Author Topic: Draining E in turns (Read 10180 times)

GODO · « **on:** August 23, 2004, 12:18:04 PM »

When you jaw left of right, the body of the aircraft (the side surface and the rudder itself) acts as an airbrake and the aircraft slows down losing energy. When you turn left or right the brake effect comes mostly from the upper surface of the aircraft and the elevators.

Assuming this is true, planes with large upper surfaces (wings and body) should lose energy quicker than smaller planes, all that being still above stall speed, that is, still having extra lift to keep flying.

But what we see is that small planes like 190s or 109s lose E much faster than much larger ones like P51s or Spits when turning well above stall speeds.

I understand than large winged planes with smaller wingloading will keep turning tighter and at slower speed than small planes with more wingload, but the small ones should lose speed at a lower rate and the others until they reach stall speeds.

Where am I wrong?

hitech · « **Reply #1 on:** August 23, 2004, 12:43:32 PM »

Quote

Assuming this is true, planes with large upper surfaces (wings and body) should lose energy quicker than smaller planes, all that being still above stall speed, that is, still having extra lift to keep flying.

This is an incorect assumption.

Planes with more wing area, also need a smaller AOA to produce the same turn rate. High wingloaded planes also suffer from more induced drag in most conditions do to the higher LCO needed for turns.

Also in your basic premiss about more side area on a plane when in a yaw condition, you are not looking at the force generated to produce a turn. And it does not neascarly follow that more side area would produce more drag in a slip. You also need to consider what the slip angle is.

dtango · « **Reply #2 on:** August 23, 2004, 12:57:31 PM »

Mandoble:

HT beat me to it already

. Size or surface area is only a single factor. You're forgetting the contribution that induced drag plays into the equation which impacts higher wing-loaded aircraft more vs. lower wing-loaded aircraft in a relative turn performance / energy-bleed comparison.

Tango, XO
412th FS Braunco Mustangs

GODO · « **Reply #3 on:** August 23, 2004, 01:01:28 PM »

Quote

Originally posted by hitech
Planes with more wing area, also need a smaller AOA to produce the same turn rate.

Certainly they will need smaller AOA for same turn rate but is that more E-consuming than the brake effect of the upper large surface? With the higher AOA a small part of the front and lower surface is generating more drag, but the whole upper side (wing and body) is the one displacing more quantity of air and probably, generating more drag? Or is the opposite?

GODO · « **Reply #4 on:** August 23, 2004, 01:12:13 PM »

I forgot to say. Im assuming both pilots are tightening the turn from a hi/medium speed as fast as possible pulling hard the stick for some seconds, not just substaning the same turn circle. This just what we do when scissoring, for example or when breaking hard left of right, not just turning in circles over and over.

hitech · « **Reply #5 on:** August 23, 2004, 01:24:44 PM »

You need to stop thinking it about area shown foward, to begin to understand the relationship between drag/weight/lift and turn radius you will need to understand the basic lift & drag equations. Then start looking at it from what happens in a given situation. It is not somthing that can be easly visulized.

HiTech

GODO · « **Reply #6 on:** August 23, 2004, 02:03:53 PM »

hitech,
Im refering to instantaneous (or short) turns, not just substained turns, sorry, I didnt clarify this well in my first post. In a substained turn where you keep turn rate, what I described as "brake effect" of the upper surface may be secondary and minimal (you are not tightering the turn very fast). But when you pull hard on the stick for few seconds things are probably different. This is the case where the wing area are really displacing a big ammount of air in a short time, in this last case I see the wings as the real big brakes (you are increasing AOA very fast, not just substaining a determined AOA).

hitech · « **Reply #7 on:** August 23, 2004, 03:23:32 PM »

Definitions of instantainious and sustained have nothing to do with AOA.

The amount of drag produced will be higher for the for higher wingloaded plane given the same speed, and same turn radius. The effect will be more pronouced the hard you pull on the stick.

What you seem to be missing is that the lower wing loaded plane dosn't need to pull as hard, to turn with the higher wingloaded plane.

