ToneFiddler
Active member
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DavidRavenMoon
New member
I hand wind pickups, but try to wind as neatly as possible. I find excessive scatter makes the pickup sound kind of lose in the low end. I don’t like it.
Neatly wound coils have a tighter low end.
All hand wound coils are random, but I see videos of people winding where they use way too much scatter. It makes each wrap of wire longer, so you end up with more wire for the same number of turns.
But either way the change is tone isn’t all that drastic.
And PAF pickups were always machine wound. No one complains about them.
Sent from my iPhone using Tapatalk
Neatly wound coils have a tighter low end.
All hand wound coils are random, but I see videos of people winding where they use way too much scatter. It makes each wrap of wire longer, so you end up with more wire for the same number of turns.
But either way the change is tone isn’t all that drastic.
And PAF pickups were always machine wound. No one complains about them.
Sent from my iPhone using Tapatalk
the fishman stuff is active which is a whole different ball game
for lower output single coils, i like more scatter and a little less tension. this does usually lead to less tightness/focus on the bottom but with a 8k turn strat neck pup, it doesnt get woofy or muddy
for lower output single coils, i like more scatter and a little less tension. this does usually lead to less tightness/focus on the bottom but with a 8k turn strat neck pup, it doesnt get woofy or muddy
eclecticsynergy
Well-known member
Remember that a number of Duncans are wound on their Leesona: all 59s, Antiquities, Pearly Gates, Seth Lovers and a few others - including many of the Custom Shop pickups.
freefrog
Well-known member
Hi,
Personally, I'd say that scatter winding theoretically diminishes distributed capacitance but also changes the shape of the coil and lowers the Q factor.
Hence a higher pitched BUT flatter and wider resonant peak, as if the tone control was lowered: the tone becomes more open but less focused.
So, in my understanding, there's no fundamental contradiction between the assumptions that you evoke.
Single coils with rod magnets have "naturally" a high Q factor (a narrow pointy resonant peak) and take advantage of scatter winding which lowers this Q factor, giving them a broader frequency response and more perceived harmonic richness, since the resonant peak less tends to hide upper harmonics.
The structure of humbuckers "naturally" favors a low Q factor (a flat round resonance) so tight machine winding gives them more tightness by increasing the Q factor of their coils. The related narrowness of the resonant peaks is perceived as an increased brightness, even if these peaks are noticeably lower pitched than those of single coils.
That being said, IME, some hand wound humbuckers exhibit a really bright tone and a single coil kind of "touch sensitivity" so my sum up above is really a gross simplification (not to mention that other variables can totally change the whole pictures).
FWIW : my 2 cents. YMMV.
ItsaBass
New member
I have these two thoughts about scatter winding.
1. If you are winding to a turn count, you will end up with ever so slightly higher resistance than that turn count would give you with neat layering.
Consider why you won't hear it, by thinking about the first "layer" of coil windings on a classic Strat pickup.
A typical Strat pickup has about 2.125 in. from the center of one E pole to the center of the other E pole. It is about 0.45 in. from the top surface of the bottom flatwork to the bottom surface of the top flatwork. The magnets are 0.188" (though other widths have been used). So one half of the circumference of each magnet is 0.295".
So one perfectly neat coil winding around the magnets is 2(2.125 in.) + 2(0.295 in.) => 4.25 in. + 0.59 in. = 4.84 in. of wire contained in one neat winding.
If "scatter" winding, the very most "extra" wire you could wind on would be a full diagonal. Use the classic geometry equation, the Pythagorean theorem, to figure out the length of the diagonal run of wire. a squared + b squared = c squared. C is the diagonal.
2.125 in. x 2.125 in. = 4.52 in. ("a squared")
0.45 in. x 0.45 in. = 0.2025 in. ("b squared")
4.52 in. + 0.2025 in. = 4.72 in. Therefore, 4.72 in. = "c squared"
Clear the square off of the c by taking the square root of both sides of the equation.
The square root of 4.72 in. = c. Therefore c = 2.17 in.
Compare to the straight run of 2.25 in. That's an extra 0.080 in. per side, or an extra 3.5 percent of wire length (therefore resistance) per side.
I don't know how to calculate the increased amount of wire when going around the magnets at an angle (like a barber pole), but let's say it's also 3.5 percent more, because it's probably close.
