Theory for the coefficient of lift

[Endercreeper01]

I am currently working on a theory for the coefficient of lift. Here is what I have so far:
C_L=Δp/q
Where Δp is the change in pressure between the top and bottom of the airfoil, and q is the dynamic pressure; q=1/2ρv²
This is because of the following:
The force of lift would also be equal to AΔp, since the force of lift depends on the change in pressure between the wings and the cross sectional area.
Because the force of lift is qAC_L, then we can write the coefficient of lift as AΔp/qA, which simplifies to Δp/q.
Now, I have to find out the value of Δp.
What do you think so far?

Edited December 6, 2013 by Endercreeper01

[Bignose]

I think it is a gross simplification of lift.

 

Lift [math]=\int_{\partial V} \mathbf{n} \, p \, ds[/math] which represents the integral of the pressure force normal over the entire surface. Replacing this integral by just a simple difference won't work in general.

 

And you have the exact same problem as I pointed out with drag, knowing the distribution of the pressure over the entire surface requires solution of the Navier-Stokes equations, which will only be analytic in some very, very special cases. You will either have to use CFD, a correlation for [math]C_L[/math], or some of the conformal mapping tricks aeronautical engineers were very, very good at (though largely supplanted by CFD today).

[Endercreeper01]

I think it is a gross simplification of lift.

 

Lift [math]=\int_{\partial V} \mathbf{n} \, p \, ds[/math] which represents the integral of the pressure force normal over the entire surface. Replacing this integral by just a simple difference won't work in general.

 

 

F=Ap, no?

Edited December 6, 2013 by Endercreeper01

[Endercreeper01]

We can also do that same integral divided by Aq.

[Bignose]

F=Ap, no?

No. 1) F is a vector quantity, and neither A nor p are vectors, so this isn't right.

 

2) You have no account for the variation of pressure over the body. That's what the integral does, it adds up the amount of pressure over the entire body. your equation would only be valid for something like a flat plate. It is a gross simplification of the true problem.

 

That is, there are times when the integral will evaluate to A*p, but that is going to be a fairly rare exception, not the rule.

 

So, you're left with, again, the tricky question of how you are going to evaluate the integral of the normal pressure over the entire body.

 

Also, I think it should be said that so far, you don't have a 'theory for the coefficient of lift'. So far, we're just using the terms as they are defined. i.e. you haven't brought anything new. If you have a new way of calculating it, then we might have something, but so far all you have is the definitions of terms...

Edited December 6, 2013 by Bignose

[Endercreeper01]

No. 1) F is a vector quantity, and neither A nor p are vectors, so this isn't right.

 

Pressure is a vector. Even hyper physics agrees that p=F/A

http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

Therefore, F=Ap

 

2) You have no account for the variation of pressure over the body. That's what the integral does, it adds up the amount of pressure over the entire body. your equation would only be valid for something like a flat plate. It is a gross simplification of the true problem.

 

Remember, it was still being developed, and I am working on that.

Another thing we can do is divide that integral by Aq, since F_L=qAC_L

For the Δp/q part, I am just trying to find the pressure change. Then, it will be a new way of calculating the lift coefficient.

Edited December 6, 2013 by Endercreeper01

[Bignose]

Pressure is NOT a vector. Pressure is a scalar.

 

The pressure force is F/A.

 

This is fluid mechanics 101, again. And a serious lack of clear communications. You can't just toss these terms around however you like.

 

For the Δp/q part, I am just trying to find the pressure change. Then, it will be a new way of calculating the lift coefficient.

That's fine. You explicitly asked what we thought about what you had so far. I gave you an answer in that so far you haven't done anything, but grossly simplifed the definition of how lift is calculated, and expressed my concern that this gross simplification would in all likelihood lead to erroneous predictions.

 

If you didn't want this feedback, maybe you shouldn't have solicited it.

 

Just like in your last thread, I really await the posting of the predictions of this forthcoming idea and comparing them to the predictions made by the best theories published today and measured experimental results. I do really hope you take some lessons learned from that previous thread and apply them here.

 

Because, it is really important that you not only have a new way of calculating the lift coefficient, but that you have an accurate way of calculation the lift coefficient. New alone doesn't mean squat if it isn't accurate. And, in my opinion, during that pressure integral into just a difference, won't be very accurate. But, prove me wrong. Show your calculations and how they compare to measured values.

[Endercreeper01]

Pressure is NOT a vector. Pressure is a scalar.

 

The pressure force is F/A.

 

This is fluid mechanics 101, again. And a serious lack of clear communications. You can't just toss these terms around however you like.

 

 

That's fine. You explicitly asked what we thought about what you had so far. I gave you an answer in that so far you haven't done anything, but grossly simplifed the definition of how lift is calculated, and expressed my concern that this gross simplification would in all likelihood lead to erroneous predictions.

