Birds fly by taking advantage of the air pressure difference created by the air velocity when the routes of air on the two side of the wing are different.
If the bird (a Golden Eagle in particular) wings bend 1 cm (0.01 m), suppose the air velocity is equal to the average velocity of the bird, 12.5 m/s and we are considering the 0.3 m air above the wing (modeled as a cylinder)
12.5 m/s * (0.15 m)2 *π=v2*(0.14 m) 2 *π
The density of air is 1.225 kg/m3, and the bird flies horizontally
P2-P1=1/2pv12 -1/2pv22 =1/2(1.225 kg/m3 )( 12.52 – 14.32 )=-30.4 Pa
The size of a birds wing is 1 m (a wing span of 2 m), and supposed the width of the wings is 0.5 m,
So the total force that the bird experiences is
F=PA=30.4 Pa*(1 m*0.5 m*2)=60.8 N.
So the bird experiences 60.8 N upward force.
To move forward, birds also lift their wings and thrust them forward. The pectoral muscle connects the humerus of the bird to the keel of sternum and its contraction causes movement of the wings.
Suppose the bird flaps its wing 45 °every 1s, and the humerus of bird is 0.2 m,
supposed the muscle attaches at 5 cm and 60 degree from the pivot and the weight of the bird is 4.05 kg,
So rFsinθ+rFsinθ=I α
If we treat the bird as a point mass with radius of the length of the body,
0.05 m*F*sin 60°+0.2 m*4.05*5%* sin45°=0.65 /s2 *4.05*(0.5 m)2 π
As a result the pectoral and supracoracoideus muscle is about 1/3 of the bird’s weight—equivalent to 60 kg of a 180 kg adult!