Theory for the coefficient of lift

[Bignose]

Why do you keep insisting on evaluating the integral as just a delta? This is going to be a terrible approximation in all but the most special of cases. This is my same objection from 20+ posts ago.

[Endercreeper01]

Why do you keep insisting on evaluating the integral as just a delta? This is going to be a terrible approximation in all but the most special of cases. This is my same objection from 20+ posts ago.4

This is why I keep using Δp: The integral is evaluated among the top and bottom of the airfoil, right? Because integrals are distributive, [latex]\int_{A}[/latex]p dA is the same as [latex]\int[/latex]Δp dA, since it is the change in the antiderivatives of pressure at the top and bottom of the wing, making it the same as the integral of the change in pressure (with respect to area). This means that the change in pressure Δp is the same as dF/dA, where F is the force of lift. Since the force of lift is also equal to qAC_L, then we need to take the derivative of this with respect to area to get the change in pressure. This gives us Δp=qC_L. Therefore, C_L=Δp/q. I am still working out what Δp will be, but I will tell you what I have so far. So far, I have that since p+q = p₀, where p₀ is the pressure when q=0 and is constant, this means that p = p₀ - q. When you have the change in pressure, then you take the change between the bottom and the top of the wing to find the pressure. If q₂is the dynamic pressure at the bottom of the wing and q₁​ is the dynamic pressure at the top of the airfoil, you get Δp = p₂ - p₁. This is the same as (p₀ - q₂) - (p₀ - q₁). P₀ cancels out on both terms, so this gives you Δp = q₁ - q₂. The coefficient of lift now becomes C_L = (q₁ - q₂)/q. What I have to do is figure out how the dynamic pressure changes depending on the geometry.

Edited December 15, 2013 by Endercreeper01

[Bignose]

Ender,

 

I have two philosophical questions for you.

 

1) If the integral can just be replaced with a delta, why do you think I and many other authors write out the integral the first place?

 

2) If pressure can be so easily replaced by dynamic pressure which is merely a function of the fluid density and velocity, why is pressure explicitly included in the lift calculation and the Navier-Stokes equation? Why wouldn't it just be written in terms of density and velocity? In particular, the N-S equations would be a lot simpler because it would eliminate a whole field of unknowns.

 

I'm not trying to be coy here, I'm trying to understand the root of your misconception, because what you are doing above is not valid, and I really don't understand why you think what you're doing is valid. I want to try to correct the error in your understanding, if you're willing.

[Endercreeper01]

Ender,

 

I have two philosophical questions for you.

 

1) If the integral can just be replaced with a delta, why do you think I and many other authors write out the integral the first place?

 

2) If pressure can be so easily replaced by dynamic pressure which is merely a function of the fluid density and velocity, why is pressure explicitly included in the lift calculation and the Navier-Stokes equation? Why wouldn't it just be written in terms of density and velocity? In particular, the N-S equations would be a lot simpler because it would eliminate a whole field of unknowns.

 

I'm not trying to be coy here, I'm trying to understand the root of your misconception, because what you are doing above is not valid, and I really don't understand why you think what you're doing is valid. I want to try to correct the error in your understanding, if you're willing.

1. I guess that would mean that the integral I had was incorrect. But what about the thing I did to find the change in pressure? I know I made a mistake somewhere in the derivation of the change in pressure as q times coefficient of lift, but I can't find it.

2. That was the change in pressure. The change in pressure is not the same as pressure. I derived it using bernoulii's principle, as shown in an above post. The pressure is not able to be replaced with dynamic pressure, but the change can be written as a change in dynamic pressure, as shown in above posts.

[Endercreeper01]

Do you know where the errors in the derivations are?

Edited December 16, 2013 by Endercreeper01

[Bignose]

I derived it using bernoulii's principle, as shown in an above post.

Here's a critical mistake, then. Bernoulli's principle only works for inviscid fluids. Navier Stokes fluids most assuredly have viscosity, and it is the solution of the Navier Stokes equations that are needed to give correct pressure and velocity distributions to give correct lift and drag calculations.

 

But what about the thing I did to find the change in pressure?

 

...you get Δp = p2 - p1. This is the same as (p0 - q2) - (p0 - q1).

