July 4, 2014 at 6:36 am #142349
I have recently been curious about what makes a harp string feel like it does when played in terms of its properties and being me, have applied some simple physics to the problem. I have come up with some interesting explanations and results that I’d like to share with everyone because I think they might be useful to people (as well as interesting)!
[All derivations are in square brackets and italics so you can skip them if you’re not interested.]
So, the first thing is: What governs how much work you have to do to pull a string? We often refer to this as the “tension” of the string, but actually, it isn’t quite that. Imagine you have a harp strung up all in the same string gauge (so, you buy several meters of concert gut 4th octave C or whatever and string your entire harp up in that), and you wind up all the strings to the same tension (so you put three loops on each tuning peg). Your longer strings are still going to be much easier to pluck than the shortest ones. This is because what it is that governs how hard you have to pull is not the tension on the string, it is the tension per unit length (i.e. the tension/length). So you have to wind the longer strings tighter than the shorter ones to get the same “feel”.
In terms of treating the string like a spring (remember Hooke’s law anyone? F=kx) this makes sense. You have to apply a force F to achieve a string displacement x and, your spring constant is k=T/l (T is tension, l is length), and a higher spring constant means it’s tougher to pull. So for an even feel across the whole harp, you want your k=T/l to be roughly equal for all strings. The gauge doesn’t come into it, only the tension and the length.
[why? I worked this out using energy considerations. Energy in ≈ energy out. Your energy in is Fx. Energy of a string is proportional to density per unit length “u”, frequency “f”, amplitude of string A and length in the following way:
E α ul(fA)^2
Assume amplitude of string is the same as (or at least proportional to) displacement of string: A α x (this is valid as far as I can tell but will get less good with shorter strings, as when you pluck you are making more of a a trapezium shape with the string rather than a half-sine curve but whatever). Also remember that for a perfect string fl α (T/u)^1/2 (speed of fundamental wave v = 2fl = (T/u)^1/2). Substituting out u gives you:
E α T/l x^2
Equating energy in with energy out gives F = T/l x which is the result quoted above.]
So far so good. Continuing with our imaginary harp strung up in the same gauge, imagine now that we adjust the tension on it so that the longer strings have higher tension and shorter strings have lower tension such that they all feel the same when you pluck them. So far so good. However, all our strings are sure as hell not going to be the right notes (i.e at the right frequency). How do we change the frequency to be what we want while keeping the tension per unit length (i.e. the “feel”) the same?
That’s where the gauge of the string comes in. The frequency of the string is controlled by the spring constant k (the “feel” -which we’ve fixed), the length “l” (which we’ve fixed), the density of the material “D” (which is fixed by what kind of string you choose, nylon or gut or wire or whatever) and the gauge “g” which we can change to get the required frequency:
f α (k/lD)^1/2 1/g
What does that mean for an actual harp? It means that for the same frequency of note and same “feel”, the ratio between two string gauges is inversely proportional to the square root of their density ratio times the length ratio:
g1/g2 = (D2l2/D1l1)^1/2
and for two string of the same material (i.e. D1=D2)
g1/g2 = (l2/l1)^1/2
Now I can use this formula to calculate what string gauges I need to put on my smaller practice harp so that it “feels” the same to play as my floor harp.
[why? Using the wave equation for a perfect string again fl α (T/u)^1/2, substitute in u α Dg^2 (density per unit length = density x pi x (g/2)^2) and T = kl giving f α (k/lD)^1/2 1/g as before]
One last thing: Why do thick strings sound thuddier and more dull than thinner ones? According to our formula, having a small harp with thick strings gives the same feel and should be equivalent to a longer one with thinner strings, so why do we prefer larger harps and why do they sound better?
Well, all this analysis is assuming a perfect string (infinitely long and perfectly clamped at both ends), which gives us a really good approximation but it becomes less valid as our string gets shorter relative to its thickness. Effects called “end effects” (effects from the ends of the string which are the “not perfect” bits of a string) become more and more significant as the length gets smaller. These effects cause the string to sound dull and un-perfect. So the thinner your gauge for a given note, the longer your strings are relative to the gauge and the better they sound.
Hope that was useful and/or interesting, if not at least palatable. Any questions/comments/corrections, feel free. And incidentally, has any one tried measuring or knows the density of gut, nylon or KF composite strings? Should be straightforward to find out, weigh a string and multiply mass by length by pi by (diameter/2) squared. If any one can do this, I’d be very interested!July 4, 2014 at 4:13 pm #142358AlisonParticipant
hmmm Interesting & I am wondering why it taken so long for you to post this !!!
I will print this and study it closely I would get out my maths and music book (Dermot Roaf and Robin Wilson (so of Prime Minister Wilson) and there is a table on the web somewhere with the tensions and gauges of celtic harp strings etc – all this is very important in the harp design – I wish I could read it easily – my familiarity with degree maths and physics is fading, Shall I pass this on to my maths/engineering tutor ?July 7, 2014 at 3:00 pm #142387
I am reminded I have a whole heap of notes that I started intending to be a book on the harp from a physicist’s point of view – maybe I will be inspired to finish it!
The ‘feel’ of the string is an established criteria, but (I am sorry, Mae) I do not agree with your derivation in italics (starting with energy = fx does not hold when f is variable with x).
For the problems of thicker strings the usual effect to invoke is the lateral stiffness of the string, its resistance to bending throughout its length, which affects higher harmonics more than lower ones.
For measuring the denisty of strings precisely you may want to take into account the Young’s modulus as you would be measuring them unstretched and measurements for different gut strings would be useful if the varnish thickness is significant to you.
If you want references starting off with Chris Waltham in Rossing’s book the Physics of Stringed Instruments and the other material his university has on the web is one source, Ian Firth and Alex Bell have also published. Nicolas Lynch Aird gave a talk at Edinburgh this year with some interesting data.
Hope this is useful.July 7, 2014 at 3:38 pm #142389Gretchen CoverParticipant
Please send all this information to Bow Brand International Ltd. I just restrung my harp yesterday and one of the lower octave gut strings broke already.July 8, 2014 at 7:23 am #142394
Tacye, that would be awesome, please finish it!
Ok, so maybe the explanation was too back-of-the-envelope…you’re right that E=/fx but assuming F is proportional to x (i.e. that we can treat the string as an elastic spring) integrating Fdx only adds a factor of 1/2, which is irrelevant if you’re assuming (as I did) that everything is approximately proportional:) I guess I only wanted to satisfy myself that this spring constant/feel factor makes some kind of sense, which I guess it does! Now that I’ve done some scribbling, I will actually go and do some proper research:)
Have met that book before, pre-harp, but I was more interested in harpsichords then:) Shall go back and have a re-read.
Gretchen – I feel your pain, I’ve worked my way through an entire duff string in the space of a week, highly frustrating! I think that has more to do with the manufacturing process though, and I’d be very surprised if Bow Brand didn’t already know this stuff…!July 8, 2014 at 1:50 pm #142398
Umm, Mae, Hookes law spring E = 1/2 k x^2July 9, 2014 at 12:37 pm #142406
…yes…and if F=kx then E=1/2 F x…..?
So confused. What have I missed?July 9, 2014 at 3:46 pm #142416
Right, I’ll just retire muttering about the inelegance of hiding a power I expected in a function of a function…July 10, 2014 at 5:01 am #142421
Yes, yes yes yes yes, I was just making the point that it only added a factor of 1/2 and so was largely irrelevant:)
Gosh, we are a pair, aren’t we…
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