Theory for Coefficient of drag

[Endercreeper01]

I have a new theory for calculating the coefficient of kinetic friction. The equation is:

 

μ_k=sin(θ_avg)sin(Φ_avg)

 

Where θ_avg is the average of the angles each part of the first surface makes, Φ_avg is the same but for the second surface.

This makes the force of friction be:

 

F_fric=μ_kF_N=sin(θ_avg)sin(Φ_avg)F_n

 

Where F_n 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 μ_k=sin(θ_avg)sin(Φ_avg) 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 μ_k is. The surface would be rougher when the average angle is large. Because μ_k is proportional to θ_avg and Φ_avg, this means it would be sin(θ_avg)sin(Φ_avg) since friction would be 0 when the average angle is 0 and it gets bigger and approaches 1 as θ_avg and Φ_avg approach 90

 

My theory is easier to understand visually:

 

 

What do you think?

 

[studiot]

This is the first time I have seen you present a plausible theory, so +1.

 

A few questions to help you develop it and compare it with conventional theory.

 

Why have you not posted this not in speculations since that is what it is?

 

Why have you chosen kinetic friction? surely your 'average angles' will change as the object slides?

 

What predictions does your theory make? I can immediately see a testable one, but will leave you the honour of stating it, since it is your theory.

 

How does it compare with conventional plastic flow theory of friction? The two theories are not actually incompatible.

 

I look forward to the development of this speculation with interest.

[Enthalpy]

Said theory was proposed several centuries ago and is known to be irremediably wrong.

 

Contradicted by experiments, both quantitatively and qualitatively.

 

Anyway, a solid is deformable, and a surface isn't flat.

[Endercreeper01]

Said theory was proposed several centuries ago and is known to be irremediably wrong.

 

Contradicted by experiments, both quantitatively and qualitatively.

 

Anyway, a solid is deformable, and a surface isn't flat.

Do you have any sources about that?

[Endercreeper01]

This is the first time I have seen you present a plausible theory, so +1.

 

A few questions to help you develop it and compare it with conventional theory.

 

Why have you not posted this not in speculations since that is what it is?

 

Why have you chosen kinetic friction? surely your 'average angles' will change as the object slides?

 

What predictions does your theory make? I can immediately see a testable one, but will leave you the honour of stating it, since it is your theory.

 

How does it compare with conventional plastic flow theory of friction? The two theories are not actually incompatible.

 

I look forward to the development of this speculation with interest.

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

[Endercreeper01]

This is the first time I have seen you present a plausible theory, so +1.

 

A few questions to help you develop it and compare it with conventional theory.

 

Why have you not posted this not in speculations since that is what it is?

 

Why have you chosen kinetic friction? surely your 'average angles' will change as the object slides?

 

What predictions does your theory make? I can immediately see a testable one, but will leave you the honour of stating it, since it is your theory.

 

How does it compare with conventional plastic flow theory of friction? The two theories are not actually incompatible.

 

I look forward to the development of this speculation with interest.

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 dθ_avg/dt and multiply by time. You then do θ_avg-tdθ_avg/dt for both surfaces. Then you get μ_k=sin(θ_avg-tdθ_avg/dt)sin(Φ_avg-tdΦ_avg/dt)

And sorry for posting twice, that is a glitch

Edited October 17, 2013 by Endercreeper01

[studiot]

1. I've made my point that I think this belongs in speculations.

Yes there are many ways to calculate an average.

Again I ask why kinetic friction? Perhaps I have misunderstood, but I thought your model involved considering (adding) all the side thrusts from the local irregularities pressing against each other at random angles (hence the average angle). This would obviously also work for static friction. But my point was that the meshing of the local irrecularities would vary as the two surfaces moved across each other.

This has nothing to do with conventional plastic theory. I will explain that if you wish. You can arrive at a reasonable estimate of the coefficient of static friction for metals with it and account for the drop between kinetic and static friction.

[Endercreeper01]

 

1. I've made my point that I think this belongs in speculations.
2. Yes there are many ways to calculate an average.
3. Again I ask why kinetic friction? Perhaps I have misunderstood, but I thought your model involved considering (adding) all the side thrusts from the local irregularities pressing against each other at random angles (hence the average angle). This would obviously also work for static friction. But my point was that the meshing of the local irrecularities would vary as the two surfaces moved across each other.
4. This has nothing to do with conventional plastic theory. I will explain that if you wish. You can arrive at a reasonable estimate of the coefficient of static friction for metals with it and account for the drop between kinetic and static friction.

