Ram effect
The equation for the velocity head (ram) of a fluid is: P = (p x v2)/(288 x g) where in this case, P is psi, p is the density of air, 0.076 lb./cu. ft., v is speed in ft./sec. and g is the acceleration due to gravity, 32.2 ft./sec./sec. so to get three pounds of ram pressure, even with 100% recovery of the velocity head, would require a speed of 350 ft./sec. or 239 MPH. It would also provide a power gain of ~ 22%. (Light aircraft and boat speedometers are pressure gauges calibrated on the basis of this formula.)
(Oops! I put in my $0.0174 before noticing I'd strolled into a hornet's nest, oh well, I'll let it stand now that it's written...)
My power difference was only around 5.6%
I don't know how well that follows what the equation would predict, but that's the raw data from the tests.
gain.
I wonder if the calibration of the MAP sensor was thrown off, or the gauge
had been in error.
Another thought is the original setup was so restrictive, it was creating a higher
pressure in the manifold at wide open throttle, when compared to the new
intake system?
I don't know if this 3 PSI is reference to atmosphere, or relative to the old
system?
gain.
I wonder if the calibration of the MAP sensor was thrown off, or the gauge
had been in error.
Another thought is the original setup was so restrictive, it was creating a higher
pressure in the manifold at wide open throttle, when compared to the new
intake system?
I don't know if this 3 PSI is reference to atmosphere, or relative to the old
system?
The Best V8 Stories One Small Block at Time
Sorry if you don't understand what I'm saying.
Sorry if you don't understand what I'm saying.
The equation for the velocity head (ram) of a fluid is: P = (p x v2)/(288 x g) where in this case, P is psi, p is the density of air, 0.076 lb./cu. ft., v is speed in ft./sec. and g is the acceleration due to gravity, 32.2 ft./sec./sec. so to get three pounds of ram pressure, even with 100% recovery of the velocity head, would require a speed of 350 ft./sec. or 239 MPH. It would also provide a power gain of ~ 22%. (Light aircraft and boat speedometers are pressure gauges calibrated on the basis of this formula.)
(Oops! I put in my $0.0174 before noticing I'd strolled into a hornet's nest, oh well, I'll let it stand now that it's written...)
A rain gauge with a funnel on top will fill up a whole lot faster than a straight rain gauge, but we all the straight rain gauge gives the only accurate reading.
Now I don't have any formula's to back up what I've written, but I did stay at an Holiday Inn Express last night.
As far as Helmholtz resonance (if I understand your question to Madman), this phenomenon would operate totally independent of any ram air pressure and is very limited in effect when more than four cylinders are being fed by one plenum. (there's good info in that same Vizard book, "How to Build Horsepower Vol. 2")
of about 11 square inches (at least, this is what I
recall from measuring every orifice in the tract one day
in a fit of new-owner enthusiasm). That's 0.076 sq ft.
Traveling 60MPH is 5280 ft/min. That means that the
inlet -encounters- only 400CFM of air, that is all that
would pass through a zero-air-resistance hole at speed.
So don't expect big things from it, at street speeds.
You get her up over 100MPH, maybe it starts to make
up for some suction losses in the front plumbing. Or not.
In fact, if -anyone- with a "ram air" setup can show me
a vehicle log in which MAP exceeds BARO by more than
1kPa, at WOT (a big 1%), I will be impressed.
Of course, to benefit modestly all you need is to bring
air temps down to where they do not mess with the
timing. That's a much easier mark to hit.
Is that 101kpa ?
Its not to say that at higher speeds you don't see some ram-air efect. But 99.999% of the cars on this site would never see any benefits from ram air.
Madman messes with some serious cars which have the beans to reach the velocity needed to get some effect.
Most of the gains folks see aren't from ram air, but rather from getting cooler denser air in rather than hotter air from underhood which will cost hp based simply on IAT.

A rain gauge with a funnel on top will fill up a whole lot faster than a straight rain gauge, but we all the straight rain gauge gives the only accurate reading.
Now I don't have any formula's to back up what I've written, but I did stay at an Holiday Inn Express last night.

The equation for the velocity head (ram) of a fluid is: P = (p x v2)/(288 x g) where in this case, P is psi, p is the density of air, 0.076 lb./cu. ft., v is speed in ft./sec. and g is the acceleration due to gravity, 32.2 ft./sec./sec. so to get three pounds of ram pressure, even with 100% recovery of the velocity head, would require a speed of 350 ft./sec. or 239 MPH. It would also provide a power gain of ~ 22%. (Light aircraft and boat speedometers are pressure gauges calibrated on the basis of this formula.)
(Oops! I put in my $0.0174 before noticing I'd strolled into a hornet's nest, oh well, I'll let it stand now that it's written...)
Running a manometer for relative pressure between the cabin and the ducted intake through the nose (license plate) of a C5, my results confirm the equations generally. No consideration for resonance was made.
I got no measurable increase at 60MPH and only 6 inches of water at 100mph.
about .25 PSI. But it looks neat.
I also ran temperature transducers throughout the engine compartment, at 20mph it is almost ambient everywhere, there is even air moving under the intake manifold.
So cutting your hood up for more air may be counter productive.





