towards a truly chromatic Horn
Dec. 30th, 2018 03:21 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
wrogSo here's a puzzler
The main things to keep in mind about the multiples of f (the "harmonic series"):
But if you want to do anything useful down in the lower register, you have no choice but to mess with the length of the tube somehow, the two most popular methods being
a magical device that attaches to four pieces of tubing and gives you a choice about which pair is connected at any given time (... in this case, owing to the center section that rotates 90 degrees whenever the valve lever is depressed, this is a rotary valve, the type typically used in French horns. The other main type is the piston valve typically used in trumpets, but the differences don't matter a whole lot for what I'm going to discuss, so this much will do for now...)
The standard configuration for brass instruments, settled on after a few decades of experimentation in the 19th century, is to have 3 valves, each controlling an isolated loop, as follows:
Since each loop can be included independently of the others, we get a total of 23 or 8 possible overall tube lengths, which turns out to be sufficient to get you an entire chromatic scale.
You say, "But wait, there are 12 notes in the chromatic scale. How do we get that from 8 tube lengths?"
The trick is to settle for filling in the space between 2f and 3f with half steps -- which only needs six additional tube lengths; for each tone we insert there, we get, for free, another one that's a fifth below it.
So if, following the usual convention for transposing brass instruments, we take the 3rd harmonic (3f) on the original tube length and call that note "G", the six additional tube lengths will give us the next lower F#, F, E, D#, D, C# as the respective 3rd harmonics on those lengths, each tube being roughly 5-6% longer than its predecessor. Then, having reached C#, we can, for the next half-step down, switch back to the original tube length and play the 2nd harmonic (2f) to get C, after which can re-use the same sequence of tube-lengths to continue downwards, getting B, A#, A, G#, G, and F# as the corresponding 2nd harmonics.
At this point you'll notice we now have slightly more than a whole octave from 3f down to ~1.4f, which then happily repeats itself in the higher ranges 6f-3f and 12f-6f range. Beyond that, we don't care, because
which is then enough (hahahaha) to enable you to play Every Brass Instrument, or, rather, every 3-valved brass instrument that's configured in (what is, these days) the standard way.
And yes, the 2nd valve is supposed to have the shortest valve slide; being crammed in between the 1st and 3rd, it has the least space. As for why they chose 3rd-valve-by-itself as the one combination out of the eight to not use, we'll get back to that.
Finally, we get to
So, once you decide what frequency/wavelength G (open), F# (2) and F (1) are going to be, that fixes E (1&2) to be something that is most likely going to be a bit wrong. In fact we know it's going to be wrong because if valve #2 is supposed to drop the pitch a half step and valve #1 is supposed to drop the pitch a whole step, the combination of the two is going to be adding the lengths together, whereas in order to achieve a true one-and-a-half step drop we actually need to be multiplying the length ratios.
Or, to put it in math, a half step down from L(1+b) is actually
Similarly, while we can adjust the length of the 3rd valve slide (Lc) to get D# (2&3) to be whatever wavelength we want, this unavoidably screws both D (1&3) and C# (1&2&3):
Which is why trumpets have that pinkie-ring on their 3rd valve slide (making it into a little mini-trombone) and why French Horn players still need to keep their hand in the bell to mess with the pitch, even though you'd think valves would eliminate the need for that nonsense.
Which is also why we've included 5f and 7f columns even though we originally didn't need them and you'd otherwise think they'd be useless, being either slightly (5f) or horribly (7f) flat, at least as compared with the equal tempered notes we might be trying to get to (and hence the red shading of those columns). But it so happens that, for at least a few notes, the flatness of that harmonic (mostly) cancels the sharpness of the valve combination and we get notes that are better in tune -- in particular, the preferred fingerings for C# and D once you're out of the lowest octave (...though, to be fair, I'm not sure anyone actually uses the 1&2&3 fingering for the high E...)
As for choice of which fingering not to use (3 by itself), I believe that goes something like this:
Once you've decided that 1, 2, and 1&2 are going to give you the first 3 half steps down, it's inevitable that 3,2&3,1&3,1&2&3 also gives you four notes separated by three slightly-shorter-than-halfstep intervals where we're needing only three notes separated by halfsteps. And so the fingering we toss has to one of the ones on either end of that range, i.e., either 3 or 1&2&3. Either way, the high note and the low note will differ by the length of the 1 valve (Lb, which we've already noticed is not enough), i.e., that interval will be equally messed up no matter which fingering we drop.
