Originally posted by howardbradley
Consider a slightly simpler problem: two strings are the same length and tuned in unison. To get a perfect fifth we need the "stopped" lengths to be in the ratio 2:3 (that's a perfect fifth not an equal temperament fifth). So if the strings are separated by distance s and the first string is stopped at length L, the other string will be stopped at ...[text shortened]... vation that the nearer bridge you get, the more perpendicular to the strings your slide got.
Ah, I see a problem with the ratios, the strings are not the same diameter or tension, the 2:3 ration would be for strings of the same mass, diameter and strength and length, identical strings in other words. The string sizes are 0.43 mm for the second string and 0.86mm for the 4th string but not even that tells the whole story. The fourth string is a double wound one with a 0.36 mm core and wound with a 0.25 mm string. It adds up to 0.86mm or twice the diameter of the 2nd string but it can't be the same as a string= to #2 since there are air spaces between the windings so the mass doesn't just double.
The big hint however is the string lengths, the 2nd string 2nd fret to the bridge is 533 mm and the 4th string 5th fret is 476 mm to the bridge so the ratio is 1: 0.893 or 1:1.119 depending on which inversion you look at.
2:3 would be 1.5:1 or 0.6666:1 whichever way you want to call it.
So all things being equal, the 4th string 5th fret at 476 mm would want a length of about 317 mm to give you a perfect fifth, which in fact is just about what I get measuring it ON THE SAME STRING.
Its interesting however, the same note on a smaller diameter string is now 533 mm, just the opposite of what an equal mass/diameter string would be. So how does that effect the function of that angle?
The angle of the bar that makes that perfect fifth.