# Why do aircraft inner wings lose lift when turning?

first question here, so please be gentle!

I'm reading an entry-level engineering course book and am currently up to discussion of aircraft design.

There's one particular statement that is unclear to me:

Turning with the wing level is aerodynamically inefficient and the inner wing loses a lot of lift because it is moving relatively slowly through the air.

Does this mean that, for maximum lift efficiency, a wing has to be horizontally level? I understand that the upward pressure beneath the aerofoil is greater than that above it pressing down, so how does the upward pressure change beneath the aerofoil change when an aircraft is turning? Does downward pressure increase?

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""Turning with the wing level is aerodynamically inefficient and the inner wing loses a lot of lift because it is moving relatively slowly through the air."" That is not really true! Lacking some keel or centreboard as ships have, a aircraft will not turn with level wings (almost). In any case the inner wing has lower airspeed (simple geometry!) and thus lower lift, if the aircraft turns. Prerequisite is a certain amount of roll. – Georg Nov 26 '11 at 12:24
In normal turns, whether banked or not, the speed differential between inner and outer wing is negligible. To get noticeable difference in lift between inner and outer wing, you either need to 1) have very long wings, like a glider, or 2) be moving slowly so as to have a small turning radius. – Mike Dunlavey Dec 28 '11 at 22:52

Lift is a function of the speed of the air from the leading edge to the trailing edge. In a flat turn, the inner wing is moving slower than the outer wing therefore there will be a difference in the amount of lift produced.

But in fact, an airplane can not change direction by flat turning this way. Rolling into the turn by the use of the ailerons is the way a plane turns. This causes part of the lift to be directed into the turn and used to pull the plane around the turn. Think vectors.

The slowed inner wing will still produce less lift and the rudder is used to compensate against the tendency to slip into the turn further. The elevator is used to raise the angle of attack again to compensate for decreased vertical lift in the the turn. Angle of attack is the angle that the wing makes relative to the air flow. A higher angle of attack at the same speed creates more lift. To a point.

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Thanks - I had to read up on aerodynamics to try to understand parts of this answer, hence the delay! – James Dec 12 '11 at 0:27

There's a beautiful book on the subject of aerodynamics, Stick and Rudder.

Either the book you are reading is not very smart about how airplanes work, or you are mis-reading it.

Normally, the way you turn an airplane is the same way you turn a bicycle, motorcycle, or high-speed boat. You bank or tilt the vehicle in the direction you want to go. Since you always have a lift vector of force coming up through the center of the vehicle, by tilting the lift vector you are using some of it to accelerate you to one side. That puts you into a circular path. (You see this in movies about flight, where the aces are constantly twisting and turning.)

You can turn an airplane without banking it. If you keep the wings level and simply give it left rudder, that swings the nose to the left, which presents the right side of the plane to the wind. You feel this as a lateral force. That lateral force does cause the plane to travel in an arc to the left. However, that force is much weaker than if you simply bank the plane. In fact that maneuver is called a skid, just like skidding an automobile.

There are violent maneuvers that require advanced training in which parts of the wing surface can be stalled, or some parts stalled more than others, such as spins and snap rolls.

Normal utility aircraft (non-aerobatic) have wings slightly angled upward (Dihedral angle) In such a plane, if you simply apply left rudder and nothing else, that swings the nose left, and swings the right wingtip forward into the relative wind. Since it is angled upward, the wind gets "under it" and pushes it up, thus putting you into a bank. So if you simply apply left rudder, the plane will bank itself, which is the better way to turn. (Also, a certain amount of sweep-back angle to the wings will do that.)

(Oddly enough, if you are in a negative-lift situation, like an outside loop or inverted flight, to turn left, in your personal reference frame, you need to bank right, which makes sense when you understand that the lift vector is reversed.)

