Moving the bridge hack: Difference between revisions

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If you have a [[~~12edo|~~12edo]] guitar, or other fretted [[~~string_instruments|~~string instrument]], and you want to play in an EDO that is numerically near 12 (e.g. [[~~11edo|~~11edo]] or [[~~13edo|~~13edo]]), then rather than redoing the whole fretboard, you might be tempted simply to move the bridge. If you move the bridge so that the 13th fret is now precisely 2/1, the frets will play precisely 13edo, right?

If you have a [[12edo]] guitar, or other fretted [[string instrument]], and you want to play in an EDO that is numerically near 12 (e.g. [[11edo]] or [[13edo]]), then rather than redoing the whole fretboard, you might be tempted simply to move the bridge. If you move the bridge so that the 13th fret is now precisely 2/1, the frets will play precisely 13edo, right?

...well, actually, no. The frets form a geometric series of lengths that converges at a specific point, which is where the bridge ought to be. (That's what an EDO ~~is - a~~ geometric sequence of frequencies, corresponding to a geometric sequence of string lengths.) If you move the bridge, the new string lengths no longer form a mathematically correct geometric sequence. However, depending on what range of the fretboard you want to be usable, and what accuracy you desire, a moving-the-bridge solution may be possible.

...well, actually, no. The frets form a geometric series of lengths that converges at a specific point, which is where the bridge ought to be. (That's what an EDO is—a geometric sequence of frequencies, corresponding to a geometric sequence of string lengths.) If you move the bridge, the new string lengths no longer form a mathematically correct geometric sequence. However, depending on what range of the fretboard you want to be usable, and what accuracy you desire, a moving-the-bridge solution may be possible.

== Derivation of the resulting scale ==

Let the EDO number of the original instrument be N (so very often N=12). Let the original scale length of the instrument (distance from bridge to nut) be 1. In other words we're measuring all lengths relative to the original scale length. Then the playable string lengths of the unmodified instrument are

Let the EDO number of the original instrument be ''N'' (so very often {{nowrap|''N'' {{=}} 12}}). Let the original scale length of the instrument (distance from bridge to nut) be 1. In other words we're measuring all lengths relative to the original scale length. Then the playable string lengths of the unmodified instrument are

If the bridge is moved so that the new scale length is x, this adds (x-1) to all string lengths, so the new string lengths are simply

If the bridge is moved so that the new scale length is ''x'', this adds ({{nowrap|''x'' − 1}}) to all string lengths, so the new string lengths are simply

The frequencies are inversely proportional to the string lengths. If we plug in i=0 to the above formula, we get x, so the frequency ratios relative to the open string are

The frequencies are inversely proportional to the string lengths. If we plug in {{nowrap|''i'' {{=}} 0}} to the above formula, we get ''x'', so the frequency ratios relative to the open string are

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-If you have a [[12edo|12edo]] guitar, or other fretted [[string_instruments|string instrument]], and you want to play in an EDO that is numerically near 12 (e.g. [[11edo|11edo]] or [[13edo|13edo]]), then rather than redoing the whole fretboard, you might be tempted simply to move the bridge. If you move the bridge so that the 13th fret is now precisely 2/1, the frets will play precisely 13edo, right?
+If you have a [[12edo]] guitar, or other fretted [[string instrument]], and you want to play in an EDO that is numerically near 12 (e.g. [[11edo]] or [[13edo]]), then rather than redoing the whole fretboard, you might be tempted simply to move the bridge. If you move the bridge so that the 13th fret is now precisely 2/1, the frets will play precisely 13edo, right?
-...well, actually, no. The frets form a geometric series of lengths that converges at a specific point, which is where the bridge ought to be. (That's what an EDO is - a geometric sequence of frequencies, corresponding to a geometric sequence of string lengths.) If you move the bridge, the new string lengths no longer form a mathematically correct geometric sequence. However, depending on what range of the fretboard you want to be usable, and what accuracy you desire, a moving-the-bridge solution may be possible.
+...well, actually, no. The frets form a geometric series of lengths that converges at a specific point, which is where the bridge ought to be. (That's what an EDO is—a geometric sequence of frequencies, corresponding to a geometric sequence of string lengths.) If you move the bridge, the new string lengths no longer form a mathematically correct geometric sequence. However, depending on what range of the fretboard you want to be usable, and what accuracy you desire, a moving-the-bridge solution may be possible.
 == Derivation of the resulting scale ==
-Let the EDO number of the original instrument be N (so very often N=12). Let the original scale length of the instrument (distance from bridge to nut) be 1. In other words we're measuring all lengths relative to the original scale length. Then the playable string lengths of the unmodified instrument are
+Let the EDO number of the original instrument be ''N'' (so very often {{nowrap|''N'' {{=}} 12}}). Let the original scale length of the instrument (distance from bridge to nut) be 1. In other words we're measuring all lengths relative to the original scale length. Then the playable string lengths of the unmodified instrument are
 <math>2^{-i/N} \text{ for } i = 0, 1, 2, 3\dots</math>
-If the bridge is moved so that the new scale length is x, this adds (x-1) to all string lengths, so the new string lengths are simply
+If the bridge is moved so that the new scale length is ''x'', this adds ({{nowrap|''x'' − 1}}) to all string lengths, so the new string lengths are simply
 <math>2^{-i/N} + x - 1 \text{ for } i = 0, 1, 2, 3\dots</math>
-The frequencies are inversely proportional to the string lengths. If we plug in i=0 to the above formula, we get x, so the frequency ratios relative to the open string are
+The frequencies are inversely proportional to the string lengths. If we plug in {{nowrap|''i'' {{=}} 0}} to the above formula, we get ''x'', so the frequency ratios relative to the open string are
 <math>\frac{x}{2^{-i/N} + x - 1} \text{ for } i = 0, 1, 2, 3\dots</math>