Endercreeper01

Because the shape of a shock wave is a cone, you can also write the area A of the shockwave as πrs+πr2, where r is the radius and s is the length of the side of the cone

Do you think there is anything I need to change about my theory? Another equation for the coefficient of drag is that Cd=p/q, where p is the pressure and q is the dynamic pressure.

I think that it would be D/cosθavg when it is concave because when it is concave, the drag increases when θavg is bigger, and the drag approaches infinity as θavg approaches 90. It is also equal to Dcosθavg when θavg is 0. Since it is dependent on θavg in the opposite way then as in when it is convex, this means it would be D/cosθavg, making the equation 2D/cosθavg - D/cosΦavg only when the shape is concave

Also, for the form drag, there is a region behind the object where there is low pressure. This adds to the drag. According to my theory, the coefficient of flow drag would actually be: Cf=2Dcosθavg-DcosΦavg Where Φavg is the same as θavg, but the axis is at the back rather then at the front. This is why I think this: First, we have the term Dcosθavg. You need to add something to it, and that is the effect from the pressure difference. The force from the pressure difference is: F=AΔp Where A is the surface area and Δp is the change in pressure between them. Now, you need to find the coefficient of drag for this. You get it by: Cd=2Fd/Aρv2=Fd/qA In this case, Fd is AΔp, so we can rewrite this as: AΔp/qA=Δp/q But, what is Δp? Δp is p2-p1, where p1 is the pressure in the low pressure region and p2 is the pressure in the front. Sine F=Ap, this means p would be qCd In this case, Cd is just Cf not including the effects from the low pressure area. This would mean it is Dcosθavg. Therefore: p=qDcosθavg And so: Δp=qDcosθavg - qDcosΦavg This means Δp/q is: qDcosθavg - qDcosΦavg/q=Dcosθavg-DcosΦavg But also with the form drag, if the object is concave instead of convex, then it is D/cosθavg instead of Dcosθavg, and the same with Φavg

And there is also the interference drag. According to my theory, the interference drag effects would be in the velocity. The velocity would not be v=v1+v2, but v=(v12+v22)1/2

Remember my previous theory about the coefficient of drag? Now, I have a new theory on the coefficient of drag. My theory is for the coefficients that affect the coefficient of drag. First, let's start off with the coefficients themselves. The equation for Cd in terms of other coefficients is: Cd=Cf+Cw+Cs+Cl2/eπAR First, lets start with Cl2/eπAR, where e is the efficiency factor and AR is the value of the span squared over the area, s2/A. That is the equation for induced drag. Cl is the lift coefficient, and my theory does not cover that. A new theory I am still developing does, but I need to finish a separate theory in order to have an equation for that. Now, lets go on to the parasitic drag. Let's start with the form drag coefficient. The form drag coefficient is: Cf=Dcosθavg Where D is the drag coefficient it would have if it was a flat 2D plate and is a function of Reynold's Number. θavg is the average of the angles that each surface makes. The angles are relative to the axis perpendicular to the velocity. The reason that I think this is the equation is because the form drag goes around the object. Let's say you had two 2D plates joined together at one side and they were both tilted at an average angle of θavg. Then they would now have a form drag coefficient of Dcosθavg. This can be applied to all shapes. The form drag is moving around the object. But we do not count angles that are greater then 90 degrees in the average. The drag from that is covered in the interference drag. The skin friction drag coefficient Cs already has an equation for it that I did not come up with. The equation is: Cs=2dΘ/dx where Θ is the momentum thickness. Finally, let's talk about the wave drag. The wave drag is the drag from shock waves and only has an affect for supersonic or transonic velocities. The equation for that is: Cw=Av/vs Where A is the area of the shock wave, v is the velocity, and vs is the speed of sound. v/vs is a ratio called the Mach number. It is the ratio of the velocity to the speed of sound. I think this because the more area the shock wave covers, then the more dag it would make, and A=0 when v<vs, and this means there is no drag from the shock waves when there is none, which is correct. I put the Mach number into the equation because when you have a bigger Mach number, you have stronger shock waves, and so more drag. This makes the equation for drag coefficient: Cd=Dcosθavg+Av/vs+2dΘ/dx+Cl2/eπAR What do you think?

Oh. But also plastic theory would work and I don't think it would just be an approximation because stress is force over area, and this means that Ss/Sy is also equal to ASs/ASy , since both of the A's cancel out. Because S=F/A, we can rewrite this as Fs/Fy. Since the yield strength would be the same as the normal force, and the shear force would just be the frictional force, this would prove the plastic theory.

It is a good approximation. I think that the equation for that would be μs=sin(θavg)sin(Φavg)Ss/Sy , and that would make it the coefficient of static friction. Would the plastic theory be for static friction? If so, then that is what I think the equation might be because sin(θavg)sin(Φavg) is μk. That would make the equation also μs=μkSs/Sy. I think that the coefficient of kinetic and static friction are related in some way, and the factor would be Ss/Sy from plastic theory.

Why don't you think my theory will work?

It's because of the ambiguity of square roots.

But that means a/i equals ai and -ai

He said there was ambiguity In fact, you yourself said so.

Because of that, then the equation is the same, but it can be represented 2 ways. That way is an approximation, so I have to figure out the exact solution for that.

1. Ok, then someone move it 3. That is correct. It is also the same for static friction 4. Yes, can you explain?

So therefore, a/I is both ai and -ai. How?

1. I don't know. 2. You can figure out the average over the whole surface in various ways, such as integrating. You would do that to find the average. 3. It predicts the coefficient of kinetic friction between 2 surfaces 4. What you do is first, you find the deformation over time dθavg/dt and multiply by time. You then do θavg-tdθavg/dt for both surfaces. Then you get μk=sin(θavg-tdθavg/dt)sin(Φavg-tdΦavg/dt) And sorry for posting twice, that is a glitch

1. I don't know. 2. You can figure out the average over the whole surface in various ways, such as integrating. You would do that to find the average. 3. It predicts the coefficient of kinetic friction between 2 surfaces

Do you have any sources about that?

That's like saying 2=-2 just because 22 and (-2)2 are the same

The principle square root, or the square root most people mean when they say square root, can be. I did not say that (-1)3 doesn't mean -13. I said that root((-1)^3) # (root(-1))^3

I have a new theory for calculating the coefficient of kinetic friction. The equation is: μk=sin(θavg)sin(Φavg) Where θavg is the average of the angles each part of the first surface makes, Φavg is the same but for the second surface. This makes the force of friction be: Ffric=μkFN=sin(θavg)sin(Φavg)Fn Where Fn is the normal force. To find the average angle, you would need a graph to show all the angles and find the average. I made the equation for my theory μk=sin(θavg)sin(Φavg) because you have different parts of the two surfaces pressing against each other, and this makes a force that acts against the velocity. That's friction. The rougher the surface, then the bigger μk is. The surface would be rougher when the average angle is large. Because μk is proportional to θavg and Φavg, this means it would be sin(θavg)sin(Φavg) since friction would be 0 when the average angle is 0 and it gets bigger and approaches 1 as θavg and Φavg approach 90 My theory is easier to understand visually: What do you think?

What was that error?