- The mathematical quest for the perfect brass instrument
- 28 Feb 2019 08:45:55 am
- Last edited by Sam on 15 Mar 2021 02:06:36 am; edited 1 time in total
The calculations made within this post were calculated to 12 digit floating point, but for the sake of readability I am rounding all numbers to the thousandths. This means math done with these numbers will not be perfectly accurate.
To understand this post, it is required to know some prerequisite jargon, basic sound science, and the construction of valve-based brass instruments. If you're well versed in music theory and brass instruments, you can probably skip the next paragraph.
Brass instruments (in this case, a trumpet) are constructed by making a long tube (usually about 57 inches) and placing three valves along the pipe which redirect the flow of the air in the instrument to longer tubing. The first valve is tuned to lower the instrument two semitones, the second valve one semitone, and the third valve three semitones. These valves are not mutually exclusive, as engaging all three valves will lower the instrument's pitch approximately six semitones. I will denote which valves are engaged by referring to valves as 1, 2, and 3 and simply placing them next to each other. When no valves are pressed, I will call that an open fingering. For example, when I am pressing the first and third valves, I would call that fingering 13. If I am pressing all three valves, that would be fingering 123.
Brass instruments, as well as most wind instruments, produce sound by blowing/buzzing on the end of a tube and producing a sound wave, the trough/peak of which is exactly at the exit of the tube. (This is in very general terms.) Since the speed of sound is relatively constant, the frequency of the dominant note is determined by the length of the tube being played. The wavelength of the note being played is therefore twice the length of the instrument. It's a little more complicated with partials (overtones), but we'll talk about that later.
To help explain, the functionality of brass instruments is actually very similar to a guitar in construction. When you strum the E string on a guitar, the guitar plays an E. If you place your finger on the middle fret of the E string (thereby effectively halving the length of the string) the string, when strummed, will also play an E, but this time an octave up. Well, how do we get all the notes in between those two Es? There are 11 frets between the open E and the octave up E, which of course play the chromatic scale between those two notes, but notice that the distances between frets gets larger as they approach the headstock. this is because to lower a note one half step, you decrease the length of a string (or tube!) proportionally, not linearly. I could go over all the math, but just know that to raise the pitch of a trumpet by one half step, you must multiply the length of the tube by about 0.944. Lowering the pitch requires multiplying the length by about 1.059. Remember, you must increase and decrease the length proportionally, not linearly, so as the trumpet gets longer, you begin requiring increasingly more tubing to lower it one half step. This is the root of the issue.
Brass instruments can not be, as of now, perfectly in tune. On a hypothetical perfectly manufactured instrument, tubing would be as follows, rounded to the nearest thousandth:
You should immediately notice that the third valve, which lowers the instrument three semitones, is notably more than thrice as long as the second valve, which lowers it one semitone. This is because the length of tubing required to lower a brass instrument by n semitones is not simply length+3.368n, it is actually more like length((1.059)^n)-length. By this math, we find that the valves are actually tuned perfectly, because each valve follows this formula. For instance, the third valve should be 56.633((1.059)^3)-56.633, which is 10.715.
This is where the tuning inaccuracies come into play. If the first and third valves are played, for instance, it will raise the length by 17.650 inches, which is 1.313 inches less than the 18.966 inches required to lower the instrument the full five semitones. You can see now why tuning on-the-fly is crucial to remaining in tune. A perfectly tuned trumpet will still be out of tune on every fingering that requires more than one valve. Here's a table describing the tube lengths:
Indeed, as more valves are pressed down, the tuning inaccuracy becomes drastic. On our hypothetical perfectly manufactured trumpet, fingering 123 played with its root note required a slide to be pulled out a full 1.220 inches to be in tune. Wow. How do we fix this?
Well, the solution needs to be a mechanism which extends a slide/extra valves depending on what combinations of valves are pressed. This means we need mechanical logic gates. And this is where I appeal to the infinite creativity of you guys. How would I do this? This is my working design as of now. A slide attached to all three valves would move different distances depending on what is pressed. This is just a model, and I have not yet crunched any numbers, because holy cow it's complicated, but that's for another post. Look at the diagrams on the bottom left to understand what this thing is supposed to do. Do you guys have any ideas?
To understand this post, it is required to know some prerequisite jargon, basic sound science, and the construction of valve-based brass instruments. If you're well versed in music theory and brass instruments, you can probably skip the next paragraph.
