# Piano string tuning dilemma

1. Dec 21, 2004

### physicsisphirst

ok certain notes have 3 strings.
the rightmost string of a note loops around a peg and becomes the leftmost string of the next higher note.
so if you increase the tension on the first, you have to increase the tension on the adjacent. right?

well i thought so too, yet i've had 2 piano tuners tell me that this isn't so - that the tightening the rightmost string on a B does not affect the tension on the adjacent C's leftmost string.

they laugh at me saying that if it were so then it would be impossible to tune the piano.

i tell them that the tuning can come from the length of the delimiters on the string and that may be the 'shared' string idea was done to actually speed up tuning - ie by tuning the rightmost string on the B, you automatically end up with the correct tension for the adjacent C's leftmost string (the delimiters end up giving you the correct pitch when the tension is correct).

i figured my understanding needs to be ratified or corrected so i brought my question here. besides, i don't want to be laughed at anymore.

in friendship,

2. Dec 21, 2004

### rcgldr

I'm guessing that the thickness and tension properties of the wire isn't consistent enough to accurately tune a piano. If it is just one continuous wire, somehow there's something to take up the slack between adjacent runs of the wire.

3. Dec 21, 2004

### NateTG

Depending on the way that the string is looped, you might find that the amount of friction is so large that you could consider it to be attached in a fixed fashion instead.

It may also be that the two strings should be at the same note since there is some black magic with blending to make the well-tempered klavier sound right.

4. Dec 21, 2004

### physicsisphirst

the string simply goes around a smooth metal peg and we are looking at tensions of around 150 lbs and over from what i read on the net.

even if the string were looped around the peg and even if the peg had rough contact points, i don't see how the friction could be great enough to not cause even distribution of the tension between the 2 segments of string.

in friendship,

5. Dec 21, 2004

### rcgldr

Have you looked at the other (non-adjustable) end of the frame holding the strings? Maybe the setup there prevents the strings from sliding.

6. Dec 21, 2004

### physicsisphirst

well the string goes from the tuning key around a peg and then to another tuning key.

here is the order for a sequence:

leftmost string for second highest G wraps around the peg and becomes middle string for the same note. then the rightmost string for that G wraps around the next peg and becomes the leftmost string for G#.

the middle string for G# wraps around a peg and becomes the rightmost string for the same G#.

then we start the sequence again with A's leftmost string becoming the same as its middle string. but A's rightmost string goes around the peg and becomes A#'s leftmost string.

the pegs are simply smooth pieces of metal.

in friendship,

7. Dec 21, 2004

### rcgldr

I did a web search on this, and some strings do affect two adjacent notes. I'm not sure how this is handled. Note that the center strings don't have this issue, and these are tuned first.

8. Dec 21, 2004

### physicsisphirst

thanks for looking, jeff.
i think the delimiters that end up determining the actual length of the struck string decide the frequency it will vibrate at. so you can have 2 strings at the same tension, but because the delimiters create different lengths, you can get different notes.

in friendship,

9. Dec 22, 2004

### T.Roc

the "delimiter" that you mentioned creates a "center node", that in effect, creates 2 separate strings (standing wave). each one is a different length, mass, and tension, which are the 3 inputs you need to determine frequency. f=1/L x sqrt T/(m/L). as long as the center node creates the right ratio of lengths, the right ratio of frequencies will be produced. to go down an octave and a step (G' to G#), the length of the shorter run would be 1/2 the longer run /1.05946 (the 12the root of 2). the "slack" in the longer can be played with (tuned) to a great degree (more than nec. for tuning) before the shorter will slip off the node. now, i've never seen the inside of a piano, but i doubt you'll get a better answer from the piano tuners.

