I am trying to understand the derivation of some simple equations to calculate the neccecary takeoff speed for an airplane (under ideal conditions).

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Which is taken from 15. Takeoff and Landing. My main issue is the derivation of the horizontal equation

To get from the first to the second line my guess is that we use

R = W - L   and    m = W / g

My main problem is understanding the details between what happens from line 2-3 and 3-4. I would love if someone could give me a brief explenation of the symbols used, what does S denote?

  • 2
    $\begingroup$ What don't you understand from 2-3? It's just rearranging the algebraic terms. #4 replaces lift and drag with the lift and drag coefficients. $\endgroup$ – JZYL Jan 20 at 13:43

We want to reduce (or simplify) equation 2, to get get dV/dt by itself, so we multiply both sides by g/W. On the right hand side, this reduces to dV/dt. On the left hand side it reduces to what is given in equation 3, but let me show you how: 1. First term is T x g/W 2. Second term is -D x g/W 3. Third term is -u x g/W (W - L). By the distributive property, we have -[u x g/W x W] + [u g/W x L]. The first part of this term simplifies to u x g.

Finally, I want to group like terms together.

  1. The first term combined with the first part of the third term give me: T x g/W - ug = g(T/w - u).
  2. The second term combined with the second part of the third term give me: -D x g/W + u g/W L = g/W(u L - D).

This yields equation 3: g(T/w - u) + g/W(u L - D) = dV/dt.

Finally, recall that L = 1/2 p S V2 Cl, with a similar equation for Drag, and we can substitute these directly into L and D to get equation 4.

S is the the wing planform area. Other sources use A, like this one: NASA - Lift equation.

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