Jumble Jumble
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Original Floyd Rose comes with a shim under the four middle saddles that bumps it up to a 10" radius. Remove that shim and it's a 12" radius.Street Avenger said:I keep seeing people state that the Floyd is a 10" radius. It's not. It is a 12" radius, and the locking nut is a 10" radius.
Sorry but that's completely wrong. If you have a nut radius of 10", and then you set your bridge radius to 12", then the string radius at the 12th fret -- halfway along the string -- will be 11", not 12". Think about it - the radius will definitely be 10" at the nut. So your claim, that setting the bridge radius to 12" would result in a radius of 12" at the 12th fret, makes no sense. When does the radius "switch" from 10" to 12"? Or if it's gradually changing from 10 to 12 and then staying there, how does that work? The strings are all straight lines, and going from one radius to another and staying there would require them to have a corner in them. Plus how would they know that they're meant to hit 12" by the 12th fret?I also have a really hard time buying the "math" described here for radius.
Anything past the last fret is irrelevant as far as I'm concerned. I adjust my action @ the 12th fret, so if the radius (compound) at the 12th is 12" or 14", that's what my bridge radius needs to be to make the string height even for all 6 strings over the 12th fret. Not 18" or 19". If I wanted to set my action at the 22nd fret, then my bridge would need to be set (or shimmed as the case may be) to 16". If the fretboard continued to the bridge, and you wanted to set your action at the last fret, you would need to set it at 18" or 19", but that is never the case.
I understand that you don't get it, but I can assure you that it is right. Just give it a good ol' think-about.
Here's the workings-out:
http://www.jumbleguitar.com/2012/04/17/bridge-logic-revisited/