Theory for Coefficient of drag

[Bignose]

1. I know how to create graphs, but not how to put them on top of each other

Fine. I'm going to reply to this, though I am loothe to actually need to do it.

 

Sometimes, not all of science is glamour and exciting. Sometimes, it is re-transcribing plotted data into another format.

 

Sometimes, that means printing out a graph at as high a resolution as possible and using a ruler and re-typing all the x & y points. That way, you can plot someone else's graph on the same graph as your own.

 

So, now that that is out of the way, any chance you'll address some of the other criticisms I've had? Like the one where you just changed the functional form for no apparent reason (the rotation may explain why you can go from sine to cosine, but it does NOT explain why you can change from the sum of two different cosines to a single sine), or why all the other terms but Cf go to zero on the lone data set you've considered despite the functional forms you've posted showing that there is no way that that should happen?

[Endercreeper01]

The functional form changed because what you have is an angle of attack from the front and back, angle a and angle b, making it 2C_f0sin(a)-C_f0sin(b)

In the angle of attack, a=b, and simplifying makes C_f0sin(a)

It is most likely not 0 at all, actually. It may seem to go to 0, but there is a slight y intercept that is the other effects.

Edited November 20, 2013 by Endercreeper01

[Bignose]

but there is a slight y intercept that is the other effects.

from what you described the other effects as, I don't see how they can be described as 'slight'. And it sure is convenient that a=b, isn't it? Why would you use two different variables in the first place if a = b?

 

This hand-waving dismissing of my concerns is getting quite annoying and is a major reason I kind of stopped writing back the first time...

[Endercreeper01]

The other terms can be described as slight because first, the wave drag coefficient is only for supersonic speeds. Second, there is most likely little, if any lift. Third, this leaves us with the skin friction and for drag, and the skin friction coefficient should not be very high.

And I used 2 different variables to prove a point

[Bignose]

Bull****. Little to no lift for airfoil?!?! You're just making this up now, I won't be back until you say something meaningfull.

[Endercreeper01]

Lift drag, I meant. The lift drag, or induced drag, would be small because of the other factors that go into it.

And also, what about the other graph? There is an intercept for that. I don't even think it shows the intercept on the other graph.

[Endercreeper01]

I have something to change about my theory, and this is the skin friction coefficient, which is now (according to my theory):

 

C_df=(μ/q)(du/dx)

 

Where μ is the viscosity coefficient, du/dx is the shear rate, and q is the dynamic pressure.

This can be proven, since:

 

C_df=τ_w/q=μ(du/dx)/q=(μ/q)(du/dx)

Edited November 23, 2013 by Endercreeper01

[Endercreeper01]

I wasn't paying attention to the fact that it was an airfoil. If that is the case, then the induced drag must be small because of the small intercept.

And as I said before, pay attention to the other graph I posted on the previous page. I can't post pictures anymore for some reason, so take a look at it. It has a y intercept, doesn't it?

There is certainly other effects that you would add together for the drag coefficient. I am trying very hard to explain the small y intercept. If I cannot explain the y intercept, then you tell me, because even if my theory is incorrect, it is certain that you have to add all of the other effects to get a y intercept.

[Bignose]

Thanks to whomever is giving me downvotes... stay classy out there.

 

I have something to change about my theory, and this is the skin friction coefficient, which is now (according to my theory):

 

C_df=(μ/q)(du/dx)

 

Where μ is the viscosity coefficient, du/dx is the shear rate, and q is the dynamic pressure.

This can be proven, since:

 

C_df=τ_w/q=μ(du/dx)/q=(μ/q)(du/dx)

So... you're going to set the portion of the drag coefficient due to friction equal to the exact value of the viscous stresses at the surface. AND include something about some undefined pressure.

 

I have no idea why you are just clumping these together willy nilly. Total drag is usually the sum of the friction drag and the form drag (that due to the pressure change). You're going to have to demonstrate that this product makes sense.

 

Also, the drag coefficient, CD, is the dimensionless drag per unit length of a 2-D body (per unit area of a 3-D body). CD = (drag)/(denisty*characteristic velocity^2 * characteristic length).

 

drag = friction drag + form drag

 

friction drag is defined to be the shear stress at the wall, or viscosity*velocity gradient normal to the wall for a Newtonian fluid... on other words, what you set your coefficient to.

 

which doesn't make sense.

 

Furthermore, it doesn't really help to define your coefficient that way, since usually it is not very easy to know what the shear stress at a boundary is. That's why drag coefficients are measured in the first place... to get around the difficulties in calculating the shear stress at the wall.

