What is RAM AIR?

[F4UDOA]

Specifically as it relates to the PW R-2800.

I am familier with the concept of compressed air in an engine giving it more power. However shouldn't this affect manifold pressure?

Here is a quote from Corkey Meyer Grumman test pilot on why the F4U was faster at sea level than the F6F.

The reason that the Corsair was faster in the Main stage blower was because the engine and carburator were provided with Ram Air coming in directly from the forward facing airduct in the wing, where as the Hellcat had the carburator air coming in from the accessory compartment of the fusalauge just behind the engine, with no Ram Air affect. Our airplane was getting carburator air at the same pressure as if it were motionless on the ground and the Corsair was getting supercharged air from the speed of the aircraft giving it more power (speed) in the main stage blower.

It sounds like it should mean a higher MAP but both engines are rated the same in all blower stages. 52.5" MAP

So what gives?

[funkedup]

If everything else (including MAP) is equal, then the ram air would allow the Corsair to generate the same MAP with less supercharger input shaft power.  This would mean more power at the crankshaft.

[Box1]

a male sheep with gas

[Shiva]

Ram air just means using a forward-facing air intake to gain some extra intake pressure. I'm sure that, back when you were a kid, you stuck your hand out the window of the car while a parent was driving and felt the pressure of moving air on your hand. When moving air is brought smoothly to rest, the energy of its motion is converted into pressure. Motorcycles went through a "ram-air" period in the early 1990s, during which street bikes were equipped with the forward-facing "rocket-launcher" engine air intakes seen on many road-racing machines.

Where vehicle speeds are very high, gains from ram air are significant. This was discovered by Rolls-Royce in the late 1920s as the company developed its R Schneider Trophy air racing engine. At speeds above 300 mph, it was noticed that the R’s fuel mixture leaned out enough to cause backfiring. When the mixture was corrected for ram-air pressure gain, the engineers realized they had a "free" source of power. At 350 mph the gain from ram air is almost 15 percent. Similar mixture correction is necessary when ram air is used on drag-race and Bonneville cars and bikes.

Intuition suggests that a forward-facing intake made in the form of a funnel, large end foremost, should somehow multiply the pressure of the air, resulting in a much larger pressure gain at the small end. Sadly, intuition is wrong. In order to convert velocity energy into pressure, the air has to be slowed down, and this requires a duct that widens rather than narrows. Next time you fly on a commercial airliner, note that its engine intakes widen as the airflow approaches the compressor face. Such widening passages are called diffusers, and they are universally used in the conversion of velocity into pressure. (The diffuser is also used, with supersonic jet aircraft, to slow the incoming air below supersonic speed so that combustion can be maintained; sustained combustion in a supersonic airflow is only now being demonstrated in the first scramjets.)

Open up an aftermarket car add-on catalog, and you can find all kinds of ram air intakes for most sports cars, although for the speeds at which street cars operate, what you get is not ram air but a tuned resonance between the carburetor/throttle body and the airbox fed by the ram vents, so that at a particular point on the power curve, the high-pressure point of the oscillation of the air in the airbox hits at the same time as the suction pulses from the carburetor, with the low-pressure point serving to help feed more air into the airbox. This is commonly used to smooth out flat spots in the torque curve, although most street-rocket modifications put the resonant point at the high end of the power curve to increase high-end power (10%-15% is a good boost).

One use that you will see on modern combat aircraft is the RAT, or Ram Air Turbine. This is essentially a fan hooked up to a generator; if the aircraft takes damage that kills its engine(s), the vent to the RAT can be opened, and the RAT will generate power to operate the aircraft's hydraulic and electrical systems.

[Pongo]

From reading you description i think it refers to the way the air intakes in the forward wings of the corsair are directly in the air flow their for as the aircraft speed increases there is more air being forced into the duct.
On the hellcat the intake of air was out of the airstream so was constent. I asume that increased air gives increased performance. Although proper calibrated tests I think showed that with the same engine..the two planes were the same speed.
Something to do with the location of the pitot tube.

[F4UDOA]

Pongo,

Indeed that is what Grumman claimed in the low blow and high blower stages. However in the main blower Corkey Meyer says that the F4U was 25knots faster due to the "Ram Air" effect.

My question is how do two A/C with the same engine, same rated MAP and same rated HP at a given alt. (under 8K is the main blower stage for both)have a 25knot difference in performance?

How would high pressure air affect performance to that degree without changing rated HP, MAP or RPM?