So the only actual choice we have is about where the middle note ends up: using 3,2&3,1&3 means it'll be a 2-valve (La) below the top note, while using 2&3,1&3,1&2&3 means it'll be a 2-valve above the bottom note or L(b−a) below the top note. And we already know that La, chosen to drop the G to an F♯, will not be enough to drop the D♯ to a D. L(b-a)= L((1+a)²−(1+a))=L(a(1+a)), being longer, will be closer to what we need, and hence we most likely to go with 2&3,1&3,1&2&3.
(… Note that all of the above assumes we're in an equal-temperament-ish world where the whole step really is twice the size of the half step (L(1+b)=L(1+a)²), which I don't think I'd want to bet the farm on…)
Which then inspires:
which should serve as a good warmup for
E.g., in the case of the standard valve wiring above, we have the dependence [open] - [1] - [2] + [1&2] = L - L(1+b) - L(1+a) + L(1+a+b) always being zero, no matter how you adjust the slides.
However, looking at the wiring of the double French horn, we perhaps see another way out of the box:
Essentially a double horn is two instruments sharing the same mouthpiece and bell that are otherwise separate; a (fourth) thumb valve (T) switches back and forth between them. Typically, one instrument is an F horn (i.e., fundamental tone is an F) and the other is a Bb horn. Note that each of the four valves is actually a doubled valve, i.e., two distinct valves that are driven by the same lever -- in practice they're stacked on top of each other in the same cylindrical case so that one valve lever can rotate them both at the same time. The point of this exercise is that the 2f tones on the F horn fill in the gap between the f and 2f tones on the Bb horn, and suddenly we have a whole 'nother octave and a half available.
Since each side of the horn can provide (at least) 4 independently adjustable lengths (as the single horn can, as we've seen above), then we evidently can have at least 8 independently adjustable lengths with 4 double valves. But do we really need this much? The double horn is actually trying to solve a different problem -- getting more range rather than getting better intonation on the intermediate notes.
If we don't actually care about having a huge range, What's the minimum we need to solve the intonation problem? Can we do it with 3 double valves? Can we do it in such a way as to preserve the standard fingerings (i.e., the last two columns of the chart above) and so that the various tube segments will all be at least a certain minimum length so that this beast actually has some chance of being constructible?
Time to break out the computer and do the exhaustive search through the 459324 possible wirings for 6 valves and the 15 possible pairings for each wiring. Hey, here's one:
which is the best in terms of having the longest minimum segment-between-valves length — every segment is at least 7.45% of the base length L.
Now there is a bit of weirdness with this solution in that the mouthpiece+bell portion of the horn is less than 25% of its overall length. Which seems like it would be showstopper for any brass instrument whose bore is not cylindrical, i.e., since we need to have the tube width stay constant throughout the section where there are valves; otherwise life gets much more complicated. And in order to do that on a conical-bore instrument like the French horn, you'd need a plan where the portion between the valves is a relatively small segment of the horn.
Redoing the search ruling out the cases where the bell+mouthpiece segment isn't at least 50% of the horn, we get another solution where every segment is at least 5.6% of L and the bell+mouthpiece portion is roughly 71% of the instrument.
Note that figuring out what a,b,c,d,e,f,g need to be for any given temperment is reduced to a Small Matter of Linear Algebra.
Except it turns out that for brass instruments, this whole distinction between cylindrical and conical bore instruments — which was presented as gospel back when I was learning them — seems to be something of a lie. The French horn, in fact, has, percentage-wise, a much longer cylindrical section than the supposedly-cylindrical-bore trumpet, and that, in fact, what is really going on with both of these instruments is that the bell and mouthpiece effects drastically warp the harmonics of an otherwise-cylindrical closed tube so as to make it behave almost, but not quite, like an open cylindrical tube or, equivalently, a closed conical tube (both of which have the even multiples of f that the closed cylindrical tube lacks).
Which suggests that I have it exactly backwards as to which solution is best for which instrument (i.e., the first solution is better for the horn, and the second solution is better for the trumpet)...
There then remains the small matter of getting to actually build one of these monsters (yeah, good luck on that...). It seems like they'd be useful, at least in the sense that a whole range of intonation issues mostly go away.
On the other hand, doing the actual tuning and emptying out the condensation ("spit") will be extra-special fun.
And, of course, now I'm wondering how different the history of brass instruments would have been if Stölzel had had a personal computer available, though, to be fair, I'm guessing a lot of things would have been different if we'd had PCs back in 1814.
