Another strangled flute

[Arch_Angel]

@ George - I may only just have begun to get a handle on this myself, but you are misunderstanding the physics involved.... and yes, a pint glass is also an "amazing structure" in the same sense - it is a cylinder or close thereto - as it tapers, just as an egg is an awkward shape, doing an effective snug or tight whipping may be tricky given a tendency for the thread to slip off - not a problem on a true cylinder... Crushing anything round is mechanically very difficult, especially if you apply pressure evenly all around the circumference. Do you understand how an arch bears a load? A ring or cylinder is like a never-ending arch, transferring and sharing any load around itself. Its strength is limited by properties of the material it is made from and its own dimensions (diameter, wall thickness) in relation to those.

Wouldn't the fact that the tenon is initially round qualify it as an "amazing structure"?
With the "arch load" distributing the tension across the load, instead of multiplying the tension by 2pi shouldn't it be divided by pi (or 2pi)?

On another thought it sounds like some people are saying stacked threads multiply the pressure. Wouldn't that only be added, if not only adding a factor of?

Can somebody get or build a gauge that can measure a internal pressure? I was thinking something like
(O) where the () represent pressure plates and the O is the spring gauge (bar/psi).

[talasiga]

What I cannot wrap my own head around is how anyone could read the physics report as provided by Terry and still maintain that it's not possible to apply thread in such a way as to exert a crushing force.

.......

I don't have the technical know how to dispute the facts of thread pressure and I am not doing that. I accept Terry's credentials in terms of practical experience and the info he provides. However,I do have a brain and and I cannot find it in my pants or anywhere else so I presume it's still behind my eyes and therefore I THINK.

What I THINK is that one cannot conclusively say its the thread pressure ALONE (whatever it may be) that damages the wood unless one can show that happening on an untenoned wooden tube as a "control".

If the thing only happens with tenon in socket wood then the question of concomitant causes arises or an alternative exclusive cause to the thread pressure arises.

[Terry McGee]

Actually, this may be of assistance. It requires a confession first!

I've been playing a mopane flute for a few years now, a 6 key of my Rudall 5088 model. You're probably aware that I, like most makers I'm sure, recommend always pulling the flute apart after playing. But, in my work, it's handy to have a flute always available, so the 5088 stands beside my computer monitor, fully assembled and always at the ready. I use a heavy desk-top microphone stand, with the vertical tube replaced by a wooden dowel, turned down a little to fit in the foot bore. The tenons are corked in my usual way. I suddenly realised it might have something to tell us in terms of whether cork-lapped flutes suffer any signs of strangulation. I don't know why I didn't think of it earlier.

So, I just pulled it apart and went hunting with the telescoping gauge for any sign of strangulation. None whatsoever, in any of the three tenons. In all three, the taper remains smoothly negative throughout the tenon area, whereas the strangled flute and the Camp flute both have easily found negative minima in the centre of their three tenons, the strangled flute being much more pronounced.

I ran the 5088 reamer through the LH joint - it took a tiny shaving off the lower end in the area of the tenon, but nothing at the top.

The joints on the 5088 are pretty tight (indeed, I had to grunt a bit getting the top one off, as I hadn't greased it lately), so I think we can confidently say that the pressure exerted through cork by the socket is not enough to cause problems. Indeed, as Casey pointed out, we know the opposite, that if overpacked with cork, it will be the socket that fails first.

I think, by inference, that the same will apply to threaded tenons - the thread is capable of much more force or resistance than the socket.

I just tried a few more old flutes with threaded tenons, and all but one tenon on one flute showed the "local minimum". I had previously used the existence of a local minimum to define strangulation, but, if we do, most old threaded flutes are strangled. So, perhaps we need to tighten up our descriptors? Possibilities:

Bore compression - when externally applied force (eg tight lapping) compresses some section of the bore to a diameter lower than when originally made? (Incurs the difficulty of knowing what was there originally!)

Severe bore compression - when excessive externally applied force compresses a section of bore to the extent that a section of taper is cancelled? (at least easy to test for)

Strangulation - when sufficient externally applied force causes collapse of the flute wall in a section of the bore ....
- reducing the bore diameter by <insert % here>?
- actually reversing the intended direction of taper in a section of bore?
- sufficient to cause visible narrowing of the bottom of the thread trough?
- sufficient to cause noticeable tuning issues (define noticeable!)
- other suggested criteria?

It would also be good to come up with measurement technique that would enable any concerned owners to check for signs of bore compression. A cheap 3/4" telescoping gauge is all I can think of, but maybe someone has a better idea?