 

1. Ok, then someone move it

3. That is correct. It is also the same for static friction

4. Yes, can you explain?

[studiot]

OK, plastic theory works like this.

 

Take an irregular (bumpy) horizontal surface.

Because it is bumpy it must have a highest point

 

Now place a second perfectly smooth flat surface onto the bumpy one. I am only using a flat second surface for simplicity of explanation the pricciple is the same for an irregular surface.

 

Clearly the flat surface will touch the highest point of the bumpy one first.

 

Now the stress equals the contact force (weight of the second object) divided by the area of contact. But this area is very small.

So the high point will deform plastically until contact is made with the second highest point.

And so on to the third, fourth etc highest points

Until the area of contact is sufficient to support the contact force without further deformation.

 

This plastic deformation joins (bonds) the two objects across the area of contact. Breaking this bond requires a shear force equal to the frictional force calculated from the coefficient of friction.

 

I do not have the time tonight but you can derive that

[math]\mu [/math] [math] \approx [/math] [math]\frac{{{S_s}}}{{{S_y}}}[/math]

That is the coefficient of friction is approximately equal to the ratio of the shear strength to the yield strength.

Edited October 17, 2013 by studiot

[Endercreeper01]

OK, plastic theory works like this.

 

Take an irregular (bumpy) horizontal surface.

Because it is bumpy it must have a highest point

 

Now place a second perfectly smooth flat surface onto the bumpy one. I am only using a flat second surface for simplicity of explanation the pricciple is the same for an irregular surface.

 

Clearly the flat surface will touch the highest point of the bumpy one first.

 

Now the stress equals the contact force (weight of the second object) divided by the area of contact. But this area is very small.

So the high point will deform plastically until contact is made with the second highest point.

And so on to the third, fourth etc highest points

Until the area of contact is sufficient to support the contact force without further deformation.

 

This plastic deformation joins (bonds) the two objects across the area of contact. Breaking this bond requires a shear force equal to the frictional force calculated from the coefficient of friction.

 

I do not have the time tonight but you can derive that

[math]\mu [/math] [math] \approx [/math] [math]\frac{{{S_s}}}{{{S_y}}}[/math]

 

That is the coefficient of friction is approximately equal to the ratio of the shear strength to the yield strength.

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.

Edited October 18, 2013 by Endercreeper01

[Enthalpy]

OK, plastic theory works like this.

[...]

That is the coefficient of friction is approximately equal to the ratio of the shear strength to the yield strength.

And the coefficient of friction differs radically from the shear-to-yield strength ratio, which tells that the plastic theory does not work at all.

 

Presently there are half a dozen theories often cited, none works. Theoreticians try to mix all these ad explanations and, with enough tuning coefficients, claim to match one observation under one set of conditions.

 

Mechanical designers refer to experimental values and keep in mind that these are not reliable nor reproducible.

 

We lack a working theory. But the one with contact angles (was it from Newton?) is not the solution.

[Endercreeper01]

And the coefficient of friction differs radically from the shear-to-yield strength ratio, which tells that the plastic theory does not work at all.

 

Presently there are half a dozen theories often cited, none works. Theoreticians try to mix all these ad explanations and, with enough tuning coefficients, claim to match one observation under one set of conditions.

 

Mechanical designers refer to experimental values and keep in mind that these are not reliable nor reproducible.

 

We lack a working theory. But the one with contact angles (was it from Newton?) is not the solution.

Why don't you think my theory will work?

Edited October 18, 2013 by Endercreeper01

[studiot]

 

which tells that the plastic theory does not work at all.

 

Really?

 

For steel, Youngs modulus is 2 x 10¹¹ N/m² Modulus of rigidity is 0.8 x 10¹¹ N/m²

 

Thus the plastic approximation to the coefficient of friction is 0.8/2 = 0.4.

 

Pretty close I'd say.

[Endercreeper01]

Really?

 

For steel, Youngs modulus is 2 x 10¹¹ N/m² Modulus of rigidity is 0.8 x 10¹¹ N/m²

 

Thus the plastic approximation to the coefficient of friction is 0.8/2 = 0.4.

 

Pretty close I'd say.

It is a good approximation.