 

BTW, a lot of what I just wrote can be found in Leal's text referenced above. Had you read it, or maybe any of the other texts I posted, you would have know this. My recommendation would be to quit just flinging things around that you do not appear to have a good understanding of in the hopes of getting something to stick, and actually start on learning what we know today. At the very least, you need to know what is published today so that you can demonstrate that your idea is actually better... something you haven't done at all so far, BTW. Fluid mechanics has a lot of unknowns out there still, it is a very active area of research, but there are a heck of a lot of things we do know fairly well. And drag is one of those things. I am not saying that there isn't a lot more to learn about drag, because there is; but, what you're doing here conflicts with a great deal of the very well established knowledge we have so far. And you've provided zero evidence comparing your idea to what is known, and zero evidence showing us your idea is better. Either start providing that evidence, or this thread really ought to be closed. And please do yourself a favor and start working on understanding what we know today. I'll even go so far as to offer to answer questions about the texts I referenced above, when I have time... I have all of them, and have studied all of them in depth.

[Endercreeper01]

I set the friction drag to that value. It makes sense because I have proven it mathematically. The coefficient I set my value to is a quotient, not a product. Notice how you divide the viscosity by the dynamic pressure? You add the form drag, the wave drag, then the friction drag, and finally the induced drag. I was just stating another form of the skin friction coefficient

Also, I am not just sticking things around, and have had proofs. See above proofs.

What is it that my work violates?

And how much does the book cost?
Also, do you understand my hypothesis? (not theory, since according to you, I have no evidence)

[Bignose]

I set the friction drag to that value.

Well then, that's wrong, because friction drag is already defined. See Leal or others. Friction drag does not include pressure in any way shape or form.

 

t makes sense because I have proven it mathematically.

You haven't 'proven' anything. You just set one thing equal to another, without any given rhyme or reason. And especially without support from the knowledge we have of fluid mechanics today, because your definition is at odds with the very commonly used defintion today.

 

The coefficient I set my value to is a quotient, not a product. Notice how you divide the viscosity by the dynamic pressure?

Fine. Quibble over a small language inaccuracy. (You do realize that quotient and product are tightly related, right? I was only trying to make it clear that it wasn't a sum as it is commonly represented.) Arguing on this is like arranging the silverware on the Titanic, but whatever.

 

You add the form drag, the wave drag, then the friction drag, and finally the induced drag. I was just stating another form of the skin friction coefficient

See, you're all over the map here again. You gave an equation for Cf. I take Cf to be the skin friction coefficient. But you set it equal to some equation that has a large chunk of the skin friction itself in it. NOT the coefficient. You seem to be mixing the drag coefficients and the drag itself. You must be extremely careful in being very explicit about what you mean.

 

On top that the equation for skin friction coefficient or not has a pressure term, which, by definition, has nothing to do with the friction drag.

 

Also, I am not just sticking things around, and have had proofs. See above proofs.

What is it that my work violates?

No, you don't have proofs. You just assume forms for terms. That is not proof. If you want to talk proof, you would derive the forms for these equations directly from the Navier Stokes equations. Just like they do -- and I'm sorry I sound like a broken mp3 player here -- in the fluid mechanics texts.

 

And how much does the book cost?

I believe you can log on to Amazon.com as well as I can.

 

Also, do you understand my hypothesis? (not theory, since according to you, I have no evidence)

No, I don't understand it because it seems very haphazard and flung together without any substantial substance behind it. I understand the current fluid mechanics pretty well. Your hypothesis has a long way to go.

 

Lastly, in accordance with silverware straightening on the Titanic, it cracks me up that you didn't bother to address any of the real issues I posted. Namely, the fact that using the shear stress at the wall for your calculation is troublesome because that is not very easily known. And that you still have posted very, very little supporting evidence -- and what you have posted is suspect to me at least. But, no let's keep arguing over product v. quotients, and your supposed proofs. In my mind, this is just more evidence of the lack of your background in this area, because I don't think you can recognize what the really important issues at hand are. That is, you think it is more important to correct quotient v. product rather than address the serious definitions that are at odds with the well verified fluids knowledge we have today.

 

My point is, if you aren't going to bother using your calculations to compare to experimentally reported values, then what is the point? I don't get what satisfaction you get from them slapping together some functional forms and calling it a day. I'll give you one thing -- you are ahead of the vast majority of posters with a new 'idea' in that you are actually doing math. But, you haven't demonstrated in the least the value of that mathematics. No derivation it comes from first principles, no source given at all, and no demonstration that math actually makes useful predictions. So I guess -- again -- I don't have a lot of interest because you just aren't saying anything meaningful, and you don't seem to take my criticisms seriously.

