String tension -- physics of length and afterlength - created 04-14-2011

Wybierala, Mark - 04/14/2011.09:29:40
Full Time Instrument Tech

Is string tension cumulative along the length of a string?

Model: A string is anchored at its ball end. From the anchor point, it travels to the right, 24 inches to a pully, over the pully, and downward toward the floor. A clip is attached to the free hanging end of the string and weight is connected to this clip until the pitch of the string equals A-440hz.

At this point, the clip is removed and the free length of the string past the pully is cut is half and the clip reattached. The distance from the anchor to the pully has not changed.

Will the same weight cause the string to reach A-440hz?

Is the same pull force at the anchor?

The answer seems to me to be yes to both questions but every once in a while I doubt my conclusion. The process relates to the distance of a guitar tailpiece to the bridge and the distance to from the nut to the tuner post.

One point of clarification: The vibrating length is between the ball end and the pulley, yes?

The answer is actually no, the pitch will have shifted higher, as the compliance of the string has been changed. The string is elastic, and you have changed the ratio of elasticity to tension across the ENTIRE string, not just the vibrating length. The tension across the vibrating length has therefore changed, and so it vibrates at a different pitch.

A better way to think about this is in regards to complaince as it relates to afterlength. If you have a two strings of the same gauge tuned to the same pitch, but with different afterlengths, the one with a longer afterlength will have greater compliance. It would be easier to "fret" this string. This string will also have required more tension to bring to the same pitch as the other.

However, examining the vibrating length in isolation of the afterlength, will show that the tension across the vibrating length itself, will be the same between the two examples. So, the tension that causes the pitch will be exactly the same. The tension on the entire string has changed, because we have changed the length of the entire string, but the tension and the pitch of the vibrating length has stayed the same.

Hope I haven't confused you more...

The length does not matter. Only the cross sectional area of the string determines the tension, and this can be verified by looking at the engineering formula to calculate tension. And only the vibrating length of the string determines the pitch. Any after length will make no difference on the pitch. The elasticity (what was referred to as compliance) will make string bending a little softer but that is a non-related issue.

Mark and I had a big discussion on this many moons ago. Should be in the library somewhere.

Sounds like someone should do this experiment instead of just talking about it.

Sort of like proving gravity?

If I understand what Greg is saying, I have to disagree. The longer afterlength does not require more tension. The tension on the entire length and the tension on the vibrating length are the same (assuming a frictionless pulley), so the last statement cannot be correct, unless we're considering the weight of the length of string removed, which would require the weight on the clip to be increased by that amount to get the same tension.

Anyway, the answer to both questions is YES.

I agree, answer to both is "yes".

The tension on the string is the force exerted by the weight (mass x acceleration(g))

The vibrating length does not change.

Therefore the pitch does not change.

If the after-length mattered, we would not be able to calculate fret position purely on vibrating length.

Thank you folks for the discussion. The model eliminates one of the two free lengths that normally exist on every guitar. I just want this point of real science not to bother me any more. Its funny how when you write something out like this you see the answer from another perspective. To put things a third way, regardless of the free length, the force on the ball-end is going to remain the same at the anchor point so the tension across to the pully will also remain the same because tension equals pitch.

Another thing that is true that Greg illudes to is that the longer free length will be easier to fret and feel easier to bend the string across the length of the fret (lateral movement). For example, If you took a Left-Handed Fender neck (six in-line tuners) and put it on a right handed guitar. You would find that the High E-String "feels" tighter and "feels" harder to bend. Actually, it is only the lateral movement that causes this change in "feel". Due to the shorter free length from nut to tuning post, the elevation in pitch is greater per measurement of lateral deviation -- a whole step bend would take less deviation. The feeling that bending notes is easier with a longer free length is actually an illusion because you need to bend further across the fretboard to achieve the intended changing in pitch.

Thank you everyone -- its this kinda stuff that keeps me awake at night. Hopefully I have the science put away securely this time.

Ok, this comes up all the time, and there's never a consensus. I don't have the materials to do a demonstrative experiment right now, but I think I might just have to do so.

For those who disagree with my conclusions, explain this to me: if you take two instruments of different scale lengths, string them both with the same gauge string, and tune those to the same pitch, why does the longer scale length require more tension?

The reason is that you have changed the tension per unit length of the string.

Let me try an illustrate it in a different way.

You have two strings of the same gauge but different lengths suspended from the ceiling. You attach the same weight to both strings. The strings now vibrate at different pitches because the "scale length" is different. Does anyone disagree with that point?

Now you could increase the weight on the longer string to bring it up to the same pitch as the shorter string. Anyone disagree with that?

