Theory for the coefficient of lift

[Endercreeper01]

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.

Oh, then that would mean I am wrong on the gradient.

But also, in a previous post in the beginning I wrote the integral as [latex]\int_A[/latex]np dA. When there was the normal vector, you said it was correct. I forgot the normal vector in my post. Would it be correct if I didn't leave out the normal vector?

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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?

 

Would it be okay if I posted another equation about the gradient that is different then what I said before, while providing a mathematical argument?

Edited December 21, 2013 by Endercreeper01

[imatfaal]

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Moderator Note

If you have taken on board the comments made and thus have revised your ideas or are gonna produce a reasoned argument against the comments - then please do so. But please make sure it is a serious attempt - ie it takes known ideas and works on them in a simple and arguable manner.

[Endercreeper01]

Because [latex]\nabla[/latex]p = i dp/dx + j dp/dy + k dp/dz, would this mean that p = [latex]\int[/latex] (dp/dx) dx + [latex]\int[/latex](dp/dy) dy + [latex]\int[/latex] (dp/dz) dz, since you would need to integrate over each term to get p? I also didn't include i, j, and k since that would make p a vector when p is a scalar, so I left out i, j, and k.

[Bignose]

Ender,

 

you should be able to test this yourself.

 

Let p = p(x,y,z), some arbitrary function. I.e. let [math]p(x,y,z) = x + y^2 + \exp(z) + xyz[/math] or any other nontrivial function.

 

Now, calculate [math]\nabla p[/math]

 

Then from that calculation, compute the integrals you posted above.

 

If you don't recover your original p(x,y,z), you'll know something is wrong.

 

THIS is the kind of stuff someone who was actually serious about the work would do... check themselves. Not just toss something off-the-cuff and hope that it works. It shouldn't be up to -->us<-- to catch your mistakes. You should be checking yourself before posting it.

 

Also, you need to quit using 'd' when the partial derivative is called for. They are two different symbols, and have two different meanings. This is part of the clear communication that I've also been requesting of you. It would also be nice to denote the vector quantities in some way, such as bolding the symbol or an overarrow. The clearer the communication, the better.

Edited December 21, 2013 by Bignose

[Endercreeper01]

I am checking myself. I just wanted to see if it was correct in order to proceed to the next step.

I want to make sure I don't get any more errors in my calculations.

[Bignose]

I am checking myself.

I have doubts about this.

 

post the results of the check I listed above, and see for yourself.

Edited December 21, 2013 by Bignose

[Endercreeper01]

It actually would be correct. The pressure p has components, and p is the sum of those components. This gives you p = p_x + p_y + p_z. If you work out each integral, you get the components of p for the terms. This is because you are taking the integral of dp/dx dx. This is the same as the integral of dp_x.We can now write the term as p_x. This is the same with y and z.

Edited December 21, 2013 by Endercreeper01

[Bignose]

No...

 

Let's go through the steps in the check I requested.

 

[math]p(x,y,z) = x + y^2 + \exp(z) + xyz[/math]

 

Then

 

[math]\nabla p = (1 + yz)\boldsymbol{\delta_x} + (2y + xz)\boldsymbol{\delta_y} + (\exp(z) + xy)\boldsymbol{\delta_z}[/math]

 

Now, let's do the integrations you listed.

 

[math] p = \int (1 + yz) dx + \int (2y + xz) dy + \int (\exp(z) + xy) dz[/math]

 

this would be

 

[math] p = (x + xyz) + (y^2 + xyz) + (\exp(z) + xyz) = x + y^2 + \exp(z) + 3xyz [/math] (ignoring the constants of integration)

 

not the same as the original function we started with! Your formula can't be right... you should recover the original function if your steps were valid. We didn't, therefore your steps are wrong.

 

Again, you should have done this before you posted it. Checking your own work is a necessary step. It would prevent us from having the opinion that you're just throwing formulas off-the-cuff in the hope that something will stick. Because you'll find your own formulas that don't stick, you won't post them, and have us point out your errors.

 

Look, everyone makes mistakes. All of us do. All of us will continue to make mistakes. But, if you develop the skill of catching your own mistakes, you'll be much, much better. And you won't be reliant on others to catch your mistakes. I consider this part of the mathematics skills that you need to develop.

The pressure p has components, and p is the sum of those components. This gives you p = p_x + p_y + p_z.

This needs to be addressed, too. Because, as I explained before, pressure is a scalar. It doesn't have components in the x, y, z direction. You can of course break any sum up if you want to, but there is not a pressure in the x direction, in the y direction, in the direction, because the pressure at each point applies pressure equally in all directions at that point.

 

This is more unclear communication that is at odds with what we know how pressure behaves today, that needs to be cleaned up.

[Endercreeper01]

Your right. I would have to rethink that.

But, if p is a function of x, y, and z, shouldn't there be components in the x, y, and z directions?

For example, in the function , you can have p_x = x + xyz/3, p_y = x² + xyz/3, and p_z= e^z + xyz/3. The sum of this would give you p.

[Bignose]

But your decomposition isn't unique, or even necessary. Why break the sum into 3 parts? Those 3 parts don't mean anything.

 

And most importantly, it still doesn't fix the problem of your formula with the integrals being wrong.

 

Just because something is a function of 3 variables, doesn't mean it has to be broken into 3 variables.

 

Temperature is also a function of x, y, and z in a 3-D problem. But it is like pressure, there isn't any temperature in the x direction that is different than in the y direction, and so on. And there isn't any reason to break its sum into x, y, z parts necessarily.

[Endercreeper01]

We could fix the integral by doing [latex]\int[/latex]dp_x/dx dx + [latex]\int[/latex]dp_y/dy dy + [latex]\int[/latex]dp_z/dz dz. This would fix the problem.

In the above case, this would mean that

[latex]p = \int (1 + yz/3) dx + \int (2y + xz/3) dy + \int (\exp(z) + xy/3) dz [/latex]

giving you the original function.

[Bignose]

We could fix the integral by doing [latex]\int[/latex]dp_x/dx dx + [latex]\int[/latex]dp_y/dy dy + [latex]\int[/latex]dp_z/dz dz. This would fix the problem.

In the above case, this would mean that

[latex]p = \int (1 + yz/3) dx + \int (2y + xz/3) dy + \int (\exp(z) + xy/3) dz [/latex]

giving you the original function.

You can't do this, though. How the heck are you supposed to know what goes into px, py, and pz?

 

what if [math]p(x,y,z) = \sin(x + y + z)[/math]? Or what if [math]p(x,y,z) = \Gamma(2xy) + J_2(\arctan{(x^3y^4z^5)})[/math]? There is no rational way to just split pressure into components, and no physical meaning behind it.

 

There is no such thing as an x component of pressure, any more than there is an x-component of temperature or an x-component of a dollar.

 

You're just making stuff up and hoping it sticks. This is absurd. I thought you agreed to stop this. All you did was conveniently define terms so that your flawed equation would work in one situation.

 

How about just accepting your equation is wrong. And actually learning about fluid mechanics instead of just making crap up.

[hypervalent_iodine]

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Moderator Note

Endercreeper, I'm going to close this thread now. You were given your chance, so please do not reopen the thread. And before you try opening another coefficient of whatever thread, please take Bignose's advice and go away and learn something first.