[Shiva]

Manifold pressure is not immediately obvious in how it affects engines.

A carbureted engine has a throttle plate that restricts the flow of air into the engine,  a venturi to speed up the air coming into the engine (the difference between the intake and venturi pressure tells how much air is moving), after which the air slows down and returns to its original pressure. Manifold pressure is this pressure, measured behind the venturi.

An engine sitting turned off obviously has a manifold pressure equal to the atmospheric pressure, even with the throttle closed (which is still open a tiny bit to maintain idle). As the engine turns over, the pistons suck air into the cylinders, lowering the manifold pressure. As the throttle is advanced, it allows more air past the throttle plate, increasing the manifold pressure. At tull throttle, there is no impediment to airflow, so the manifold pressure equals the ambient pressure.

The power an aircraft engine produces depends on both the manifold pressure and the engine RPM. If you reduce an engine's RPM from 2000 to 1,200 without changing anything else, you'd see a rise in he manifold pressure, due to the pistons pulling less air through the intake system. If you increase the RPM from 2000 to 2700 without changing anything else, you'd see a drop in the manifold pressure as the pistons try to suck in more air; there will be more fuel flow, and therefore more power.

The blower on the F6F and F4u sits between the air intake and the carburetor, and works by compressing the air coming into the engine --  increasing the manifold pressure. In order to do this, it has to tap off some of the energy produced by the engine to run the blower. At 18,000', for example, the ambient air pressure is about half what it is at sea level; boosting that pressure up to 52" represents a significant parasitic loss to the engine.

What ram air does is increase the pressure in the intake, allowing the blower to raise the pressure with less effort, so that more of the engine's power goes to the propellor. If an F6F and an F4U are at 18,000 feet, the intake blower on the F6F would have to add 38 inches of pressure to the incoming air to bring the engine to its maximum rated manifold pressure. The F4U, however, at the same speed as the F6F, would be getting, say, 15 inches of pressure from ram air, meaning that the intake blower would only have to raise the pressure 23 inches to get to the same maximum manifold pressure. This leaves more torque to be applied to the F4U's propellor.

[Pongo]

wow.
cool description. Thanks shiva. I think I acctually understand that..

[Gixer]

Ride any new sports bikes alas a GSX-R 1000 if you want to really feel what Ram Air is like. The howl of the intakes and air box along with the amazing rush of ever increasing hp numbers and speed is a mind blowing experience. I can never get use to.

In 91 I bought a new ZZR-1100 (one of the first production bikes with ram air) because it had massive HP for the time due to its ram air. And was good novelty value with your mates to get them to hold their hand against the front vents while giving the throttle a big blip. Just about sucked your hand in.  :-)



...-Gixer
The Horse Soldiers

[J_A_B]

"My question is how do two A/C with the same engine, same rated MAP and same rated HP at a given alt. (under 8K is the main blower stage for both)have a 25knot difference in performance? "


The F6F is bigger, heavier and draggier than the F4U.  Surely that would account for something.  Perhaps there other design factors which also affect top speed, like the downward tilt of the F6F's engine or such things.  


J_A_B

[Imp]

Originally posted by Shiva
What ram air does is increase the pressure in the intake, allowing the blower to raise the pressure with less effort, so that more of the engine's power goes to the propellor. If an F6F and an F4U are at 18,000 feet, the intake blower on the F6F would have to add 38 inches of pressure to the incoming air to bring the engine to its maximum rated manifold pressure. The F4U, however, at the same speed as the F6F, would be getting, say, 15 inches of pressure from ram air, meaning that the intake blower would only have to raise the pressure 23 inches to get to the same maximum manifold pressure. This leaves more torque to be applied to the F4U's propellor.

So basically it increases engine efficiency.

[F4UDOA]

Shiva,

That is a very well stated description.

Even a complete boob like myself can understand what you said with little or no need to take a nap after reading it.

Thankyou!!

[F4UDOA]

JAB,

There is method to my madness.

I don't ask questions without reason. You said yourself that the F6F is larger and draggier than the F4U. So then why does the F6F accelerate so much better than the F4U in AH??

I have been trying to prove that the F4U accelerates better than it is represented in AH for a long time but I have not been able to because of the "climb and acceleration" club on the boards. They swear that the two are one and the same.

Well I think this thread proves otherwise.

It IS possible for the F4U to acclerate better than it climb because of the RAM air affect.