Now if you were to "fret" the longer string at the same length as the short string, changing the vibrating length, but without changing the other variables, do you think the two strings would still be the same pitch?

Ok, stepping backwards now. Going back to the start of my example, we have the two strings of different lengths suspended from the ceiling, both with the same weight attached to them. They vibrate at different pitches. If we were to "fret" the longer string at the same length as the shorter, do you think they will both vibrate at the same frequency? The answer is no, because we have changed the ratio of tension (weight) to elasticity per unit length of the string.

do you think they will both vibrate at the same frequency?

Yes they will. Why do you think they would vibrate at different frequency?

In the long scale length vs short scale length question, it comes down to "restoring force" and mass. The longer string has more mass between the stop points, so more restoring force (tension) is needed to cause it to vibrate at the same frequency.

Greg, do you reject my simple proof that if after-length mattered we would not be able to use simple geometry to calculate fret locations?? There would have to be a "after-length" term in the equation

ut there is not.

The changes in distance on a FB correspond directly to the changes in frequency (wavelength)

ndependent

of after-length - which is behind your fretting finger.

Do the experiment of the hanging weights on wires - you will prove yourself wrong.

Ok Stephen, then why do we require adjustment (compensation) of the (theoretically correct and perfect) scale length?

You're failing to take string stiffness into account.

The error is small, yes, but we are talking string physics here.

I'm off to the hardware store to get some bits for the experiment.

I'll do the experiment -- exactly as I described in the model unless anyone has any other sugggestions. I have a decent place at my bench to set this up. It'll force me to clean my bench which is long over due.

Greg, the very same thoughts that are causing you to arrive to your conclusion are what were driving me nuts. I'll get to work in an couple of hours and recheck this post before I proceed.

I believe that string compensation has nothing to do with this. In the model, the stiffness of the string remains constant.

If a pretty girl with blonde hair who weighed 110lbs fell over a cliff near Big Sur in California and was being rescued by a local luthier using a rope and a quickly constructed pully system at the edge of the cliff -- lets say he is standing about 20ft from the edge of the cliff, it would not matter if she was either 10ft below the edge of the cliff or 100ft below. The weight that the luthier would bear would be the same (with the exception of the weight of the extra 90ft of rope). The weight that the luthier would bear and the weight on the quickly fabricated pulley at the edge of the cliff would be the same and thus the tension of the rope between the luthier and his quickly constructed pulley would remain the same too.

In the end, the luthier discovered that the pretty girl was actually a skinny surfer dude so they didn't hook up or anything.

The End.

Good luck with the experiment Mark. The hardware store didn't have the fish scale I was intending to use for my experiment, and neither did two sports goods stores.

If I could suggest trying to exaggerate the results as much as possible, by radically changing the amount of afterlength, even making the afterlength much longer than the vibrating length if possible, as I believe the error would only be slight.

The example I like to use it this:

Imagine having a long spool of rope and a 50 pound dead weight tied to the end. Lets say you initially have 100 feet of rope unwound from the spool. You pick up the spool and pull until you overcome the friction and drag the weight across the ground. If you unroll more rope would the force increase? If you think the answer is yes, then theoretically, you should have zero force if you wound the spool up so there was no free rope left. But this is just not the case. The force stays the same no matter how long the rope is.

I understand that this is not intuitive, but it is scientifically true.

Also, bringing in string compensation or string bending are completely different issues that will only serve to confuse.

Greg, there is no such thing as "tension per unit length of the string". As I said, it is totally dependent upon the cross section area of the material and this is backed up by basic physics and engineering.

Barry, Exactly. Its the concept of tension per unit of length that gets into my head and its something that doesn't exist in this particular universe.

I have to wonder if the hypothetical scenarios of pulleys and weights, though perhaps worthy of use, may add more confusion than it clears for some.

Let's look at the example on a guitar. You have the string connected to a tuner, going over the nut, to the bridge, and connected to a tailpiece 1" past the bridge. Simple enough, and clear that the speaking length we are plucking is between the nut and bridge, right?

Now lets say you got a really long string, set up your test bed in the salt flats of Utah, and moved the tailpiece from one inch to, let's say two miles behind the bridge. If you tuned it to the exact same pitch as before, the total tension on the string would be exactly the same. The main difference would be that you would end up with a whole lot of wraps on the tuner post before you got it up to pitch because there is a lot more net elasticity in the string, which would be cumulative - i.e., if a string will stretch .1% in length for each kg of tension applied, by the time you reach 10 kg of tension, 1% of two miles (plus the scale length and distance from nut to tuner) will take up a lot more room on a tuner post than 1% of the distance from tuner to tailpiece with a 1" back tailpiece. Once up to the desired pitch though, the tension on the string will be the same.

