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Why do we Lose Airspeed in a Turn and What Causes This?

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Why do we Lose Airspeed in a Turn and What Causes This?

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Old 11th May 2007, 20:19
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Question Why do we Lose Airspeed in a Turn and What Causes This?

This is a question i have been asked to find out by my 'educated' (polite way of saying 'old') instructor who also happens to be Ex RAF.

Many think you need to include a vast variety of scientific words and avionic lingo, however, he always says the shortest answer is often the simplist - logically! Or so he says! Plus he hates people making up bulls**t for the sake of speaking or pretending to know the answer!

His exact question is and i quote; "In one clear and concise paragraph, explain why we lose airspeed in a turn, and what causes this?"

Please assist, as this question is doing my head in - i kind of know the answer but just not how to phrase it simply, or to his standard.

If you could include some references or sources where you got your answers from i would be most grateful.

Many thanks in advance.

Al
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Old 11th May 2007, 20:31
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As soon as you start a turn you have to pull back on the yoke to maintain your altitude. This increases the drag and therefore you lose speed if you do not increase power.
In shallow turns normally there is no noticeable difference, but with bank angles more than 30 deg you will definitely have to increase pitch and power to maintain altitude and speed.
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Old 11th May 2007, 20:49
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When in a turn, you have to produce more lift, because a portion of that lift is used to turn you.
More life=more drag
More drag=less speed
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Old 11th May 2007, 20:51
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The above is the most correct answer but an additional factor is that if you are flying into a reasonably brisk wind and then rapidly turn downwind you will tend to temporarily lose airspeed. This is due to the momentum of the aircraft being related to groundspeed whilst the windspeed goes from a positive to a negative value. A large groundspeed acceleration is required to recover the 'equilibrium' airspeed.

Some people will argue that this effect does not exist, saying that the airspeed will not vary (only true in light winds with gentle turns). If this effect did not exist, neither would windshear.

Doing tight low level field circuits in a heavy helo with a strong wind, the effect is very noticeable, requiring significant nose down and power to recover airspeed on turning downwind.

Edited to add: If turning from downwind into wind your airspeed will temporarily increase.

Last edited by Mike Oxmels; 11th May 2007 at 21:03.
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Old 11th May 2007, 20:54
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In a balanced turn the lift generated by the wings is no longer just acting vertically to maintain level flight but is also acting horizontally to make the turn. To maintain level flight the total lift generated by the wing must therefore be increased by increasing the angle of attack resulting in more drag and thus speed loss. Same answer, another way of saying it.
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Old 11th May 2007, 21:02
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Ignoring the indirect aerodynamic effects and just using phyics.
If you're flying at 350kts stabilised, then initiate a turn through 180, you've effectively accelerated in velocity terms by 700kts. It's always going to need thrust above that required for straight & level flight.
It's incidental that for an aeroplane, the direction change (relative acceleration) requires lift which leads to drag.
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Old 11th May 2007, 21:10
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Sorry LP. I disagree with that explanation. In still air (unlike my example) the acceleration is lateral, not in the direction of flight (mechanics of circular motion). Imagine any object rotating on perfect bearings (no friction or drag losses); it will rotate forever - it will not decelerate because of the apparent change of speed you describe.

In still air it's purely about the inclination of the lift vector and the increased drag that results. With wind the momentum effect, I mentioned also comes into play.
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Old 11th May 2007, 21:21
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Don't forget in your own words I'm sure your experienced instructor looks on here as well?
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Old 11th May 2007, 21:23
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yeah but no but yeah but no but...

an aeroplane isn't an object rotating on a frictionless bearing, it's a body with mass travelling with a velocity of +350kts which in my example is accelerated to -350kts, deriving this acceleration from the properties of the atmosphere around it.

If it was a vehicle in frictionless space, it would fire a rocket motor, still consuming energy.

However, I've had a bottle and a half of wine now and sense I'm going to lose this one.
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Old 11th May 2007, 21:29
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But that rocket motor would fire sideways to get the centripetal acceleration (plus a couple of puffer jets to establish rotation) and circular motion. You would not need any thrust in the direction of flight.

If your vehicle travelled in 1 dimension only, along a line, coming to a stop and then going back the other way, you would be correct. For circular motion I regret, Sir, that you are not.
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Old 11th May 2007, 21:34
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It's Friday night and LP is drinking vino

What the hell am I doing talking physics on Pprune. Time to open a bottle myself!
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Old 11th May 2007, 21:37
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Disagree.
We aren't talking about just rotating an aircraft about its axis without changing the direction of travel. A turn will always reduce the velocity on the original heading and increase velocity on the reciprocal (relatively).
My spacecraft analogy was on the same lines ie don't want to go to the moon now, want to turn and come home. First an RCS burn to rotate, then a main engine burn (don't know the acronymn) to decel/accel.
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Old 11th May 2007, 21:39
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"When in a turn, you have to produce more lift, because a portion of that lift is used to turn you."

