But the cool has it'* benefits.

125mph is the number to be effective

For the ram-air effect maybe, but 30mph is the magic number for cooling effects.

i dunno, my highway butt dyno says 60 mph.

well damnit! ya wnana go and

Possibly, if you mean a ram hood, where the inlets are in the boundary layer air.
But for a higher scoop, up out of the boundary layer, or for a scoop mounted inside the lower grille, the magic number comes way down. Remember that if you mount the inlet in the grille you take advantage of air that is slightly pressurized as it hits the front of the car, and is directed into the scoop, whereas a ram air hood takes in air that is flowing over the car at high speed and is therefore less dense(as in the case of a wing, which uses that to produce lift). Not to mention that most "ram-air" hood scoops are angled back and have a screen that forces most of the air to flow OVER them.
There is also cowl induction, which takes advantage of a high pressure area in the angle of the hood and the windshield to push air into the intake.

"Hmmm I was in a coma like state earlier today ( sitting through a four hour training lecture at work). Anyway I was wondering how fast the air in a 3" inlet pipe to the turbo would be traveling if 800 cfm of air was going into the turbo. So when I came up with my anwser it woke me up enough to ask my buddies at work to calculate it for them selves and see what they came up with. I didnt tell them how I went about calculating it to avoid getting them started on the wrong foot incase I was wrong. They came up with the same results as I did. When I asked them how they went about calculating it they had approached it the same way I did.
here is how I did it:

1) I'am going to use standard temp and press for the air

2) area of 3" pipe = pi * radius squared
3.14 * 1.5in * 1.5in = 7.07 in2

3) Inside volume of pipe = area * length
I wanted to know how long a piece of 3" pipe had to be to half an internal volume of 1 cubic foot. I rearanged the equation:
length = volume / area
length = 1 ft3/ 7.07 in2 = 1 ft3 * 144in2 / 7.07in2 = 20.4 ft

4) OK now I know that 1 ft3 is the volume of a 20.4 ft long piece of 3" pipe. I also know that Iam looking for 800 ft3/min. so this is next:
Speed of the air in the pipe (mph) = (800 * 20.4 * 60) / 5280
800 is cfm (ft3/min)
20.4 (ft of pipe for one cfm)
60 (convert from minutes to hours)
5280 (convert from feet to miles)

ANWSER: 185.45 mph!!! Air speed in a 3" pipe with 800 cfm of air moving through it. Makes me shake my head but it appears to be right. I also calculated for a 4" pipe and the speed dropped to 104 mph.

There is no intended point of this. I was just kind of suprized by my answere and thought I would pass it along. BUT....it does make me wonder how much ram effect some of these late model cars get at 60 mph or even 80 mph with the little factory snorkels. Maybe they're just fancy looking cold air induction set ups? it does not ram air.

This is why I don't think ram air works. it might be a cold air intake but it is not ramming the air in the engine"
Courtesy of Hypsi87 on Automotiveforums.com
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Plugging our 460cfm (ish) into the formula comes up with 106mph. So you would need to go 106.1mph to get ANY "ram" air effect. Some time back I rigged up a very crude cooling system, it was simply a tube to bring air from the front onto the bottom of the filter. Even driving out of my parking lot at ~10mph reversed the IAT readings very quickly.

Cold air = good and achievable with this.
Ram air = not so much.

at what speed does a ramjet engine need to be going to produce power? ~400mph
here check this out:
http://www.aardvark.co.nz/pjet/ramjets.htm

case in point even the scoop on a alcohol funny car doesnt add power it merely directs cool air to the engine inlet.
but the debate will still go on.....