Why is the velocity $v$ squared in the lift equation? $$L=C_L\cdot\frac{1}{2}\cdot\rho\cdot v^2\cdot S$$

  • $\begingroup$ Can you share the formula you are asking about? $\endgroup$
    – CrossRoads
    May 10 '18 at 12:52
  • 3
    $\begingroup$ At a very high level of abstraction, it's because you're hitting more air per unit of time (proportional to v) and you're hitting each bit of air with more momentum (also proportional to v). (At least, that's why drag is proportional to v^2; I assume it holds for lift.) $\endgroup$ May 10 '18 at 21:55

Newton's second law is:

$$F = ma$$

The $m$ass in the lift equation is of the fluid's, mass times velocity is the fluid momentum. So the equation can be written as:

$$F = {\Delta}(mv)/{\Delta}t$$

Rearrange that equation and it becomes:

$$F = \text{constant} \cdot (m/t)v$$

Mass over time is the mass flow rate:

$$m/t = {\rho}vA$$

So we end up with:

$$F = \text{constant} \cdot {\rho}vAv$$

Which is:

$$F = \text{constant} \cdot {\rho}v^2A$$

Add $A$ to the constant:

$$F = \text{constant} \cdot {\rho}v^2$$

Reference: NASA

  • $\begingroup$ It's not F=mv/t, but F=m$\Delta$v/t $\endgroup$ May 10 '18 at 17:46
  • $\begingroup$ The ½ comes from an integration step, but never mind. $\endgroup$ May 10 '18 at 19:57
  • $\begingroup$ -1, Derivation is wrong, or, at best, misleading. The A in Mass flow rate is not the same as the S in the Lift equation. The A is the cross-sectional area through which the mass is flowing, (normal to the flow), not the surface area of the airfoil. $\endgroup$ May 11 '18 at 17:45
  • $\begingroup$ @CharlesBretana - fixed the $A$ part, thanks, rest is correct backed up by the link. $\endgroup$
    – ymb1
    May 11 '18 at 19:00
  • $\begingroup$ removed my downvote, thanks for correcting that.. Also suggest you think hard about starting out with $F=ma$. I believe there's a flaw there. The idea is substantively correct, but I think (not positive), that the NASA derivation is about flow through an engine, or a defined cross-sectional area, and not for lift on an airfoil. But they both proportional to $V^2$ for the same reason - change in momentum, which is what you're getting at. $\endgroup$ May 11 '18 at 21:34

Kinetic Energy is why the V2 is there. The formula for kinetic energy is:

$$E_{kin} = \frac{1}{2}\cdot m\cdot v^2$$

and the greek letter rho is the density of the air which is mass per unit volume. So you can see the formula for kinetic energy in the lift equation. Cool huh? So what the lift equation is telling you is that the kinetic energy of the wing moving through the air is responsible for lift and we all know that you need velocity to lift a plane off the ground. This also tells you that velocity is the biggest contributing factor to the lift component so airspeed gives you lift more than anything else.


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