Math Geek Acoustics Anyone?

[kevgermany]

I think so, haven't got time to check

 

[aldevis]

Apologies for my thickness, but if I have a note produced by a conical tube long l, that will have a wavelength (circa) =2l, will the first harmonic (octave) be at l/2 (half of the length) or at (1/3 × π x 2r× l)/2 (half of the volume, where 2r is the diameter at the end of the tube)?

 

[Dave McLaughlin]

Apologies for my thickness, but if I have a note produced by a conical tube long l, that will have a wavelength (circa) =2l, will the first harmonic (octave) be at l/2 (half of the length) or at (1/3 × π x 2r× l)/2 (half of the volume, where 2r is the diameter at the end of the tube)?

The octave will be at half the wavelength of the fundamental, or l (which I think is what you meant).

One nit-picking point: I think the fundamental is usually considered to be the first harmonic, and the octave would be called the second harmonic or first overtone. But I'm sure some people would disagree with that!

 

[jbtsax]

This illustration I made a while back hopefully can help in the understanding of this topic. I posited the question that perhaps Ferron divided by 3 to find the placement of the displacement antinode (pressure node) of a higher harmonic of Ab. As you can see from the illustration, that does not provide a satisfactory answer. The website viewer cuts off the side of the illustration, but when the pdf file is downloaded you can view the complete page.

Neck Node Study

 

[aldevis]

The octave will be at half the wavelength of the fundamental, or l (which I think is what you meant).

One nit-picking point: I think the fundamental is usually considered to be the first harmonic, and the octave would be called the second harmonic or first overtone. But I'm sure some people would disagree with that!

I am fine with the wavelength. But I am not sure where the second harmonic antinode is in a conical tube.
I number harmonics with the. denominator of the wavelength.

 

[kevgermany]

First harmonic is, as Dave said, the fundamental. However the first overtone is at the octave above the fundamental. Important not to mix these things.

 

[Dave McLaughlin]

This post is to discuss specific information contained in "The Saxophone Is My Voice" by Ernest Ferron. Although the book contains several errors and some dated information nevertheless it is an excellent entry level text to learn more about saxophone acoustics.

On page 104 Ferron writes:



The basic question is why did Ferron divide by 3? What is his rationale for doing so once the distance to the antinode of the note a half step away is found by dividing (or multiplying) by 1.05946. The ramifications of this question are important since Ferron represents the "sensitive harmonic sites" in an alto neck in the illustration below copied from his book on p. 19. Note that he shows 3 octaves.

A Mark VI alto neck including the tenon measures 195 mm. Going from the antinode length of G which is 362.81 mm according to Ferron's figures and the antinode of G an octave higher 362.81 / 2 = 181.405 would indicate that the difference of 181.405 would encompass nearly the entire 195 mm length of the neck. This would indicate that there are not 3 octaves of antinode positions, but only one.

My thinking is that the illustrations of "the sensitive harmonic sites" in alto and tenor necks in Ferron's book that have been widely distributed on the internet are incorrect because of the division by 3 which does not make any sense to me. Am I missing something here? Opposing thoughts and opinions are welcome.

I find this baffling. I'm trying to follow Ferron's discussion:

Here we are interested in displacement antinodes. Piercing the neck precisely on a particular note's displacement antinode will not affect the note's (but only this note's) response. It is even possible to cut the tube here without imparing the note. It thus seems pertinent to use this peculiarity to determine the exact position of this or that note's displacement antinode ( or harmonic).

Place a 2.5 mm hole for the soprano (or a 5 or 6 mm ole for the other saxophones) on the neutral line of the neck curve. It can easily be filled in later.

Play the instrument's second register with the mouthpiece exactly in place. It is possible to play only one note, that being the one whose displacement antinode lies exactly on the hole and which will serve as a reference.

After removing the mouthpiece, measure the distance from the center of the hole to the end of the neck, and add the length of the theoretical pointed cone. This gives the harmonic's wavelength.

I've said in #3 above that I thought the distance from the apex of the cone to the displacement antinode was 0.75 of a wavelength. Actually, having thought about it further, it should be the first turning point of sin(x/λ) - see http://www.phys.unsw.edu.au/jw/pipes.html. From Wikipedia, a good approximation to that turns out to be about 0.716λ.

