Math Geek Acoustics Anyone?

[jbtsax]

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:

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. An example using G with an alto neck, shows:

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]

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

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.

 

[altissimo]

did he perhaps get mixed up with clarinets which overblow at the 12th? I'd have to go away and study the acoustics of tapered tubes for a few weeks and brush up on my maths before I could offer a better explanation - the only way would be to start at first principles and work through the maths independently and see if you end up in the same place as him.
To me the odd bit is that he seems to be calculating the nodal points of the neck alone and not as attached to the instrument - why?

 

[Dave McLaughlin]

I suspect the answer lies in the fact that the sax is conical, so that the fluctuating pressure varies as sin(x)/x, where x is the distance from the apex of the cone. If you look at the first graph under the conical pipe at http://www.phys.unsw.edu.au/jw/pipes.html, you'll see that the "motion" antinode occurs at 0.75 * wavelength, instead of the 0.5 * wavelength you'd expect.

But I'll have to ponder this further before I'm clear in my own mind!

 

[Steve Marshall]

Ferron doesn't explain things very well. It'd help if he said whether the numbers were frequency, wavelength or something else.
It still doesn't make sense to me. The initial length is the length of a harmonic rather than the fundamental so I'm not sure he's got it right.

 

[kevgermany]

Finally I had a chance to look this up and think. I did see this when I read the book and thought it strange. What I'd fogotten when I discussed with JBTsax in pm, but remembered on re-reading the passage is as follows:

Here he's talking about the volume of the cone from a point (the part replaced by the mouthpiece). And so you need to read this in the light of what he said on pages 100-102.

Remember that here he's talking about matching octaves - where you need to balance the length, with the mouthpiece volume representing the missing part of the cone the neck forms. The part you're quoting is not about tube length, but cone volume and it's effect on pitch.

 

[altissimo]

Vcone = 1/3 × pi × r2 × h

 

[Dave McLaughlin]

Hmm - I can see I'm going to have to buy this Ferron if I'm to have any chance of following this discussion!

 

[kevgermany]

Hmm - I can see I'm going to have to buy this Ferron if I'm to have any chance of following this discussion!

Hard to find, but musicmedic stock it.

 

[Dave McLaughlin]

Hard to find, but musicmedic stock it.

Thanks. Last time I went looking for it I was surprised not to find it on abebooks. I think I did eventually find it somewhere but didn't get round to buying it then.

 

[Steve Marshall]

Vcone = 1/3 × pi × r2 × h

If he's using this then why didn't he use the pi and the r2?

It seems to me he is trying to find the site of an octave pip by locating the appropriate length of a particular frequency. Then, because it is a harmonic the difference between a G and G# needs to be split over the nodes. This might work but he seems ot start in the upper register, so I'm not sure it's right.

 

[kevgermany]

He is in the upper register, otherwise the nodes would be a lot further down the sax

 

[aldevis]

Vcone = 1/3 × pi × r2 × h

In my approximate knowledge, this makes sense. If we think of strings or cylindric tubes, it should be /2 (constant section) but in a cone /3 seems to make sense.
I am not sure why clarinets skip to 12ths, though: still /3 but on the next harmonic.

 

[aldevis]

If he's using this then why didn't he use the pi and the r2?

Good point.

 

[saxplorer]

Vcone = 1/3 × pi × r2 × h

eh forgive the swerve off-topic, but I can't resist it:

A pizza base has thickness a and radius z: what is the volume of dough in the base?

 

[aldevis]

eh forgive the swerve off-topic, but I can't resist it:

A pizza base has thickness a and radius z: what is the volume of dough in the base?

You can't call it a pizza (π2za) if it has a constant thickness in the base.

Maybe "pizza pie" like in the States.

 

[saxplorer]

You can't call it a pizza (π2za) if it has a constant thickness in the base.

Is average thickness ok?

 

[kevgermany]

eh forgive the swerve off-topic, but I can't resist it:

A pizza base has thickness a and radius z: what is the volume of dough in the base?

pi.z2.a before cooking, but then it's not pizza
0, once it's cooked.

Now lets get back on topic.

 

[Dave McLaughlin]

I am not sure why clarinets skip to 12ths, though: still /3 but on the next harmonic.

A flute is essentially a cylinder, open at both ends, so it resonates when its length is a multiple of a half wavelength. Going from 1 half-wavelength to 2 goes up an octave.

A clarinet is a cylinder, open at one end and (almost) closed at the other, so it resonates when it is an odd number of quarter wavelengths. That means it sounds an octave lower than a flute of the same length. It also means that it jumps from 1 to 3 quarter-wavelengths - hence a twelfth.

A saxophone is like a clarinet in that it's open at one end and more or less closed at the other. But the fact that it's conical adds another quarter-wavelength at the pointy end, so it overblows at the octave, like a flute.

 

[Steve Marshall]

I am not sure why clarinets skip to 12ths, though: still /3 but on the next harmonic.

A clarinet can only support odd harmonics so it can't go through the usual harmonic series and when it over blows goes straight to the third mode.

[Edit; Sorry, missed Dave's post (above) where he explains this.]

 

[Steve Marshall]

He is in the upper register, otherwise the nodes would be a lot further down the sax

Ah, is this the answer?The second mode has λ2 = 4L/3, so that f2 = v/l2 = v/(4L/3) = 3v/4L or three times its fundamental frequency.