Now there seems to be two other points being discussed here, which are a somewhat separate from what I believe the original question was about. First is the amount of force it would take to deflect the string a given distance by pushing it down to the fret. This would indeed change if the string went from 1" past the bridge to the tailpiece up to two miles. As you increase the length of string beyond the speaking length, you increase the length of string over which the increased load is dispersed. In the 1" example, if we have a 25" scale and another 4" to the tuner, we have a total string length of 30". Push the string down to the 12th fret, deflecting it .1" in the center, and we will have increased the total length of the string by about 0.003%. If we deflected the string with the 2 mile tailpiece down by the same amount, then we only increased the total string length by 0.0000006%, therefore the tension will not have increased considerably.

Hopefully my numbers are right there - didn't double check my calculations, but either way, the point should be clear.

The other point that seems in question is how this affects intonation when a string is pressed down. Obviously from the previous example, it does effect how the string tension is changed when you press on a string. The question is, does this change have a significant impact on intonation? I would say a little, but not all that much. Strings need compensation for two reasons. The most often referenced is the idea that pushing a string down to the fret will increase its tension, therefore raising the pitch, and requiring an increase in speaking length to compensate for this.

Though true to some degree, I would have to argue that this is not the largest factor at play here. The other reason compensation is needed, and I believe the larger factor, is that the string has some degree of stiffness, and it's vibrating length does not fully begin until some distance beyond the nut (or fret) and saddle. Both the stiffness and the change in tension of the string undoubtedly play some role in the need for compensation, but I believe that the tension factor is much the minor compared to the stiffness factor.

There's my long-winded hypothetical scenario. I don't know if this helps to clarify or confuse, but I suppose it's my preferred example for explanation. Then again we all probably think best in different examples or analogies.

It is so nice to see a bit of reason entering into this discussion.

Takes me back to PHY 101.

An interesting factlet I gleaned from Howard Mitchell's book on Hammered Dulcimers - For a fixed string length, plain strings break at roughly the same note (frequency) regardless of gauge.

Good thing I read through the whole thread before responding - he saved me a bunch of typing. What Dave wrote is 100% correct.

I did it.

I then removed the can of lacquer thinner and reattched it 2 inches from the hook and the tuner indicated a perfect F.

So you reduced the after-length by (24/26) = 92% and it still hit the same note. Very nice - thanks for closing the door on that one Mark. Major kudos for doing the experiment - very efficiently and simply I might add.

I've actually wanted to do some testing related to this, not on tension (that's pretty well established by clear and basic physics), but rather on pitch changes resulting from increase in tension via string deflection when fretting, vs the shortened actual vibrating length resulting from string stiffness.

My plans were to set up a string on a lathe bed, tensioned between the spindle and tail stock, and make a clamping "fret" attachment for the carriage. I have some 26" digital calipers, and could easily check the string on a strobe tuner, then position the "fret" at the exact half way point (or any other arbitrary point), and clamp to restrict the string with no deflection at all. Then I could also modify the fret clamp to simulate real fretting, both by deflecting the string a given amount modeled on typical action, as well as modest deflection between two points to simulate finger pressure between frets.

My guess before starting would be that the increased sharpness with no deflection would be quite similar to sharpening with deflection when tested in the center range of the string's length. The concept that a large part of the sharpening comes from string stiffness (resulting in the true vibrating length being shorter than the total length between boundary points) is pretty clear and well established to anyone who's set up basses with taper wound strings. A low E on a bass will require much less compensation with a tapered end and the saddle, because even though the tension will be increased by deflection pretty much the same as a standard string, the difference lies in the flexibility of at least one of the end points.

The question for which I would like to get some better data is, by how much? Does 90% of the sharpening come from stiffness and 10% from tension, or is it closer to 60/40? How much greater a role does elongation and increased tension play when fretting toward the ends of the strings, both first frets and uppers, due to the increased angles at which they are bent when nearing an end point.

To some extent this can all be calculated on paper (or at least the amount of increased tension could be reasonably approximated). Still, I don't believe I've seen much data as to how much sharpening occurs within a given string length, diameter, and tension, simply from stiffness of the ends alone, even without any deflection.

One of these days. I've been wanting to do this for some time, and I have all the tools at my disposal for a clean experiment, except for of course time.

One of these days....

After reading & patialally understanding all this "String Theory"

Would it be safe to say that I could change the scale length on an archtop by simply swaping only the fretboard then adjusting the distance of the bridge from the 12th fret to same distance as nut to 12th fret?

OR, Would the tailpiece length Also, need to be adjusted to increase or decrease the legnth of strings behind bridge?