TRoc

10. Dec 22, 2004

### physicsisphirst

thank you very much troc!
i am confused about one thing though - by delimiter i mean the 2 points that establish the actual length of the struck string - i don't mean the peg that the string goes around.

i can see how the peg would create 2 strings being a node, but i don't see how the delimiters could do this:

-------------^-------------------------------^------------------------- string

the 'carats' are the delimiters and appear under the string.

here is the top view:

/ -----------^-----------------------^---------- to tuning key
|O
\ ----------^-------------------------^-------- to tuning key

the peg is the O and the string wraps around that.

my contention is that given that the tension is around 150 lbs or more and that the peg is friction-free, there is no way you can avoid that tension being equally spread between the 'two' strings (because they are in reality one string).

in friendship,

11. Dec 23, 2004

### T.Roc

i believe, in the world of physics, "friction free" may not be accurate (especially under 150 lbs tension). again, not having seen in a piano, i'm not sure, but are the strings "wound" (like a bass string)? if so, then it would be like a metal file against the peg. that failing, then consider this: the co-efficiency of friction on a polished DRY marble floor is higher than that of a textured surface because, at the microscopic level, there is more surface area to act upon. the tension on the longer string is less, and has slack. the act of finding the correct frequency could be described as logarithmic. the distance of the string is changed by the gear in the tuning peg, which changes in steps defined by the gear size. often this is too much, and you have to back off and come back to "in tune". if you marked control lines on the tuner peg and instrument, they would not be aligned from one days' tuning and the next. temperature, and age of the strings also play a role in this "floating" tuning point.

best regards,
TRoc

12. Dec 23, 2004

### physicsisphirst

i agree that it wouldn't be friction-free by any means. yet it seems to me that even if the string was wound like a bass string (and the higher ones aren't), that the increase in the coefficient of friction wouldn't be sufficiently great to offset an even distribution of the tension especically when you consider the small surface area of contact between the peg and the string. overtime (if not immediately), i would think there would have to be slipage at the peg evening out any tension discrepancies.

i suppose one way to find out is to put my own delimiters fairly near to the peg at an equal distance and see if the pitch is the same on each adjacent string.

in friendship,

13. Dec 23, 2004

### T.Roc

i think the bottom line is that, if the long side is in tune, the short side must be in tune too. the slack in the long side presents the easiest way for the tension to change. you would have to take the long side up to a full octave (or more) above the short side before the tensions on both sides were = enough to slip on the peg. if you want to experiment, use a rubber band set up in the same fashion.

TRoc

14. Dec 23, 2004

### Grogs

Is the distance from one tuner to the peg different than the distance from the other tuner to the peg? If so, the system may in fact be designed to slip.

My practical experience with strings is limited to the guitar, but I have learned a good deal of the theory along the way. Given a certain set of environmental conditions, the pitch produced by a string is a function of it's composition, thickness, tension, and length. On a standard guitar, the overall length of the strings is ~25 inches. If you shorten the string by a certain distance (by pressing down on on a fret) the pitch rises by a certain number of steps. For example, if a string is tuned to 'C' and I press down on the first fret (shortening the string by x inches), I get a C#. If I retune the string to D (increasing the tension) and press down on the first fret, I get a D#. The difference between the 2 notes remains the same.

The point I'm trying to make is that if the 2 sides have the approprate ratio of lengths and slip freely around the peg, then having one side in tune means that the other *should* be in tune. The only way I could see them not being in tune in this set up would be something physically wrong (bent tuner, damaged string, etc.)

In any case, it should be relatively easy to check whether the string slips or not. Tune 1 side to the correct pitch and check the pitch on the other side. Note the results and then tune the same side down 1 step (from D to C for example) and check both sides with the tuner. If both sides drop equally (D & D# drop to C and C#), then it's safe to say the strings slip freely over the peg. If they *don't* change relatively, then you'll obviously have to tune each side separately.

15. Dec 23, 2004

### physicsisphirst

exactly!! that's how i see it too and what makes both sides in tune is the separation of the delimiters and the fact that the tension is the same on 'both' strings.

anyway, i greatly appreciate your inputs Grogs and TRoc (as well as the others).
thank you all!

in friendship,

16. Mar 26, 2007

### graeme f

Stumbled across this (old) thread and thought I'd comment based on both my physics and piano restoration background:

There is sufficient friction on the hitch pin ("peg") to prevent slippage under normal circumstances- therefore the two halves of the string can be under different tensions. String tensions within a note are designed to be the same hence there is very little differential tension either side of the hitch pin. Tensions between adjacent notes should also be similar, although there is a gradual grading of tension across the scale with the bass section being around 20% higher than the treble. Some piano scales are more even than others. So there will typically be some variation in tension note to note although the difference is small enough not to cause slippage.
If there were to be slippage at the hitch pin then the piano would indeed be untunable - particularly with the strings that are shared between notes.

