Musical angles

I was playing slide guitar the other day when I noticed if instead of the normal way of holding the slide, perpendicular to the strings, if you bend in so the slide is over the second string 15th fret and the 17th fret of the 4th string that makes a perfect 5th, its the same thing as just fingering down the 15th fret second string and 17th fret 4th string, is a perfect 5th. So you get the same perfect 5th if you do the same thing but down one octave, now its the 2nd string 3rd fret and 4th string 5th fret, same thing but an octave down. The thing I want to show is the angle of the slide is less, approaching parallel with the strings, not quite but going in that direction. So imagine a guitar with the same string separation but the strings are a kilometer long and you can see that angle will approach but not equal being parallel to the strings and as you go up to the bridge it will go towards being perpendicular to the strings but not quite. The thing I want to know is what is the mathematical function that would describe that change of angle of the slide during the making of that perfect 5th up and down the strings? The strings are in regular tuning BTW. EADGBE.

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 length 2/3 L. This gives a right angle triangle with one side s and hypoteneus 1/3 L. Hence tan(a) = 1/3 * L/s

For strings that are not in unison the ratio 1/3 will change (for strings a 4th apart it will be 1/9), so in general we'll have:

a = arctan(some ratio * L/s).

This fits in with your observation that the nearer bridge you get, the more perpendicular to the strings your slide got.

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.

Originally posted by sonhouse
So how does that effect the function of that angle?
The angle of the bar that makes that perfect fifth.

It will just affect the "some ratio" bit in my formula. I'm assuming that the length from nut to bridge is the same for each string - an OK assumption for classical and acoustic guitars, not so good for electrics and I have no clue about slide guitars - but the string weight and tension shouldn't make a difference.

Yet again you've set a problem that is easier to explain with a diagram. 😉

Originally posted by howardbradley
It will just affect the "some ratio" bit in my formula. I'm assuming that the length from nut to bridge is the same for each string - an OK assumption for classical and acoustic guitars, not so good for electrics and I have no clue about slide guitars - but the string weight and tension shouldn't make a difference.

Yet again you've set a problem that is easier to explain with a diagram. 😉

Thats a big for sure about the diagram! I would like to see some kind of simple drawing program that we could use here for that! The guitar is an acoustic and each string is close to the same length, however most guitars have the bass string, the big E string a tad longer, in this case 635.8 mm vs. 635 mm. I assume this tends to compensate for the smaller strings changing more than the larger ones when they pick up gumpucky from your fingers, oil and acid which etches the strings and makes for porousness which gets filled with oil and dirt and bits of finger dna! The result of that is the string gets less dense but at the same time heavier so it usually ends up making a lower frequency for the same length and is no longer symmetrical as far as where the harmonics are. In a new string the harmonics work out very close to the halfway points and half of that and half of that, etc., but older strings get a differential change in density by the fact that the left hand pushing down the strings to the frets will cause the string to be slightly different on the fret side than the part by the bridge which has less corrosion and oil build up. The fret side of the strings also get mushed into the frets which tends to make the string a bit thinner there and also the fret. So its a very complex issue trying to figure all that out for sure!