# Mathematics of the Harp

After my second harp post, I thought I’d keep going with some mathematics. In particular I want to answer the question: why does a harp have that shape?

The physics of vibrating strings gives us Mersenne’s laws, which tell us that the frequency of a string of length L is (1 / 2L) √ T / μ , where T is the tension force on the string (in newtons), and μ is the density per unit length (in kg per metre).

For a string of diameter d and density ρ, we can calculate μ = A ρ, where A = π (d / 2)2 is the cross-sectional area. Nylon has a density ρ of 1150 kg/m3. For a nylon string of 1 mm diameter, we get μ = 0.0009 kg/m. Putting 0.448 m (17.6 inches) of that string under a tension of 140 newtons (31.5 pounds force), we get a frequency of (1 / 0.896) √ 140 / 0.0009  = 440 Hz. That is, the string plays the note A.

The important thing here is that the frequency is inversely proportional to the length. Over an octave the frequency doubles, which means that the string length halves. A 36-string harp covers 5 octaves, therefore if all the strings were made of the same material under the same tension, the longest string would be 32 times the length of the shortest. Doing some calculations in R for strings of diameter 0.8 mm under a tension of 140 newtons (31.5 pounds), we would get the following harp, which has strings ranging from 7 to 224 cm in length (note that the strings run from A to A, and the C strings are red):

You can see quite clearly that, starting at the treble end, there is an exponential growth in string length. That makes for a terribly unwieldy instrument, and creates all sorts of problems in playing. In practice, we make bass strings thicker (and usually of heavier material) and we vary the tension as well – although, to make life easier for the harpist, we want the properties of the strings to vary reasonably smoothly. If we re-do our calculations with string diameters varying linearly from 3 mm to 0.6 mm, and tension varying linearly from 210 newtons at the bass end to 60 newtons at the treble end, we get a much more realistic-looking harp:

We can flatten out the curve at the top a little by changing the way we vary the strings (after all, a guitar manages to span 2 octaves with all the strings being the same length). However, we cannot eliminate that curve completely – it is the inevitable result of spanning so many octaves, combined with the mathematics of exponential growth.  Left: an exponential curve (red) and a similar polynomial (dashed). Right: the quotient of the two functions (green) compared to a straight line (grey). Click images to zoom.

Mathematically speaking, the modifications to the strings have the effect of dividing an exponential function by some kind of polynomial (as shown above). Over a short range of x values, we can find a polynomial that fits the exponential well, and gives us strings of the same length. Over a wide range of x values, however, the exponential wins out. Furthermore, exponential growth is initially slow (sub-linear), so that (starting at the right of the harp), growth in string length is slower than the linear shift provided by the sloping base, which means that the top of the harp curves down. After a few octaves, growth in string length speeds up, and so the top of the harp curves up again.

A similar situation arises with the strings of a piano, although these are usually hidden from view:

And to finish, here is one of my favourite classical harpists in action:

## 9 thoughts on “Mathematics of the Harp”

1. Diana McCandless on said:

Thanks!

• Tony on said:

You’re welcome!

2. Rianne on said:

Thanks, I am using this information for my graduation project.(Of course with an acknowledgment and a link to this website) It is really helpful. I’m struggling a little bit with the explanation about the Mersennes Laws. But I’m trying to make a calculation myself but it’s really difficult.

• Tony on said:

Well, you kind of need a calculator.

The basic idea is frequency = (1 / 2L) × sqrt(T / μ) , where L is length (in metres); T is the tension force (in newtons), and μ is the density per unit length (in kg per metre).

So 2 × length ⇒ half the frequency,
4 × tension force ⇒ double the frequency,
4 × density per unit length ⇒ half the frequency.

Or feel free to ask me a question.

• James BURNS on said:

I see you have plotted this with R, Do you share this code anywhere like on github?

• Tony on said:

No, James, sorry.

There hasn’t been enough demand to justify the effort required to clean up my code and make it pretty enough to share.

• James BURNS on said:

no problem, I’m working on creating a lyre and am running some numbers, however i’m not an expert at maths. if you knew the frequency, tension, and thickness, how could you calculate the length that the string would need to be?

• James BURNS on said:

ah, got it. L = ((F/sqrt(T/μ))^-1)/2

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