When in a level turn, you have to produce more lift because only a portion of that lift now opposes the weight of the aircraft - the rest is used to turn you.

More lift=more drag
More drag=less speed
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Old 11th May 2007, 21:39
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The above is the most correct answer but an additional factor is that if you are flying into a reasonably brisk wind and then rapidly turn downwind you will tend to temporarily lose airspeed. This is due to the momentum of the aircraft being related to groundspeed whilst the windspeed goes from a positive to a negative value. A large groundspeed acceleration is required to recover the 'equilibrium' airspeed.

Some people will argue that this effect does not exist, saying that the airspeed will not vary (only true in light winds with gentle turns). If this effect did not exist, neither would windshear.

Doing tight low level field circuits in a heavy helo with a strong wind, the effect is very noticeable, requiring significant nose down and power to recover airspeed on turning downwind.



Point me at one authorititive text which confirms the "Down Wind Turn".

An aircraft flys with reference to the air around it. The fact that that air may itself be moving is completeley irrelevent.

You make the mistake because you have an erroneous idea as to what momentum is. Monentum is kinetic energy which is mass x velocity.

Ah! you will say, but velocity is relative to the earths surface.

No it isn't.

It is relative to whatever you want it to be relative to. For an aircraft in flight, the only momentum that matters is that relative to the air around it.Does a space ship in inter-steller space have to look back at the earths surface to find out it's velocity?

Look at it another way- the earth is turning at arond 600mph at the equator. Does this mean if an east-bound aircraft turn it starts flying backwards bacause of all the momentum it has due to the earths rotation?

Go flying some time on a windy day. Get some height, and by reference to the ASI alone, try and determine the wind dirsction- I assure you it won't be possible

Wind shear is a different thing. It occurs when the aircraft encounters a change in wind speed.

Last edited by PPRuNe Pop; 13th May 2007 at 13:50. Reason: Swearing
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Old 11th May 2007, 21:43
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Kinetic Energy = 1/2 mass x velocity squared
Momentum = mass x velocity

Otherwise I agree with you, wizofoz - the 'downwind turn' myth is indeed UTTER BOLLEAUX!
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Old 11th May 2007, 21:52
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4L3X I guess that's what you are looking for:-

we lose airspeed in a turn because of DRAG and what causes this is the disturbance of the airflow over the wing.
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Old 11th May 2007, 21:54
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where's Genghis?
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Old 11th May 2007, 22:13
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I see that the 'some people' I referred to have turned up.

Yes the aircraft flies with respect to its airmass, but its momentum is related to groundspeed.

My heavy helo strong wind example:

Airspeed 60kts,
Headwind 30 kts,
Groundspeed 30 kts

Turn downwind:

Airspeed 60kts
Tailwind 30 kts
Groundspeed 90 kts

A GS acceleration of 60kts is required to maintain the steady state, equilibrium airspeed. There will inevitably be a lag (more noticeable, the more aggressive the turn) resulting in a temporarily reduced airspeed.

This effect can be very noticeable in the circumstances I described, as can the opposite, turning into wind. As I acknowledged in gentle winds and gentle turns this effect will not be noticeable.

If my statement is wrong, explain to everyone how rapid changes of windspeed can affect airspeed in windshear.

Beags, Wizowaz? I'm waiting. HAVE A THINK ABOUT THE KINETIC ENERGY OF THE AIRCRAFT INTO WIND AND DOWNWIND.

LP, the speed (not velocity) is constant. No tangential acceleration.

Last edited by Mike Oxmels; 11th May 2007 at 22:28.
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Old 11th May 2007, 22:51
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Poor fella probably wishes he never asked! 4L3X, just tell your instructor about the greater lift required due to the inclination of the lift vector, and the increased drag that results.

Beags, Wiz, Deliverance: Talk to me about the kinetic energy of the aircraft in my example at 30kts GS and 90kts GS. You've got to input energy if you wish to maintain airspeed. If you don't you'll come out of the turn at a reduced airspeed.
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Old 11th May 2007, 23:03
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If you're losing speed in a turn your aircraft isn't generating enough thrust
WRT the downwind turn, can see how this applies to rotary folk:

Helicopter in the hover with a 10 kt headwind... IAS 10kts

Turns 180 out, IAS -10 kts

voila a 20 kt decelleration.

However, with the fixed wing case not so sure this applies!
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