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Good day! I am a student of aeronautical engineering and recently started studying the more natural way of flying, birds flight. After reading some scholarly articles there is still a question on my mind. Which phase of flight do birds generate more lift? My opinion is: in take-off. Knowing that birds jump in to the air without any contribution of the wings (just using leg thrust), the wings must generate enough lift to cancel the bird weight and make it climb and accelerate upwards. In the landing phase birds need to reduce the speed, and of course produce some lift to guarantee no harm to birds legs but the question is: is lift higher in landing or take-off phase? Does it depends on other aerodynamic parameters?

Emanuel Camacho
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  • For a bird to stay at some height, lift must equal weight. If it climbs, this is because lift exceeds weight, and if it loses altitude, it's because lift is lower than weight. – mins Mar 08 '17 at 00:32
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    @mins: unfortunately not quite true. If a bird is accelerating upwards (ie, vertical speed upwards is increasing) then lift exceeds weight, and if it is decelerating downwards (ie, vertical speed downwards is increasing) then weight exceeds lift. Lift is a force that provides acceleration, not velocity. – Peter Mar 08 '17 at 00:36
  • I totally agree with you. But for example, when the bird is landing, the lift is higher and the bird is losing altitude which means it is decelerating at that instant. But what I am trying to find out is f the magnitude of the aerodynamical force is higher in landing or take-off. – Emanuel Camacho Mar 08 '17 at 00:37
  • When the bird takes off, it goes from 0 -> TakeoffSpeed, the lift generated is just that needed to exceed the birds weight. As the bird lands, it goes from FullFlyingSpeed -> 0. Wouldn't there be more generated lift (negative) as the FullFlyingSpeed is without a doubt greater than TakeoffSpeed. But this also brings into question the time period. Do you allow the bird to take off to the same "FullFlyingSpeed", therefore changing the equation to 0->FullFlyingSpeed? – Prodnegel Mar 08 '17 at 00:51
  • @Prodnegel I understand your point. It's indeed true that birds go from zero to the takeoff speed but they also need to be continuously producing lift in order to keep climbing. I don't agree with your definition of landing, since there is a clear preparation (approach) where they extend their legs and the wing movements are pretty different comparing to when they are flying at the full flying speed. And yes the time period is important too. I think that my answer is somehow related with the variation of momentum during takeoff and landing. – Emanuel Camacho Mar 08 '17 at 01:05
  • When V is obtained the bird needs to keep creating lift, but in that case, Lift will have the same value as its weight. – Emanuel Camacho Mar 08 '17 at 01:15
  • @EmanuelCamacho, you need to better define your question then. What time period are you considering here? The bird's mass does not change, therefore momentum would be directly calculated from the birds velocity, which brings it back to my original question of which velocity do you allow the bird to fly to? If bird takes off from 0->X in 3 seconds, but bird lands from X->0 in 4 seconds, then it would seem that the bird creates more lift taking off. Yes? – Prodnegel Mar 08 '17 at 01:26
  • Also, it may be worth looking into gravity's effect here. If gravity helps the bird on its landing, then technically the bird would "generate" less lift by itself right? – Prodnegel Mar 08 '17 at 01:28
  • @Prodnegel I totally agree with you from the definition of momentum. And yes, the gravity effect helps the bird decreasing their height even if in the very final phase is needed some lift to guarantee a smooth landing. Thanks for referring the time factor. – Emanuel Camacho Mar 08 '17 at 01:32
  • @EmanuelCamacho, Thanks for asking! I am sure a seasoned aviator will come to our rescue soon :) – Prodnegel Mar 08 '17 at 01:32
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    @Prodnegel Thanks for bringing up those concepts. It made the problem a bit more clear, since I will try to introduce some other aerodynamically factors. – Emanuel Camacho Mar 08 '17 at 01:42
  • @EmanuelCamacho: This image depicts forces acting on an aircraft during a climb. Lift is oriented rearward, creating induced drag and providing some vertical force. Thrust is balancing drag and provides the missing vertical component to balance weight. The excess of thrust moves the aircraft forward. For a bird there is no engine, no thrust. The bird needs to orient its lift forward, and lift alone must compensate weight and and total drag, and provide some forward speed. Hence thrust is larger than weight in climb at constant rate. – mins Mar 08 '17 at 01:45
  • @mins There is thrust in birds. The difference is that in our planes, the wings represent the aerodynamics while the engine represents propulsion. In birds the wing has everything included, in other words, the wings movement generates lift and thrust, due to the inverted Von Kármán vortex street. – Emanuel Camacho Mar 08 '17 at 01:49
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    @mins I understand what you said, because the thrust production in birds and even the aerodynamics of them are totally different from these days airplanes. – Emanuel Camacho Mar 08 '17 at 01:58
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    VTR - we already have a question about bird aerodynamics and it was decided that it would be on topic: Are birds aerodynamically stable? – fooot Mar 08 '17 at 16:21
  • @Peter mins is correct in stating that for climb, lift must be greater than weight. There is an initial acceleration upwards, which leads to a vertical velocity, which leads to an aerodynamic damping force downwards, proportional to climb speed. So in climb, lift = weight + damping force, which is greater than weight. – Koyovis Jun 24 '17 at 07:03