Terry

[Jim_B1]

It would also be good to come up with measurement technique that would enable any concerned owners to check for signs of bore compression. A cheap 3/4" telescoping gauge is all I can think of, but maybe someone has a better idea?
Terry

How about a bore gauge like these:

http://www.plasticservices.com/techtips/bgauge.htm

Another question that comes to mind on this. Since it seems that it is older flutes that the compression is occurring on, I have to wonder if there are not cracks in the tenon due to the natural cycle of expansion and contraction that slide against each other even if they can't open all the way because of the string. They would still weaken the tenon and possibly shift the tension of the string to create pressure points instead of uniform pressure.

Just a thought...

[George]

The force is 6 x the tension exerted by one whole wrap - so the pressure is that force divided by the contact area of that length of thread (one wrap). If there multiple layers the forces will add - multiple windings (not layered) the pressure remains constant.

Okay please correct me where I'm going wrong -

For starters lets clear this up - 700g is way, way, way too much force and no one uses that much by hand.

Don't believe me? Wrap a single strand of sewing thread around something that weighs close to 700g and try to pick it up with just your thumb and index finger. I tried to pick up a can that weighed 425g and literally couldn't using just my thumb and index finger, not nearly enough friction. The string was close to breaking under it's weight and there's no way you could do hundreds of wraps without wrecking your arm/fingers. Next I tied a single thread around my Razr cell phone and did the same thing, it felt much closer to how much tension I'd use ... it weighs all of 95g http://en.wikipedia.org/wiki/Motorola_RAZR much like the 100g for the apple's weight in Highwood's previous example. Does anyone think they use much more than that by hand?

So the force being applied is about 6x the tension of the thread, I think a more realistic starting tension is around 100g

100g * 6 = 600g

Terry's wrap went 13 deep

600g * 13 = 7,800g (17.19lb)

Please help me out with this bit

Pressure is the force divided by the contact area. I used this website to calculate it http://www.calculatorsoup.com/calculato ... linder.php where in Terry's example the radius is 1cm and the length of the wrap on my flute is about 7/8in or 2.22cm for the upper tenon … so that gave an area of 13.94 Square Centimeters

7,800g / 13.94 sq cm = 559.54g per square cm
or
17.19lb / 2.16 sq in = 7.95psi

Okay, did I make a mess of all that?
(I'm leaving off everything past two decimal points, so "=" means "approximately")

Highwood, in your example the 2,000 wraps on an apple, they were one directly on top of the other right? That's how you got the 2,500 pounds of force mark?

Say you tripled the initial string tension ... The end result is still no where near the hundreds if not thousands of pounds that have been floating around as a figure.

300g * 6 = 1,800g
1,800g * 13 = 23,400g
23,400g / 13.94 sq cm = 1,678.62 g per sq cm
or
51.58lb / 2.16 sq in = 23.87psi

I don't know how to really put these numbers into perspective, but -

Their lightest model is 100lbs and marketed for beginners/warm-ups/youth athletes
http://www.heavygrips.com/
Here it lists 88lb as a very poor result for a man's handgrip strength
http://www.topendsports.com/testing/tests/handgrip.htm

[Rob Sharer]

Soldier #1: Where'd you get the coconuts?
Arthur: We found them.
Soldier #1: Found them? In Mercia? The coconut's tropical!
Arthur: What do you mean?
Soldier #1: Well, this is a temperate zone.
Arthur: The swallow may fly south with the sun or the house martin or the plover may seek warmer climes in winter, yet these are not strangers to our land?
Soldier #1: Are you suggesting coconuts migrate?
Arthur: Not at all. They could be carried.
Soldier #1: What? A swallow carrying a coconut?
Arthur: It could grip it by the husk!
Soldier #1: It's not a question of where he grips it! It's a simple question of weight ratios! A five ounce bird could not carry a one pound coconut.
Arthur: Well, it doesn't matter. Will you go and tell your master that Arthur from the Court of Camelot is here?
Soldier #1: Listen. In order to maintain air-speed velocity, a swallow needs to beat its wings forty-three times every second, right?
Arthur: Please!
Soldier #1: Am I right?
Arthur: I'm not interested!
Soldier #2: It could be carried by an African swallow!
Soldier #1: Oh, yeah, an African swallow maybe, but not a European swallow. That's my point.
Soldier #2: Oh, yeah, I agree with that.
Arthur: Will you ask your master if he wants to join my court at Camelot?!
Soldier #1: But then of course a-- African swallows are non-migratory.
Soldier #2: Oh, yeah...
Soldier #1: So, they couldn't bring a coconut back anyway...
[clop clop clop]
Soldier #2: Wait a minute! Supposing two swallows carried it together?
Soldier #1: No, they'd have to have it on a line.
Soldier #2: Well, simple! They'd just use a strand of creeper!
Soldier #1: What, held under the dorsal guiding feathers?
Soldier #2: Well, why not?

