Endercreeper01

This uses the geometry of the flow itself. This is also for laminar flow only. And this is to calculate the shear rate, du/dx. The shear stress would be μ(du/dx). Also, all this is for a Newtonian fluid only. For a non Newtonian fluid, we would use τ/q

Can we have more time?

Endercreeper is?

First, you need to find the shear rate, since the wall shear stress is the viscosity times the shear rate, and the shear rate is du/dx. I have found a way to calculate this: You need a function for the angle with respect to x, a(x) This function would be a(x)=arctan(dy/dx), since dy/dx would be the same as tan(a), and arctan is the inverse function The derivative of this function, da/dx, will be useful when we calculate du/dx, and the derivative of this function is (d2y/dx2)(1/(1+(dy/dx)2) Now, we must find u(a(x)), or just u(a), where a=a(x) u(a) would be the part of the velocity that is in the direction of motion, so u(a)=u0sin(a), where u0 is the velocity when x=0 We now have to take du/dx. This would be (du/da)(da/dx) Du/da would be u0cos(a), and da/dx is found above at #3 We get: du/dx = ((u0cos(a)) / (1+(dy/dx)2)) (d2y/dx2)

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

What is the model?

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 ux/x + uz/z , where ux and uz 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 duy/y. How exactly we calculate du/dxwill be something I will work on. I will tell you ASAP on what that would be.

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.

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 Cd=Cw+Cf+Cs+Cdi First, let's explain Cdi, the induced drag coefficient I did not come up with the equation for Cdi, which is Cl2/(πe(AR)) So now that we got that out of the way, let's talk about the other terms First, Cw, 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/vs Now, lets talk about the skin friction drag, Cs 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. Oh

And also, what is it with all these negs?

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.

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.

You can also make the argument that it equals both bi and -bi since -b/i=((-b)^2/i^2)^0.5=(-b^2)^(1/2), which is both bi and -bi

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)

No, it was a - bi Why does -b/i=bi?

I have a proof that proves that the conjugate of a complex number is equal to the number. Proof: You have the complex number a+bi A+bi is also equal to (i(a+bi))/i Distributing gives you (ai - b)/i Distributing the division gives you a - bi Or, in equation form: a+bi = (i(a+bi))/i = (ai - b)/i = a - bi

If you had a spinning disk inside a fluid, what would describe the motion of the fluid?

Spinning and linear motion are similar. Spinning motion is just linear motion applied angularly

I agree

Do we get a prize?

What is it with all these other science forums?

I mean the definition where you have the tangent as opposite divided by adjacent

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.

I have something to change about my theory, and this is the skin friction coefficient, which is now (according to my theory): Cdf=(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: Cdf=τw/q=μ(du/dx)/q=(μ/q)(du/dx)

I mean if you use the triangle definition for a tangent and insert that into the equation for an angle based on tangent and arctangent