harddrivin1le

Bottom line:

You THOUGHT that "Ram Air" relied solely on the "negligible" compression of air relative to the increase in vehicle speed.

BUUUUUUUZZZZZZZZZZZZZZZZZ! WRONG ANSWER!

In actuality, the MAJORITY of ram air's effect is attributale to THE BERNOULLIE EFFECT, which is what I've been saying since this began.

Skeedatl

Come on...

Do the math already. It shouldn't be hard for you.

Holy crap, you don't even know how to use that forumula do you?!?!?!?!?!?!?

HAHAHAHAHAHAHAHAHAHAHA ROTFLMAO

Skeedatl

BZZZZZ WRONG. LOL :lol1: :lol1: :lol1: :lol1: :lol1: :lol1:

EVEN GM says it's not from Bernoulli's Principle
:lol1: :lol1: :lol1: :lol1: :lol1: :lol1: :lol1: :lol1: :lol1: :lol1: :lol1: :lol1:

Bottom line...you are full of crap and still can't come up with the applied mathematics supporting your claims.

You are a joke.

Come one man...the math can't be that hard for you.

Just do it. Come on. Use pitot theory to show the HP gains from WS6 ram air as you posted that article. LOL. SHow us the applied mathematics...PROVE THAT Bernoulli... HA HA HA and Pitot theory HA HA HA result in HP gains in the WS6 HA HA HA.

But alas, you can't. HA HA HA HA HA HA HA HA.

Norse396

Most of the Ram Air (or all stock Ram Air cars for that matter) cars capable of doing 150 are running out at that speed and most don't drive at that speed and I doubt Pontiac recommends driving that fast for an insignificant increase.

Skeedatl

Come on...bring on your next stolen thought 1LE. Cut-n-paste more nonsense for us. This is fun.

The applied math can't be that hard for you since you claim to be an engineer and have taken fluid mechanics while not being able to tell the difference between Bernoulli's Equation, pitot theory and the Combine Gas Law as well as not being able to tell the difference between Centers of mass and distribution of mass...but those are just trivial errors man. Come on...DO THE APPLIED MATH FOR US. We all need a good laugh.


You REFUSING to supply the applied mathematics (that's "showing your work" to you Slomaro drivers out there) just shows us all that you don't know what you're talking about.

brahtw8

Take it from someone who has spent way too much of his life arguing with HD, it really is not worth it. I am at the point where I don't care anymore. I hope more of you can reach that point.

To HD: You were doing so well in the "AWD a reality" thread, particularly on pages 3-5:

http://www.acura-tl.com/forums/showt...t=75621&page=1

On several occasions you admitted you were wrong. You will note that I did not rub it in your face, or post it again and again, nor did the others. I thought this was a new era for you and this forum.

I don't mean to say this is entirely your fault. It takes two to tango, and as a former frequent 'dance partner' of yours I do not mean to suggest that I am somehow above such things (although I hope to be in the future).

Nevertheless, I continue to believe that you have the most control over eliminating these thread wars. Please look at the AWD thread and consider how the lessons learned in that thread can help you and this forum community.

If not for me, do it for my NSX.

Norse396

I'd try to do the math, but I'm so pathetic at math I'd be laughed off of the forum....

Donte99TL

Damn this thread should be changed to harddrivin1le gets owned :lol2: :thefinger

harddrivin1le

One doesn't have to travel @ "150 MPH" to benefit from RAM AIR (assuming a weel designed system).

That Kawasaki I showed you began making positive pressure (albiet very mild) @ 60 MPH.

TLover

Well, the moderators already said they won't ban him so we might as well have some fun with him. :lol2:

harddrivin1le

P = .5 x r x v2

http://www.sportrider.com/tech/146_9910_ram/

"The pressure build-up can be defined using the Pitot-static tube theory:

P = .5 x r x v2

We took one of Pi Research's advanced, System 3 data-acquisition systems and hooked up one of its air-pressure sensors into the airbox of eight different modern sportbikes. We also mounted a wheel-speed sensor which allowed us to precisely measure and compare roadspeed with airbox pressure. Absconding with all the motorcycles to our top-secret, high-desert test site, we then proceeded to wring the piss out of each machine and gather data from each of the top-speed runs.................KAWASAKI ZX-7R: Kawasaki was the pioneer of ram-air induction on sportbikes, and this graph shows why. Compare this graph (and the ZX-9R's) with all the others. Note that there are virtually no pressure spikes during shifts. And airbox pressure builds past ambient at approximately 60 mph, not 90 or 145 mph-just a smooth crescendo up to a peak of approximately 23mb."

Donte99TL

hell ya :lol2: :lol2:

Skeedatl

I've offered on more than one occasion to stop and explain the fluids theory (as I've done for others) if he wasn't understanding it...but of course he just gets all arrogant and continued with his flawed logic so I won't bother.

I'm always willing to stop and explain the mathematics and explain theory to someone, but they actually have to be willing to LEARN and interested in learning...which case 1LE ain't one of 'em.