HiTech

dtango · « **Reply #8 on:** August 23, 2004, 05:54:20 PM »

Quote

This is the case where the wing area are really displacing a big ammount of air in a short time, in this last case I see the wings as the real big brakes

Mandoble: You're assuming that wings have significant boundary layer separation that occur to increase profile/pressure drag while in a hard turn and that the larger the wing the more drag you would have. This doesn't happen because the wing would not be flying since massive boundary layer separation occurs at post stall. The URL below gives you a picture of this:

http://www.centennialofflight.gov/essay/Theories_of_Flight/Two_dimensional_coef/TH14G4.htm

Lift-dependent profile drag increase does occur as you increase AOA but it's contribution to the total drag of the aircraft is in addition to the induced drag (a signficant contributor to drag in a turn) of the aircraft as well.

Tango, XO
412th FS Braunco Mustangs

GODO · « **Reply #9 on:** August 23, 2004, 07:25:34 PM »

Very clear link dtango, that was what I was pointing, the larger the wing, the larger the turbulent wake producing larger pressure drag. That is the effect that should be present when you perform violent scissors, you increase the AOA as much as possible without stalling (avobe maximum lift), keep the turn for very few seconds (now reducing the AOA) and then repeat in the opposite direction. This is also what happens in a hard break turn (may be you only turn 45 degrees but probably past the stall angle, losing lift and aproaching the stall).

As the drawings show, the turbulent wake affects all the upper surface of the wing (supposedly, the larger the wing, the larger the effect for the similar AOA). Now the question is how this affects the speed of the plane.

IMO, the example playing with rudders is similar. Full rudder right and your nose starts pointing right, but not your flight path initially, similar as increasing your AOA. The larger the lateral surface of the plane and the larger the lose of speed. Keeping full right rudder further and the flight path and the nose angle difference starts to decrease, as well as the "brake" effect.

dtango · « **Reply #10 on:** August 23, 2004, 07:42:34 PM »

Hi Mandoble:

I'm afraid you're missing the point I was trying to make

.

Firstly- massive boundary layer separation occurs AFTER the stall not before.

2ndly - a higher wing-loaded aircraft will be at an higher aoa to match turn performance of a lower wing loaded aircraft therefore any pressure drag caused by some separation will affect it more than the lower wing loaded aircraft.

3rdly - surface area of the wing doesn't play a part in boundary layer separation. The shape / camber of the wing does.

4thly - not pictured here is the affect of induced drag which the higher wing-loaded aircraft will pay a higher penalty for. On top of that shorter wingspan increases the amount of induced drag as well.

Tango, XO
412th FS Braunco Mustangs

Angus · « **Reply #11 on:** August 23, 2004, 07:51:18 PM »

Guys:
Does not the Chord and even spanloading twist into this?

GODO · « **Reply #12 on:** August 23, 2004, 08:09:39 PM »

dtango,
I'm not talking about plane A trying to match the turn of plane B, but doing a similar movement, just imagine both reaching similar AOA and keeping it for few seconds.

Lets suppose a 190 is bounced by a 262 and breaks hard to the left to negate guns solution. In that case, the defender just turned few degrees while keeping a very high AOA. But in the process a noticeable amount of speed was lost. Now the bounced is a P51, it breaks hard also with very high AOA but the decrease of speed was not so noticeable.

In these examples, both planes increased AOA very fast well above the needed to keep a substained turn and probably reaching similar distance between separation points in the wings. But the turbulent wake generated by the P51 will affect a larger section of the wing area.

dtango · « **Reply #13 on:** August 23, 2004, 09:38:07 PM »

Hi Mandoble:

Critical aoa for a specific aircraft is the same no matter the type of turn. You can't have an aoa above what is needed for a sustained turn because the maximum aoa for a sustained turn is the same as that for an instantaneous turn.

Comparing aoa in a turn of a 190 vs. P-51 is comparing apples and oranges. The example you've given doesn't mean much unless you have specific data spelled out you're working with - flight config, conditions, specific type of plane, alt, speeds etc.

You're "greater wing surface equals more drag" theory is ignoring the aerodynamics. The condition you're thinking about primarily exists after the wing is stalled and no longer flying.

Tango, XO
412th FS Braunc Mustangs

hitech · « **Reply #14 on:** August 24, 2004, 09:03:42 AM »

Mandoble: IF you are asking which has more drag in a stalled condition. Note this is not turning but falling, Then in most cases I would say the bigger wing does.

You also seem to have a missunderstanding of what AOA is, just simply because you use statments like rapidly increasing AOA. And turns little but keeps a High AOA.

For this discusion just think of AOA as how far you are holding the stick back.

HiTech