
CFM = ( Cubic Inches x RPMs / 3456 ) x VE%
http://www.4secondsflat.com/Carb_CFM_Calculator.html
-a Javascript carb calculator that automatically doubles displacement to accomidate piston engines
http://golenengineservice.com/calc/calccarb.htm
-another of the same
http://www.rdrop.com/~/larry/download/formulas.pdf
Engine size (cid) x maximum RPM / 3456 = CMF
http://www.nightrider.com/biketech/c...mulas_page.htm
-for bikes; racing carb cfm = rpm x displacement / 3456 x 1.1
http://users.bigpond.net.au/webstuff...selection.html
(CID x RPM) / 3456 = max CFM
Originally Posted by "How to Build & Power Tune Holley Carburetors", by Des Hammill
CFM CALCULATION
1 - Multiply half the cubic inches of the engine's total capacity by the maximum rpm to be used.
2 - Divid this answer by 1728.
3 - Multiply this answer by the efficiency rating of the engine (e.g. - 75% effiency meaning that .075 is the multiplication factor)
4 - The answer is the cubic feet of air per minute (CFM)that the engine can flow, and the choice of carburetor is based on this amount of airflow.
If you're wondering why just half the engine's capacity is used in the calculation, it's because we're dealing with four stroke (otto cycle) engines, and only every second revolution of the engine per cylider is to do with drawing air into the engine. The next part of the calculation divides the answer by 1728 because 1728 is the number of cubic inches in a cubic foot. The final part of the basic calculation is multiplying this answer by the volumetric efficiency rating of the engine and this is where the gross errors are likely to creep in. The efficiency of an engine should be realistic, not over rated.
1 - Multiply half the cubic inches of the engine's total capacity by the maximum rpm to be used.
2 - Divid this answer by 1728.
3 - Multiply this answer by the efficiency rating of the engine (e.g. - 75% effiency meaning that .075 is the multiplication factor)
4 - The answer is the cubic feet of air per minute (CFM)that the engine can flow, and the choice of carburetor is based on this amount of airflow.
If you're wondering why just half the engine's capacity is used in the calculation, it's because we're dealing with four stroke (otto cycle) engines, and only every second revolution of the engine per cylider is to do with drawing air into the engine. The next part of the calculation divides the answer by 1728 because 1728 is the number of cubic inches in a cubic foot. The final part of the basic calculation is multiplying this answer by the volumetric efficiency rating of the engine and this is where the gross errors are likely to creep in. The efficiency of an engine should be realistic, not over rated.
Here, for those who still aren't quite on board...
http://web.archive.org/web/200005261...r/12ptdyn.html
This is the dyno for street ported 12a with a Yaw Super Street Nikki.
Note the peak VE, and the peak CFM, as well.
What all this illustrates is that a Holley 600 CFM is absolutely, inarguably, too damned big for a 12a or a 13b rotary engine.
G'ahead and do the math for a bridge-ported 13b with pinned rotors that spins to 10,000 RPM and a VE of 110%, and then the Holley 600 makes sense.
For more realistic rotary configurations, using an oversized Holley, where the power band cannot be tuned by the user, it may perform particularly well in a narrow band up high, but it will downright suck everywhere else. You may be able to learn how to "drive around" those short comings and exploit the narrow performance band a Holley gives by changing your style, but putting an oversized carb on an engine is a guaranty that there will be very little low end power.
My point here is not to sell carbs. Go get a Holley, I don't give a **** who you get your carb from. But at least get the right damned size, and stop telling newbies who are confused about rotary engine displacement to get carburetors sized for piston engines.
02-23-08, 09:43 AM
#21
Nikki-Modder Rex-Rodder
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e=mc2
e = Energy
m = Mass
c = The speed of light (299,792,458 meters per second)
...but no, I can't really explain how it works.
e = Energy
m = Mass
c = The speed of light (299,792,458 meters per second)
...but no, I can't really explain how it works.
02-23-08, 10:28 AM
#24
wow, one of the best info that I've read on the internet.
I do have a question that might be relevant to this whole thing concerning 2 barrel carbs/ ITB. How would you calculate what diameters would be good for what engine porting. I don't really see much CFM numbers for 2 barrels.
I do have a question that might be relevant to this whole thing concerning 2 barrel carbs/ ITB. How would you calculate what diameters would be good for what engine porting. I don't really see much CFM numbers for 2 barrels.
Last edited by dj55b; 02-23-08 at 10:35 AM.
02-23-08, 10:42 AM
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Hmmm well if you're looking for a general equation, let's think:
Output = Work - friction
So, power in Kilowatts is going to be a function of the combustion energy minus the drivetrain and internal friction loss of the engine.
Let's assume all of the work loss is the same over different engines, and you have a function relating output to combustion energy.
the equation for work of a shaft is 2*pi*(rotations per second)*(torque in KiloNewton meters)*(reference time in seconds)
So the enthalpy (energy) in joules per meter of the system is equal to (internal energy in joules) plus pressure*volume
You can see that increasing pressure can increase output energy, such as turbocharging.
However, increasing the air/fuel mix will increase internal energy of the system by adding to potential combustion energy (again in joules)
Energy contained in gasoline is .032 megajoules per millileter..
Energy of added oxygen gas is, let's say 6.82 Kilojoules per mol...
That's as far as I want to go right now..
Output = Work - friction
So, power in Kilowatts is going to be a function of the combustion energy minus the drivetrain and internal friction loss of the engine.
Let's assume all of the work loss is the same over different engines, and you have a function relating output to combustion energy.
the equation for work of a shaft is 2*pi*(rotations per second)*(torque in KiloNewton meters)*(reference time in seconds)
So the enthalpy (energy) in joules per meter of the system is equal to (internal energy in joules) plus pressure*volume
You can see that increasing pressure can increase output energy, such as turbocharging.
However, increasing the air/fuel mix will increase internal energy of the system by adding to potential combustion energy (again in joules)
Energy contained in gasoline is .032 megajoules per millileter..
Energy of added oxygen gas is, let's say 6.82 Kilojoules per mol...
That's as far as I want to go right now..