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Gibson Burstbucker Question


wyldbil

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I really don't mean to sound ignorant, but what difference does it make?  :D

There are three distinct models ofthe BBucker-

#1= slightly underwound (lower output)

#2= wound to the "average" late 50s PAF (average output)

#3= slightly overwound (hotter output)

As you can see, these specs make a big difference as to where the optimal positioning of each pickup would be, ie, given two different levels of "hotness", the hotter by the bridge, the less hot by the neck.

Makes a BIG difference, especially since I prefer to use a hotter pickup in all applications.

As it turns out, the two I have are stilll in their mounting rings, one clearly neck, one bridge. The person I inherited them from assures me the neck is a #1 and the bridge a #2 which makes perfect sense.

Problem solved!

If the "hotter by the bridge, milder by the neck" principle still does not make sense to you, go to the DiMarzio website where they explain it in more technical terms yet very clear. B)

Edited by wyldbil
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So, instead of simply making a bridge and neck model, they make 3 different pickups with increasing output levels, and leave it up to the user to figure out how to use them. Sometimes I'd like to be a fly on the wall in their marketing department! :D

:D And I know I used the word "ignorant", but I really do understand that the string induces less signal strength in the bridge pickup due to the more limited string motion. Anyway, thanks for the pointer to the Dimarzio site. B) Glad you solved your problem.

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:D Glad you solved your problem.

Well, not quite solved. B) I would really like something hotter (like DiMarzio X2N or Super Distortion) for this particular project. BUT, the Gibsons were donated (aka: GRATIS!) by a fellow Hamer enthusiast, so the price was right for a pair of high quality pickups.

Rock on, Loveraft! :D

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lol yeh, prolyl not the best place to psot this, but do you happen to know why you get induced voltage int he first place? i get the idea of steel wire moving through the field can give a charge to the wire, but no why as the copper only carries a EMF when powered. just wondering thats all, my physics teacher said he couldn't explain it :S

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