I don't see why p2 - p1 should be equal to q2 - q1. How is p2 - p1 the same as (p0 - q2) - (p0 - q1)? There is no real reason a change in pressure, [math]\nabla p[/math], should be directly proportional to a change in the 'dynamic pressure', [math]\nabla q = \nabla \rho u^2 [/math].

 

The pressure change and the velocity change are not so directly related. That's exactly what the Navier Stokes equations say:

 

[math]\rho \frac{D \mathbf{u}}{D t} = \rho \left( \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} \right) = \nabla p + \nabla \cdot \mathbf{T} + \mathbf{b}[/math]

 

if you rearrange that for just [math]\nabla p[/math], you get a whole bunch of other stuff besides just [math]\nabla \rho \mathbf{u} \cdot \mathbf{u} = \nabla \rho u^2[/math].

[Endercreeper01]

Here's a critical mistake, then. Bernoulli's principle only works for inviscid fluids. Navier Stokes fluids most assuredly have viscosity, and it is the solution of the Navier Stokes equations that are needed to give correct pressure and velocity distributions to give correct lift and drag calculations.

 

 

I don't see why p2 - p1 should be equal to q2 - q1. How is p2 - p1 the same as (p0 - q2) - (p0 - q1)? There is no real reason a change in pressure, [math]\nabla p[/math], should be directly proportional to a change in the 'dynamic pressure', [math]\nabla q = \nabla \rho u^2 [/math].

 

The pressure change and the velocity change are not so directly related. That's exactly what the Navier Stokes equations say:

 

[math]\rho \frac{D \mathbf{u}}{D t} = \rho \left( \frac{\partial \mathbf{u}}{\partial t} + \mathbf{u} \cdot \nabla \mathbf{u} \right) = \nabla p + \nabla \cdot \mathbf{T} + \mathbf{b}[/math]

 

if you rearrange that for just [math]\nabla p[/math], you get a whole bunch of other stuff besides just [math]\nabla \rho \mathbf{u} \cdot \mathbf{u} = \nabla \rho u^2[/math].

In this case, do you mean change in pressure or gradient for [latex]\nabla[/latex]p? And also, the thing I derived earlier using Bernoulli's principle was based on the fact that p+q is constant, and this constant is the total pressure. Edited December 17, 2013 by Endercreeper01

[Bignose]

Why would p + q necessarily be constant? Again, that is true for an inviscid fluid, but that's not what we're talking about here.

 

and, you do realize that [math]\nabla p[/math] is the change in pressure, right? It is how pressure changes with position. Your delta of the pressure of the top and bottom is a (exceptionally poor) approximation of this change in pressure.

[Endercreeper01]

Would [latex]\nabla p[/latex] be constant among the airfoil?

Also, what about the post where I showed that [latex]\nabla[/latex]p=qC_L?

[Bignose]

Would [latex]\nabla p[/latex] be constant among the airfoil?

Only in very special cases, so really, no.

 

Also, what about the post where I showed that [latex]\nabla[/latex]p=qC_L?

where was that? no one even used the nabla symbol until I did, so I don't see where you showed this at all. Also, this equation as it is posted isn't dimensionally sound or tensor rank sound.

 

I know I am repeating myself here, but you really are just symbol pushing, and giving little to no thought about what the math is actually saying.

 

You kept trying to turn an integral, the sum of a whole bunch of forces over very small areas, into a difference. Now you're asking if the change in pressure, as you travel along an airfoil, would be a constant. What in the mathematics would suggest that the change -- in every direction mind you since you just wrote [math]\nabla p[/math], not its projection in a certain direction -- would be the same. If you had given that some thought, you'd realize how unlikely that really would be.

 

And this whole aside of using Bernoulli's equation where we are clearly discussing a viscous flow. This seems like you found an equation somewhere that said p + q = a constant. But, you obviously were missing under what conditions that was valid.

 

Again, this smacks of symbol pushing. You're seemingly just trying to get it out of the way (and ever insistently into some kind of delta form), and not thinking about what the mathematics you are writing is actually saying. To do this work (and really, most of science) correctly, you need to take the time and work on gaining an understanding of how the math is actually describing the physics.

Edited December 17, 2013 by Bignose

[Endercreeper01]

Only in very special cases, so really, no.

