# What is the math behind how keys/holes on woodwind instruments affect the pitch?

To make this question more specific, I'll provide an example of what I mean.

Let's say you have an instrument that's just a tube you blow into, and it has one hole exactly halfway down the tube (I know that's not an "instrument" but I'm speaking theoretically). If one blows into the instrument without covering the hole, let's say it produces a B flat 4. If one were to cover the hole in the center and blow, how would that affect the sound? Would it change the Bb's octave from 4 to 3 or 5 since the hole is exactly halfway down the tube?

https://en.wikipedia.org/wiki/Music_and_mathematics

In this article under "Frequency and Harmony" there is a chart with frequencies and stuff, but I don't understand it, and how it would apply to a wind instruments. I understand the "Common Name" column and the "Example name Hz", but the other ones confuse me greatly. I'm not sure if this would help me understand what I asked more, but it seems like it.

Hopefully this makes some sort of sense.

• Just to be clear: this theory applies to the woodwinds, where a reed or a "pocket" (in the case of the flute) is the source of vibrations. Brass instruments, where the source (lips) is essentially a square wave, sound awful if you try to change the pitch via holes in the tube. – Carl Witthoft Nov 20 '15 at 13:12
• @CarlWitthoft It's completely fine. I shouldn't have to specify every single little thing about this question. As long as it's understandable. Chris Erwin was kind enough to answer my thoroughly, proving that he understood it. This question is more so about sound waves and instruments, not about how "good" a brass instrument's sound is. – Sam Nov 20 '15 at 20:17

Covering the hole would bring the note down to Bb 3. With the hole covered (or no hole at all) you would get a standing wave with a wavelength of twice the length of the instrument as the fundamental pitch.

The reason for this is because the instrument is a tube capable of holding pressure in the middle, but not capable of holding pressure at either end (let's assume a flute with open ends). When you visualize a wave, remember that the wavelength is from 0, up to the highest peak, back to zero, down to the lowest peak, and back to zero again. Line two of those zeros up at either end of the tube and you now have half the wave length inside the instrument, and this is the longest wave that fits in the instrument.

When you open the hole at exactly the center, think about a wave that has 0 at either end and 0 at the hole. The longest wave capable of this has a wavelength of 1L, where L is the length of the instrument.

Basically, whatever combination of holes is open, the only notes that will resonate are ones where the wavelengths work out so the pressure nodes line up at the holes.

Now let's go back to having all the holes closed. Don't forget that you can blow harder on a flute (and more importantly, brass instruments) to raise the frequency. The only notes that will resonate are those that have nodes (0 pressure) at either end of the instrument. Since Bb3 is the fundamental pitch, you could potentially play Bb4, F5, Bb5, D6, F6, Ab6, and so on.

That's where we get into that chart on the wikipedia page. It shows A2 having a frequency of 110Hz. To get A3, we double that (220). To get A4, we need to double THAT, right? so 440. But what about 3x 110 (330)? That's where you get E4.

The "ratio within octave" means the ratio between the note given and the A below it (since that chart is based around A). So, in regards to frequency, E4 is 3x A2, but 1.5x A3. C#5 is 5x A2 (the fundamental) but 1.25x the closest A, A4. There are 1200 cents per octave (100 per note), so that column is just giving how many out of 1200 increments it is above the A below it.

That's all confusing, but think about tuning a piano. You start by tuning A440. Now you play that A440 with the A below it. If it's out of tune, the frequency won't be exactly double. If it was, the waves would line up and it would sound great. When the waves don't quite line up, you get 'beating' where the waves cycle between adding their sound together and slightly canceling each other out. You tweak the A3 until the beats stop, meaning it's at exactly 220.

You can then tune the E to that. Since E4 is 1.5x A3, the waves also line up when they're in tune. When they're out of tune, you hear the beats, and you can adjust until they line up. That's why that 1.5x is important.

Which brings us to the circle of 5ths. One you tune that E4 to A3, you can tune all the Es. And then you can tune the Bs because they're 1.5x the nearest E. Then the F#s, and so on. 12 notes later and you're back at A and you've tuned the piano. I won't get into temperament though.

