- - Ram Air explanation?
||02-07-2009 11:26 PM
Ram Air explanation?
Alright, I know this is FI forum, but this is regarding the intake system so maybe you guys can help me. I was looking at the Ram Air theory and I came up with this on team-integra.com.
So, what's my confusion about?
Our pressure wave travels at the speed of sound. A good estimate for the speed of sound is 1,125 feet per second.
For the F & G camshaft, the intake valve is open for 268(deg) of crank rotation.
The engine rotates twice (720[deg]) for the intake to open once.
720(deg) minus 268(deg) = 452(deg) of crank rotation that the intake is closed.
At 2,800 RPM, 452(deg) = .0268 seconds. (See below)
2800 rev/minute divided by 60 seconds/minute = 46.66 rev/second
46.66 rev/second X 360(deg)/rev = 16,8000/second
452*deg) / 16,800(deg) per second = .0268 seconds.
This .0268 seconds is the critical time factor. During this .0268 seconds that the intake valve is closed, the pressure wave is moving at 1,125 feet/second and travels 30.15 feet.
At resonant conditions, the pressure wave has to travel 30.15 feet to arrive at the intake valve when it is open. Since the pressure wave spends this time going up the runner AND going back down the runner, the runner length is actually only half of 30.15 feet, or 15.075 feet, which is equal to 181 inches.
(where deg=degrees, or that little 0 mark at the top)
Where did he get the 30.15 feet from? I tried 1,125/.0268 and I came up with 41977.61194
I tried vice versa, .0268/1,125 and I came up with 2.3822(repeating dec)
Maybe someone has a better explanation of the Ram Air technology?
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