[Endercreeper01]

I meant skin friction drag coefficient

I mean proof as in rewriting the equations since a value(s) are equal to something else, so I rewrite them, and I consider that as proof

Ex: the skin friction drag coefficient is defined as τ/q, where q is the dynamic pressure and τ is the shear stress. Since τ=μ(du/dx), we can rewrite it as (μ(du/dx))/q, which we can also rewrite as (μ/q)(du/dx)

So, you want me to derive the drag coefficient from the navies stokes equations instead of what I am doing?

And I would try to compare to other expirental graphs, but I can't post links or pictures anymore.

What I will now do is find a good source involving form drag and use a different computer to post it. And also, if you don't understand my theory, then how can you criticize it? If you want me to explain it more clearly, just ask.

[Bignose]

I mean proof as in rewriting the equations since a value(s) are equal to something else, so I rewrite them, and I consider that as proof

Ex: the skin friction drag coefficient is defined as τ/q, where q is the dynamic pressure and τ is the shear stress. Since τ=μ(du/dx), we can rewrite it as (μ(du/dx))/q, which we can also rewrite as (μ/q)(du/dx)

But, if you start from a wrong place, e.g. "the skin friction drag coefficient is defined as τ/q", then re-writing it doesn't matter. Because the statement in quotes isn't justified in the first place. What you've written here is NOT the definition used by, well, everyone else. THAT's what I want you to justify. THAT's what I want to see derived from first principles or the Navier Stokes equations.

 

And also, if you don't understand my theory, then how can you criticize it? If you want me to explain it more clearly, just ask.

I've been asking! You haven't been answering! Once again, you decide not to address the large fundamental flaws, and nitpick on word choice! (Which, by the way, if you're going to nitpick on word choice, you probably need to make sure your words are right. It is Navier Stokes equations, not 'navies stokes'. It is named after two famous researchers, so it would be respectful to capitalize and spell their names correctly.) What am I supposed to think about this!?!? I'm at the point where I just want to give up because you seem very unresponsive to the questions I ask. But, I also want to help make sure you understand what I see as major issues with what you've posted to date, so I'm still on here -- but losing hope and interest fast.

Edited November 25, 2013 by Bignose

[Endercreeper01]

But, if you start from a wrong place, e.g. "the skin friction drag coefficient is defined as τ/q", then re-writing it doesn't matter. Because the statement in quotes isn't justified in the first place. What you've written here is NOT the definition used by, well, everyone else. THAT's what I want you to justify. THAT's what I want to see derived from first principles or the Navier Stokes equations.

 

 

I've been asking! You haven't been answering! Once again, you decide not to address the large fundamental flaws, and nitpick on word choice! (Which, by the way, if you're going to nitpick on word choice, you probably need to make sure your words are right. It is Navier Stokes equations, not 'navies stokes'. It is named after two famous researchers, so it would be respectful to capitalize and spell their names correctly.) What am I supposed to think about this!?!? I'm at the point where I just want to give up because you seem very unresponsive to the questions I ask. But, I also want to help make sure you understand what I see as major issues with what you've posted to date, so I'm still on here -- but losing hope and interest fast.

First, according to the Wikipedia article, then it would be τ/q. Third, I am typing from my phone and have autocorrect on, and it is autocorrect's fault for saying navies. I meant Navier Stokes. Fourth, I will show you and work on solutions for the Navier stokes equations and try to prove it that way, but not now. Tomorrow since I have time, I will throughly explain everything, provide a graph, and then I will try to solve the Navier Stokes equations for my theory.

Edited November 26, 2013 by Endercreeper01

[Endercreeper01]

And also, what is it with all these negs?

[imatfaal]

And also, what is it with all these negs?

 

Possibly it was because you had a First, a Third, and a Fourth - but no Second; these things really grate! Or maybe it is because too many of your posts are effortlessly condescending - even when it is quite clear that another member is spending a fair amount of time trying to communicate.

 

From a quick look at the wikipedia page on parasitic drag and the other on dynamic pressure I would kinda agree with you - as long as the velocity designated as [latex]U_\infty[/latex] is the same in every way as that shown as [latex]v[/latex]

[Bignose]

First, according to the Wikipedia article, then it would be τ/q.

