Endercreeper01

And now, I will talk about what I have done so far. First, you have FL= [latex]\int_{A}[/latex] np dA. To get the coefficient of lift, then you must divide this by [latex]qA[/latex]. This gives you CL=[latex]\int_{A}[/latex] [latex]{np}/{qA}[/latex] [latex]{ dA}[/latex], which is equal to [latex]\int_{A}[/latex] [latex]{n}/{q}[/latex] [latex]{ dp}[/latex] Using integration by parts, you find that: latex]\int_{A}[/latex] [latex]{n}/{q}[/latex] [latex]{ dp}[/latex] = nΔp/q - [latex]\int_{A}[/latex] p/q dn For the first term, then it is change in pressure because it is evaluated among the top and bottom areas, and so, you get the change in pressure. I just need to find what this change would be. For the second term, then I need to make this integral in terms of A, so I changed dn to dA dn/dA, and this makes it an integral with respect to A. Now, I need to find dn/dA, along with the change in pressure.

I can't find the sticky, though. Do you know which forum it is in?

Could we also write the integral as integralA(np dA), where integralA would mean the integral over the area?

Would it be considered a personal attack if you called someone a crackpot?

Also, why is pressure a scalar if force is a vector?

If there is no angular acceleration, then there will be no tangential force, but there will be a centripetal force. The centripetal force is mv2/r, where r is radius, m is mass, and v is velocity.

A spinning object would gain mass, and this would then cause more gravity. In General Relativity, the solution for a rotating spherical object with no charge would be called the kerr metric, and you can learn about it at http://www.roma1.infn.it/teongrav/leonardo/bh/bhcap3.pdf

If you go to the last page(s) of forums, then you will see threads with -1 replies, and they are locked. Why is this?

I agree. Many humans waste resources. We are already seeing people suffering. Farmers produce enough food for 12 billion people, but 1 billion people are starving. Humans waste vast amounts of food, and also animals on farms. Animals living on farms eat much more food then they produce from being killed, and this consumes a lot of our food.

In a vacuum, the path of light is bent by gravity, and this is called gravitational lensing. The angle of deflection is 4GM/(rc2), where c is the speed of light, r is the distance from the mass, G is the gravitational constant, and M is mass.

Because p=dF/dA, then would it be F=integralA( n dp)? If not, what would F be in terms of A? For n, I meant the vector normal to the area.And also, for n, I meant something else then what I said in the other post. I have trouble communicating sometimes, and this sometimes makes it hard to explain things. Sorry for the communication errors. And, I'm still waiting for my order to come.

Wouldn't angular motion in a fluid create lift?

Because p is not a vector, then I will change it to F=Apn, where n is the unit vector that the pressure is pointing in. And also for this, then it would be FAΔnΔp. Would that work? And also, once I finish my theory, then I will show you the calculations. This would also change it to CL=ΔnΔp/q. Or, if that doesn't work, then it will then be that integral that you posted divided by qA. I also forgot about how p=dF/dA. Could you also write it as F=integral(p) dA, since p=dF/dA? And if so, would that make CL=(integralA(p) dA)/(qA), giving you (integralA dp)/q?

Pressure is a vector. Even hyper physics agrees that p=F/A http://hyperphysics.phy-astr.gsu.edu/hbase/press.html Therefore, F=Ap Remember, it was still being developed, and I am working on that. Another thing we can do is divide that integral by Aq, since FL=qACL For the Δp/q part, I am just trying to find the pressure change. Then, it will be a new way of calculating the lift coefficient.

We can also do that same integral divided by Aq.

F=Ap, no?

What about debates?

Yes

I am currently working on a theory for the coefficient of lift. Here is what I have so far: CL=Δp/q Where Δp is the change in pressure between the top and bottom of the airfoil, and q is the dynamic pressure; q=1/2ρv2 This is because of the following: The force of lift would also be equal to AΔp, since the force of lift depends on the change in pressure between the wings and the cross sectional area. Because the force of lift is qACL, then we can write the coefficient of lift as AΔp/qA, which simplifies to Δp/q. Now, I have to find out the value of Δp. What do you think so far?

Try www.lumosity.com It's a brain training website.

I was just wondering.

Does (a+b)1/2 equal a1/2 + b1/2 - (2ab)1/2? If you have ( (a+b)2)1/2, then because of the binomial theorem, then (a+b)2=a2+b2+2ab. Inserting this into the equation, then you get (a2+b2+2ab)1/2, which gives you a+b, since a2+b2+2ab=(a+b)2. If you set x=a2 and y=b2, then you have (x+y+2x1/2y1/2)1/2, which gives you x1/2+y1/2. This is also equal to x1/2 + y1/2 + (2x1/2y1/2)1/2 - (2x1/2y1/2)1/2, since (2x1/2y1/2)1/2 and -(2x1/2y1/2)1/2 cancel out. What if you removed 2x1/2y1/2 from the square root? Then, would you get x1/2 + y1/2 - (2x1/2y1/2)1/2?

There is angular force, actually. The torque is τ=Fr. Torque is also the moment of inertia I times the angular acceleration α, T=Iα. The moment of inertia is the sum of mr2 for each of the masses in the system, written as Σmr2. Since there is a force term in torque for the first equation, we can make the angular force F=Iα/r. Since F=ma, we can also write torque as τ=mar. So yes, there is angular force.

The angular displacement is (rcos(a), rsin(a)) where a is the angle displacement