Skeedatl

Still waiting for the math there guy.

http://www.vetteguru.com/ramair/

http://www.snowgoercanada.com/tech_ram_air.shtml

What's the problem. You said you're an engineer and have taken a year of fluid mechanics.

You having problems with applied pitot theory? It's not that hard for someone who is and engineer and has taken a year of fluids. Use P = .5 x r x v2 and show us the increases in pressure and temperature from compression, show us the losses from the bend and diameter changes and come up with the result. It's not hard...come on sport. Just do it.

You simply can't. Surprise surprise.

Donte99TL

He should change his name to "cut and paste." :lol2: :lol2:

harddrivin1le

You were ignoring this simple equation and how it relates to RAM AIR, yet you're an expert on fluids:

P = .5 x r x v2

Skeedatl

That formula doesn't mean anything with the application. It doesn't SHOW anything other than your ignornace.

You don't even know how/when to apply it. Use P = .5 x r x v2 and show us the increases in pressure and temperature from compression, show us the losses from the bend and diameter changes and come up with the result.

We've all been 4 hours now waiting for you to do applied math using that formula to show the HP gains and for 4 hours you have dodged and weaved while we all laugh at you.

Of course that's 'cause you don't understand pitot theory or how to use the formula.

I ignore P = .5 x r x v2 for a number of reasons. It doesn't cover all the variables in compressions of gasses. It doesn't take into account anything other than the shape of a tube, then relies on the temperature being constant, the diameter being true and the "tube" straight.

NONE of which occur in ram air applications. But of course you wouldn't know that 'cause you don't understand pitot theory and what it's used for.

TLover

You claim to be an engineer yet won't show us the math. Show us the math not the formula.

brahtw8

I've said my peace. One of you will have to step up and be the big man, thereby ending this debate without an attempt to one-up the other or get in a parting shot. Otherwise, this will apparently go on forever.

And with that, I am hitting unsubscribe on this thread, since the issue about Ram Air was beat to death 8 pages ago, if not several threads ago.

Enjoy.

Donte99TL

harddrivin1le you are getting called out homie. You best step up and show them what you are made of.

Skeedatl

Well I didn't take Swat Dude's advice but I'll take yours. If 1LE wants to live in ignorance fine...there has been enough said here so that anyone reading this will get the truth that automotive ram air is a myth. If 1LE is truly interested in learning WHY pitot theory hasn't an application in ram air, I will take the time to explain it to him.

But the pissing contest is over. 1LE is soaked and is beginning to stink. As I bow out of this thread...I'm just waiting for the "ricer flyby" and will laugh rather than respond.

Norse396

Where do I state you do? The benefit is cooler fresh air getting into the engine. I never dispute that, what I dispute is any useful form of Ram Air taking place, I find it very hard to believe and apparently Pontiac agrees as they make no claim to any kind of Raming of air taking place.

I think people agree that at very high speeds (and 150 is very high speed) some effect can take place. At 60 it's so negligable to be laughable. Who cares when the major benefit is the cooler air having more direct access to the engine than normal intake tracts? A Ram Air system is great in name but doesn't really Ram air into the engine, it's just a cool name to help sell cars.

harddrivin1le

His original claim was that RAM AIR does nothing below Mach 0.5 (~ 350 MPH). The speeds that we're discussing are MUCH lower than that.

Additionally, you are ignoring the Bernoulli effect.

harddrivin1le

Don Vogel read it and he disagrees with you.

That's because he knows this:

P = .5 x r x v2

And he also knows that 90 MPH in a LS1 powered 1LE is NOTHING.

Norse396

I'm not talking about what he said, I'm talking to you. Ram Air does nothing of value that is worth jumping up and down over, I don't care what effect or equation you use. I think everybody will agree that there is a benefit to the use of such a system, but not because of any Air getting Rammed into the engine. If measuring insignificant values is more important to you than the real gains provided by these systems, well just say so and it's over.

90mph is nothing to many cars, so? I would rather speak of real gains than insignificant ones gained at much higher speeds or very insignificant gains at lower speeds.

harddrivin1le

I've never said that Ram Air produces gains that "are worth jumping up and down over." My original claim was that it begins to produce "quantifiable gains" in HP @ ~ 90 MPH... (or ~10 seconds after my 1LE leaves the light.)

The Bernoulli effect is what produces POSITIVE GAINS in STATIC PRESSURE, which yields added power for any given atmopheric condition (on Earth).

Minimizing inlet temps (to maximize air density) is also important as is ensuring a "fresh charge." I agree with you on that.

Skeedatl

:lol1: -coughullsh!t-

TLover

Huh?

Skeedatl

I know...it's really funny.

Skeedatl

LOL, no he hasn't. He asked how I came to my conclusion and I showed him the math. He hasn't posted here since. The only one who doesn't get it is you.

Skeedatl

OKAY...

Just to set the record straight since 1LE is continuing to molest Bernoulli's Principle.

