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Old 19th Dec 2006, 03:27   #1 (permalink)
 
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Formula for lift: Question

So, why is it 1/2 x rho (air density) in the formula?

I understand that it is dynamic pressure but why the one-half?
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Old 19th Dec 2006, 04:13   #2 (permalink)
 
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Quote:
So, why is it 1/2 x rho (air density) in the formula?
Correct me if I'm wrong, but I thought the 1/2 applied to the V squared part of the formula?

or

The Baffler

Last edited by baffler15; 19th Dec 2006 at 04:16. Reason: spel cheking
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Old 19th Dec 2006, 05:15   #3 (permalink)
 
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Quote:
Originally Posted by baffler15 View Post
Correct me if I'm wrong, but I thought the 1/2 applied to the V squared part of the formula?

or

The Baffler
Whatever! But why the 1/2?!
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Old 19th Dec 2006, 05:18   #4 (permalink)
 
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It all goes back to the initial experiments with lift. They were able to measure the lift, and were trying to devise a formula that gave them the answer they already had.

All related to Kinetic Energy, which is 1/2 x mass x V squared (obtained by integration from momentum, mV)
To find mass, you need air density (rho) x surface area
Thus for airflow, you got 1/2 x (rho x s) x V squared

Plus the variable with angle of attack, the coefficient of lift.

Or near enuf .....
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Old 19th Dec 2006, 05:29   #5 (permalink)
 
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Great. Thanks DD, it's been bugging me!
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Old 19th Dec 2006, 05:42   #6 (permalink)
 
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Not sure if this is really a serious question, but here goes a serious answer.

The 1/2 rho v squared is essentially the kinetic energy density or kinetic energy per unit volume. Think of rho as mass over volume and the formula can be made to look like 1/2 mass v squared over volume. Of course 1/2 m v squared is the kinetic energy formula you would have learned during basic physics.

So where does the 1/2 come from in 1/2 m v^2? Try a simple derivation for an arbitrary mass that is initially at rest. If you move the mass over a distance, x, using a force, F. Then the work you perform is W=Fx. Substitute F=ma and recognize that without friction all the work you do increases the kinetic energy of the mass and you get KE=max. Now you assume constant acceleration over the whole distance and apply some basic calculus (or just borrow the answers from a textbook) and you kind that v^2=2ax, where v is the final velocity after moving distance x. Substitute back into the equation and you get KE = 1/2 m v^2.

It comes into the lift equation because that is developed using basic principles of physics. When the kinetic energy term (or more accurately KE density) is required a 1/2 shows up. It isn't associated specifically with the rho nor the v^2 but with the entire KE term.

A word of caution: with all physics you'll find that there are multiple ways of describing the same thing. It may be that you find a derivation where the 1/2 comes from somewhere else. That derivation may not be wrong, just different, but as long as the final answer is the same and the concepts are consistent then the alternate derivation is valid.
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Old 19th Dec 2006, 06:45   #7 (permalink)

 
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And I always thought that the ½ was there to represent the average of a body’s velocity between the start and end of its travels. Oh well.

Phil
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Old 19th Dec 2006, 09:10   #8 (permalink)
 
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Basic calculus (if there is such a thing !!!)
Let me try for a simpler version of Matt's answer, and a more correct version of DD's answer.
The lift depends on the kinetic energy of the free stream
From physics we have:
momentum = mass x velocity, and
kinetic energy = the derivative of momentum
From any calculus 101 textbook:
If momentum = mv, then
kinetic energy = 1/2 mv^2
Now substitute rho for m, and you get the answer as it applies to free stream aerodynamics.
Does that help ? Dunno !! It's still simplified (a lot) but it's more accurate.
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Old 19th Dec 2006, 10:01   #9 (permalink)
 
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Paco, you are correct, 1/2 is the average velocity. Don't let the Calc fool you, since the relationship of momentum to velocity is linear, the 1/2 is just the average!
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Old 19th Dec 2006, 11:55   #10 (permalink)
 
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Another explanation

Another way of looking at it is that the term 1/2 * rho * V^2 is the dynamic pressure. The 1/2 is there to ensure that the term is comparable with the static pressure.

Along any streamline the static pressure and the dynamic pressure sum to a constant. So where the velocity is higher, the dynamic pressure is higher and hence the static pressure is lower - Bernoulli's principle as we all know it.

If we didn't have the 1/2 term, then this relationship would not hold.

Of course, this relates to what Mathew Parsons was saying:

Quote:
The 1/2 rho v squared is essentially the kinetic energy density or kinetic energy per unit volume.
Static pressure is the the kinetic energy density associated with the random thermal motion of the air molecules. Dynamic pressure is the kinetic energy density associated with the non-random motion of the air molecules.

Daniel
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Old 19th Dec 2006, 14:45   #11 (permalink)
 
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Actually the 1/2 doesn't make an average velocity because the average is v/2 and when you square that you get v^2/4.

However, you may be able to come up with a derivation that uses both the average velocity, v/2, and the final velocity, v.

Matthew.

Last edited by Matthew Parsons; 21st Dec 2006 at 21:55.
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Old 20th Dec 2006, 00:20   #12 (permalink)
 
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Quote:
Originally Posted by Matthew Parsons View Post
Not sure if this is really a serious question, but here goes a serious answer.
Yes, it was a serious question and thanks for the serious answers!
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Old 20th Dec 2006, 01:16   #13 (permalink)
 
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The half has nothing to do with averages.