Wrt the discussion on "delimiters" - the critical length is the so-called speaking length which is defined by the "v-bar" up near the tuning pins and the bridge pins - ie the pins on the wooden strip that connects to the soundboard. This is the only length that is intended to vibrate. The frequency of vibration is defined by this length, as well of course by tension and the wire mass. The short length between the bridge and the hitch pin would tend to vibrate at a higher frequency than the speaking length. This is an unwanted vibration that is suppressed by a felt strip or similar means ( although in some grand pianos (such as Steinway) this length is tuned by moveable terminations and allowed to vibrate symathetically - the so-called Duplex scale)

Note that each (half) string is under uniform tension over its entire length from the tuning pin to the hitch pin.

Looped stringing is common to most pianos. It simplifies the stringing process - avoiding the need for a lot of hitch pins and the need to terminate each string in an eye.

A few top brands use non-looped stringing but it is the exception rather than the rule. The main advantage is that if a string breaks then you only lose one, rather than two (complete loss of tension on one half will indeed cause slippage!! )

Hope this gives some insight!

Cheers,

Graeme

17. Mar 26, 2007

### AlephZero

I suggest you actually look at a piano carefully, rather than reading about them on the web.

You will see that the way the string is held is DESIGNED to create a lot of friction! It doesn't just go round the tuning pins and the hitch pin at the other end. There are also pins (more than one at each end) where it goes over the bridges, and there are metal plates (called "agraffes") that press down on the non-playing parts of the strings to create more friction.

Your piano tuners are quite right, in practice the two parts of the string can be tuned completely independently. However there is some skill involved in getting all the parts of ONE half of the string to be at the same tension, because of the amount of friction. It ain't as simple as just turning the tuning pin. That's the usual reason for a piano to go out of tune quickly - i.e. it wasn't tuned properly in the first place.

Piano wires are usually stressed right up to the limit where they begin to go plastic. The average tension in each "string" of a grand piano is of the order of 150 to 200 pounds force. The forces in guitar strings are way lower than that, so any analogies about tuning, or friction, are most likely to be misleading.

Last edited: Mar 26, 2007
18. Mar 27, 2007

### graeme f

In truth the only parts designed to provide friction on a piano string are the hitch pin and the tuning pin. Any other points of friction are unintentional. The agraffes or v-bar and the bridge pins are most definitely not intentionally designed to provide friction - rather they purely provide the termination points that define the speaking length of the string. In fact many bridges are coated with graphite to improve the slippage of the string over them.
The string tension equalizes over its length. During tuning, when the tension of the string is changed, there is a transient differential in tension - the tension between the tuning pin and the v-bar (or agraffe) is slightly higher than the remainder of the string, if the string is being pulled up to pitch, due to friction at the v-bar . This is the reason that the tuner strikes the note loudly - the mechanical shock helps to release the binding at the v-bar and hence equalize the tension across the string.
If this is not done then the string can subsequently go out of tune. The other main cause of the piano going out of tune shortly after tuning is if the torsion on the tuning pin introduced during tuning isn't neutralized by applying a slight reverse torque on the pin as the final stage of tuning.

19. Mar 27, 2007

### AlephZero

The maximum difference in tension on the two sides of the hitch pin can be estimated by the standard "capstan equation" which is derived from Coulomb friction for a rope wrapped round a circular object. The ratio of tensions between the two ends of the rope when slipping occurs is
$$T_2 = T_1 e^{\mu \theta}$$
where mu is the friction coefficient and theta is the wrap angle (in radians).

For a hitchpin the angle is pi radians. For a typical metal-on-metal friction coefficient of 0.1 that means the possible difference in tensions on each side is 37%. A friction coefficient of about 0.036 would hold the 12% difference in tension needed to tune two equal length strings a semitone apart (6% difference in pitch). In practice the strings are not equal length, so an even lower friction coefficent would be enough to do the job.

Last edited: Mar 27, 2007
20. Mar 27, 2007

### Andrew Mason

I agree with the last two posts. Having tuned many pianos, it is certainly possible to tune two halves of the same string at significantly different pitches. The only time you have a problem is when there is you let the tension in one string go too low, as this will lower the tension of the adjacent half that you may have just tuned. If you are raising the pitch of an old piano, you have to do a rough tuneup to raise the pitch and then retune. It can take several tunings to become stable.

The best grand pianos have a single strings and avoid this problem entirely.

AM