basic layout of 2.8 engine
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basic layout of 2.8 engine
How many times does the crank turn per firing order? I know most v-8's turn 2 revs per firing order, what about v-6's?
(edited to remove wrong statment concerning I-4 engines)
(edited to remove wrong statment concerning I-4 engines)
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thanks...
but additional question then, why do inductive tach's have a switch for the # of cylinders?
The goal of all this mental masturbation is to determine in CFM the air requirements at a given engine speed, any info would be appreciated.
The goal of all this mental masturbation is to determine in CFM the air requirements at a given engine speed, any info would be appreciated.
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must be trying to read ALL cylinders, not just one
Tachs and dwell meters (what?) that are driven from the low side of the coil see all voltage pulses for all 4,6,8 cylinders and so must know how many there are in order to divide the input properly.
I am not sure where your inductive tach hooks up, but if it is between the electronic ignition and coil, or between coil and distributor it would be the same.
I am not sure where your inductive tach hooks up, but if it is between the electronic ignition and coil, or between coil and distributor it would be the same.
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CFM needs
The formula for a four stroke engine would be:
(RPM )
(------) times engine capacity (cubic inches) = CFM, assuming 100% VE
(3456 )
0R: (RPM X engine capacity)/3456
If you only have engine capacity in CCs, divide by 16.4 to get cubic inches.
As a by the way, no street engine will have 100% volumetric efficiency at either peak power or peak rpm. *Some* (but not many) street engines can get close to 100% volumetric efficiency at peak torque.
A good rule of thumb is to figure around 85% VE at peak power (or 90% if you must, but absolutely no more than that), so your calculated cfm should be reduced appropriately.
Therefore, a 255 cubic inch Audi engine (the 4.2), will look like this at redline:
6500
------ = 1.88076, times 255 = 480 CFM, times 85% (assumed VE) = 408 CFM
3456
(RPM )
(------) times engine capacity (cubic inches) = CFM, assuming 100% VE
(3456 )
0R: (RPM X engine capacity)/3456
If you only have engine capacity in CCs, divide by 16.4 to get cubic inches.
As a by the way, no street engine will have 100% volumetric efficiency at either peak power or peak rpm. *Some* (but not many) street engines can get close to 100% volumetric efficiency at peak torque.
A good rule of thumb is to figure around 85% VE at peak power (or 90% if you must, but absolutely no more than that), so your calculated cfm should be reduced appropriately.
Therefore, a 255 cubic inch Audi engine (the 4.2), will look like this at redline:
6500
------ = 1.88076, times 255 = 480 CFM, times 85% (assumed VE) = 408 CFM
3456
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Re: CFM needs
Well... you should clarify here. Virtually no _naturally aspirated_ street engine will have 100% volumetric efficiency. A turbo/supercharged engine does though... and I'm guessing some of the bikes with ram air are around there when they're running 150mph+...
-Craig
-Craig
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