It has everything to do with integral calculus.

Start with momentum, MV. Integrate it once (with respect to V), you get 1/2 MV squared. If you integrated it again, you get 1/6 MV cubed, but this item means nothing to us, so we don't do it. (There are a few constants thrown in as well, but we won't worry about them either.) Purely mathematics, no averages.
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Old 20th Dec 2006, 02:07   #14 (permalink)
 
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It does not matter any way as it is incorrect theory, (and it was only a theory after all), before you start typing have a look at NASA's website and type "incorrect theory" in the search engine.

Only 2% of lift is generated by the bernouli theory and the rest is Coanda effect.

Don't argue with me argue with NASA, (you'll lose)
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Old 20th Dec 2006, 07:44   #15 (permalink)
 
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Punto - I think you'll find Deemar mentioned Bernoulli first and since the argument is about a formula used in his theorem (or a derivation of it) then it seems fair to mention him. However, since his theorem concerns ideal, incompressible fluids and ignores friction it is not surprising that it had to be fudged to explain what happens in the real world (isn't that what scientists and mathematicians do?)
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Old 20th Dec 2006, 07:49   #16 (permalink)
 
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HaHaHa. Deemar, Deeper it's all the same to me. Apologies to Deeper for the slur ! Post deleted.

Last edited by puntosaurus; 20th Dec 2006 at 08:23.
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Old 20th Dec 2006, 11:18   #17 (permalink)
 
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I know I posted it on another thread---but I believe this thread deserves it too..........Sometimes my brain hurts ! ! !

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Old 20th Dec 2006, 14:36   #18 (permalink)
 
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Why would any pilot bother with where the lift formula comes from? You'll never need it. Unless you want to be a designer later don't bother.
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Old 20th Dec 2006, 14:58   #19 (permalink)
 
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Agreed Helibee, It makes no difference to a working pilots daily life. Lack power, dont lift to a higher DA/temp etc. so have a day off, easy as that, but they dont teach that stuff.

Last edited by Heliringer; 20th Dec 2006 at 15:09.
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Old 21st Dec 2006, 13:23   #20 (permalink)
 
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airfoils and the oneness of zen

Quote:
Originally Posted by deeper View Post
It does not matter any way as it is incorrect theory, (and it was only a theory after all), before you start typing have a look at NASA's website and type "incorrect theory" in the search engine.

Only 2% of lift is generated by the bernouli theory and the rest is Coanda effect.

Don't argue with me argue with NASA, (you'll lose)
Well, a quick search of the nasa site under incorrect theory doesn't show anything on the ongoing coanda/bernoulli/deflected airstream debate (well at least on the first page of results).

Nevertheless, it amazes me that people continue with the misconception that the three explanations are different. They are all different ways of looking at the same effect. It goes as follows if you decide to start with the momentum change associated with deflecting the airstream. You can, of course, start from coanda or bernoulli and get the same conclusions. It is the same physical effect.

1) In order to produce lift we need to produce a change in momentum (F = dp/dt, where p is momentum). This is achieved by deflecting the airstream passing over the airfoil.

2) In order to deflect an airstream, you need to have a pressure gradient across the airstream. To deflect the airstream downwards the pressure above the airstream needs to be higher than the pressure below.

3) For the air going over the top of the airfoil, the pressure above it is just the free stream pressure, therefore in order for there to be lower pressure below this stream, the pressure at the top of the airfoil needs to be lower than the free stream pressure.

4) Conversely, for the air going underneath the airfoil, the pressure below it is just the free stream pressure, so the pressure just below the airfoil needs to be higher than the free stream pressure.

5) The high pressure below the airfoil and the low pressure above it results in the airfoil being pushed up. This is the mechanism by which the force due to the change in momentum of the airstream is conveyed to the airfoil.

6) Because the pressure above the airfoil is lower, the air velocity needs to be higher to comply with bernoulli's law (P + 1/2*rho*V^2 = constant). Conversely, because the pressure below the airfoil is higher, the velocity needs to be lower.

7) Circulation is defined as the integral of air velocity around a closed path encircling the airfoil. Basically, you can think of the faster upper surface velocity and the slower lower surface velocity as being comprised of an equal velocity above and below superimposed with a weaker circulating flow that goes from the back of airfoil to the front along the lower surface, and then comes back along the top surface. The flow circulating flow along the top adds to the base velocity leading to a higher velocity over the top of the airfoil. The circulating flow is subtracted from the flow underneath the airfoil, resulting in a lower velocity under the airfoil. Note very carefully that the circulation is really just a mathematical tool to describe the difference in speeds above and below the airfoil.

8) The coanda effect basically states that the lift of the airfoil is proportional to the circulation times the free stream velocity. But remember, we only have a circulation because there is a difference in speeds above and below the airfoil, and we only have the difference of speeds because bernoulli requires it to generate the difference in pressures, and we only have the difference in pressures because we are deflecting the airstream.

So, you can't claim that only 2% of the lift is due to the "coanda effect", it is all due to the coanda effect, but then it is also all due to the deflection of the airstream, etc... They are all part of one and the same physical phenomenon.

The theory of circulation (coanda effect) is important mainly from a theoretical point of view, it provides a mechanism for calculating the strengths of downwashes and trailing vorticies. All very useful when Nick Lappos and co want to work out how to make a more efficient, quieter rotor.

Questions?

Daniel
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