 

 

where was that? no one even used the nabla symbol until I did, so I don't see where you showed this at all. Also, this equation as it is posted isn't dimensionally sound or tensor rank sound.

 

I know I am repeating myself here, but you really are just symbol pushing, and giving little to no thought about what the math is actually saying.

 

You kept trying to turn an integral, the sum of a whole bunch of forces over very small areas, into a difference. Now you're asking if the change in pressure, as you travel along an airfoil, would be a constant. What in the mathematics would suggest that the change -- in every direction mind you since you just wrote [math]\nabla p[/math], not its projection in a certain direction -- would be the same. If you had given that some thought, you'd realize how unlikely that really would be.

 

And this whole aside of using Bernoulli's equation where we are clearly discussing a viscous flow. This seems like you found an equation somewhere that said p + q = a constant. But, you obviously were missing under what conditions that was valid.

 

Again, this smacks of symbol pushing. You're seemingly just trying to get it out of the way (and ever insistently into some kind of delta form), and not thinking about what the mathematics you are writing is actually saying. To do this work (and really, most of science) correctly, you need to take the time and work on gaining an understanding of how the math is actually describing the physics.

Just wondering.

I have made an equation for the average velocity among the top and bottom of the wing based on the free stream velocity v₀. θ_avg is the average angle for the top of the airfoil and φ_avg is the average for the bottom. All angles are measured with respect to the axis in the direction of motion. First, we know that the velocity at any angle θ would be the component of v in that direction. This would mean it is v₀cos(θ). The average of this among the top of the airfoil will therefore be v₀cos(θ_avg). This would be the same for the bottom of the airfoil, making it be v₀cos(φ_avg) for the bottom of the airfoil. Would this be correct?

Also, is their an equation for the force on lift based on [latex]\nabla[/latex]p?

BTW: I'm not just "symbol pushing" whatever I say. You can't just assume I am.

[Bignose]

Ender, I'm going to start at the bottom.

 

As I wrote above, you have demonstrated very little to no knowledge of what the mathematics is actually representing. And, for that matter, little to no knowledge of mathematics itself. You've continuously posted equations that aren't dimensionally consistent, and tensor rank consistent. And it isn't like once or twice, but over and over.

 

What exactly am I supposed to think? At the very, very best, you're demonstrating that you are incredibly sloppy with the mathematics. Or, as I've written above, that you are just symbol pushing. Pulling any equation from any source, taking shots in the dark hoping that the step is valid, and trying to turn something you don't understand into something you do.

 

Sure, this assumption could be wrong. But you haven't demonstrated it at all.

 

And, really, this latest post is just another example of this.

 

You say you are calculating the "average velocity among [sic, I think you mean along] the top and bottom of the wing..."

 

The problem is that it is a basic fluid mechanics assumption that the fluid velocity right at a solid boundary is equal to the velocity of that boundary. This assumption is only invalid if the fluid is exceptionally thin (the molecular density is such that the molecules can travel a very large distance before colliding with another molecule) or that the solid boundary is porous in some way. These are not valid assumptions for typical use of an airfoil.

 

Therefore, there is no need to do any kind of complicated and frankly wrong calculation of the fluid velocity along the airfoil. Because that average fluid velocity is just the airfoil velocity.

 

So, as I wrote above, either you are very sloppy here, or you are demonstrating lack of a rather elementary basic assumption of fluid mechanics.

 

I just don't know what else to tell you except to repeat what I've written several times before that you need to put some time in and understand the basic mathematics and equations of fluid mechanics. It is my strongest assumption that you don't have enough background for it yet. Your elementary mathematics mistakes exhibit this. If you don't agree, then quite simply, stop posting some many mistakes.

 

The last thing I want to say is that I hope you don't take this as a personal attack, because that is not how I intend it. We are all at different abilities and knowledge levels. The question you need to take back is: are you going to put in the time and work to gain the mathematical maturity to work through a modern fluid mechanics text and gain the knowledge and intuition of what the mathematics is describing? As before, I am willing to occasionally assist when I have time if you want to start on this path. But, I no longer want to assist in you fumbling, bumbling, and praying that you stumble on something correct. I've pointed out this lack of knowledge many times and the mistakes many times, and I see very little acknowledgement and correction of those mistakes. If you're not going to bother learning from these mistakes, I'm not going to bother pointing them out any more, because I don't think anyone is getting anything out of this.