Brief review of the physics of wind instruments
Your standard wind instrument is a pipe attached to some kind of sound source (lip or reed). Being of a certain length it resonates at particular frequencies and suppresses the others, ulimately being capable of producing tones that are multiples of a certain fundamental frequency f, 2f being an octave higher than f, 3f being a fifth above that and so on.The main things to keep in mind about the multiples of f (the "harmonic series"):
- going up by a particular interval means multiplying the frequency (or dividing the wavelength) by a particular ratio (2 ↔ octave, 3/2 ↔ perfect fifth, 5/4 ↔ major third, etc.…), and
- intervalwise, the notes are getting closer together the higher you go.
But if you want to do anything useful down in the lower register, you have no choice but to mess with the length of the tube somehow, the two most popular methods being
- (the woodwind solution) poke holes in the tube (and cover them with removable keys)
- (the brass solution) insert valves or slides in the tube to change its actual length
So how do valves actually get used?
Let's first get clear what a valve is: |  |
a magical device that attaches to four pieces of tubing and gives you a choice about which pair is connected at any given time (... in this case, owing to the center section that rotates 90 degrees whenever the valve lever is depressed, this is a rotary valve, the type typically used in French horns. The other main type is the piston valve typically used in trumpets, but the differences don't matter a whole lot for what I'm going to discuss, so this much will do for now...)
The standard configuration for brass instruments, settled on after a few decades of experimentation in the 19th century, is to have 3 valves, each controlling an isolated loop, as follows:
Since each loop can be included independently of the others, we get a total of 23 or 8 possible overall tube lengths, which turns out to be sufficient to get you an entire chromatic scale.
You say, "But wait, there are 12 notes in the chromatic scale. How do we get that from 8 tube lengths?"
The trick is to settle for filling in the space between 2f and 3f with half steps -- which only needs six additional tube lengths; for each tone we insert there, we get, for free, another one that's a fifth below it.
So if, following the usual convention for transposing brass instruments, we take the 3rd harmonic (3f) on the original tube length and call that note "G", the six additional tube lengths will give us the next lower F#, F, E, D#, D, C# as the respective 3rd harmonics on those lengths, each tube being roughly 5-6% longer than its predecessor. Then, having reached C#, we can, for the next half-step down, switch back to the original tube length and play the 2nd harmonic (2f) to get C, after which can re-use the same sequence of tube-lengths to continue downwards, getting B, A#, A, G#, G, and F# as the corresponding 2nd harmonics.
At this point you'll notice we now have slightly more than a whole octave from 3f down to ~1.4f, which then happily repeats itself in the higher ranges 6f-3f and 12f-6f range. Beyond that, we don't care, because
- by that point that harmonics are so close together you hardly need valves anymore
- unless you're Maynard Ferguson, your lip can't get there anyway, and
- the tube itself isn't being that much of a resonator any more because it's so large as compared with the wavelength of sound being pushed through it -- you might as well be yelling into a sewer pipe.
| 9f | 8f | 7f | 6f | 5f | 4f | 3f | 2f | fingering |
|---|---|---|---|---|---|---|---|---|
| D | C | A#/Bb | G | E | C | G | C | open (no valves depressed) |
| C#/Db | B | A | F#/Gb | D#/Eb | B | F#/Gb | B | 2 (i.e., 2nd valve) |
| A#/Bb | G#/Ab | F | D | A#/Bb | F | A#/Bb | 1 | |
| A | G | E | C#/Db | A | E | A | 1&2 | |
| G#/Ab | F#/Gb | D#/Eb | C | G#/Ab | D#/Eb | G#/Ab | 2&3 | |
| F | D | B | G | D | G | 1&3 | ||
| E | C#/Db | F#/Gb | 1&2&3 |
which is then enough (hahahaha) to enable you to play Every Brass Instrument, or, rather, every 3-valved brass instrument that's configured in (what is, these days) the standard way.
And yes, the 2nd valve is supposed to have the shortest valve slide; being crammed in between the 1st and 3rd, it has the least space. As for why they chose 3rd-valve-by-itself as the one combination out of the eight to not use, we'll get back to that.
Finally, we get to
The Problem
which makes itself apparent in the blue-shaded rows:Having seven distinct tube lengths available does not mean they can all be tuned independently.On a typical brass instrument, each valve segment is adjustable, and there's a tuning slide for the whole rest of the instrument, but that's all you get, i.e., just four (4) adjustments that are possible.