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+1 for the recommendation of Stick and Rudder, it's one of the best books I've ever read for this sort of question. – Daniel Chisholm Jan 4 '12 at 16:31

Something to keep in mind is the difference between changing the direction that the aircraft is pointing, versus changing the direction that the aircraft is moving. In straight and level flight you can stomp on the rudder and fairly quickly cause the nose to point 5 degrees or 10 degrees away from where it was. But (at least initially) you will not have changed the direction that the aircraft is moving. Forces have to be applied to the aircraft over a period of time to cause its direction of motion to change. By far the largest force that the pilot can control is the amount of lift produced by the wing, which is why turns are normally done using the lift of the wing.

With respect to the difference in speed over the wings, let's run some numbers and see:

• small aircraft flying at 100knots/100mph (50m/s in round terms)
• A comfortable turn rate - 180 degrees in one minute
• wingspan 33'/10m (so 5m from a/c centreline to each wing tip)

Turning 3 degrees per second means that the wingtips have a speed difference of tan(3 degrees)/sec * 10m, which is 0.52m/s or 1%. Even with the fact that lift is proportional to the square of the airspeed (so the lift at the wingtips differs by 2%), this is still a pretty small and pretty negligible effect in cruising-like maneuvers like this.

Where thing get more interesting is with lower speeds and higher turn rates (for example during an approach to landing, with the last 90 degree turn onto final approach). If the airspeed is 60knots (0.6X the above) and the turn rate 90 degrees in 6s (so 15deg/s or 5X the above), then the difference in airspeed between wingtips will be about 8.3 times as much or about 4.4% Based on a the lift being a quadratic function of speed this indicates a 9% difference in lift, and in fact closer to the stall speed like this the lift-vs-angle-of-attack curve is nonlinear and increases the effect further.

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I suppose the hypothetical turn mechanism is to turn off the engine on the inner wing, and blast the engine on the outer wing full throttle. This would work, in principle, although it is a really silly way to turn an airplane. You will produce a torque on the body, because the thrust on the two wings is unbalanced, and you will produce this torque even if the airplane is perfectly level.

This method doesn't require the airplane to slow down, and the lift in the inner wing is not significantly smaller, so you can't answer why it is smaller. But this is still a silly way to turn an airplane.

When you tilt an aircraft, and use the lift vector horizontal component to turn the airplane, the vertical component decreases very slightly, and it is best to think that it is not changed at all. The reason is that rotations are a lot like relativity--- rotation by small angles is nearly indistinguishable from a tilt.

If you have a vertical bar of length 10 meters, and you tilt it so that it leans at the top by 10cm, or 1%. How much shorter is the bar? The answer is by .00005 percent, or half the square of the fractional horizontal change. This is zero for all intents and purposes, in the same way that the Lorentz contraction, or time dilation is negligible for moving at 1% of the speed of light.

This means that you can turn an airplane with essentially no loss of lift, simply by taking the vertical lift vector and taking some of it and turning it into a horizontal turning pull with hardly change in the vertical component. There is no conservation law for the sum of vector components, (only for the squares).

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What is the downvote for? This is what they are talking about. It's the book's silly turn mechanism. – Ron Maimon Nov 27 '11 at 22:05
I'd vote your answer up, but I don't have enough rep points yet! I appreciate your answer, anyway :) – James Dec 10 '11 at 12:49
The extra lift you need to pull with the elevators to maintain altitude in a turn is $1/cos(\theta)$. So for a 60 degree bank angle, you need to pull 2 Gs. Needless to say, the stall speed is elevated, so you need sufficient extra speed when you do it. – Mike Dunlavey Dec 28 '11 at 22:41
@Mike: the bank angle is usually 1 degree. That's the small angle limit I was talking about. There is no loss of lift for 1 degree turns, and the loss is negligible even for 10 degrees, only becoming a significant 40% correction at around 45 degrees, the doubling point is 60 degrees, while the infinity-point is 90 degrees! This is how slowly rotation kicks in. – Ron Maimon Dec 29 '11 at 14:20
@Ron: You're right about the small-angle approximation, of course. I haven't flown for a couple years now, but as a rule of thumb, 15 degrees is a safe bank in a small plane. We practice 45 degree banks in steep turns. Since at a given bank angle, turning radius is quadratic in speed, jets have to bank steeper in approach patterns. – Mike Dunlavey Dec 29 '11 at 14:58