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The easy answer is that they create more lift during take-off because the bird is heavier. Like aircraft, birds shed some mass during flight when their metabolism converts food to energy. Unless the bird picks up some prey during the flight, it will weigh less during the following landing and will also create less lift. But from now on I will interpret your question as if you had asked for the higher lift coefficient.

You cannot calculate bird flight with stationary aerodynamics (which is usually good enough for airplane aerodynamics). The flapping motion of the wings adds instationary effects wich dominate flow in case of birds.

Therefore, there is not one lift coefficient of a bird's wing, but a variation over flapping angle and angle of attack change, the integral of which keeps the bird in the air. If you want to get an idea how much lift can vary with the rate at which the angle of attack changes, please read this NACA report from 1951 (NACA TN 2525). Another hint: Birds with large wingspan, which prevents effective flapping while close to the ground, must run to reach sufficient flight speed, just like airplanes do. Watch albatrosses take off: They have to restrict flapping so instationary effects can contribute less to increase overall lift. Pidgeons, on the other hand, which flap their wings during take-off so much that their wingtips meet below their belly (and create the typical clapping sound of a pigeon taking off), can fly instantaneously when hopping into the air.

Now to your question: I have no hard data, only observations. They lead me to think that overall the lift coefficient during landing is greater. The bird uses full tail feathers down deflection to add more lift and moves the wings forward to compensate the pitching moment. This makes the whole configuration completely instable, but birds fly with "artificial" stability anyway. Only in this case they max out the lifting potential of all surfaces, fully knowing that inertia will not allow their bodies to become uncontrollable before their feet are on the ground or perch.

Peter Kämpf
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  • The only way the bird's mass is reduced during flight is if it "sheds" some poo. Although this is a quite likely event particularly for flights over humans this is not always the case. – mezzanaccio Mar 08 '17 at 10:10
  • First of all, thanks for your answer. Second, I was not even considering the mass loss during flight, and to be honest I didn't even think on that at all. Your conclusion, based on observations was that the lift coefficient is greater during landing, because of all the surfaces movement, but I have a question. For example, during a takeoff (escape takeoff is an interesting case too) does not the frequency of flapping increase the lift? The lift that you referred during landing can not be the drag, that helps the bird reduce its horizontal speed during approach? – Emanuel Camacho Mar 08 '17 at 13:40
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    @mezzanaccio: Nonono! It inhales oxygen and exhales the same oxygen as carbon dioxide and water. Guess where that carbon and hydrogen come from! Same happens with your body between meals. – Peter Kämpf Mar 08 '17 at 14:27
  • @EmanuelCamacho: Right, the flapping intensity should contribute more lift and I did not consider that aspect. Maybe take-off does create a higher lift coefficient after all. However, it could also be that the flapping intensity goes into forward acceleration, and less into lift production. – Peter Kämpf Mar 08 '17 at 14:30
  • @PeterKämpf Isn't acceleration in this case also lift (in the sense that a propeller generates lift, or perhaps a helicopter rotor)? – Steve Mar 08 '17 at 14:36
  • @Steve: No, during take-off the acceleration is mostly horizontal. Lift should be mostly vertical. – Peter Kämpf Mar 08 '17 at 14:37