I will probably argue this in another post so as to be more annoying to Pyro and HT

[HoHun]

Hi F4UDOA,

>I have been trying to prove that the F4U accelerates better than it is represented in AH for a long time but I have not been able to because of the "climb and acceleration" club on the boards. They swear that the two are one and the same.

That's Newton's fault only :-) We've actually mathematically proven the equivalence of climb and acceleration to you, though of course only because we failed at explaining it well enough for you to understand it so you can verify it for yourself.

>It IS possible for the F4U to acclerate better than it climb because of the RAM air affect.

The difference in "climb" and "high-speed" power is real, so you've really discovered a significant design difference between the F6F and the F4U here.

However, this doesn't mean there's an internal contradiction in the  equivalence of climb and acceleration. The important thing to remember is that both are equivalent in any specific flight condition, but not (of course) for different flight conditions.

Imagine an F6F and an F4U flying next to each other at the F6F's top speed. By definition, the F6F can neither accelerate nor climb at top speed - all of its power is consumed just overcoming air resistance.

The F4U, on the other hand, has a power reserve at this speed. If the pilot advances the throttle to full power now, his aircraft will start to accelerate since the F4U has a higher top speed than the F6F. However, he has another option: He can pull up a little and keep going at the same speed as the F6F, but climb away from it with his extra power.

Again, climb and acceleration prove to be interchangable :-)

How's the situation at low speed? Cruising next to each other, F6F and F4U advance their throttles to the firewall. Enjoying greater specific excess power at low speed, the F6F will outclimb the F4U if the planes go for a climbing contest, or leave it behind if they accelerate level.

Climb and acceleration are interchangable here, too :-)

But: While the aircraft accelerate, they leave the flight situation they started at, and traverse a whole lot of other flight situations. At the starting situation, the F6F had more specific excess power, somewhere in the medium speed range is a point where both have equal excess power, and at high speeds the F4U is superior - as we've already seen.

Now what's going to happen if we give the aircraft a bit more time for a race?

At frst, the F6F pulls ahead, faster than the F4U all of the time. It also reaches the point of equality before the F4U does, so it keeps outrunning the F4U even beyond that point. The F4U, having better high-speed aerodynamics (and additional ram power) steadily decreases the speed difference though. While it's far behind when it matches the F6F's speed for the first time, it out-accelerates it for the rest of race, catching up with and finally overtaking the F6F.

I guess the pitfall in understanding the equivalence of climb and acceleration is just that they're equivalent for each snapshot only, for a single frame in a long film, even though each frame in the film is different.

For the race, you can't put a single number on the acceleration capability as it varies with speed. The same is true for the climb rate - though you usually read a single number for that in the books, the truth is that this number is valid for the speed of best climb only. It's the same with acceleration - it's highest at some low speed, and zero at top speed.

Since climb and acceleration are equivalent, the highest acceleration actually is reached at the speed of best climb :-)

Well, I hope you enjoyed my attempt to show the difference between simple situations, for which it's easy to give accurate statements, and complex situations that can give people headaches :-)

Regards,

Henning (HoHun)

[F4UDOA]

HoHun,

I think what is so frustrating here is the fact that from an engineering point of view the math is clear and the logic follows one hundred percent of the time. However I already know that things are not equal so I am presenting an abstract point of view.

A+B may always equal C but not when A and B are not the same.

Case and point. You mentioned that you thought the F6F would always accelerate better at it's climb speed. I disagree.

Maybe not with the result just that you can say that as a blanket statement.

How is it possible to say that the acceleration is constant with the climb if one A/C is getting not excess power but to full power faster? At some point the F4U becomes faster because of a parasitic drag advantage for sure but what is the result of the sustained affect of Ram Air through different speed ranges?

If a F4U and F6F are flying in formation at 100MPH at sea level in main blower and both apply full power the F4U will reach full power faster and with greater torque applied to the prop. How can you say which will reach 150MPH first? Then 200MPH, 250MPH etc.?


I'll step back from my own explanation and use one from a Aeronautical Engineer who wrote America's Hundred Thousand.

He rates acceleration in his calculation of the following fighters in this order.

1.P38L
2. P-47M
3. P-51D
4. P-39Q
5/6. Tie P-63A and F4U-4

The order of Climb would be from the AHT numbers

1. P-63A
2/3. Tie F4U-4 and P-38L
4. P-47M
5/6. tie P-51D and P-39Q  

These results are not close.

Why? What calculation is he using?