An example using G with an alto neck, shows:

A second register G on an alto is a concert A# at 466 Hz - I think!

Length from the center of the hole to the end of the neck=157. Length of the theoretical pointed cone=205.81
Wavelength of the G harmonic (*) 157 + 205.81 = 362.81 [mm]

If we assume the speed of sound is 340 m/s, the wavelength of a second register G on an alto is 729 mm. So already, I think Ferron's an octave out.

To find the position of G#, divide this length by the 12th root of 2 (1.05946),

No disagreement there; that's a semitone.

deduct this figure from the preceding length and divide by 3. [emphasis added]

For example: 362.81/1.05946 = 342.448, 362.81 - 342.448 = 20.362

20.362/3 = 6.78 mm G# is 6.78 mm from G, etc.

(*) It is in fact a quarter of a wave length (see page 9) since we are actually always working with a quarters (sic) of the total wave length.

Now he's completely lost me. I think the distance he's measured is 0.716λ. and a third of that would be roughly a quarter wavelength. But it doesn't matter whether it's a wavelength, a quarter wavelength or whatever. The key thing is that, if the distance is proportional to wavelength, then to get from a distance appropriate to G to the corresponding distance for G#, he should divide by 1.05946. That would make the G# antinode 20.362 mm from the G antinode. I don't see why he's dividing by 3.

Sorry - I had hoped to get to the bottom of this, but now I'm more baffled than ever!

 

[kevgermany]

Wish I had time as well

 

[aldevis]

If we assume the speed of sound is 340 m/s, the wavelength of a second register G on an alto is 729 mm. So already, I think Ferron's an octave out.

The tube length is about half the wavelength.
It seems to me that the placement of the first hole in the neck is empirical, and he starts from there.

 

[kevgermany]

OK, I took a close look. I agree with Ferron.(innacuracies aside)

Firstly Ferron says he's working with second register G and G#. So wavelength of this is half that of the fundamental. This is the first overtone of G/G# or second harmonic.
Starting at the mouthpiece end, and assuming a tapered cone (i.e the theoretical sax, mouthpiece doesn't come into this. )

See http://www.phys.unsw.edu.au/jw/woodwind.html Diagram under the heading "Harmonics and the different instrument bores", second row of pics.
The waveform has a point of maximum movement/lowest pressure (displacement antinode in Ferron's terms) 1/3rd of the way along the cone.
Two thirds of the way along the cone is the point of highest pressure/zero movement. (node).
At the open end of the cone is the point of reflection - where we again have maximum movement and zero pressure. It's this point that controls the wavelength or pitch.

You can see from the diagram/explanation that in the second register of the sax, the wavelength of the note is not double the length of the tube as it is in the fundamental, but it's 3/2 times the length of the sax/cone as there's are one and a half full wavelengths in the cone. Double this as the wave is reflected at the open end.

When Ferron divides by 3, it's the 3 from the 3/2 he needs to remove. And it has nothing to do with the volume of the cone. Just the shape of the waveform in the first overtone. Sorry if I confused before.

 

[jbtsax]

One of the clearest and easiest to understand explanations of standing waves I have found is this:

Standing Waves and Wind Instruments

Be sure to click the "animation" about a third of the way down the page.

Please do not be misled as I was at one time by the illustration of the pressure and displacement (air vibration) waves inside a conical instrument on the UNSW website. I even contacted Joe Wolf about it, and he indicated that using the graphics program they had at the time made it difficult to get the proper placement and proportions of the waves in that illustration.

I believe that the following sites have a more accurate illustration of the standing waves inside a conical instrument.
Colorado Physics and Standing Waves in a Conical tube. Although I will admit there are some sites that show the standing waves inside a conical tube to be the same as those inside a cylinder. Most confusing, to say the least. . .

 

[Dave McLaughlin]

OK, I took a close look. I agree with Ferron.(innacuracies aside)

Good! Maybe you can help me, because I'm obviously having a slow day. [Edit: I realise that could be read as sarcasm - it's not intended that way! I really am having trouble understanding this, but kev's previous post has helped, to some extent.]

