Tech/maintenance Saxophone Pad Acoustics Study

[aldevis]

... which seems to imply pressure variation of +/- 3% atm, (the article later describes a rarefaction wave, which presumably is of the order of 3% lesser pressure than the surrounding air)

That would be 30 mb
Almost 100 g on the aforementioned pad.

Far too much to be correct

 

[jbtsax]

Just came across this ... Clarinet - Wikipedia, the free encyclopedia

There's an interesting explanation of wave motion in the clarinet. I did also note this point particularly.

"The ‘puff of air’ or compression wave (around 3% greater pressure than the surrounding air) travels down the cylindrical tube and escapes at the point where the tube opens out. "

... which seems to imply pressure variation of +/- 3% atm, (the article later describes a rarefaction wave, which presumably is of the order of 3% lesser pressure than the surrounding air)

This is incorrect. Whoever wrote this misunderstood or misinterpreted the source that was cited for this information. The source cited actually reads:

The clarinet player provides a flow of air at a pressure above that of the atmosphere (technically, about 3 kPa or 3% of an atmosphere: applied to a water manometer, this pressure would support about a 30 cm height difference). This is the source of power input to the instrument, but it is a source of continuous rather than vibratory power.

 

[saxplorer]

This is incorrect. Whoever wrote this misunderstood or misinterpreted the source that was cited for this information. The source cited actually reads:

Ah, I think I understand: the blowing pressure might be 3% atm, but that doesn't mean at all, that the compression wave will have that pressure.... right?

 

[jbtsax]

Here is an update to the thread showing the data from the various tests conducted. The first image is the "porosity" measurements of the various pads tested. The Magnehelic measures "differential pressure" between a perfectly air tight inner chamber and the instrument being measured. A zero differential is a perfect score indicating no pressure is lost in the object being measured. A score of .2 means that there is a constant .2in. H2O pressure loss at the input pressure indicated. The input pressures were 2in. H2O and 4in. H2O. It is also important to remember that the maximum measured pressure inside an alto saxophone is 130 dB which converts to just .25in. H2O.

 

[jbtsax]

These are the results of the "acoustic absorption coefficients" tests conducted by Dr. Pauline Eveno. It came as a complete surprise to both of us how most of the pads were "all over the map"---especially at the higher frequencies. At this point I am not sure any conclusions can be drawn from the data except that more acoustic tests need to be conducted in this area. One can note on the graph that the "winners" of the porosity tests, those pads that tested zero at both pressures also did the best in Dr. Eveno's test of absorption. It is possible that the "hard" more reflective surface of those pads contributed to their low absorption readings along with their non-porosity. The "cutoff frequency" marked on the chart is the frequency above which the harmonics go straight out the bell and do not cooperate with the fundamental and lower harmonics in sharing energy. This raises the question that if the frequencies above "cutoff" don't "see the open toneholes" [Benade's venacular] then what effect do the pads in the closed toneholes have upon their energy?

 

[aldevis]

How was the sound generated?

The "cutoff frequency" marked on the chart is the frequency above which the harmonics go straight out the bell and do not cooperate with the fundamental and lower harmonics in sharing energy. This raises the question that if the frequencies above "cutoff" don't "see the open toneholes" [Benade's venacular] then what effect do the pads in the closed toneholes have upon their energy?

Is it possible that the concept of "cutoff frequency" needs to be revised?

 

[jbtsax]

How was the sound generated?

I'll need to review the description of the apparatus used in Dr. Eveno's resonator study and get back to you on that.

Is it possible that the concept of "cutoff frequency" needs to be revised?

In a word no. It is a well established principle of woodwind acoustics. You can read more about it here.

 

[kevgermany]

A few things strike me.

The acoustic losses are mostly going to affect the sound generating part of the sax - i.e. from mouthpiece to first open tone hole.

Cut off frequency is above the fundamental frequency of most of the notes we use.

There's work to be done to establish why absorption varies with frequency, but it looks from the graphs as if one could derive a formula for each pad type. From the graph shape, there's a lot of consistency over the first couple of peaks, but the interval differences confuse things.

