kevgermany
ex Landrover Nut
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I think so, haven't got time to check
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)?
I am fine with the wavelength. But I am not sure where the second harmonic antinode is in a conical 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!
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.
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.
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.
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 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.
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.
np, this is interesting.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.]
yes (466.16 really )Do you agree with me that a second register G on an alto is a concert A# at 466 Hz?
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.
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.
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.
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 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.