How will the pitch of a string change when I stretch it? (sound)

[Paul Colby]

True temperament frets, elliptical fretboard, fanned scalloped neck.

I'm having a hard time with the magnitude of the fret corrections you're calculating. They seem disproportionately large and discontinuous fret to fret.

 

[Paul Colby]

I don't get it. If you lower the tension won't the string be out of tune?

We're talking about the change in pitch added by the tension incurred by pressing the string to the 12'th fret. Initially the bridge is adjusted such that the twelfth harmonic matches in pitch with the pitch obtained when the string is pressed into the 12'th fret. The strings age and wear. The wear reduces the string material at each fret as time passes. The stiffness at these wear points is reduced. Now the string looks like many springs of slightly differing stiffness connected in series.

Years pass (I recall 20 years or so if I read the OP correctly)

The guitar is then tuned before testing. The open string now vibrates at the nominal pitch. The harmonic is sounded. Works great because the wavelength is exactly halved the open string value. Now the string is depressed. Since the stiffness has been reduced at the wear points, the tension increase by pressing into the fret is now less than it was initially when the guitar intonation was set. This causes the note to be flat.

 

[JT Smith]

Since the stiffness has been reduced at the wear points, the tension increase by pressing into the fret is now less than it was initially when the guitar intonation was set. This causes the note to be flat.

Thanks, that makes sense. But I don't think it works. When you fret a string (gently) you form a kind of triangle between nut, fret, and bridge. The string elongates by only a tiny amount, assuming the action isn't crazy. For a twelfth fret action of 2mm you'd expect the frequency at the twelfth fret to increase by less than 1hz. So it would be impossible to lose a few cents. If the action were higher, say 5mm, the stiffness would have to decrease by about 25% to flatten by that much, which seems kind of unlikely. I think you'd notice that when winding the string.

So unless the action is quite high, or maybe with non-metal strings, it doesn't seem to me that a loss of stiffness explains it.

 

[bobob]

I was trying to search but found no results.
Is there a way to calculate how frequency will change if I stretch a string by certain amount (0.14 mm in my case)?[

Yes.

I found out I can measure its frequency once stretched, but no results as of how to estimate new frequency ahead of the time. I know a speed of string can be calculated from its weight, length and tension, but that doesn't seem to get me anywhere:

Use the formula for the frequency of a string under tension. Several things to consider. As you stretch the string, the mass density will decrease slightly. Is this important for your purpose? If so, look up Poisson's ratio.
You can find a wealth of string data at the D'Addario website to compare your results with. They have data for all of their strings giving pitches at different tensions based on a 25.5 in scale length. Beware of the formulae they use. They've written things in a way that combines factors to give weird dimensionless constants so that people who aren't physicists won't go, "WTF?" and get confused. Instead, physicists will look at it and go "WTF?" Their formulae are correct, but the "magic numbers" like 386.4 might not be obvious. Do a google search for D'addario string data.[/quote]

 

[bobob]

The properties of strings change with time. I find it takes about a day for a new set to stop stretching. My guess is they lose stiffness as the age.

The metal work hardens, plust oxidation plus grime at random spots. My strings always feel really slinky before stretching them out and get a bit stiffer to bend to the same pitch. Supposedly, boiling them will somewhat restore them, which I assume is some sort of annealing, but it seems too low a temperature for that. I've personally never tried that, but that's the lore. Maybe it just removes the grime and oxidation well.

 

[bobob]

Why do you think that a reduced spring constant for the upper half of the string would cause the lower part to go flat?

There are different lengths to consider. One is the scale length, which is the distance from the nut to the bridge. The other is the total length of the string, since the tension in the entire string changes, so to figure out what are younge in tension you need, you have to consider the length of the string from the tuners to the tailpiece.

I am catching the last minutes of a weekend, I've got much work next week. Here is what I have so far. First picture shows ellipse, if you look closely. Upper spikes of teeth count. Second picture is 3d view. True temperament frets, elliptical fretboard, fanned scalloped neck.

I hope you have the proper tools for this. If you are using a scalloped neck, the reason for scalloping is that you do not push the string to fretboard. If you try to set it up by pushing the string all the way to the fretboard, you will never get it in tune across the fretboard. With scalloping, the idea is to just press lightly enough for the string to make contact with the fret. You have to have a very light touch to play that in tune.

As far a temperment goes, if you aren't using equal temperment, playing in any but the key it's tempered for is a problem. That's the issue Bach settled with equal tempered tuning. Every key is a few cents out of tune, but you can play in any key. The fanned frets just give you different scale lengths for different strings, presumably to make the tension across the strings uniform. (This problem and others were solved by Ned Steinberger with calibrated double ball strings. BTW, Ned's father was a Nobel Prize winner in physics for discovering the muon neutrino.)

 

[sophiecentaur]

The string elongates by only a tiny amount,

It's not the increase in length he's discussing; it the change in string modulus, which is affected by both sections of the string. Your finger pressure may have a small effect on that because a metal clamp would isolate the two sections but your finger is not a Capo.

