Lift formula
 

Hi all,
Is there any easy way to explain why V is squared in the lift formula ?
I'm not entirely sure why it is squared.
Thanks.

It's really just describing the Kinetic Energy (Ek=1/2 m v^2) of the air mass, which is then combined with the other parts of the fomula's multipliers eg surface area. Ek increases with the square of the speed. Are you familiar with the idea that having a head on accident in your car at 4mph is nowhere near as bad as the same accident at 8mph? The speed is doubled but the *energy to be dissipated* has quadrupled!

If you double your TAS then you hit twice as much air per unit time.

You also hit it twice as hard.

You can show that it has to be that way if you reduce it to dimensional analysis. All units resolve to one of three dimensions - kilogrammes (mass), metres (distance) or seconds (time).

So lift force has dimensions of N. One Newton is the force to accelerate 1kg at 1m/s/s. So the dimension is kg.m.s^-2

Now look at the basic equation of lift, L=0.5 Rho V^2 S Cl

Rho, density, is kg.m^-3
V, speed, is m.s^-1
S, area is m^2

So take the right hand side of the equation and you get

0.5 x kg.m^-3 x (m.s^-1)^2 x m^2 x Cl

You can actually ignore the 0.5 and the Cl at the moment as we're only interested in dimensions. So, ignore those, and multiply out, we get the dimensions for the right hand side of the equation as...

kg.m^-3.m^2.s^-2.m^2

Which is kg.m.s^-2

Which has the same dimensions as force, hence the dimensions of the equation work out. If it was ony V on the right hand side, it wouldn't.

Not for the first time, I wish that Pprune would let me type equations like I can in MS Word !


This is actually a pretty standard physicists trick. If you know that one term must be a function of some other terms, in this case...

Lift =F(speed, air density, aeroplane size, something to do with shape)

You reduce speed, density and size to dimensions, and throw in a coefficient, which here we call coefficient of lift nad has no dimensions. Then doing it this way you can work out what you *think* the dimensions look like, and from there you throw in a load of experimental data and start trying to see (a) if you are right, or need to go back and re-think the whole thing, and (b) what values your coefficient has. The same basic approach was used to work out the dimensions, and thus units, for things like Reynolds Number, viscosity, Strouhall number and dozens of others.

G

Absolutely fantastic, thank you for the replies.

This question has intrigued me for years. I didn't really get from the answers why it is proportional to V squared and not just V. Same question applies to braking distances and drag.

I enjoyed the dimensional analysis but that started assuming v^2 is a given (I think) Is there a non formulaic explanation why doubling the speed quadruples the lift, drag, braking distances?

I think that if you want to find a non-mathematical explanation for a mathematical formula, you are somewhat doomed to failure.

Half-Rho-V^2 exists in multiple places as a term to define a force. It's in the lift formula, drag formula, Bernoulli's equation, rolling friction...

Basically I think that either you need to accept it "just is", or learn and work through enough maths to follow the derivation of kinetic energy, Bernoulli's equation, and then the lift force equation from basic principles.

Sorry. If you do want to give that a go, I can recommend some books and appropriate chapters, but realistically this is A level maths and Physics, then a chunk of first year degree level engineering.

G

The lift formula is in fact a definition. It's the definition of the coefficient of lift. There is no knowledge implied in a definition. You may define whatever your want.
The knowledge comes when you make experiments in a wind tunnel. Then you can see that the coefficient of lift, at usual GA aviation speed & pressure, does not depend on speed, nor on temperature, nor on pressure. The coefficient of lift only depends on the angle of attack and on the shape & surface of the airfoil. That's an experimental knowledge, and it's enough to built safe aircrafts.

Most of the time in history, knowledge comes first with experience, then a mathematical model is built and some theoretical knowledge can be built on the basis of the previous experience.
Most explanations of lift in pilot's manual don't explain anything (see https://www.grc.nasa.gov/www/k-12/airplane/wrong1.html, https://www.grc.nasa.gov/www/k-12/airplane/wrong2.html, etc.).

The best explanation, to my point of view, is to explain the actual experimental protocol that led to the properties of the coefficient of lift.

Why is this being over complicated?
Lift = airflow over the wing.
Airflow is determined by speed so “V” is the most important.
Hence V squared in the lift formula.
Private pilot pre solo student explanation.


* I’ll get my coat :}

It isn't.


The coefficient of lift is the ratio of lift pressure to dynamic pressure.

Yes... my earlier response for lift and drag.

Let me expand a bit.

Remember, although we tend to draw wings as stationary with air flowing around them it is equally valid to consider the air as stationary and the wing moving though it. It's just more difficult to draw in a simple graphic!

If you double your TAS you hit twice as much air per unit time.

You also hit it twice as hard.

Twice the amount, twice as hard = 4 times the force.

Does that help for an intuitive rather than formulaic explanation?



For braking energy there isn't really a good non formula method, at least not that I've found. But like most mechanics, it really isn't that complicated as long as your basic science is actually understood. Unfortunately far too few people actually understand basic mechanics but rather know it by rote. And there are far too few Physics graduates teaching science (and far too many biologists).

However if you can get your head around the idea that work = force x distance then you may be able to visualise that when an object is already moving it travels further per unit time. Thus the faster you go, the more energy you need to put in for the same change in velocity.

½ρV² is simply the result of first order integration of ρV with respect to V...

Study the derivation of Bernoulli's result, considering incompressible fluids and the continuity equation for inviscid fluid flow and you'll find that dp=-ρVdv. Simple integration will then yield p= - ½ρV² + const, meaning that the sum of static and dynamic pressure remains constant for the steady flow of an ideal fluid. Since p is static pressure (normally written as S), ½ρV² can only be the dynamic pressure, normally written as Q. Thus S+Q=const!

If you don't understand the level of schoolboy calculus which I was taught at the age of 15, then just accept that L=CL½ρV²S....

Calculus alone will not lead to the given lift equation which, as indicated in earlier posts, is simply a definition.