 

Look. This kind of curiosity and creativeness is needed in sciences, including fluid mechanics. It is craved for. There is lots of active research in fluid mechanics, lots of neat questions that need further study, including drag and lift. But, you can't just toss things out there and hope they are right. You need to understand the mathematics and what it is saying in order to contribute something meaningful. I hope you will put the time in to learn the math and the current state of fluid mechanics so that you can then apply this creativity. But, you really can't skip over that. It will take some time and effort. I really hope that you will take that time and make that effort.

[Endercreeper01]

I know I might have some errors, but I've tried to correct them.

What do you think I should do to demonstrate my knowledge on fluid mechanics?

[Endercreeper01]

Is there an equation for force of lit based on [latex]\nabla[/latex]p?

[Bignose]

Is there an equation for force of lit based on [latex]\nabla[/latex]p?

This is a question that, per my above comment, I think that if you had a better grasp of fluid mechanics, you could answer it yourself. Considering what you've tried to do with the answers I've provided before -- specifically performing incorrect mathematics operations and misinterpretations of what the math actually says -- I don't think I want to provide an answer to this equation.

[Endercreeper01]

Seriously?! You just brush aside my question?!

And you still haven't told me how to demonstrate my knowledge on fluid mechanics.

Edited December 19, 2013 by Endercreeper01

[Bignose]

Seriously?! You just brush aside my question?!

And you still haven't told me how to demonstrate my knowledge on fluid mechanics.

Yep. I'm not sure sure your indignity here is warranted since between your thread on drag and this thread, you've brushed aside plenty of my comments.

 

If you actually want to demonstrate your fluid mechanics and mathematics knowledge, you could answer this question yourself.

[Endercreeper01]

I didn't brush away any comments, you just didn't accept them. You instantly reject everything I say!

Edited December 19, 2013 by Endercreeper01

[Bignose]

Yeah. I instantly reject equations that are dimensionally unsound or tensor rank unsound. Simply because there has never been a successful dimensionally unsound or tensor rank unsound.

 

If you are seriously asking me to believe that you've found some of the first ones, then you need to present extraordinary evidence to support this really extraordinary claim. Evidence, by the way, that hasn't been forthcoming in either thread despite my asking for it.

 

I also reject things when they are gross simplifications of the equations, and are at odds to what is already known. Again, if you really think that's wrong, you need to be presenting overwhelming extraordinary evidence that you're right and the body of knowledge in fluid mechanics is wrong.

 

I reject elementary mathematical errors. Do I really need to defend this?

 

And, lastly, I am getting really tired of writing out explanations for things -- which I guess in my opinion is not 'instantly' rejecting things because I am trying to take the time to explain where you made errors. But, I don't see you caring. So, I'm done. You demonstrate zero effort to understand what I write, so I'm not going to waste my time and write it anymore; I honestly am not sure I should have wasted my time to write this... If you really, really cared, you'd take the time to do some research and read about the great wealth of data and theories was have our there. So, I don't know what you're getting out of this. If you really cared about the subject matter, you sure as hell aren't showing it.

Edited December 20, 2013 by Bignose

[hypervalent_iodine]

!

Moderator Note

Once again, Endercreeper, you need to back your threads up with something a little more intellectually sound. I've let this thread go on without moderator notes for as long as Bignose's patience was intact, but since that has begun to diminish I am going to have to ask you to provide evidence for your claims and stop waiving off Bignose's rather rigorous and detailed criticisms or this thread will be closed. Given that we've been here so many times before, I am giving you only one chance to do this in this thread.

[Endercreeper01]

Fine, I will try to provide evidence. (Try, since big nose will deny them instantly) when I have time.

BTW: I mean when I fix them and no matter what I do, you reject.

You didn't even tell me what to do to show my knowledge, so I don't know how. Therefore, you think I have no knowledge. If you actually wanted to see if I had knowledge, you would show me how to do so. I could argue all day with you on some of your criticisms, but you will instantly reject what I say, and it will be a waste of my time.

I'm done. It is just a waste of time talking to bignose.

[Bignose]

BTW: I mean when I fix them and no matter what I do, you reject.

