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
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    $\begingroup$ 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.) $\endgroup$ – Russell Borogove 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

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  • $\begingroup$ It's not F=mv/t, but F=m$\Delta$v/t $\endgroup$ – Acccumulation May 10 '18 at 17:46
  • $\begingroup$ The ½ comes from an integration step, but never mind. $\endgroup$ – Peter Kämpf 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. $\endgroup$ – Charles Bretana 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. $\endgroup$ – Charles Bretana 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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The correct answer depends on having a correct understanding of what causes lift. Strangely engineers still debate this. For example, Navier-Stokes equations are all about airflow, and provide no insight into lift; So cannot answer this question. But if lift is due to the wings pushing air down, the 'equal & opposite' force pushes the airplane up. Then Newtonian view of lift can answer the question..... As Force (lift) = ma; then lift is proportional to the mass of air flown through by the wings and pushed down ('m'), and the acceleration of this air ('a') downwards. So, if aircraft velocity doubles; Then the wings fly through x2 as much air each second. The 'm' (mass of air) doubles. Also, as the airplanes is going x2 as fast, then it hits each air molecule x2 harder, and this air flown through is accelerated down x2 as fast as before. Therefore, in summary, if aircraft velocity doubles, then lift quadruples: Force (LIFT) x 4 = (m x 2) x ( a x 2). Simple.

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  • $\begingroup$ This is actually a very sensible answer as velocity need only compared to amount of air the area strikes and the momentum transferred to each air molecule. Both increase x 2. Notice transfer of energy is mv, not 1/2mv2. The 1/2 does not seem to belong in the equation, at best it is combined with coefficient of lift constant? $\endgroup$ – Robert DiGiovanni Aug 9 '19 at 13:25
  • $\begingroup$ I did a 4 minute video explaining this see: youtu.be/hIx_hRjWTA8 or buoyancy-explains-flight.com/a-physics-of-lift This helps to resolve the debate on how planes fly; like NASA says, lift is explained by just applying Newtons laws of motion (not Navier-Stokes equations). $\endgroup$ – Nick Landell Aug 14 '19 at 8:18

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