 

If you didn't want this feedback, maybe you shouldn't have solicited it.

 

Just like in your last thread, I really await the posting of the predictions of this forthcoming idea and comparing them to the predictions made by the best theories published today and measured experimental results. I do really hope you take some lessons learned from that previous thread and apply them here.

 

Because, it is really important that you not only have a new way of calculating the lift coefficient, but that you have an accurate way of calculation the lift coefficient. New alone doesn't mean squat if it isn't accurate. And, in my opinion, during that pressure integral into just a difference, won't be very accurate. But, prove me wrong. Show your calculations and how they compare to measured values.

 

Because p is not a vector, then I will change it to F=Apn, where n is the unit vector that the pressure is pointing in. And also for this, then it would be FAΔnΔp. Would that work?

And also, once I finish my theory, then I will show you the calculations.

This would also change it to C_L=ΔnΔp/q. Or, if that doesn't work, then it will then be that integral that you posted divided by qA.

I also forgot about how p=dF/dA. Could you also write it as F=integral(p) dA, since p=dF/dA? And if so, would that make C_L=(integral_A(p) dA)/(qA), giving you (integral_A dp)/q?

Edited December 7, 2013 by Endercreeper01

[Bignose]

Because p is not a vector, then I will change it to F=Apn, where n is the unit vector that the pressure is pointing in.

I'm sorry, but this just still isn't right.

 

Pressure doesn't point in a direction. Pressure actually applies equally in every direction. This is actually a big reason why pressure is a scalar.

 

The normal vector points in the direction normal to the area.

 

I am hesitant to actually address anything more, because, again this is fluid mechanics 101 basic stuff. I really feel like you are just symbol pushing, and not understanding at all what the equations you are writing actually mean. And, again, I'm sorry, but this isn't any way to actually form a meaningful idea. You need to actually understand what you're doing, and writing things like "the pressure is pointing in" shows you have very fundamental misunderstandings.

 

Have you gotten yourself a fluid mechanics text? Like the one I suggested in the other thread? You really, really need to be reading through that to correct these wrong ideas you have.

 

Before you try to tackle something as complicated as lift, I'd like to see that you have some of the basics down first.

[Endercreeper01]

I'm sorry, but this just still isn't right.

 

Pressure doesn't point in a direction. Pressure actually applies equally in every direction. This is actually a big reason why pressure is a scalar.

 

The normal vector points in the direction normal to the area.

 

I am hesitant to actually address anything more, because, again this is fluid mechanics 101 basic stuff. I really feel like you are just symbol pushing, and not understanding at all what the equations you are writing actually mean. And, again, I'm sorry, but this isn't any way to actually form a meaningful idea. You need to actually understand what you're doing, and writing things like "the pressure is pointing in" shows you have very fundamental misunderstandings.

 

Have you gotten yourself a fluid mechanics text? Like the one I suggested in the other thread? You really, really need to be reading through that to correct these wrong ideas you have.

 

Before you try to tackle something as complicated as lift, I'd like to see that you have some of the basics down first.

Because p=dF/dA, then would it be F=integral_A( n dp)? If not, what would F be in terms of A? For n, I meant the vector normal to the area.And also, for n, I meant something else then what I said in the other post. I have trouble communicating sometimes, and this sometimes makes it hard to explain things. Sorry for the communication errors.

And, I'm still waiting for my order to come.

Edited December 7, 2013 by Endercreeper01

[Endercreeper01]

Also, why is pressure a scalar if force is a vector?

Edited December 9, 2013 by Endercreeper01

[Bignose]

this isn't a perfect definition of what a vector is, but it'll be good enough for these purposes: a vector is a quantity that has a magnitude and a direction.

 

Force does this, right? You have a magnitude -- how much force there is, and a direction -- the direction the force is applied in.

 

But pressure doesn't have this. It has how much pressure there is... it has the magnitude. But it doesn't have a direction. Pressure applies, well, pressure equally in every direction.

 

Hence it is not a vector. It doesn't have a direction.

 

That's where the normal vector comes in. If you want to find out how much force the pressure applies on a spot on a surface, the pressure, over that infintesimal area, is applied normal to that area.

 

And that's where the integral comes in. To find the total force on the object, you have to sum up all the infintesimal bits. Hence the integral I wrote in post #2. Note the form of that. The LHS is a vector, the lift force. The RHS is a vector because of the normal vector in there. The pressure in the integral on the RHS is a scalar, only n is a vector quantity on the RHS.

 

Just like units, it is important to keep track of what is and isn't a vector quantity. The tensor rank of each side of the equation has to be equal, just like the units on each side of the equation has to be equal. Scalars have tensor rank 0, vectors have tensor rank 1, and tensors have rank 2 or greater.

[Endercreeper01]

Could we also write the integral as integral_A(np dA), where integral_A would mean the integral over the area?

