max cfm of a 968 @ 8 psi?
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10-23-2006 | 11:42 PM
#10
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From: seattle
I wish it were that easy!
right now I'm trying to turn thrust into psi
thrust = density x cfm sq
----------------
Pi x R sq
density of air at 122 degrees F = .06
the cfm is the 530
and my radius is .155 feet
(95 mm = .311 foot dia = .155 foot radius)
.06 x 530 / 3.14 x .155sq = 421
so my lb force of thrust is 421
I do know that 1 lb force/ inch squared = 1 psi
so now I have a 95mm tube, which has a 11" area using Pi R squared
1.8 X 1.8 X Pi = 11
and divide from my 421
421/11= 38.27
38 psi?
WTF?
right now I'm trying to turn thrust into psi
thrust = density x cfm sq
----------------
Pi x R sq
density of air at 122 degrees F = .06
the cfm is the 530
and my radius is .155 feet
(95 mm = .311 foot dia = .155 foot radius)
.06 x 530 / 3.14 x .155sq = 421
so my lb force of thrust is 421
I do know that 1 lb force/ inch squared = 1 psi
so now I have a 95mm tube, which has a 11" area using Pi R squared
1.8 X 1.8 X Pi = 11
and divide from my 421
421/11= 38.27
38 psi?
WTF?
Last edited by emwporsche; 10-24-2006 at 02:29 AM.
10-24-2006 | 12:06 AM
#11
Rennlist Member
"I used the black magic way of Rennlist - facts combined with mathematical calculations"
Yep, and that would equal about 375 bhp at sea level.
And using those same facts and mathematical calculations, one could question the accuracy of the almighty "dyno".
Yep, and that would equal about 375 bhp at sea level.
And using those same facts and mathematical calculations, one could question the accuracy of the almighty "dyno".