So we can say theoretically, considering one turn, that you're adding, at the absolute maximum, 3.5 percent more resistance per turn. Since we are assuming that we have the room to wind a specific number of turns, we can say that the end result is 3.5 percent extra resistance, if every single turn goes "corner to corner."
A hypothetical 6.0 KOhm Strat pickup becomes a 6.21 KOhm Strat pickup at the theoretical max. But remember, this is simply extra total resistance at the same number of turns. It's not the the reason we normally see added resistance in a pickup (due to extra turns). So would you hear it?
And as we know, when scatter winding, you don't, and indeed, you can't, go corner to corner on every single wind. It's impossible to layer a coil that way and still get anywhere close to the same number of turns on the bobbin. So the physical reality of how the wire must be layered to get the right number of turns means that you can't get anywhere near adding 3.5 percent extra wire length.
I would guess, out of my ass, that you're closer to a quarter of that amount of extra wire, at most – maybe even less. Let's say 1 percent more wire, though. So you're 6.0 KOhm Strat pickup becomes a 6.06 KOhm Strat pickup, and it has the same number of windings as the neatly layered one. Can you hear 0.06 KOhm more resistance on a coil, when the turn count is the same?
2. If you are winding until the bobbin is full, you will end up with fewer turns on the pickup. Whether you hear this or not depends on how many fewer turns.
You take 8,000 pieces of rope, all of the same length, and layer them across each other at various angles within the boundaries of a long, rectangular walled area. Measure how high the stack is.
Then, take those same 8,000 lengths of rope and lay them out neatly in flat layers within the same walled area. Measure the height of the stack.
Common mechanical sense tells you that the pile laid out neatly is going to be lower. Or as a better way of thinking about it for our purposes, if you neatly lay out ropes until you reach the same height as the randomly stacked pile, then you end up using more than 8,000 ropes to get to the same height.
Unfortunately, with my limited mechanical knowledge, I have no idea how to calculate how many turns you could lose by scatter winding a Strat coil! So I can't really say based on theory.
But I can say that just screwing around and experimenting at home, I have wound a humbucker that had one neatly wound coil and one scatter wound coil, both until full (using AWG 40 wire). The scatter wound coil took 2,850 turns. The neatly wound coil took 3,350 turns. That's 15 percent less wire on the scatter wound bobbin!
If that number is a fair representation of scattered vs. neatly layered, AND you are winding bobbins till completely full, then that would be a very large and very audible difference in the number of windings you end up with.
So, my untested theory on the matter is that if you are scatter winding to a turn count, you won't hear a difference. But if you are scatter winding until your bobbins are full, then you will.
But that theory is formulated only knowing what I know...which isn't much! There are other things that could be at play, e.g. does the angle at which the wires run affect anything about the magnetic field, or the production of current inside the coil? Does capacitance notably change? Etc.
1. If you are winding to a turn count, you will end up with ever so slightly higher resistance than that turn count would give you with neat layering.
Consider why you won't hear it, by thinking about the first "layer" of coil windings on a classic Strat pickup.
A typical Strat pickup has about 2.125 in. from the center of one E pole to the center of the other E pole. It is about 0.45 in. from the top surface of the bottom flatwork to the bottom surface of the top flatwork. The magnets are 0.188" (though other widths have been used). So one half of the circumference of each magnet is 0.295".
So one perfectly neat coil winding around the magnets is 2(2.125 in.) + 2(0.295 in.) => 4.25 in. + 0.59 in. = 4.84 in. of wire contained in one neat winding.
If "scatter" winding, the very most "extra" wire you could wind on would be a full diagonal. Use the classic geometry equation, the Pythagorean theorem, to figure out the length of the diagonal run of wire. a squared + b squared = c squared. C is the diagonal.
2.125 in. x 2.125 in. = 4.52 in. ("a squared")
0.45 in. x 0.45 in. = 0.2025 in. ("b squared")
4.52 in. + 0.2025 in. = 4.72 in. Therefore, 4.72 in. = "c squared"
Clear the square off of the c by taking the square root of both sides of the equation.
The square root of 4.72 in. = c. Therefore c = 2.17 in.
Compare to the straight run of 2.25 in. That's an extra 0.080 in. per side, or an extra 3.5 percent of wire length (therefore resistance) per side.
I don't know how to calculate the increased amount of wire when going around the magnets at an angle (like a barber pole), but let's say it's also 3.5 percent more, because it's probably close.