Brass instruments (in this case, a trumpet) are constructed by making a long tube (usually about 57 inches) and placing three valves along the pipe which redirect the flow of the air in the instrument to longer tubing. The first valve is tuned to lower the instrument two semitones, the second valve one semitone, and the third valve three semitones. These valves are not mutually exclusive, as engaging all three valves will lower the instrument's pitch approximately six semitones. I will denote which valves are engaged by referring to valves as 1, 2, and 3 and simply placing them next to each other. When no valves are pressed, I will call that an open fingering. For example, when I am pressing the first and third valves, I would call that fingering 13. If I am pressing all three valves, that would be fingering 123.
Brass instruments, as well as most wind instruments, produce sound by blowing/buzzing on the end of a tube and producing a sound wave, the trough/peak of which is exactly at the exit of the tube. (This is in very general terms.) Since the speed of sound is relatively constant, the frequency of the dominant note is determined by the length of the tube being played. The wavelength of the note being played is therefore twice the length of the instrument. It's a little more complicated with partials (overtones), but we'll talk about that later.
To help explain, the functionality of brass instruments is actually very similar to a guitar in construction. When you strum the E string on a guitar, the guitar plays an E. If you place your finger on the middle fret of the E string (thereby effectively halving the length of the string) the string, when strummed, will also play an E, but this time an octave up. Well, how do we get all the notes in between those two Es? There are 11 frets between the open E and the octave up E, which of course play the chromatic scale between those two notes, but notice that the distances between frets gets larger as they approach the headstock. this is because to lower a note one half step, you decrease the length of a string (or tube!) proportionally, not linearly. I could go over all the math, but just know that to raise the pitch of a trumpet by one half step, you must multiply the length of the tube by about 0.944. Lowering the pitch requires multiplying the length by about 1.059. Remember, you must increase and decrease the length proportionally, not linearly, so as the trumpet gets longer, you begin requiring increasingly more tubing to lower it one half step. This is the root of the issue.
Brass instruments can not be, as of now, perfectly in tune. On a hypothetical perfectly manufactured instrument, tubing would be as follows, rounded to the nearest thousandth:
- Trumpet Length: 56.633 in.
First Valve Length: 6.935 in.
Second Valve Length: 3.368 in.
Third Valve Length: 10.715 in.
You should immediately notice that the third valve, which lowers the instrument three semitones, is notably more than thrice as long as the second valve, which lowers it one semitone. This is because the length of tubing required to lower a brass instrument by n semitones is not simply length+3.368n, it is actually more like length((1.059)^n)-length. By this math, we find that the valves are actually tuned perfectly, because each valve follows this formula. For instance, the third valve should be 56.633((1.059)^3)-56.633, which is 10.715.
This is where the tuning inaccuracies come into play. If the first and third valves are played, for instance, it will raise the length by 17.650 inches, which is 1.313 inches less than the 18.966 inches required to lower the instrument the full five semitones. You can see now why tuning on-the-fly is crucial to remaining in tune. A perfectly tuned trumpet will still be out of tune on every fingering that requires more than one valve. Here's a table describing the tube lengths:
- Fingering 12:
- Current tube length: 10.303 in.
Ideal tube length: 10.717 in.
Length deficit: 0.413 in.
Fingering 13:
- Current tube length: 17.650 in.
Ideal tube length: 18.966 in.
Length deficit: 1.313 in.
Fingering 23:
- Current tube length: 14.083 in.
Ideal tube length: 14.722 in.
Length deficit: 0.637 in.
Fingering 123:
- Current tube length: 21.018 in.
Ideal tube length: 23.462 in.
Length deficit: 2.440 in.
Indeed, as more valves are pressed down, the tuning inaccuracy becomes drastic. On our hypothetical perfectly manufactured trumpet, fingering 123 played with its root note required a slide to be pulled out a full 1.220 inches to be in tune. Wow. How do we fix this?
Well, the solution needs to be a mechanism which extends a slide/extra valves depending on what combinations of valves are pressed. This means we need mechanical logic gates. And this is where I appeal to the infinite creativity of you guys. How would I do this? This is my working design as of now. A slide attached to all three valves would move different distances depending on what is pressed. This is just a model, and I have not yet crunched any numbers, because holy cow it's complicated, but that's for another post. Look at the diagrams on the bottom left to understand what this thing is supposed to do. Do you guys have any ideas?