Last modified: 2025-08-17 17:30:05
< 2025-07-30 2025-08-19 >We came across a fly-tipped piano at the side of the road, and stopped to rescue the keys:
There are 85 keys, I was expecting 88. Helpfully most of them have a number stamped into them so it is easy to put them in the right order.
I'm thinking it might be fun to make some sort of clavichord or keyed monochord with these keys. Obviously not all 85.
One of the keys is stamped "SHENSTONE" and "BRITISH MANUFACTURE", and the two on the left-hand edge are marked 69331, although these two do not have stamps for "1" and "2", so potentially are replacements.
You can see the keys mostly have a dog-leg in them, if repurposing for a keyed monochord you could pick a subset of keys that have appropriate dog-legs to position the strikers at the right place on the string.
The white keys are 22.5mm wide, black keys are 12mm wide, and the "tails" are 13mm wide.
Dog-legs available are, from left to right:
You can easily get several mm (maybe +/- 6 mm) of adjustment at the end of the tail based on where you position the tangent, so a close solution would be good enough.
But how many keys do you need? A single octave? (which is 12, including black keys). You could make the other end of the string easily adjustable so that you could change the "key". Maybe. Maybe you'd actually want a different string for different keys.
ChatGPT tells me that the formula for the position of the tangent from the "nut" for the "nth semitone" is L0/2^(n/12).
So I need to find a string length that is going to allow me to position the keys in workable positions. Annoyingly it is looking like it might need to be quite a short string, which means very high pitch.
I think we can imagine an extremely long string with hundreds of keys on it, work out where they would all be, and then just take the 12 or 24 keys from somewhere in the middle that seem most appropriate.
Actually I think we know that in the course of 12 notes you halve/double the string length (depending on which way you're counting).
If I put 12 keys together then they would span let's say 12.5 * 12 = 150 mm on the keyboard. And the tangents (if we use the extremes
of the +47 to +28 mm group) would span a range of 150 + 19 = 169mm. So the string would be twice as long as that, 338 mm. Is that
too short? We can maybe add 12 mm to the span of the tangents, if the left one is bent all the way left and the right one all the
way right, we get an extra 24 mm on the string, 362 mm.
I have rigged up a "string" out of a piece of the spring steel wire I bought for the watch mainspring. I can put a pair of "bridges" about 362 mm apart, and half that, and see if it sounds like a reasonable range.
A couple of observations:
Here is a taut string, and a key with a pivot, and a "tangent" made out of a nail. The paper towels are for damping.
This does not work very well at all, it is almost totally inaudible. Maybe this wire is too thick and there's not enough tension in it? You can see the smaller wire towards the top right, that one also sounded too quiet. I may give up on this idea.
OK, what about this: it's basically like a keyed monochord, except just after the tangent touches the string, a plectrum plucks it like on a harpsichord? So then it would still only need one string, and the note is defined by the tangent, but you get more energy into the string. You also get the opportunity when transitioning from one note to another to choose whether or not you press far enough to pluck. If not, the note will kind of smoothly change as it decays, if you pluck again then it will sound like a separate note.
Also do you really need a plectrum on each key? Is it possible to rig up the keys so they all actuate the same plectrum?
And for that matter, do you need a plectrum at all? What if you had a proper piano hammer mechanism, but all keys operate the same hammer on the same string?
< 2025-07-30 2025-08-19 >