Mudcat Café message #754226 The Mudcat Café TM
Thread #49771   Message #754226
Posted By: JohnInKansas
25-Jul-02 - 02:18 AM
Thread Name: Guitar string tension v gauge
Subject: RE: Guitar string tension v guage
With very red face....

I have just discovered that the (mostly reliable) source where I first found that the string frequency equations can be rearranged to give the results based "only on the string weight" neglected a "minor" square root dependency. I hadn't needed the formula recently and (hanging head in shame) didn't check back to "original sources."

For a simple string, meaning basically an unwound single wire, the relationship between frequency (f), tension (T), length of the string between bridge and nut (L) and total weight of the string (W) can be written as:

f = SQRT(T/kWL)

Another way of putting it is that for a given string ( f x f x W x L / T ) is a constant.

If you use all of the right "units," the constant k is 3.14159.... but the important thing is the relationship between the factors included.

For purposes of my previous comments, the relationship can be stated as:

"The square of the frequency is proportional to Tension, divided by weight and length."

At 23-Jul-02 - 03:42 AM: the italicized words should be inserted:

23-Jul-02 - 03:42 AM
The square of the frequency is directly proportional to the tension, and inversely proportional to the weight.

The ratio in frequency going from C to E (4 semitones) is 1.05946^4 (1.05946 x 1.05946 x 1.05946 x 1.05946 = 1.2599), so for a given string, tuning from C up to E requires a tension at E equal to the square root of] 1.26 ( = 1.122) times whatever the tension was at C

Working backwards - tuning down, if you know the tension for the heavier string tuned to E, you can divide by 1.26 1.122 to get the tension for the same string at C.

Dividing by 1.26 1.122 is the same as multiplying by 0.7937 0.891, so you can estimate that the string tuned to C has about 80 90% of whatever tension it would take to tune the same string to E.

At 23-Jul-02 - 10:39 AM:

Regardless of any details of how the string is put together, for any one string, the tension required to tune to E is the square root of 1.26 (which is 1.122) times the tension required to tune the same string to C. That's because the frequency of an "E" is 1.26 times the frequency of the next lower "C" and the only thing you're changing when you retune a particular string is the tension.
...
Because only the ratio T/W appears in the equation, the following paragraph is still correct:

For practical purposes though, the tension at a given pitch depends only on the total weight of a length of the string equal to the distance between bridge and nut. If one string weighs 1.26 times as much as another, then the heavier string will require 1.26 times the tension of the lighter one to bring them both to the same pitch.

But:

Since you are tuning down from E to C, the ratio of the frequencies is 1.26. If you use a string that is 1.26 x 1.26 = 1.59 times as heavy as your original "lightweight" E, it will have the same tension at C that the original string had at E. If your "heavy" string weighs less than 1.26 1.59 times what the same length of your lightweight string weighs, you will not have a higher string tension with the heavy string at C than you had with the light one at E.

If you attempt to tune an "80%" string up by the same 4 semitones - E to G# - it will probably break (.80 times 1.26 = 1 and you've reached the yield point).

You should be able to increase an "80%" string's tension by a factor of 1.20 to get to the "yield point". An increase in tension by 1.2 increases frequency by the square root of 1.2, or 1.095, which is between one and two semitones. The "fact" that you can, in practice, tune more than a semitone up may or may not mean that you started with a stress level of 80% of yield. When you reach yield, the string will continue to stretch by reducing its diameter (permanently) - hence the pitch can continue to go up because the weight decreases - for a little ways until the string breaks.

Again, my apologies for the error in the equation - particularly to those who spent time using it to look at something.

John