Here's a great article on flute accoustics that has helpful images: https://newt.phys.unsw.edu.au/jw/fluteacoustics.html

• Thank you so much! It makes sense now, but I cannot open the article because my computer, and my cell phone also, block it. – Sam Nov 19 '15 at 13:53
• Not sure why you can't open the article. Maybe the https version is causing issues? Try this link: newt.phys.unsw.edu.au/jw/fluteacoustics.html – Chris Erwin Nov 19 '15 at 19:23
• Thank you! However, I want to ask you one more thing. From what the article says, the pitch depends on the frequency of a wave from the end of the instrument to the first open hole (I'm probably wrong). What if you were to have a sound wave in between two open holes? What "pitch" would play? Sorry if this doesn't make sense. – Sam Nov 20 '15 at 20:21
• Totally missed your comment and you'll probably never see this, but the short answer is that, if you have two holes, and two open ends, the instrument will play the note where each end and the two holes line up on pressure nodes (0 pressure). It's not just the distance between the two holes, it's the least common multiple of the distance between the end and the first hole, the first hole and the second hole, and the second hole and the other end. – Chris Erwin Aug 10 '16 at 20:52

The other answer is quite misleading. Holes do not just force nodes in the resonating waves. Consider you cover all holes but the bottom one, you'll hear a note 1 or 2 semitones higher than the lowest note (depending on the flute design), meaning the wavelength is 6-12% shorter. This is definitely not a harmonic selected from the original set. Also factor in that when making hand-made flutes, they are tuned by drilling the holes small and gradually expanding them starting from the bottom until each note is correct. The hole placement is much less important to tuning!

You have to consider how standing waves are created in the first place: reflections. A flute has three main types of reflections: hard boundary (at the mouth end) which reflects a wave inverted to the original, soft boundary (the open end) which reflects a wave matching the phase of the original, and mixed boundary (open holes of varying sizes), which reflect waves of varying phase to the original depending on how much the hole size acts like an open boundary.

In the simplest case with all holes closed, only the first two reflections are at play, and the result is a standing quarter-wave, 4 times the tube length.

In the next simplest case with one hole open, you have an additional reflection. An interesting property of sine waves is that the sum of any two sine waves of the same frequency is a single sine wave of that frequency with some phase offset. So the reflection from the hole and the reflection from the open end of the tube are effectively equivalent to a single reflection from some virtual end-point in between. Considering that, the same mathematical result as above applies and you'll get a standing quarter wave, 4 times the wavelength as that virtual reflection point. The hole size controls the amplitude and phase of the hole's reflection, which in turn controls the location of the virtual reflection point (it's similar to a weighted average, a larger hole brings the virtual reflection closer to the hole, while a smaller hole lets it go back towards the real tube end).

In the multiple open holes case, the same principles apply and approximately an average of tube end and each open hole's location weighted by the hole's size gives the functional end point of the resonant tube. (Weighted average here, really referring to how vector sums in the complex plane have resultant phases that resemble the average phases of inputs weighted by their amplitudes)

There are some pages that go into this in some detail. Clarinets are here: https://newt.phys.unsw.edu.au/jw/clarinetacoustics.html That page has links to others.

As well as the mathematics, the pitch of a note on a practical instrument is affected by several factors. The recorder is close to your "theoretical instrument" and is one I am very familiar with, so I will use it as an example.

Recorders have a 2 octave and a tone range. The bottom octave is the fundamental notes and the upper octave notes are "overblown" at the octave. A thumb hole is used to introduce a controllable leak that encourages the instrument to resonate at twice the frequency. Most notes have the same fingering in both octaves - only the thumb hole is partially opened (pinched) to play the upper octave note.

So what sets the pitch of a note?

1. The basic pitch of a note is set by the size and position of the first open hole below the closed holes (e.g. On a descant or tenor recorder, the fingering for G has the thumb hole plus holes 1, 2 & 3 covered, so hole 4 is the main tuning hole for G). Moving a hole up the tube or making it bigger sharpens the note.
2. Bottom octave notes are more sensitive to the distance down the tube than hole size.
3. The upper octave notes are more sensitive to hole size than distance down the tube.
4. The internal shape of the tube matters. Early recorders were cylindrical, baroque recorders are conical, so the hole spacing is different. On larger instruments, makers often vary the bore shape to move holes closer together to reduce the need for keys.
5. The hole internal profile has a small effect. Undercutting the hole produces approximately the same result as enlarging it.
6. A thick-walled instrument will lower the pitch of each hole slightly as the air column becomes a little longer.
7. On small instruments the two small holes on the foot joint are usually bored at at an angle. This moves them downstream closer to the correct acoustic position in the bore, but keeps them in a comfortable physical distance of your right-hand pinky.
8. On larger instruments the key pads shade their open holes, flattening the note slightly.

So the process to tune a recorder involves:

1. Turning the body exterior to a cone shape, and using a reamer to ream the internal cone.
2. Adjusting hole size and position of each hole to get both octaves in tune. This step will be repeated many times.
3. Possibly modifying the bore taper to fix tuning problems.

For anyone who would like to delve deeper into the acoustics of the recorder, this book is invaluable and reasonably priced.