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)

v1A1=v2A2

12.5 m/s * (0.15 m)

^{2 }*π=v2*(0.14 m)^{ 2 }*π
v2=14.3 m/s

The density of air is 1.225 kg/m

^{3}, and the bird flies horizontally
P1+1/2pv

_{1}^{2}+pgy1= P2+1/2pv_{2}^{2}+pgy2
P2-P1=1/2pv

_{1}^{2 }-1/2pv_{2}^{2 }=1/2(1.225 kg/m^{3 })( 12.5^{2 }– 14.3^{2 })=-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,

α=0.65/s

^{2}
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 /s

^{2 }*4.05*(0.5 m)^{2 }π
F=118 N

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!

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