You mean this page? http://en.wikipedia.org/wiki/Parasitic_drag

 

Where it says

 

[math]C_f = \frac{\tau_w}{\frac{1}{2}\rho U^2_\infty}[/math]?

 

The equation that doesn't include pressure at all?

 

The equation that is similar to the nondimensionalization I posted above?

 

You're missing the bigger point here.

 

Total drag = skin drag + form drag + possible others.

 

Skin drag is due entirely to surface friction effects. That is why it contains the shear stress at the wall.

 

Form drag is due entirely to pressure effects. That is why it contains the pressure difference.

 

I have never seen anyone turn that above sum into some kind of product. That is, the surface terms and the pressure terms are kept separately and then added together -- not multiplied. And if you want to propose blending there two terms together, then I am going to request extraordinary evidence to support this.

 

Note that I'm not saying that the two effects don't interact, because I have little doubt that they do. But the first order approximation has the terms separate. And given the difficulties in measuring the terms directly as above, usually the first order terms are sufficient. Teasing out any higher order effects would require significant evidence and experimental verification.

 

And, I'd like to see some very compelling reason to change the well established and accurate 1st order terms that are already in common use.

 

Lastly, a meta-comment. In clicking on the above wikipedia page and following some links there, I suspect very strongly that most of this work is based on those wikipedia pages. (In the future, you ought to get in the habit of citing the references you use.). This is a case where wikipedia isn't bad, but it really isn't all that good. If you really want an understanding of fluid mechanics, you are not going to learn it from wikipedia. You are going to need to put in some time with some good books.

 

Back when I taught the class, I used to tell the students that fluid mechanics was one of the first classes where just doing the homework and memorizing the text wasn't going to guarantee them a good grade. Because, there are so many different problems in fluid mechanics, that I would write exams different enough from all the others that the only way to get a good grade was to actually understand fluid mechanics. This was a shock to some students that otherwise had very good marks to date. Because they were very good at symbol pushing and memorizing equations, but no real understanding of those equations. Fluid mechanics -- if taught well -- can expose those flaws.

 

This is why I keep repeating myself, but I cannot recommend strongly enough that you put in some time with some good textbooks. If you don't want to buy new books, there are many good used texts out there -- and the basics of fluid mechanics haven't changed in the last 20 years, so even a good text from 1990 can be valuable. I don't know of any web resource that can supplant a good text. And wikipedia as I viewed it today certainly can't.

Edited November 26, 2013 by Bignose

[Endercreeper01]

You mean this page? http://en.wikipedia.org/wiki/Parasitic_drag

 

Where it says

 

[math]C_f = \frac{\tau_w}{\frac{1}{2}\rho U^2_\infty}[/math]?

 

The equation that doesn't include pressure at all?

 

The equation that is similar to the nondimensionalization I posted above?

 

You're missing the bigger point here.

 

Total drag = skin drag + form drag + possible others.

 

Skin drag is due entirely to surface friction effects. That is why it contains the shear stress at the wall.

 

Form drag is due entirely to pressure effects. That is why it contains the pressure difference.

 

I have never seen anyone turn that above sum into some kind of product. That is, the surface terms and the pressure terms are kept separately and then added together -- not multiplied. And if you want to propose blending there two terms together, then I am going to request extraordinary evidence to support this.

 

Note that I'm not saying that the two effects don't interact, because I have little doubt that they do. But the first order approximation has the terms separate. And given the difficulties in measuring the terms directly as above, usually the first order terms are sufficient. Teasing out any higher order effects would require significant evidence and experimental verification.

 

And, I'd like to see some very compelling reason to change the well established and accurate 1st order terms that are already in common use.

 

Lastly, a meta-comment. In clicking on the above wikipedia page and following some links there, I suspect very strongly that most of this work is based on those wikipedia pages. (In the future, you ought to get in the habit of citing the references you use.). This is a case where wikipedia isn't bad, but it really isn't all that good. If you really want an understanding of fluid mechanics, you are not going to learn it from wikipedia. You are going to need to put in some time with some good books.

 

Back when I taught the class, I used to tell the students that fluid mechanics was one of the first classes where just doing the homework and memorizing the text wasn't going to guarantee them a good grade. Because, there are so many different problems in fluid mechanics, that I would write exams different enough from all the others that the only way to get a good grade was to actually understand fluid mechanics. This was a shock to some students that otherwise had very good marks to date. Because they were very good at symbol pushing and memorizing equations, but no real understanding of those equations. Fluid mechanics -- if taught well -- can expose those flaws.