Bernoulli's Principle states simply that moving air has less pressure than still air. BERNOULLI'S PRINCIPLE DOES NOT SHOW ANY AIR VOLUME HAVING HIGHER PRESSURE THAN THAT OF STILL AIR. PERIOD. Anyone who says differently is a retard and can't read the equation. IOW, to those who aren't following this, if you have 2 columns of air, 1 going 50 the other going 100, if you slow them both to 25, they both have the SAME FINAL PRESSURE. The fact that the one going 100 increases more doesn't mean anything...simply from the fact that it was a lower pressure to begin with. So through Bernoulli's principle a STANDARD air box would give you the SAME static pressure since the velocity of the air is FIXED BY THE THROTTLE BODY. A FIXED AIR VELOCITY THROUGH A GIVEN SPACE IN BERNOULLI'S PRINCIPLE HAS A FIXED STATIC PRESSURE...PERIOD.

Pitot theory DOES NOT EXPLAIN RAM AIR. It does not contain all of the variables necessary. Pitot Theory only applies to the variables involved...which is a fixed radius tube...WHICH DOES NOT EXIST IN RAM AIR.

These 2 formulas DO.
For computing the temperature change of the air as it's compressed
Tramair = ((Vcar^2/2gc)/cp) + Tair
And once finding that, computing the pressure change.
Pram = Pair (Tram/Tair)^(k/(k-1))

With Cpair = 1005J/KgK and K = 1.4; the Cpair and K are constants for air.

Bernoulli's Equation WILL NEVER RESULT in higher pressure than that of STILL AIR AROUND THE CAR. Hense the B.S. response from me regarding 1LE's quote "which yields added power for any given atmopheric condition (on Earth).".



To use Bernoulli's Equation to explain ram air is IDIOCY. To use pitot theory alone to explain ram air is IDIOCY.

Anyone who does it BETTER APPLY THOSE FORMULAS AND GENERATE SOME FRIGGIN' APPLIED RESULTS or STFU as they have ZERO clue what they're doing.

1LE...I find your posts amusing because you are obviously a retard...but others unfamiliar with fluids as you obviously are may be confused by your LIES.

Since you obviously don't know what the F you're talking about...how about you STFU beforeyou cost someone some money.

Anyone disagreeing with the above step up and DO THE APPLIED MATHEMATICS PROVING ME WRONG.

Don't be a b!tch and just repeat the same WRONGLY APPLIED FORMULAS or flawed experiments again.

DO THE ACTUAL APPLIED MATHEMATICS or STFU.

Norse396

Now I know why I hated math, I got a headache reading that, why, because it parted my hair on the way over my head.....

TLover

Yeah, someone better wake up Daniel Bernoulli and tell him he was wrong.

Skeedatl

It's idiots like him that write these retarded articles in magazines. They're all full of crap...just making it up as they go. Ask anyone in the PSCA what they think of joke mags like "Hot Rod". They don't understand what they're doing and just print any lie as fact...and dumbass toilet racers (those who sit on the toilet reading Hot Rod instead of actually building motors and drag racing) believe it as gospel.


Here's your racecar 1LE with a nice padded race bucket... just flip open Hot Rod and start racin'. The plunger is good for 10 HP at 150MPH!!!!! Put a few more plungers in there and you'll have yourself a nice Lenco.




This is the extent of your racing expertise stupid ass.

TLover

So when I fart, is that considered ram air?

Skeedatl

No, that would be a new LS1.

harddrivin1le

Pstatic = .5 x r x Velocity^2

harddrivin1le

http://www.sportrider.com/tech/146_9910_ram/


"P = .5 x r x v2

Pressure (P) is force divided by an area. In the English system of measurement the units of pressure are (lb - force)/in2 which translates to psi. Density (r) is mass divided by volume. The units of density in the English system are (lb - mass)/in3. Velocity (v) is air speed, with units ft/sec. Plotting pressure vs. speed gives a graph that has theoretical pressure rising with the square of speed, and this is why ram air has much more effect at greater speeds."

harddrivin1le

You've been telling us that gases are "incompressible below Mach 0.5." So why would CARE about "compression of gases" in vehicles that are traveling MUCH slower than that? :grenade:

You DO know what the basis of that equation is...Right?

TLover

You're such a joke. You quote only those parts of the article that support your statement.

"For a speed of 150 mph, the resulting maximum theoretical pressure would be about 27mb (approximately .4 psi). Millibar (mb) is a metric unit for pressure. We used millibar instead of psi to give more workable numbers."

"Without going into a lot of theory, suffice it to say that efficiency depends on the relative areas of the fairing intake, airbox intake and the airbox itself; even with a good system the loss can be as much as 70 percent."

So a maximum THEORETICAL pressure increase of .4 psi @ 150 mph. Now consider the second quote of a 70 percent loss of efficiency in real-world applications.

And why aren't these jokers measuring the difference in air temperatures in the air box? For instance, on a similar bike without air ram and a "ram air" bike?