Assuming that reflectors work by cutting absorption losses, there's confirmation that they'll significantly affect the sound, and that the bigger the reflector, the more effect.

It goes towards explaining why flutes have a high, clear sound. Every tone hole has, in sax terms, a large reflector.


So if we want to move towards the big lush sound of the earlier recordings - say 70/80 years ago - we need nice absorptive pads and no reflectors. For a purer, more classical french horn tone we need big reflectors.

 

[peterpick]

that's a very interesting deduction, kev. so in general this study means that roo pads would give a less 'bright' sound? i wonder if any of us could tell in a blind test? i must say i have never noticed the difference even with reflectors / no reflectors on the pads in a given sax, but then there are so many other variables.....

 

[Dibbs]

Assuming less absorption and less leakage is good then the gold pads appear to win hands down. Who'd have thought? I'd imagined it was all snake oil.

It might be interesting to see the absorption coefficient curve for a piece of brass on that graph.

Do you have any idea why they all have a well defined peak around 500Hz then most have a less well defined one around 1KHz? If I'd made those measurements that would be bothering me.

 

[aldevis]

Do you have any idea why they all have a well defined peak around 500Hz then most have a less well defined one around 1KHz? If I'd made those measurements that would be bothering me.

If the test has been done playing a Bb....
Did I win?

 

[Morgan Fry]

Very interesting,John. So there seems top be a correlation between porosity and acoustic dampening. I think frequencies above the cutoff may matter here, but I'm not sure. The Y axis is absorption coefficient, how does that affect the

These are the results of the "acoustic absorption coefficients" tests conducted by Dr. Pauline Eveno. It came as a complete surprise to both of us how most of the pads were "all over the map"---especially at the higher frequencies. At this point I am not sure any conclusions can be drawn from the data except that more acoustic tests need to be conducted in this area. One can note on the graph that the "winners" of the porosity tests, those pads that tested zero at both pressures also did the best in Dr. Eveno's test of absorption. It is possible that the "hard" more reflective surface of those pads contributed to their low absorption readings along with their non-porosity. The "cutoff frequency" marked on the chart is the frequency above which the harmonics go straight out the bell and do not cooperate with the fundamental and lower harmonics in sharing energy. This raises the question that if the frequencies above "cutoff" don't "see the open toneholes" [Benade's venacular] then what effect do the pads in the closed toneholes have upon their energy?

Very interesting, thanks for posting this. It appears that there is some correlation between leakage and acoustic absorption. Did these all have the same reflectors in? It would be useful IMO to scatter plot Absorption at the cutoff frequency against leakage so you can quantify the correlation. I think we can safely ignore any effects above the cutoff frequency.

 

[Dibbs]

If the test has been done playing a Bb....
Did I win?

Wouldn't you use a sine wave sweeping through all frequencies of interest? I'm thinking maybe a room resonance around 750 causing apparent peaks either side or something.

 

[aldevis]

Wouldn't you use a sine wave sweeping through all frequencies of interest? I'm thinking room resonances or the like.

I wouldn't. I would use the richest wave.
Also I would see if possible to reduce many notes to the same "imaginary" fundamental, mathematically.
Waiting to know about the tone generator, but many thoughts are emerging.

 

[abadcliche]

This is not surprising to me. It matches with my experience, though people have been telling me for years I'm crazy. You can even change the response of a horn just by coating the pads with a layer of lacquer! Pad porosity is a leak. You might like the "darkening" of the sound that comes as a result (though personally I'd call it "deadening") and that's your choice, but it IS a leak.

As far as why frequencies above the cutoff are also responding, I would assume its because they are harmonics and the fundamental that created them is being drained of energy. Cut the bottom 2 inches off a ladder, the top rung is also 2 inches shorter.

I wonder about the pressure conversion from decibels to inches of water. If a low C# can be vibrated open easily on almost any leak-free horn, then the pressure must be enough to do so, and I'd wager that is more than the weight of a 1/4" column of water the size of the tonehole. And its easier to vibrate open a key with a subtone, which isn't as loud, so perhaps decibels aren't the do-all measurement for what pads undergo during playing.