So, once you decide what frequency/wavelength G (open), F# (2) and F (1) are going to be, that fixes E (1&2) to be something that is most likely going to be a bit wrong. In fact we know it's going to be wrong because if valve #2 is supposed to drop the pitch a half step and valve #1 is supposed to drop the pitch a whole step, the combination of the two is going to be adding the lengths together, whereas in order to achieve a true one-and-a-half step drop we actually need to be multiplying the length ratios.
Or, to put it in math, a half step down from L(1+b) is actually
L(1+b)(1+a) = L(1+a+b+ab) > L(1+a+b)where L is the length of the original tube, and La, Lb are the respective lengths of the 2nd and 1st valve slides. Our E is thus doomed to always be too sharp, hence the blue shading of that row.
Similarly, while we can adjust the length of the 3rd valve slide (Lc) to get D# (2&3) to be whatever wavelength we want, this unavoidably screws both D (1&3) and C# (1&2&3):
L(1+a+c)(1+a) = L(1+b+c+ac) > L(1+b+c)with (1&2&3) coming out particularly badly (the error being L(a+c)b ≈ 4.4Lab as compared with the error for (1&3), Lac ≈ 1.6Lab), hence the darker blue for that row.
L(1+a+c)(1+b) = L(1+a+b+c+(a+c)b) > L(1+a+b+c)
Which is why trumpets have that pinkie-ring on their 3rd valve slide (making it into a little mini-trombone) and why French Horn players still need to keep their hand in the bell to mess with the pitch, even though you'd think valves would eliminate the need for that nonsense.
Which is also why we've included 5f and 7f columns even though we originally didn't need them and you'd otherwise think they'd be useless, being either slightly (5f) or horribly (7f) flat, at least as compared with the equal tempered notes we might be trying to get to (and hence the red shading of those columns). But it so happens that, for at least a few notes, the flatness of that harmonic (mostly) cancels the sharpness of the valve combination and we get notes that are better in tune -- in particular, the preferred fingerings for C# and D once you're out of the lowest octave (...though, to be fair, I'm not sure anyone actually uses the 1&2&3 fingering for the high E...)
As for choice of which fingering not to use (3 by itself), I believe that goes something like this:
Once you've decided that 1, 2, and 1&2 are going to give you the first 3 half steps down, it's inevitable that 3,2&3,1&3,1&2&3 also gives you four notes separated by three slightly-shorter-than-halfstep intervals where we're needing only three notes separated by halfsteps. And so the fingering we toss has to one of the ones on either end of that range, i.e., either 3 or 1&2&3. Either way, the high note and the low note will differ by the length of the 1 valve (Lb, which we've already noticed is not enough), i.e., that interval will be equally messed up no matter which fingering we drop.
So the only actual choice we have is about where the middle note ends up: using 3,2&3,1&3 means it'll be a 2-valve (La) below the top note, while using 2&3,1&3,1&2&3 means it'll be a 2-valve above the bottom note or L(b−a) below the top note. And we already know that La, chosen to drop the G to an F♯, will not be enough to drop the D♯ to a D. L(b-a)= L((1+a)²−(1+a))=L(a(1+a)), being longer, will be closer to what we need, and hence we most likely to go with 2&3,1&3,1&2&3.
(… Note that all of the above assumes we're in an equal-temperament-ish world where the whole step really is twice the size of the half step (L(1+b)=L(1+a)²), which I don't think I'd want to bet the farm on…)
Which then inspires:
Question:the point of this, by the way, being not necessarily to achieve Actual Equal Temperment — arguably a dubious goal, since there's a not unreasonable argument that, in fact, equal temperment sucks, — but rather to be able to do a particular temperment at all. Four adjustable slides just isn't enough.
What would it take to build a truly chromatic horn where we have seven (or twelve!) independently adjustable pitch families?
And now the fun starts
Clearly, what we need at the very least is to have more inter-valve segments available. So let's just scramble things around and see what happens:Exercise #1:
Consider this arrangement of valves:
Now we have six segments whose lengths can be independently adjusted.
Running through all possible valve-pressing combinations, how many distinct lengths do we actually get?
How many of these are actually independent?
Answers: 5 and 4, respectively.
which should serve as a good warmup for
Exercise #2:Just to be clear about what we mean by "independent": For every fingering there is a length. A set of fingerings then has a linear dependence if there is some nonzero combination of their lengths that always sums to zero regardless of how one might adjust the individual edge lengths, and is linearly independent if there is no such combination.
