Ram effect
There is far more to driving than the 1/4 mile.

Edit: anybody got times on the 1/4 mile with and without an air filter?
Last edited by ConnClark; Apr 5, 2006 at 06:13 PM.
The average sea level pressure is 29.92" of mercury, which equates to just under 14.7 psi. I would consider a .07 psi increase as pretty insignificant ...
Humm ... connclark could you clarify this for us?
There is far more to driving than the 1/4 mile.

Edit: anybody got times on the 1/4 mile with and without an air filter?
I made 3 runs with the air filter, then pulled it for the last run to see if I would be faster.
The trap speed was almost identical. Didn't seem to make a difference.
However, on the dyno. Pulling the airfilter, lifting up the airbox, and running a fan got me about 12 rwhp!
Humm ... connclark could you clarify this for us?
Detroit has spent money for metal on making horns on their air cleaners like the one above. As this just draws hot air from under the hood and has no ram air inlet that mates with it, they have to have some reason. Oh yes.... I belive its call Bernoulli's Principle. The horn serves no other purpose than raising the air pressure before it goes through the air filter to reduce the pressure drop across it, I would consider a .07psi increase signifigant by comparison.
After you consider how much money they could save on metal and labor making their air cleaners like the one below this must have some real bennefit. I mean they have done this for decades and these guys are so cheap that they wouldn't spend a penny to keep their mothers from being thrown out in the street. (Note to self, don't raise children to be auto execs)
Conclusion....
If detroit will spend money on getting minor gains with a horn on an air cleaner then ram air must be well worth it.
If you think your time would be better spent doing something rice like putting a leaf blower before your intake feel free to do so.

Edit: Note this is a picture of a 66 t-bird air cleaner. My 63 doesn't have a horn. A 66 t-birds engine noise is no quieter than a 63 .
Last edited by ConnClark; Apr 5, 2006 at 09:09 PM.
Detroit has spent money for metal on making horns on their air cleaners like the one above. As this just draws hot air from under the hood and has no ram air inlet that mates with it, they have to have some reason. Oh yes.... I belive its call Bernoulli's Principle. The horn serves no other purpose than raising the air pressure before it goes through the air filter to reduce the pressure drop across it, I would consider a .07psi increase signifigant by comparison.
After you consider how much money they could save on metal and labor making their air cleaners like the one below this must have some real bennefit. I mean they have done this for decades and these guys are so cheap that they wouldn't spend a penny to keep their mothers from being thrown out in the street. (Note to self, don't raise children to be auto execs)
Conclusion....
If detroit will spend money on getting minor gains with a horn on an air cleaner then ram air must be well worth it.
If you think your time would be better spent doing something rice like putting a leaf blower before your intake feel free to do so.
So because there was a "horn on the air cleaner" back in the 60's, it means that "ram-air" (completely different concept) works?
Ram-air = putting the air intake INTO the air stream, and claiming that it effectively "RAMs" the air into the intake, compressing the air and acting as a sort of supercharger. Making a larger air box to increase the pressure through the filter is a completely seperate topic, since that air still has increase in velocity to go through the intake tubing (and will decrease in pressure as a result).
A .07psi increase is not large at all. If your intake increases the standard atmospheric pressure of 14.7psi by .07 psi, to 14.77psi, that is not even a Half of a single percent increase in air pressure. I consider that an extremely Insignificant gain.
To put that in perspective, you would find more of a pressure difference in a 20 foot change of elevation.
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Ram-air = putting the air intake INTO the air stream, and claiming that it effectively "RAMs" the air into the intake, compressing the air and acting as a sort of supercharger. Making a larger air box to increase the pressure through the filter is a completely seperate topic, since that air still has increase in velocity to go through the intake tubing (and will decrease in pressure as a result).
A .07psi increase is not large at all. If your intake increases the standard atmospheric pressure of 14.7psi by .07 psi, to 14.77psi, that is not even a Half of a single percent increase in air pressure. I consider that an extremely Insignificant gain.
To put that in perspective, you would find more of a pressure difference in a 20 foot change of elevation.
What's the air pressure inside the airbox when an engine is sucking air?
It has to be LESS than 14.7 psi since the air is moving INTO the engine.
Let's say it's 13.7 for example. Going from 13.7 to 14.77 is a decent difference. It'll probably get you 10-15hp, which is basically what ram air claims.
What's the air pressure inside the airbox when an engine is sucking air?
It has to be LESS than 14.7 psi since the air is moving INTO the engine.
Let's say it's 13.7 for example. Going from 13.7 to 14.77 is a decent difference. It'll probably get you 10-15hp, which is basically what ram air claims.
The MAP sensor measures the relative change of air pressure inside the manifold itself and is normally used on forced induction cars to measure the boost PSI. I’m sure someone with tuning software could most certainly go see if there is any change in internal manifold pressure (map sensor voltage) with ram-air, and w/o ram-air. I would guarantee that it would barely be picked up by the map sensor at all. Of course you would have to account for any increase from the colder & denser air it would be picking up, and remove that from the equation. I would estimate that you would have to hit at least 100mph before you started to see an appreciable gain in pressure.
Anyways, I'm not saying that ram-air doesn't work .... it most certainly does, and as I said in a previous post it is used extensively on many aircraft engines. HOWEVER, at the speeds a normal street car travels, the gains from ram-air would be insignificant at best. A standard cold air intake would provide much more power by just bringing in colder air then a ram-air system would. If you're driving around town at 150mph, then by all means, strap a ram-air system to your car.

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Good discussion, lots of numbers and formulae- I like both the 'hard science' and the seat-of-the-pants approach
Measuring 10HP at the track with a timeslip difference is a crapshoot as well. A tiny better launch, a tad more traction, a better shift, a small tailwind - any of those could change your times.
Jim
There is far more to driving than the 1/4 mile.

Edit: anybody got times on the 1/4 mile with and without an air filter?
Just my low tech personal experience in this high tech topic.
Last edited by Steve Gunn; Apr 9, 2006 at 08:31 AM.
Measuring 10HP at the track with a timeslip difference is a crapshoot as well. A tiny better launch, a tad more traction, a better shift, a small tailwind - any of those could change your times.
Jim
I was wondering how he figured a "10rwhp gain" via a track time, and how he can prove that entire 10rwhp is from the "ram-air" effect and not just colder air, less heat soak, track conditions, etc ... but after some thought I figured I'd just leave it alone.
Imagine a sliding resonance box with variable volume. The box is controlled by an actuator programmed to choose its position based upon vehicle speed and rpm.
It would require a lot of testing, either in a combination wind tunnel/dyno or in a very consistant car on a very consistant track. Multiple runs at several box positions and airspeeds.
Alternately, would there be some group of sensors that would help compute an optimal airbox size? Vacuum, speed, and rpm combined?
Once you get past that you are then qualified to run a performance shop.
Jim
If you take Ram Air Effect (RAC) and multiply it by 14.7 psi (pressure at sea level) and divide it by S.W.A.G it equals net Horse Power. It looks like this (rac*14.7/S.W.A.G.=HP) Oh, I forgot S.W.A.G.=scientific wild *** guess!
So when I use this model it gives me a lot more HP! lol
Last edited by mvonada; Apr 11, 2006 at 01:55 PM. Reason: wrong word





