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.

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

Seriously?! You just brush aside my question?! And you still haven't told me how to demonstrate my knowledge on fluid mechanics.

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

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?

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. θavg is the average angle for the top of the airfoil and φavg 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(θavg). This would be the same for the bottom of the airfoil, making it be v0cos(φavg) 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.

Would [latex]\nabla p[/latex] be constant among the airfoil? Also, what about the post where I showed that [latex]\nabla[/latex]p=qCL?

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.

Do you know where the errors in the derivations are?

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.

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 qACL, then we need to take the derivative of this with respect to area to get the change in pressure. This gives us Δp=qCL. Therefore, CL=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 CL = (q1 - q2)/q. What I have to do is figure out how the dynamic pressure changes depending on the geometry.

If both of the above posts are true, then CL=Δp/q, which is the same as (q1-q)/q, giving you CL = [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.

You have provided no scientific evidence for your claims. People on this forum are looking for proof. If you don't have any, they won't believe you.

You forgot to multiply the left sides of each equation by 2. 1+1 does not equal four, but 2(1+1)=4.

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?

Also, how do we distinguish trolls from other people? Sometimes, people are wrongly accused of trolling (as with me). We must be able to distinguish.

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

There are many trolls everywhere. Sadly, there is always going to be internet trolls.

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

There is no scientific evidence for creationism, but there is scientific evidence for evolution.

right in by skrillex

K I just accepted your request

Would this work?

I think that calling an idea crackpot is generally not the best way of explaining why it is not a good idea. Also, if you mean crackpot idea as in a creative, outside the box idea, then it may get confused with it being a crazy idea.

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: CL = 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 CL = 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.