The wind beneath our wings

Since the air molecules above and below the wing need the same time to reach the end of it, their speeds ,assuming they are constant, are related by the following equation:

$\frac{u_{t}}{u_{b}}$ = $\frac{s_{t}}{s_{b}}$ ,

where u represents flow speeds and s the corresponding arc lengths. We have $s_{b}=7 m$ and $s_{t}=7.1852 m$ .

$s_{t}$ can be calculated as follows: The circle whose part is the wing arc is unique. The arclength and radius can be obtained by solving the following set of equations,all derived from simple trigonometry applied to the circle:

$s_{t}=2φ\cdot{R}$

$sinφ=\frac{s_{b}}{2R}$

$R^2=(R-h)^2+\frac{s_{b}^2}{4}$

Here φ is the central angle corresponding to the given arc and h the maximum distance between the top and bottom of the wing, which is equal to 0.7 m. The solution is

$R=\frac{4h^2+l^2}{8h}$

$s_{t}=\frac{4h^2+l^2}{4h}\arcsin\frac{4hl}{4h^2+l^2}$ .

It's reasonable to assume that at the bottom of the wing the speed of the air flow is the same as that of the plane. That means $u_{b}=250 m/s$ and from the previous results $u_{t}=256.6 m/s$ . Now we can solve for the pressure difference in Bernoulli's equation and obtain:

$p_{b}-p_{t}=ρg(h_{t}-h_{b})+\frac{1}{2}ρ(u_{t}^2-u_{b}^2)$

These equations are approximate,even if the first term is considered negligible, due to the fact that $u_{t}$ is not constant but instead varies because the top of the wing is not level. The exact solution requires integral calculus which takes into account the varying height of the wing and consequently the varying air flow velocity on top of it.

We first need the distance over the top of the airfoil. From the Pythagorean theorem and the figure, we can find the radius r of the circle for which the top of the wing is an arc,

$r^2=3.5^2+(r-0.7)^2\rightarrow r=9.1~m$ .

Using $L=2rsin\theta$ allows us to solve for $\theta=0.395~rad$ and so the length of the top part of the wing is $d=2r\theta=7.185~m$ . The ratio of the velocity on the top to the velocity on the bottom is $v_t/v_b=7.185/7=1.026$ . We can now use Bernoulli's equation, slightly rewritten:

$p_b-p_t=\frac{1}{2} \rho(v_t^2-v_b^2)=\frac{1}{2} \rho 0.053 v_b^2=1987.5~Pa$ .