When the fixes don't actually fix the problem, or sometimes make it worse... then, yeah, I continue to reject them. I'll stop rejecting them when they actually, you know, work.

 

Demonstrate something that works, and I promise I won't reject it. I think that's supremely fair, and it is how I try my best to treat everybody.

[Endercreeper01]

Demonstrate something that works, and I promise I won't reject it. I think that's supremely fair, and it is how I try my best to treat everybody.

Would this work? :

Since [latex]\nabla[/latex]p=dp/dA and F_L=[latex]\int_A[/latex] p dA (evidence), this means that F_L=[latex]\int_A \int_A[/latex] [latex]\nabla[/latex]p d²A. Rearranging the Navier Stokes equations for [latex]\nabla[/latex]p (more evidence), we get (where * represents dot product):

[latex]\nabla[/latex]p=ρ(du/dt + u * [latex]\nabla[/latex]u) - [latex]\nabla[/latex] * T - B

Inserting this into the integral gives us

F_L= [latex]\int_A \int_A[/latex] (ρ(du/dt + u * [latex]\nabla[/latex]u) - [latex]\nabla[/latex] * T - B) d²A

This means the force of lift is the double integral over the area of the pressure divergence with respect to the area.

I would now have to work out this integral, then divide by qA to get the lift coefficient.

[Bignose]

Would this work? :

Since [latex]\nabla[/latex]p=dp/dA and F_L=[latex]\int_A[/latex] p dA (evidence), this means that F_L=[latex]\int_A \int_A[/latex] [latex]\nabla[/latex]p d²A. Rearranging the Navier Stokes equations for [latex]\nabla[/latex]p (more evidence), we get (where * represents dot product):

[latex]\nabla[/latex]p=ρ(du/dt + u * [latex]\nabla[/latex]u) - [latex]\nabla[/latex] * T - B[/size]

 

Inserting this into the integral gives us[/size]

F_L= [/size][latex]\int_A \int_A[/latex] (ρ(du/dt + u * [latex]\nabla[/latex]u) - [latex]\nabla[/latex] * T - B) [/size]d²A

This means the force of lift is the double integral over the area of the pressure divergence with respect to the area.

I would now have to work out this integral, then divide by qA to get the lift coefficient.

Well, let's see.

 

[math]\nabla p \ne \frac{dp}{dA}[/math]. What is true is: [math]\nabla p = \frac{\partial p}{\partial x} \boldsymbol{\delta_x} + \frac{\partial p}{\partial y} \boldsymbol{\delta_y} + \frac{\partial p}{\partial z} \boldsymbol{\delta_z}[/math] in a Cartesian coordinate system. It will have different definitions depending on the coordinate system picked.

 

And

 

the correct equation is:[math]\mathbf{F_L} = \int_{\partial V} \mathbf{n} \, p \, ds[/math], not what you wrote. You left out the normal vector on the RHS.

 

both of these two equations as you wrote them are tensor rank incorrect. Both are vectors on the LHS, and you have only written scalars on the RHS.

 

And then even if these two equations were not wrong, your third certainly wouldn't follow.

 

This is just another example of what I've been railing on above. If this is demonstrating anything it is demonstrating a lack of knowledge. Again, what am I supposed to think from your posting these just plain incorrect equations?

 

What I think is that you don't even know what the grad operator does (since you got its definition and tensor rank wrong) AND you can't even copy what I wrote all the way at the beginning of this thread! Please, please, please take the time to actually go and learn the mathematical tools used in fluid mechanics before you keep posting dead wrong stuff.

 

So, yeah, I'm rejecting this, again, because it's wrong.

Edited December 21, 2013 by Bignose

[imatfaal]

!

Moderator Note

 

Endercreeper

 

You are on thin ice in this thread having been warned on a few occasions. This thread will not be locked because of a failure to understand the science or maths - but it may be locked if you do not engage actively with the well-explained flaws in your ideas that members have raised. As a guide - the previous post made two comments about your assertions of equality (a gradient equalling a simple derivative rather than an orthogonal set of partials , and a mis-formed integral); these points should be answered either through a mathematical argument showing you are correct or an acknowledgement that you are mistaken. Do NOT just throw together another equation and ask what about this?