I think that the equation for that would be μ_s=sin(θ_avg)sin(Φ_avg)S_s/S_y , 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(θ_avg)sin(Φ_avg) is μ_k. That would make the equation also μ_s=μ_kS_s/S_y. I think that the coefficient of kinetic and static friction are related in some way, and the factor would be S_s/S_y from plastic theory.

[studiot]

 

I think that the equation for that would be μ_s=sin(θ_avg)sin(Φ_avg)S_s/S_y

 

 

Not really

 

You have the basic equation [math]F = \mu R[/math]

It doesn't mater whether we are talking static or kinetic here.

 

Your attempt to account for the mu factor works on summing all the contributions from F.

 

The plastic theory works by summing all the contributions from R.

 

You do not need to do both, just one or the other.

 

Plastic theory suggests that [math]{\mu _s} > {\mu _k}[/math]

 

 

simply because the bonds don't have time to fully form before the surfaces move on.

 

You should also note that sinx is never greater than 1 so the product of two sines is smaller than either.

Edited October 18, 2013 by studiot

[Endercreeper01]

 

 

Not really

 

You have the basic equation [math]F = \mu R[/math]

It doesn't mater whether we are talking static or kinetic here.

 

Your attempt to account for the mu factor works on summing all the contributions from F.

 

The plastic theory works by summing all the contributions from R.

 

You do not need to do both, just one or the other.

 

Plastic theory suggests that [math]{\mu _s} > {\mu _k}[/math]

 

 

simply because the bonds don't have time to fully form before the surfaces move on.

 

You should also note that sinx is never greater than 1 so the product of two sines is smaller than either.

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 S_s/S_y is also equal to AS_s/AS_y , since both of the A's cancel out. Because S=F/A, we can rewrite this as F_s/F_y. 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.

[Endercreeper01]

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 C_d in terms of other coefficients is:

C_d=C_f+C_w+C_s+C_l²/eπAR

First, lets start with C_l²/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. C_l 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:

C_f=Dcosθ_avg

Where D is the drag coefficient it would have if it was a flat 2D plate and is a function of Reynold's Number. θ_avg 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 θ_avg. Then they would now have a form drag coefficient of Dcosθ_avg. 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 C_s already has an equation for it that I did not come up with. The equation is:

 

C_s=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:

C_w=Av/v_s

Where A is the area of the shock wave, v is the velocity, and v_s is the speed of sound. v/v_s 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<v_s, 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:

C_d=Dcosθ_avg+Av/v_s+2dΘ/dx+C_l²/eπAR

 

What do you think?

[hypervalent_iodine]

!

Moderator Note

One thread per topic. Closed.

[hypervalent_iodine]

!

Moderator Note

Similar topics merged. Endcreeper01, please stop opening multiple topics on this.

Edit: I originally closed the repeat topic and then decided to merge the two threads, only to then merge the incorrect ones. I have now rectified the situation. Thanks to studiot for pointing it out.

[Endercreeper01]

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

[Endercreeper01]

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:

 

C_f=2Dcosθ_avg-DcosΦ_avg

 

Where Φ_avg is the same as θ_avg, but the axis is at the back rather then at the front. This is why I think this:

First, we have the term Dcosθ_avg. 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:

 

C_d=2F_d/Aρv²=F_d/qA

 

In this case, F_d 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 qC_d

In this case, C_d is just C_f not including the effects from the low pressure area. This would mean it is Dcosθ_avg. Therefore:

 

p=qDcosθ_avg

 

And so:

 

Δp=qDcosθ_avg - qDcosΦ_avg

 

This means Δp/q is:

 

qDcosθ_avg - qDcosΦ_avg/q=Dcosθ_avg-DcosΦ_avg

But also with the form drag, if the object is concave instead of convex, then it is D/cosθ_avg instead of Dcosθ_avg, and the same with Φ_avg

Edited October 20, 2013 by Endercreeper01

[Endercreeper01]

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

Edited October 21, 2013 by Endercreeper01

[Endercreeper01]

Do you think there is anything I need to change about my theory?

Another equation for the coefficient of drag is that C_d=p/q, where p is the pressure and q is the dynamic pressure.

[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

[Endercreeper01]

There is also the drag from the wind If you have a wind speed of v_w and an initial velocity of v₀, then the total velocity is v=(v_w²+v₀²)^1/2