[Bignose]

if you want, so long as you define the terms. It looks like it is just a re-write of what I put in post #2. I used [math]\partial V[/math] for the area, and ds. You are using A's. So long as they mean the same thing, it's, well, the same thing.

 

I'd suggest you use the forum's LaTeX capabilities, rather than writing "integralA", though. The symbols help make it a lot clearer. There is a LaTeX primer thread stickied somewhere if you need help.

Edited December 10, 2013 by Bignose

[Endercreeper01]

I can't find the sticky, though. Do you know which forum it is in?

Edited December 10, 2013 by Endercreeper01

[hypervalent_iodine]

http://www.scienceforums.net/topic/3751-quick-latex-tutorial/

[Endercreeper01]

And now, I will talk about what I have done so far.

First, you have F_L= [latex]\int_{A}[/latex] np dA. To get the coefficient of lift, then you must divide this by [latex]qA[/latex]. This gives you C_L=[latex]\int_{A}[/latex] [latex]{np}/{qA}[/latex] [latex]{ dA}[/latex], which is equal to [latex]\int_{A}[/latex] [latex]{n}/{q}[/latex] [latex]{ dp}[/latex]

Using integration by parts, you find that:

 

latex]\int_{A}[/latex] [latex]{n}/{q}[/latex] [latex]{ dp}[/latex] = nΔp/q - [latex]\int_{A}[/latex] p/q dn

 

For the first term, then it is change in pressure because it is evaluated among the top and bottom areas, and so, you get the change in pressure. I just need to find what this change would be.

For the second term, then I need to make this integral in terms of A, so I changed dn to dA dn/dA, and this makes it an integral with respect to A. Now, I need to find dn/dA, along with the change in pressure.

 

 

 

[Bignose]

And now, I will talk about what I have done so far.

First, you have F_L= [latex]\int_{A}[/latex] np dA. To get the coefficient of lift, then you must divide this by [latex]qA[/latex]. This gives you C_L=[latex]\int_{A}[/latex] [latex]{np}/{qA}[/latex] [latex]{ dA}[/latex],

Really? You're integrating over the area, and you just move the area into the denominator of the integrand like this? That isn't valid. You can't just willy nilly change the integrand on the variable you're actually integrating on. This needs to be corrected.

[Endercreeper01]

Really? You're integrating over the area, and you just move the area into the denominator of the integrand like this? That isn't valid. You can't just willy nilly change the integrand on the variable you're actually integrating on. This needs to be corrected.

Oh yes, I need to change that. This integral [latex]\int_{A}[/latex][latex]np/qA[/latex] dA is the same as n/q[latex]\int_{A}[/latex][latex]p/A[/latex] dA, since n and q are not functions of A. We then get:

C_L = n/q[latex]\int_{A}[/latex][latex]p/A[/latex] dA

Using u substitution, we can set du as du=(1/A)dA, making the integral C_L = n/q[latex]\int_{A}[/latex][latex]p[/latex] du.

Using integration bu parts, we find that n/q[latex]\int_{A}[/latex][latex]p[/latex] du = n/q(up - [latex]\int_{A}[/latex]u dp)

Because du=dA/A, and so u = -1/A², then we can rewrite this as n/q( [latex]\int_{A}[/latex]1/A² dp - p/A² ), or [latex]\int_{A}[/latex]n/qA² dp/dA dA - nΔp/qA². There is a change in pressure for the second term since it is evaluated among both areas, the top and bottom, giving us the change in pressure. I now need to evaluate the integral and find the change in pressure.

[Endercreeper01]

Would this work?

[Bignose]

Oh yes, I need to change that. This integral [latex]\int_{A}[/latex][latex]np/qA[/latex] dA

No, you are missing my point. My point is that you can't bring the 'A' into the integrand like this because you are integrating over A. This invalidates all the other steps.

 

[math]\frac{\int_A \mathbf{n} p \, dA}{qA}[/math] is what it really looks like and is not the same as what you wrote.

[Endercreeper01]

This would mean I would have to work out the integral, and then I will divide by qA

[Endercreeper01]

If [latex]{\int_A}[/latex]np dA = F_L = qAC_L, then would this mean that Δp=qC_L, since dF/dA=Δp, and the derivative works out to be qC_L?

Edited December 14, 2013 by Endercreeper01

[Endercreeper01]

Also, since p=p₀-q, does this mean that Δp=q₁ - q (where q is the dynamic pressure at the bottom of airfoil and q₁ is the dynamic pressure at the top of the airfoil), since Δp=(p₀ - q) - (p₀ - q₁), giving you Δp=q₁- q?

[Endercreeper01]

If both of the above posts are true, then C_L=Δp/q, which is the same as (q₁-q)/q, giving you C_L = [latex]\frac{q_1}{q}[/latex] - 1. What I would now have to do is find the velocity at the top of the wing compared to the velocity at the bottom of the wing.