Are you a nerd? I am. But don’t worry, it’s alright. Hey, if you were normal you wouldn’t be either playing the dulcimer, or reading this web page in the first place, would you? You’d be listening to a top 40’s station on commercial radio while drinking a cold frosty Pabst on the chaise lounge in the back yard while your 2.3 children play basketball.
But no, you’re different. I can tell these things.
Admit it, you own a pocket protector, don’t you? Uh-huh, I thought so. You have the presets in your car radio set for every NPR radio station within a hundred mile radius of your house. A thrill runs down your spine when the announcer says, “Coming up next, The Thistle and Shamrock with Fiona Richie…” If you’re over 40 you have a slide rule and still know how to use it. Your kids take dancing and karate lessons, and play in the school orchestra. And you only have 1.8 of them in the first place.
Well, this is the web page you’ve been looking for. I’ll bet you’re tired of choosing your strings by trial and error, aren’t you? You wanted to try to make your McSpadden into a Baritone dulcimer, but didn’t know what weight strings to use, and ended up collapsing the sound box, didn’t you? Tsk, tsk, such a pity.
But never again!
Use the equations below, and with a little physics, you can quickly calculate the proper string diameter to use for a given pitch, string length and tension. So, get your slide rule and reading glasses (the ones held together at the bridge with tape), pull your favorite pen out of your plastic pocket protector, get a fresh pad of quad-ruled paper, and let’s begin.
Your freshman college physics text will tell you that the basic equation for the frequency of a vibrating string is
where f is the frequency of vibration,
F is the tension on the string (think “Force”)
L is the length of the string,
and the greek “mu” (looks like a “u” with a descending tail in front) is the mass of the string per unit length. “Mu” is a somewhat awkward unit, since it changes depending on the string diameter, which is how we buy strings.
We can replace mu with its expression in terms of the string diameter, d, and the density of the string material, rho (Another greek letter, us techno-nerds love them. This one looks alot like a lower case p.)
“g” here is the acceleration of gravity (386.4 in/sec/sec). We need this to make the units come out right because F is in pounds-force, while m is in pounds-mass, and no, they’re NOT the same. The greek pi is the familiar ratio of a circle’s circumference to its diameter, 3.14159265358979…. (we techno-nerds like lots of digits of precision.)
Inserting the expression for mu in the first equation gives the result: which gives us the frequency of vibration for a string of a given material, length, diameter, and tension. Normally, however, we know the frequency we want, it’s the diameter we’re interested in, and, as we will see, the tension. Solving this equation for d gives the following:
and solving for F gives:
So, how to apply the above?
To begin with, you need to know the density of the string material. For dulcimers this will usually be steel. It has a typical density of 0.283 lbm/in^3 (pounds-mass per cubic inch). We also have to know the frequency of the pitch of interest in units of “Hertz” (vibrations per second.) Most western music is based on “concert pitch” of an A above middle C being 440 Hz. The frequency of a note a half step up or down from that differs from the A by a ratio of the 12th root of 2, which is equal to 1.05946… This is only for equal temperament. Treatment of mean temperament or just intonation is “beyond the scope of this article.” (Code words that mean the author doesn’t know how to do it.)
So for a B-flat the frequency is 440 * 1.059… = 466.16 Hz, and for a B-natural the frequency is 440 * (1.059)^2 = 493.88 Hz. Going lower is the same, except you divide by the factor instead of multiplying. So for an A-flat the frequency is 440 / 1.059 = 415.30 Hz, and a D-natural (seven half-steps below the A) the frequency is 440 /(1.059)^7 = 293.66 Hz. Also, notes that differ by a whole octave in pitch differ in frequency by a factor of 2. (Count’em, 12 half steps in an octave) Therefore an A an octave below the “standard” A has a frequency of 440/2=220 Hz.
To make an intelligent choice of the tension, you need to know some starting combination of string weight, length and pitch that you are comfortable with. For example, say you customarily use a 12 gauge string (d=0.012 in) on your Rockwell Kentucky hourglass dulcimer, which has a 28 inch string length, when tuned to D. Plugging all this into the equation for F above gives:
F=(293.66)^2*28^2*0.012^2*0.283*3.1416 / 386.4 = 22.4 lbf (pounds-force)
If this is comfortable for you, but you want to restring your dulcimer for different pitches, it probably makes sense to try to keep the tension of the string (F) constant by changing the gauge of the string. Let’s say you want to retune your dulcimer lower, to a B-natural (246.94 Hz). If you leave the same strings on your instrument they will probably be a little loose and floppy resulting in a “wimpy” sound (to use the technical term). The tension on a 12 gauge string of this length at B-natural is 15.8 lbf. But if you want to keep the same tension on the string that you’re used to in the key of D, 22.4 lbf, you need to increase the string diameter. Substitute the values into the equation above for d,
with F = 22.4 lbf, and the result is about 0.0143 inches. Try a 14 gauge string instead. To get the other strings of the dulcimer when tuned to B F# B, you just plug in the appropriate values in for the frequency (f) and flail away with your slide rule. For the F#, (f=185.00 Hz) the appropriate diameter for the string is 0.0192 (might be hard to find, you may have to use a 20 gauge string, or maybe an 18). For the Bass string (f=123.47) the diameter would be 0.0287.
Hmm… I’d definitely use a wound string on the bass, and maybe on the middle string as well. This, however, presents a teensy problem. The physics and math above are only strictly valid for plain gauge strings.
Still, no problem is so large that it can’t be ignored, er, dealt with.
Wound strings have a relatively thin flexible core of steel wire to bear the tension, and are wrapped with (usually) either a nickel wire or phosphor bronze wire to add mass to the string without sacrificing flexibility (much). Both nickel and bronze are heavier per unit volume (denser) than steel, but the precise amount more is “beyond the scope of this article.” (Remember what those code words mean?) Also, since the cover wire is wrapped around the core, the exterior of the wire is not smooth, which changes the mass per unit length calculation in some ill-defined manner.
This situation cries out for what an engineer would call a “simplifying assumption.” That’s what an engineer does when he doesn’t know how to solve the problem he wants to solve. He solves some other problem instead, that by some stretch of the imagination he thinks is similar to the “real” problem.
In this case, we will assume that the material missing between the wraps of the cover wire is exactly balanced by the amount that the cover wire is denser than the steel. Also, we assume that the combination of a steel core wrapped with a cover wire is sufficiently similar to the case of the plain wire that the frequency equation at the top of this section still applies.
There, wasn’t that easy? Now we can use the same equations above for either plain gauge or wound strings. Engineers go to college for YEARS so they can do this.
An interesting additional result of this analysis is that the stress in a string of a given length (stress is the tension divided by the area) is a function only of the frequency that it vibrates. The string diameter divides out of stress calculation. This means that the pitch at which a string will break as you keep tightening it is not related to the diameter of the wire. That is, you can’t tune a 10 gauge string any higher than a 14 gauge string before it breaks. (Plain strings only, now. The simplifying assumption doesn’t work here.)
The fact that this does not seem to match my experience may mean either that thinner strings are more “work hardened” by having been drawn through smaller dies, and are therefore stonger than the steel in larger diameter strings, or that this analysis is all wrong.
(P.S., I must admit, with some embarrassment, that I used an electronic calculator on the numerical examples above, instead of using my slide rule. I’d left my good slide rule at work (a Pickett model 110 ES circular), when I sat down at home to write this. Honestly, and in all seriousness, I COULD have done the calculations with it without much sacrifice in time or accuracy.)
I welcome questions. I’ll try to answer them on a time permitting basis,
so bear with me if it takes a bit to get back to you.