 

[sophiecentaur]

The elliptical theory suggests that open string has no tension, which is not true

There a few basics that you need to sort out. The 'ellipse' theory only predicts the incremental increase in length (= extension) and the tension is incremental too.

Also, you really need to understand the way that equations can be re-arranged and variables eliminated. This is the point I was making at the top of the thread. The best way to approach a problem is to make the experimentation as simple as possible. Only when the simple approach becomes untenable do you go deeper. At the moment, everything you seem to want i.e. detecting vibrato, as a frequency variation, and applying a suitable model to what's happening. That can be done with string length measurements and frequency - if you can come to terms with what the Maths is telling you. What sort of balance do you have that will tell you anything about mass per unit length to an appropriate accuracy? Frequency and incremental length measurements are potentially much easier and much more precise. Otoh, tension and linear density are harder.

 

[JT Smith]

It's not the increase in length he's discussing; it the change in string modulus, which is affected by both sections of the string. Your finger pressure may have a small effect on that because a metal clamp would isolate the two sections but your finger is not a Capo.

When you fret a string it elongates slightly and this results in an increase in tension and hence pitch. Paul Colby was suggesting that the average string constant of the string decreases with use and this results in less of an increase in pitch when the string is fretted.

What I'm saying is that because the elongation is so small the resultant increase in pitch is too small for this to make sense. The string constant would have to change by far more than is likely.

 

[Paul Colby]

What I'm saying is that because the elongation is so small the resultant increase in pitch is too small for this to make sense. The string constant would have to change by far more than is likely.

It’s the tension change per unit length that is reduced. The length change due to fretting remains constant.

On my electric guitars the bridge position (string length adjustment) depends quite a bit on which string. I attribute this to the variation of the tension delta on fretting depending on wire gauge and string height. What’s your explanation?

 

[JT Smith]

It’s the tension change per unit length that is reduced. The length change due to fretting remains constant.

Yes, N/m. That's the string constant.

A concrete example: a 0.010 steel string tuned to E4 on a guitar with a 25.5" scale length. If the action is 2mm the increase in string length is about 12 microns. The spring constant of steel "10" on a guitar of that scale length is roughly 14kN/m. So the increase in tension due to fretting is about 0.2N. That corresponds to a pitch increase of around 0.7Hz, which is only about 2 cents. So it's basically impossible in this example for the tension to reduce by "a few cents". If you increase the action to 5mm then the pitch goes up about 11 cents meaning to lose a few cents the spring constant would have to diminish by more than 25%. Does that seem likely? Only if the action is way too high does it start to seem plausible.

On my electric guitars the bridge position (string length adjustment) depends quite a bit on which string. I attribute this to the variation of the tension delta on fretting depending on wire gauge and string height. What’s your explanation?

From what I have read the bridge adjustment is because of the finite diameter of the string. The larger the string the less free it is to move near the nut and bridge which makes the string effectively shorter. So larger strings need to be longer than an infinitely thin string would have to be. The wound strings start over because the core diameter is what matters.

Edit: You're right that fretting the strings and their gauge also has significant bearing on the bridge adjustments. I think this paper does a good job of describing the problem: https://arxiv.org/pdf/0906.0127.pdf

As for why strings go flat with age I don't have an explanation.

 

[sophiecentaur]

There's been confusion about the effect of the 'dead end' of the guitar string (the section between finger and nut). When you stop a guitar string with a finger, you aren't providing a 'hard' boundary. (It's not a node)_We all know about pressing particularly hard to get a better note when needed. This is because your finger allows the string to move underneath it. For a fundamental or low overtones, the string is only around a half (or very few) wavelength long so the termination is shared by the finger and also by the tension in the dead end. The modulus of the dead bit is actually relevant and an old string will have a lower modulus and it will lower the note. So I think it's not down to the change in length (which is tiny).
End effects are very relevant in musical instruments and are responsible for the interesting sounds they produce. Many of the 'imperfections' in a guitar will make it sound nice. Small frequency errors can be subconsciously corrected for by a good player. I imagine that it's easier to cope with nylon strings. Electric guitar techniques use the long sustain and feedback so that the available notes are on an almost continuous scale.

 

[Paul Colby]

You're right that fretting the strings and their gauge also has significant bearing on the bridge adjustments.

I assume I can take this as a partial vindication. Nice paper BTW.

 

[JT Smith]

There's been confusion about the effect of the 'dead end' of the guitar string (the section between finger and nut). When you stop a guitar string with a finger, you aren't providing a 'hard' boundary. (It's not a node)_We all know about pressing particularly hard to get a better note when needed. This is because your finger allows the string to move underneath it.

Yes. Even with a capo the string is not fixed. For that matter any string above a normal nut or below the bridge also has the effect of reducing the spring constant. A floating bridge does the same.

 

[JT Smith]

I assume I can take this as a partial vindication. Nice paper BTW.

Okay then, you're partially vindicated! :-)