Rope wrapped around a rod - belt friction

[Lotto]

TL;DR Summary

I have a rope wrapped let's say twice around a rod and have two bodies attached to its two ends. I know this equation $T_2=T_1 \mathrm e^{2\pi N \mu}$, where N is in our case 2. I understand the equation, but I think it is valid only when the rope is wrapped around the rod in a perfect circle, but that is not possible, so my qustion is: is this formula only an approximation?

Ideally, it should look circa like this:

But in reality, it looks often like this:

Is the formula valid also in this case?

 

[Baluncore]

Everything is only an approximation.
The rope ends need to be at 90° to the axis of the capstan. In the first picture, there are no ends, in the second you do not show the ends, so it is invalid in both your depicted situations. Other than that, the equation works well.

 

[Lotto]

Everything is only an approximation.
The rope ends need to be at 90° to the axis of the capstan. In the first picture, there are no ends, in the second you do not show the ends, so it is invalid in both your depicted situations. Other than that, the equation works well.

Of course the rope ends are at 90° to the axis of the rod, I didn't show it in the pictures because I thought it was clear. I showed the pictures to show that the rope isn't often wrapped around the rod in the perfect circle. But as I said, I have two weights at the ends of the rope.

However, my intention was to find out whether the formula is correct both for the first picture and for the second one (when there are two weights at the rope's ends). Isn't there a measurement error for the second picture?

 

[Baluncore]

However, my intention was to find out whether the formula is correct both for the first picture and for the second one (when there are two weights at the rope's ends).

The open helix of the second example would close if the ends were at 90° to the axis of the capstan. It is the number of turns about the axis that is important, not the length of the rope.

 

[Lotto]

The open helix of the second example would close if the ends were at 90° to the axis of the capstan. It is the number of turns about the axis that is important, not the length of the rope.

OK, my system should look like this:

(although there is a second weight at the end of the rope, at the place where we can see that arrow). So, when the turns aren't perfect circles as shown here (I consider "the perfect circles" turns shown in the very first picture of this thread):

can we use the formula without any measurement errors? Is the formula valid or is it a condition to use it when the turns are very close together (like in the first picture)?

 

[Baluncore]

can we use the formula without any measurement errors? Is the formula valid or is it a condition to use it when the turns are very close together (like in the first picture)?

If the radius of the capstan is not part of the equation, then the equation should also hold for an open helical wrap. The length of rope per turn increases, the virtual radius increases, while the rope pressure on the capstan is reduced, which compensates for the longer rope per turn.

 

[Lotto]

If the radius of the capstan is not part of the equation, then the equation should also hold for an open helical wrap. The length of rope per turn increases, the virtual radius increases, while the rope pressure on the capstan is reduced, which compensates for the longer rope per turn.

So when the lenght of the rope per turn changes, but the weight and turns are the same, the force $F$ in the picture above doesn't change at all? The lenght of the rope per turn is totally irrelevant and what only matters is the number of turns. So when the distances between two turns are always different, then the equation work as well. Do I understand it well?

 

[Dale]

can we use the formula without any measurement errors?

You cannot use any formula in physics without any measurement errors. Every measurement intrinsically has uncertainty.

In something like this there are lots of sources of error. The angle is not exact, the rope is not flexible, the coefficient of friction is inaccurate, the tension is uncertain, the formula may be inexact.

 

[Baluncore]

A physicist would accept that the equation is correct in what it relates, but knows it lacks many of the parameters that would be needed to compute an accurate value in the real world.

In the absence of a design code, an engineer would happily use the equation, but would then double or triple the value, to get a wider safety margin.

A mathematician would consider the topology as a knot, then collapse it to a non-knot, an imaginary loop of useless rope.

Do I understand it well?

You understand it like an engineer. Give a bit, take a bit.
If you want to understand it better, you will need to go back to its derivation, to look at the assumptions and approximations that were made in the original formulation.

 

[Lotto]

A physicist would accept that the equation is correct in what it relates, but knows it lacks many of the parameters that would be needed to compute an accurate value in the real world.

In the absence of a design code, an engineer would happily use the equation, but would then double or triple the value, to get a wider safety margin.

A mathematician would consider the topology as a knot, then collapse it to a non-knot, an imaginary loop of useless rope.You understand it like an engineer. Give a bit, take a bit.
If you want to understand it better, you will need to go back to its derivation, to look at the assumptions and approximations that were made in the original formulation.

And when the number of turns and the weight are constant, then (no matter how distant are single turns) the force $F$ from the picture above should be constant as well? It is only important for the rope turns not to touch each other, then friction between the rope itself would appear. Am I right?

 

[Baluncore]

It is only important for the rope turns not to touch each other, then friction between the rope itself would appear. Am I right?

I do not see a problem with that since the rope-rope contact is normal to the capstan surface, and the differential rope velocity is zero.
The only requirement I see is that the rope wrap slides continuously on the surface of the capstan. If it switched from sliding to static friction, that would cause hysteresis.