 

[abadcliche]

...
So if we want to move towards the big lush sound of the earlier recordings - say 70/80 years ago - we need nice absorptive pads and no reflectors. For a purer, more classical french horn tone we need big reflectors.

Well, the graph above shows a *decrease* in the richness of the tone the more absorptive the pads got.

Old saxes typically came out of the factory with pads that had been treated- most had been sprayed with lacquer, some even had metal foil behind the pad skins, and by and large the skins were not very porous (even after all these years, especially ones that had a lacquer coating). You can take an original pads 10M (or VI, or whatever) and put new pads with large resonators in them and find that the original setup had a fuller sound, while the large resos mostly turn up the volume of the higher partials. The lush sound of older saxes I think comes primarily from the ideals of the player, mouthpiece selection at the time, then from the design of the horn, THEN from resonator choice. I think the result of taking a modern mouthpiece on a modern horn and put no resos and porous pads in it would be ending up with something that just plays dead. Conversely, you could take a Buescher True Tone with non-porous pads and the stock snap-on resonators (which are quite small) and put a S80 C* on there and sound pretty modern. If you want a full, lush tone, use a mouthpiece with more internal volume and less baffle, have your horn be leak-free, and don't go nuts with oversized/overdomed resonators. Probably an older horn would help too, but that part is not essential.

Unless you want to sound like you are playing from behind a closed door, you don't want porous pads.

 

[abadcliche]

And cool study, JBT. Thanks for sharing. Maybe people will start believing that pad porosity matters now.

 

[aldevis]

As far as why frequencies above the cutoff are also responding, I would assume its because they are harmonics and the fundamental that created them is being drained of energy. Cut the bottom 2 inches off a ladder, the top rung is also 2 inches shorter.

This sums up what I had in mind.

When we talk of "higher partials" we are talking about analytic composition of a complex wave.
What is happening above the cutoff frequency is actually the effect of the "shape" of the fundamental wave.
Personally is where the actual "tone" gets defined.

So pads seem to affect a wave shape (= high partials) more than I expected.

It would be useful IMO to scatter plot Absorption at the cutoff frequency against leakage so you can quantify the correlation.

+1

 

[abadcliche]

Personally is where the actual "tone" gets defined.

So pads seem to affect a wave shape (= high partials) more than I expected.

Yes! The graph stops around 3000hz, which is really not all that high. What we are looking at in that graph is a clear change across the main tonal spectrum of the saxophone.

This is 3000hz:

View: https://www.youtube.com/watch?v=I-QPhMIv4pM

Major changes start to be evident around 1500hz in the graph- here is 1500hz:

View: https://www.youtube.com/watch?v=NIbwpFosXxw

 

[jbtsax]

Assuming that reflectors work by cutting absorption losses, there's confirmation that they'll significantly affect the sound, and that the bigger the reflector, the more effect.
It goes towards explaining why flutes have a high, clear sound. Every tone hole has, in sax terms, a large reflector.

Except for open hole flutes where each open hole has a fat spongy finger. 🙂

Assuming less absorption and less leakage is good then the gold pads appear to win hands down. Who'd have thought? I'd imagined it was all snake oil.

Correct, but there is lots of anecdotal evidence that saxophones padded with gold pads sound too "harsh" for most players.

It appears that there is some correlation between leakage and acoustic absorption. Did these all have the same reflectors in? It would be useful IMO to scatter plot Absorption at the cutoff frequency against leakage so you can quantify the correlation. I think we can safely ignore any effects above the cutoff frequency.

At the higher frequencies there is definitely a correlation. If you look at the beginning of this thread, all pads used had no reflector (resonator).

Wouldn't you use a sine wave sweeping through all frequencies of interest? I'm thinking maybe a room resonance around 750 causing apparent peaks either side or something.

I believe you are correct that a sine wave sweeping through all the frequencies was used. Actually the measurement was of the "acoustic impedance" of a brass tube with a loudspeaker on one end and the pad at the other. The "acoustic absorption ratio" is then deduced from the impedance measurements as far as I can tell.