 

This is why I keep repeating myself, but I cannot recommend strongly enough that you put in some time with some good textbooks. If you don't want to buy new books, there are many good used texts out there -- and the basics of fluid mechanics haven't changed in the last 20 years, so even a good text from 1990 can be valuable. I don't know of any web resource that can supplant a good text. And wikipedia as I viewed it today certainly can't.

I never multiplied them. The form drag coefficient and the skin drag coefficient were never multiplied, they were added together in my theory. It is just that there are 2 terms. I never said I multiplied them. I think you may have gotten confused on what I was saying. Also, most of my knowledge on areodynamics actually comes from Areospace and NASA, not just Wikipedia

Now, I will clearly explain my theory again.

 

We have the sum C_d=C_w+C_f+C_s+C_di

First, let's explain C_di, the induced drag coefficient

I did not come up with the equation for C_di, which is C_l²/(πe(AR))

So now that we got that out of the way, let's talk about the other terms

First, C_w, which is the wave drag coefficient

Since it depends on the area of the shock wave and the mach number, we can write it as Av/v_s

Now, lets talk about the skin friction drag, C_s

This is equal to τ/q, according to the wiki article.

Since the shear stress is equal to the viscosity times the shear rate, we can write it as μ(du/dx)/q, which is also equal to (μ/q)(du/dx)

I have already talked about the form drag on the 1st and 2nd pages, so go see that for info on form drag.

A graph from http://www.aerospaceweb.org/question/aerodynamics/drag/drag-cone.jpg has the following image:

In this case, the average angle is ε - 90 because of the way ε is measured. This is also equivalent to the sin in this case.

 

Possibly it was because you had a First, a Third, and a Fourth - but no Second; these things really grate! Or maybe it is because too many of your posts are effortlessly condescending - even when it is quite clear that another member is spending a fair amount of time trying to communicate.

 

From a quick look at the wikipedia page on parasitic drag and the other on dynamic pressure I would kinda agree with you - as long as the velocity designated as [latex]U_\infty[/latex] is the same in every way as that shown as [latex]v[/latex]

Oh

[Bignose]

This is equal to τ/q, according to the wiki article. [/size]

Where do you see this?! I posted exactly the same formula from wikipedia! There is no pressure in the denominator!

 

[math]C_f = \frac{\tau_w}{\frac{1}{2}\rho U^2_\infty}[/math]

 

You even quoted this in my reply! Where is the pressure in this equation!?!?!

Edited November 26, 2013 by Bignose

[Endercreeper01]

q, the dynamic pressure, is equal to one half of the density times the velocity squared. I'm sorry. I should have clarified this. I also should have called it q₀, representing the dynamic pressure using free stream velocity.

[Bignose]

So, your q is not an actual pressure, but the pressure they use for nondimensionalization, like with the Euler number.

 

e.g. [math]q=\frac{1}{2}\rho U^2_\infty[/math]

 

Sheesh. Clear communication would have helped here tremendously.

 

ok. Let's address the next issue then. How are you going to get the shear stress at the boundary, then?

 

And you should be adding a qualifier that [math]\tau_w = \mu \frac{\partial u}{\partial x}[/math] for a Newtonian fluid only. But, it is the calculation of [math]\frac{\partial u}{\partial x}[/math] that is very difficult to do. How are you proposing to do it?

[Endercreeper01]

In this case, you can do 2 things. If you are simply taking shear rate as u/x, where u is the free stream velocity, then it would be just be u/x in 2 dimensions, where x would be the axis perpendicular to the velocity. In 3 dimensions, it would be u_x/x + u_z/z , where u_x and u_z is the x and z parts of u, and x and z are the 2 axis' perpendicular to the velocity. So basically, it would be the divergence of u minus du_y/y. How exactly we calculate du/dxwill be something I will work on. I will tell you ASAP on what that would be.

[Bignose]

gotta work on that communication again. It is NOT ux/x or u/x. It is [math]\frac{\partial u}{\partial x}[/math] can't just drop the partial signs willy nilly here. Must be very, very clear about getting the math right. This has been a cause of a lot of the problems to date.

Edited November 26, 2013 by Bignose

[Endercreeper01]

Should I make a new thread for the shear rate or use this thread?

[Bignose]

I'd put it here... you're explaining how you intend to calculate the wall shear stress (NOT exactly the same at the shear rate... gotta get these right, my friend) in order to calculate your Cf. That's a natural continuation of this thread.

Edited November 27, 2013 by Bignose