Do a new pickup need to activate or run in.?

[RedHouse]

...With the oblong shape, the force, such as it is, is applied to the ends and tends to crush the pick up along its length. So in a Fender single coil, the force tends to collapse the two outermost magnets together by buckling the flat work...

In my experience it's not the pole-to-pole aspect that's causing the flatwork to buckle, it's outward pressure on the flatwork at the ends of the bobbin. Measuring a actual example, the pole-to-pole distance doesn't reveal a crushing-in effect, while the flatwork expansion at the ends is measurable.

This another area where one can over-think the problem, really should go by observation instead.

[Mike Sulzer]

It does buckle the flatwork. By a few thousandths of an inch. This (and the relaxation of the copper itself, as mentioned by other posters) is why the winding doesn't collapse into a little ball.

The key is that copper is far stiffer a material than forbon, so it takes very little crushing to reduce the force to whatever the forbon can sustain.

This is not a correct description of the physics. Wind one turn on and pull to a certain tension. The bobbin or flat work deforms a small amount and is held there by the tension of the wire. Wind a second turn on and hold the same tension. The flatwork deforms a bit more. How much depends on whether the wire can slide around the bobbin. If it can, the tension remains in the first turn, and there is now twice as much force on the bobbin, and so it deforms twice as much as with one turn. If the wire cannot slide around, at least some of the force due to the first turn on the bobbin is relieved as the bobbin deforms, and so it deforms less than if the wire can slide. Even if two turns can slide, as one continues to wind more turns on, eventually the wire cannot slide all the way around and the tension on the first turns is at least partly relieved as the bobbin deforms slightly more. But the forces tending to deform the bobbin increase very slowly from this point on.

[RedHouse]

This is not a correct description of the physics. Wind one turn on and pull to a certain tension. The bobbin or flat work deforms a small amount and is held there by the tension of the wire. Wind a second turn on and hold the same tension. The flatwork deforms a bit more. How much depends on whether the wire can slide around the bobbin. If it can, the tension remains in the first turn, and there is now twice as much force on the bobbin, and so it deforms twice as much as with one turn. If the wire cannot slide around, at least some of the force due to the first turn on the bobbin is relieved as the bobbin deforms, and so it deforms less than if the wire can slide. Even if two turns can slide, as one continues to wind more turns on, eventually the wire cannot slide all the way around and the tension on the first turns is at least partly relieved as the bobbin deforms slightly more. But the forces tending to deform the bobbin increase very slowly from this point on.

Mike you should wind a few bobbins. What you're describing is some thought you're comming up with, not really what actually happens. Wind a few and observe.

[Mike Sulzer]

Mike you should wind a few bobbins. What you're describing is some thought you're comming up with, not really what actually happens. Wind a few and observe.

The fact that the wire tends to separate the flat work does not contradict what I am saying.

[Joe Gwinn]

This is not a correct description of the physics. Wind one turn on and pull to a certain tension. The bobbin or flat work deforms a small amount and is held there by the tension of the wire. Wind a second turn on and hold the same tension. The flatwork deforms a bit more. How much depends on whether the wire can slide around the bobbin. If it can, the tension remains in the first turn, and there is now twice as much force on the bobbin, and so it deforms twice as much as with one turn. If the wire cannot slide around, at least some of the force due to the first turn on the bobbin is relieved as the bobbin deforms, and so it deforms less than if the wire can slide. Even if two turns can slide, as one continues to wind more turns on, eventually the wire cannot slide all the way around and the tension on the first turns is at least partly relieved as the bobbin deforms slightly more. But the forces tending to deform the bobbin increase very slowly from this point on.

I think it will be a bit more complicated than that. Certainly the force crushing the flatwork will grow slowly as one winds, as you describe. But only the current top windings will be able to slide, even as those same current top windings provide the clamping force to prevent the growing mass of inner windings from sliding. The closer to the core, the greater the total clamping force, the less the sliding that's possible.

What's important is the static coefficient of friction, this being used to compute the force required to start sliding. (The dynamic coefficient of friction is used to compute the forse required to keep on sliding.) Friction - Wikipedia, the free encyclopedia

I don't offhand know the friction coefficients for magnet wire, but it's low, so let's assume that the static coefficient is 0.1. This means that if the winding tension is 20 grams, that a clamping force of 200 grams will prevent sliding of that strand. In round numbers, this means that ten or twenty layers of wire above the test strand will prevent sliding of the test strand and all strands closer to the core.

Now, soft plastics (like wire enamel) deform slightly on such pressure, raising the coefficient of static friction, which complicates things a bit.

It's possible to mathematically analyze this, at least numerically (versus symbolically), but it won't be simple.

[Mike Sulzer]

I think it will be a bit more complicated than that. Certainly the force crushing the flatwork will grow slowly as one winds, as you describe. But only the current top windings will be able to slide, even as those same current top windings provide the clamping force to prevent the growing mass of inner windings from sliding. The closer to the core, the greater the total clamping force, the less the sliding that's possible.

What's important is the static coefficient of friction, this being used to compute the force required to start sliding. (The dynamic coefficient of friction is used to compute the forse required to keep on sliding.) Friction - Wikipedia, the free encyclopedia

I don't offhand know the friction coefficients for magnet wire, but it's low, so let's assume that the static coefficient is 0.1. This means that if the winding tension is 20 grams, that a clamping force of 200 grams will prevent sliding of that strand. In round numbers, this means that ten or twenty layers of wire above the test strand will prevent sliding of the test strand and all strands closer to the core.

Now, soft plastics (like wire enamel) deform slightly on such pressure, raising the coefficient of static friction, which complicates things a bit.

It's possible to mathematically analyze this, at least numerically (versus symbolically), but it won't be simple.

I am glad to see that we are converging on the physics. However, experience shows (yes, Brad) that it does not take many turns to stop sliding, much less than a full layer. I believe that a friction coefficient of .1 implies this.

Have a good Thanksgiving. (back to cooking)

[RedHouse]

Sounds like cerebral rectitus, but just wind and observe (yes, Mike).