Firstly Ferron says he's working with second register G and G#. So wavelength of this is half that of the fundamental. This is the first overtone of G/G# or second harmonic.

Do you agree with me that a second register G on an alto is a concert A# at 466 Hz?

Starting at the mouthpiece end, and assuming a tapered cone (i.e the theoretical sax, mouthpiece doesn't come into this. )

See How Do Woodwind Instruments Work? Diagram under the heading "Harmonics and the different instrument bores", second row of pics.
The waveform has a point of maximum movement/lowest pressure (displacement antinode in Ferron's terms) 1/3rd of the way along the cone.
Two thirds of the way along the cone is the point of highest pressure/zero movement. (node).
At the open end of the cone is the point of reflection - where we again have maximum movement and zero pressure. It's this point that controls the wavelength or pitch.

Ah! The penny drops! I had been assuming the relevant length was the from the apex of the cone to the new hole in the neck and we were looking for the fundamental mode of that truncated cone.

You can see from the diagram/explanation that in the second register of the sax, the wavelength of the note is not double the length of the tube as it is in the fundamental, but it's 3/2 times the length of the sax/cone as there's are one and a half full wavelengths in the cone. Double this as the wave is reflected at the open end.

Here I disagree. Whether you're looking at the UNSW page or the one jbtsax pointed to (Standing Waves in a Conical tube), the pressure wave looks, at first glance, like 3/4 of a wavelength. But look more closely. (I note that jbtsax has some reservations about these illustrations, but this looks right to me.) The first 1/4 wave takes up half the length of the cone. The other 1/2 wave takes up the other half. How can 1/4 wave be as long as 1/2 a wave? Remember, in a conical tube, the pressure varies, not as sin(x), but as sin(x)/x. As x tends towards 0, that tends to a maximum of 1. So that first 1/4 wave is actually half a wavelength. The wavelength is the same as the length of the cone.

When Ferron divides by 3, it's the 3 from the 3/2 he needs to remove. And it has nothing to do with the volume of the cone. Just the shape of the waveform in the first overtone. Sorry if I confused before.

I still don't get it. I still think, if the distance is proportional to wavelength, then to get from a distance appropriate to G to the corresponding distance for G#, he should divide by 1.05946. It doesn't matter how many wavelengths it is.

 

[kevgermany]

Good! Maybe you can help me, because I'm obviously having a slow day. [Edit: I realise that could be read as sarcasm - it's not intended that way! I really am having trouble understanding this, but kev's previous post has helped, to some extent.]

np, this is interesting.

Do you agree with me that a second register G on an alto is a concert A# at 466 Hz?

yes (466.16 really )

Here I disagree. Whether you're looking at the UNSW page or the one jbtsax pointed to, the pressure wave looks, at first glance, like 3/4 of a wavelength. But look more closely. (I note that jbtsax has some reservations about these illustrations, but this looks right to me.) The first 1/4 wave takes up half the length of the cone. The other 1/2 wave takes up the other half. How can 1/4 wave be as long as 1/2 a wave? Remember, in a conical tube, the pressure varies, not as sin(x), but as sin(x)/x. As x tends towards 0, that tends to a maximum of 1. So that first 1/4 wave is actually half a wavelength. The wavelength is the same as the length of the cone.

Looks like I was wrong. Sorry.

Assuming you're talking about the first overtone.
I understand a full wave as going from 0 to full positive pressure, back to zero and then to full negative pressure and back to zero again. (Or from any starting point on the same pattern, until it's completed the cycle). So the starting from max pressure, dropping to 0 is the quarter wave. Then the wave drops to lowest pressure, returning to 0 at the end of the cone - a half wave, then the return gives us the second half wave, followed by the second quarter wave. So one and a half cycles in the length of the cone. What the line shows is the pressure distribution on the OU graph. If you go back to the UNSW graph, it also shows the pressure being zero half way along L. But look at amplitude - the blue lines. This does not coincide with pressure, but shows (roughly) a 2:1 relationship between the lengths of the half and quarter cycles. Position of the 0 pressure point on the OU and UNSW graphs doesn't seem right, but may be explained by the pressure drop along the length of the cone. Not sure. As the angle is narrower than illustrated, maybe not.