So we can say theoretically, considering one turn, that you're adding, at the absolute maximum, 3.5 percent more resistance per turn. Since we are assuming that we have the room to wind a specific number of turns, we can say that the end result is 3.5 percent extra resistance, if every single turn goes "corner to corner."
A hypothetical 6.0 KOhm Strat pickup becomes a 6.21 KOhm Strat pickup at the theoretical max. But remember, this is simply extra total resistance at the same number of turns. It's not the the reason we normally see added resistance in a pickup (due to extra turns). So would you hear it?
And as we know, when scatter winding, you don't, and indeed, you can't, go corner to corner on every single wind. It's impossible to layer a coil that way and still get anywhere close to the same number of turns on the bobbin. So the physical reality of how the wire must be layered to get the right number of turns means that you can't get anywhere near adding 3.5 percent extra wire length.
I would guess, out of my ass, that you're closer to a quarter of that amount of extra wire, at most – maybe even less. Let's say 1 percent more wire, though. So you're 6.0 KOhm Strat pickup becomes a 6.06 KOhm Strat pickup, and it has the same number of windings as the neatly layered one. Can you hear 0.06 KOhm more resistance on a coil, when the turn count is the same?
2. If you are winding until the bobbin is full, you will end up with fewer turns on the pickup. Whether you hear this or not depends on how many fewer turns.
You take 8,000 pieces of rope, all of the same length, and layer them across each other at various angles within the boundaries of a long, rectangular walled area. Measure how high the stack is.
Then, take those same 8,000 lengths of rope and lay them out neatly in flat layers within the same walled area. Measure the height of the stack.
Common mechanical sense tells you that the pile laid out neatly is going to be lower. Or as a better way of thinking about it for our purposes, if you neatly lay out ropes until you reach the same height as the randomly stacked pile, then you end up using more than 8,000 ropes to get to the same height.
Unfortunately, with my limited mechanical knowledge, I have no idea how to calculate how many turns you could lose by scatter winding a Strat coil! So I can't really say based on theory.
But I can say that just screwing around and experimenting at home, I have wound a humbucker that had one neatly wound coil and one scatter wound coil, both until full (using AWG 40 wire). The scatter wound coil took 2,850 turns. The neatly wound coil took 3,350 turns. That's 15 percent less wire on the scatter wound bobbin!
If that number is a fair representation of scattered vs. neatly layered, AND you are winding bobbins till completely full, then that would be a very large and very audible difference in the number of windings you end up with.
So, my untested theory on the matter is that if you are scatter winding to a turn count, you won't hear a difference. But if you are scatter winding until your bobbins are full, then you will.
But that theory is formulated only knowing what I know...which isn't much! There are other things that could be at play, e.g. does the angle at which the wires run affect anything about the magnetic field, or the production of current inside the coil? Does capacitance notably change? Etc.
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ItsaBass
New member
Hi,
Personally, I'd say that scatter winding theoretically diminishes distributed capacitance but also changes the shape of the coil and lowers the Q factor.
Hence a higher pitched BUT flatter and wider resonant peak, as if the tone control was lowered: the tone becomes more open but less focused.
So, in my understanding, there's no fundamental contradiction between the assumptions that you evoke.
Single coils with rod magnets have "naturally" a high Q factor (a narrow pointy resonant peak) and take advantage of scatter winding which lowers this Q factor, giving them a broader frequency response and more perceived harmonic richness, since the resonant peak less tends to hide upper harmonics.
The structure of humbuckers "naturally" favors a low Q factor (a flat round resonance) so tight machine winding gives them more tightness by increasing the Q factor of their coils. The related narrowness of the resonant peaks is perceived as an increased brightness, even if these peaks are noticeably lower pitched than those of single coils.
That being said, IME, some hand wound humbuckers exhibit a really bright tone and a single coil kind of "touch sensitivity" so my sum up above is really a gross simplification (not to mention that other variables can totally change the whole pictures).
FWIW : my 2 cents. YMMV.
Assuming I actually did understand the electronic theory behind this, the question would be: Are things changed by enough to make a tonal difference?
I'm not asking you to explain the various forms of capacitance to me. I'm just curious whether you think the effects of what you talked about would be major enough to cause a real world tonal change, in two pickups with the same number of coil windings – one neatly wound, one randomly wound.
Just to clarify, the coil dimensions thing...I get. I talked about it myself, in a way. The capacitance thing, I don't know about.
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freefrog
Well-known member
My answer will be that I don't "think" it to be the case: I've witnessed it.
More later if time permits (am way too busy right now).
