We know that the lift formula is:

$Y= Cy0.5 ρv^2s$

$ρ$-air density

$s$-wing area

$Cy$-lift coefficient

$v$-airflow velocity

We all know that.

$F = ma$ (1)

Let the aircraft take off with uniform acceleration, zero initial speed, $a$ acceleration, $v$ final speed and $x$ distance

According to the formula of uniformly accelerated motion,

$a = 0.5v ^ 2/x$ (2)

Substitute Formula (2) into Formula (1) and get

$F = 0.5 mv ^ 2/x$ (3)

Multiplication of formula (3) with $s/s$ is obtained.

$F = 0.5mv ^ 2s/xs$ (4)

xs is volume, so formula (4) is

$F = 0.5 ρv ^ 2s $ (5)

Since the wing has an angle of attack, it is necessary to modify the (5) mode.

$F=Cy0.5 ρv^2s$ (6)

And formula (6) is the lift formula.

  • $\begingroup$ In the last line, $C_y$ is not taking care of angle of attack - this is included in the swept-through area $s$ - but is rather taking other factors, such as surface interactions, internal fluid activity, turbulence and other air flow, etc. That drag coefficient simply covers all unknowns or other unpredictabilities or oddities. $\endgroup$ – Steeven Jun 4 at 8:46
  • 2
    $\begingroup$ The acceleration is horizontal, the lift force is vertical. This looks like nonsense to me. $\endgroup$ – alephzero Jun 4 at 8:54
  • $\begingroup$ @alephzero Because the wing has an angle of attack, the horizontal acceleration of course can produce vertical component. $\endgroup$ – enbin zheng Jun 4 at 11:42

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