I still don't get it. I still think, if the distance is proportional to wavelength, then to get from a distance appropriate to G to the corresponding distance for G#, he should divide by 1.05946. It doesn't matter how many wavelengths it is.

If the position of the of the point on the amplitude graph is right it makes sense to divide by 3 I think. He's already done the division by root 12, now needs to correct for the waveform in the cone. But I need to sleep on it again.

 

[jbtsax]

I started this topic thinking I would have time to discuss the in's and out's, but then I got hit with a wave of repair work and haven't been able to do any research or think through the responses. Nevertheless, let me try to bring the question back to its simplest form putting aside the issue of trying to figure out the placement of nodes and antinodes in a conical instrument.

To give everyone the same starting point, I suggest we use Frequencies of Musical Notes that is based upon speed of sound at 345 m/s. G1 on the alto sax corresponds to Bb3 on the keyboard. G2 on the alto corresponds to Bb4 on the keyboard.

Using this reference for frequencies and wavelengths, one can multiply any given frequency by 1.05946 and get the frequency of the note 1/2 step higher. One can also multiply any wavelength by 1.05946 and get the wavelength of the note 1/2 step higher.

As I understand it, the same would be true for the frequency and wavelength of each harmonic of any given note as well as the wavelength positions of nodes or antinodes. All bear the same square root of 2 relationship to its chromatic neighbor.
Dividing any frequency by 3 will produce the frequency of a note an octave and a 5th lower, whereas dividing any wavelength by 3 will produce the wavelength of a note an octave and a fifth higher.

Ferron's division by 3 makes no sense whatever.

 

[old git]

Thanks guys. All this wavelength stuff and resonance is helpful in my radio and internal combustion engine exhaust pipe studies, especially two strokes. If only Walter Kaaden had become a saxophone manufacturer, it would be a fight between his and Jim Schmidt's products.

 

[jbtsax]

You may be interested in Curt Altarac's motorbike. He plans to add keys to it to "tune" the exhaust or whatever.

 

[ProfJames]

Aprilia did this some time ago! Great sound!

 

[Steve Marshall]

Using this reference for frequencies and wavelengths, one can multiply any given frequency by 1.05946 and get the frequency of the note 1/2 step higher. One can also multiply any wavelength by 1.05946 and get the wavelength of the note 1/2 step higher.

As I understand it, the same would be true for the frequency and wavelength of each harmonic of any given note as well as the wavelength positions of nodes or antinodes. All bear the same square root of 2 relationship to its chromatic neighbor.
Dividing any frequency by 3 will produce the frequency of a note an octave and a 5th lower, whereas dividing any wavelength by 3 will produce the wavelength of a note an octave and a fifth higher.

Ferron's division by 3 makes no sense whatever.

I'm not convinced by dividing by 3. he seems to base things on the clarinet a lot and the diagrams on wave formations (p10 etc) are based on clarinet tubes. Harmonic series given on p14 is for stopped cylindrical tubes (clarinet), so I think he might have confused clarinet modes with saxophone ones.

 

[Morgan Fry]

My thinking is that the illustrations of "the sensitive harmonic sites" in alto and tenor necks in Ferron's book that have been widely distributed on the internet are incorrect because of the division by 3 which does not make any sense to me. Am I missing something here? Opposing thoughts and opinions are welcome.

I don't know whether you have missed someting or Ferron has. But it seems crazy to me to spend much time thinking about something so easy to test. Just take a neck you don't mind drilling two or three holes in and see which calculation is correct.

 

[aldevis]

I don't know whether you have missed something or Ferron has. But it seems crazy to me to spend much time thinking about something so easy to test. Just take a neck you don't mind drilling two or three holes in and see which calculation is correct.

And kill the conversation?
This reminds me of Achilles's tortoise and other similar paradoxes from the Sophists. The actual mathematical explanation arrived only about 2000 years later, when the concept of "limit" was introduced.
In the meantime the tortoise kept losing.