Yep, had a revelation last night. I was wrong. Sorry for the confusion I brought to the discussion. This and the idea that increased break angle increases tension comes up so often that I got myself confused. Oh well, at least I simulated some discussion!

Mark, while you're at it, could you change the angle of the vibrating length after the pulley to demonstrate that changing the break angle doesn't change the tension? It changes the direction of force obviously, but does not change the tension.

Oh, and sorry Chuck, just realised I called you Stephen in a previous post!

Howard, Stephen, it's all the same. Chuck's being called a lot of things these days.

Mostly by you Klay.

Jeff, What the experiment shows is that yes, you could change the fretboard, move the bridge and all would be good. The length of the strings from the bridge to the tailpiece is irrelevant to tuning.

Of course, moving the bridge drastically could bring up other issues like string action and structural considerations.

I'm late to the party, but reading through I don't think I see anyone mentioning the actual elephant in the room: the formula.

Being personal friends with google, did I get this right? - Frequency = sqrt(t/m)/2L

t is tension, m is mass, L is vibrating length.

So doubling the length lowers the pitch by an octave, but one must quadruple the mass to get an octave lower, or quadruple the tension to get an octave higher.

Which supposedely Pythagorus proved back in the day in the apocryphal story about him listening to the sound of the blacksmith's hammer on different anvils.

That's the formula for the vibrating length of the string. The focus of the question here however was related to the length of string beyond the boundary points, outside of the speaking length, to which this does not apply.

Hi,

String length beyond the span 'L' is not a factor.

I don't get why you say it doesn't apply, I thought that was the point of the conversation - that nothing in the formula has to do with anything beyond the vibrating portion of the string, therefore the length beyond the nut or saddle are not variables.

If I wanted to do a real world example of how the formula works on a real string, for example for the A string to vibrate at 110 Herz would my units be newtons for tension, meters for length, and kilograms per meter for mass?

Doesn't matter Bill, as long as you use internally consistent units.

Frequency is [1/s] (one over seconds)

e.g. in SI

F[1/s] = sqrt(t[N] / d[kg/m]) / (2L[m])

A Newton is really [N] = [kg][m] / [s^2] (kilogram*meter/second_squared)

So the original equation is:

F[1/s] = sqrt(t[kg*m/s^2] / d[kg/m]) / (2L[m])

All the [kg] and [m] cancel and you are left with a sqrt(1/[s^2]) ~ which is [1/s] or Frequency.

And, yes cycles/second is the same is [1/s]

That makes sense, thanks.

Hi again,

I think the 11 inches was the vibrating length. All else remains equal, if you double that you lower the pitch one octave. I think he was trying to prove that the length of the string outside of the vibrating length was irrelevent to the pitch.

My thinking is that, while there is one string, there are in effect two 'L's.

No, we just disproved that thinking. There is only

one

L with regard to the pitch a string will sound at.

Sorry that I didn't explain very clearly.

In Mark's experiment, he used two hangers 11" apart. I'll call that 11" span L1. He fixed one end of a string to one hanger, and ran the string over the second hanger. He then hung a weight on the free end of the string and adjusted the weight so that the string at L1 when plucked would vibrate at an F, I don't know what octave... let's just say F1. Whether the length of string between the second hanger and the weight was 2" or about 15" that string would sound an F1.

Suppose that the length of string between the second hanger and the weight is set to exactly half of L1. You'll know when you're at exactly half because, if you pluck this section of string, the pitch will be F2. Let's call length of string between the second hook and the weight L2.

Now you have one section, L1, that will sound at F1, its fundamental, along with its harmonics: F2, F3, F4...

You also have one section, L2, that will sound at F2 and it's harmonics.

L1 and L2 are strongly coupled (same string), and both will sustain an F2 or harmonics. What happens now when you pluck L1?

I don't think they are coupled, since the saddle or in this case the hook would be a node. L2 would probably vibrate sympathetically when it hears L1, but so would the B string of the guitar propped up in the corner.

Sorry, I'm not getting it.

You have reduced the distance between the hooks to L2 = L1 / 2

Then when you say:

"Now you have one section, L1... You also have one section, L2"

I say, no, you only have L2 now. You shortened the distance between the hooks.

Picture??

Let's call length of string between the second hook and the weight L2.

He said L2 is the free part that the weight is on and that gets set to half of L1 (the span between the hooks).

L1 is still the distance between the hooks and doesn't change.

Sympathetic vibration strays beyond the original topic.

Hi Chuck,

sympathetic vibration implies coupling.

I agree.

What was the topic?

Asymmetric aura observed around Punxsutawney Phil during cold winters.

Oh.....

groundhogs. got it

They are using Fishman pickups to monitor hibernating groundhogs? What will they think of next!