Given an arbitrary valve wiring, which is to say a connected graph on n nodes such thatwe will have, for each possible combination (i.e., "fingering") of node states, a unique loop that contains the special edge and only traverses connected pairs at each node, and said loop will have a length that is the sum of the lengths of some subset of the edges.
- each node has out-degree four (i.e., four edges coming out of it),
- each node has designated for it a pair of possible states, each of which partitions the outgoing edges into pairs that are "connected" (i.e., each node is a valve)
- one edge is marked special (this being the one that we can take a pair of scissors to and attach a mouthpiece and bell to the resulting cut ends),
What is the maximum number of linearly independent fingerings we can get from an n node valve-wiring?
E.g., in the case of the standard valve wiring above, we have the dependence [open] - [1] - [2] + [1&2] = L - L(1+b) - L(1+a) + L(1+a+b) always being zero, no matter how you adjust the slides.
Answer: n+1. (and I'm still looking for a more elegant proof of this)Which means that 3 valves are just not enough to get 7 independent fingerings, no matter how you wire things up.
However, looking at the wiring of the double French horn, we perhaps see another way out of the box:
Essentially a double horn is two instruments sharing the same mouthpiece and bell that are otherwise separate; a (fourth) thumb valve (T) switches back and forth between them. Typically, one instrument is an F horn (i.e., fundamental tone is an F) and the other is a Bb horn. Note that each of the four valves is actually a doubled valve, i.e., two distinct valves that are driven by the same lever -- in practice they're stacked on top of each other in the same cylindrical case so that one valve lever can rotate them both at the same time. The point of this exercise is that the 2f tones on the F horn fill in the gap between the f and 2f tones on the Bb horn, and suddenly we have a whole 'nother octave and a half available.
Since each side of the horn can provide (at least) 4 independently adjustable lengths (as the single horn can, as we've seen above), then we evidently can have at least 8 independently adjustable lengths with 4 double valves. But do we really need this much? The double horn is actually trying to solve a different problem -- getting more range rather than getting better intonation on the intermediate notes.
If we don't actually care about having a huge range, What's the minimum we need to solve the intonation problem? Can we do it with 3 double valves? Can we do it in such a way as to preserve the standard fingerings (i.e., the last two columns of the chart above) and so that the various tube segments will all be at least a certain minimum length so that this beast actually has some chance of being constructible?
Time to break out the computer and do the exhaustive search through the 459324 possible wirings for 6 valves and the 15 possible pairings for each wiring. Hey, here's one:
which is the best in terms of having the longest minimum segment-between-valves length — every segment is at least 7.45% of the base length L.
Now there is a bit of weirdness with this solution in that the mouthpiece+bell portion of the horn is less than 25% of its overall length. Which seems like it would be showstopper for any brass instrument whose bore is not cylindrical, i.e., since we need to have the tube width stay constant throughout the section where there are valves; otherwise life gets much more complicated. And in order to do that on a conical-bore instrument like the French horn, you'd need a plan where the portion between the valves is a relatively small segment of the horn.
Redoing the search ruling out the cases where the bell+mouthpiece segment isn't at least 50% of the horn, we get another solution where every segment is at least 5.6% of L and the bell+mouthpiece portion is roughly 71% of the instrument.
Note that figuring out what a,b,c,d,e,f,g need to be for any given temperment is reduced to a Small Matter of Linear Algebra.
Except it turns out that for brass instruments, this whole distinction between cylindrical and conical bore instruments — which was presented as gospel back when I was learning them — seems to be something of a lie. The French horn, in fact, has, percentage-wise, a much longer cylindrical section than the supposedly-cylindrical-bore trumpet, and that, in fact, what is really going on with both of these instruments is that the bell and mouthpiece effects drastically warp the harmonics of an otherwise-cylindrical closed tube so as to make it behave almost, but not quite, like an open cylindrical tube or, equivalently, a closed conical tube (both of which have the even multiples of f that the closed cylindrical tube lacks).
Which suggests that I have it exactly backwards as to which solution is best for which instrument (i.e., the first solution is better for the horn, and the second solution is better for the trumpet)...
There then remains the small matter of getting to actually build one of these monsters (yeah, good luck on that...). It seems like they'd be useful, at least in the sense that a whole range of intonation issues mostly go away.
On the other hand, doing the actual tuning and emptying out the condensation ("spit") will be extra-special fun.
And, of course, now I'm wondering how different the history of brass instruments would have been if Stölzel had had a personal computer available, though, to be fair, I'm guessing a lot of things would have been different if we'd had PCs back in 1814.