[Denny]

[david_h]

per turn, not per layer

[highwood]

I was not wrapping an apple with thread...

I did calculations with a tension equal to the weight of an apple - 100g weight or 1 newton

The apple is a fairly common object with which many are familiar and hence have 'a feel' for its weight, as opposed to 100g or more correctly 1N

The force is per wrap...

According to Terry a ball park number is about 2000 wraps...

Disclaimer:
I have never wrapped a flute tenon - my oboe has cork, and the whistles I make use brass tube or turned plastic.

I have wrapped fishing rods, tool handles and various other random things, including Harp strings

[George]

The force is per wrap...

According to Terry a ball park number is about 2000 wraps...

Okay, so you're saying it's each individual loop that needs to be added together?

So then after the first layer of 150 loops it's 6*Tension*150 as the force being applied?
100g*6=600
600*150 = 90,000g (198lb)
90,000g / 13.94 sq cm = 6,474g sq cm
198lb / 2.16 sq in = 91.6psi (like three times the psi of my front tire)

You're saying it's that much force/pressure after the first layer?

The force is 6 x the tension exerted by one whole wrap - so the pressure is that force divided by the contact area of that length of thread (one wrap). If there multiple layers the forces will add - multiple windings (not layered) the pressure remains constant.

[david_h]

Try putting a few turns on a bicycle inner tube then pump it and check the pressure when you think some damage is about to be done.

That, by the way, is a way of working towards a way of measuring pressures -put a 'bladder' inside the tenon - but you will need a lot more that a car foot pump if you go for a full set of turns.

[George]

Now if there are about 2000 wraps (150 x 13 = 1950 but who's counting) that is 12000 N of force applied inwards which converts to over 2500 lbs or well over 1 ton.

The resulting pressure would be easy to figure out and is left as an exercise for the reader...

Okay, I'm just trying to get a hold on things here ...

Using your method from the apple example -

Tension applied on the string is multiplied by 2pi (about 6)

100g * 6 = 600g

2,000 wraps * 600 = 1,200,000g or 2,645.54lb of force

Pressure is the force divided by the area over which it's applied, flute tenon is about 13.94 sq cm

1,200,000g of force / 13.94 sq cm = 86,083g per square cm

or

2,645lb / 2.16 sq in = 1,224 psi

That's the same psi as being 2,723ft under water or 829m below the sea

http://www.calctool.org/CALC/other/games/depth_press

Am I wrong to suggest that if you believe a thread wrap is putting 2,645lb of force over a tenon which has an area of about 2.16 sq in and has the same psi as being 2,723ft under water that maybe you should start building budget submarines out of flute tenons?

The numbers you're suggesting just seem so outrageous. That's why I asked if you meant 2,000 loops directly on top of one another.

[Rob Sharer]

The natural world is a amazin' thing.


Rob

[david_h]

Review the turn count ? But would 200 turns change the astonishment factor significantly ?

(ask jem what length of thread he normally uses)

[jemtheflute]

I don't doubt the basic formulae, but I suspect there are still aspects of their contextually correct application we have not yet worked out/dealt with/had adequately clearly explained (in clear language for maths dunces like me) - and I also somewhat doubt the point of going any further until we have some real (measured) rather than speculative applied tension figures. I think all these numbers getting bandied about are more confusing than helpful at this stage.

Responding to david-h, I don't know off-hand how much thread I would use on average for the largest tenon. In any case, so much depends on how deep the lapping bed (trough) is and the size of the socket in relation to the tenon. Well made older flutes often don't need more than 2-3 full layers (of the thread I use) after filling the combings (usually one or maybe two layers just in the combings). However, less well made old flutes, also those that may have lost their (more-or-less) original dimensions for whatever reason and some modern ones (like e.g. the Glenluce type subcontinental ones, relatively crudely made) can need quite a few more layers.

Here's another angle: the skeins of (mercerised cotton) embroidery thread I use (heavily cork-greased during application) are nominally 8m in length, consisting of six medium spun strands, each of which is two-ply. Those two ply strands are probably close to the same gauge as sewing thread, maybe a little thinner, and certainly looser spun, and the two strands composing each of them are also not particularly tight spun - less so than sewing thread. I always split the six-strand skein into two groups of three strands, thus producing a quite loosely spun working thread that is roughly twice as thick as ordinary sewing thread, but much more"squashable" because, being loosely spun, the strands can open up: the breaking strain will also be much less (than a similar gauge but harder spun thread of the same material) because of the loose spin - fibres can more easily pull apart longitudinally in the thread under strain because they are less locked in by the twist. It is a rare (and very loose-in-its-socket and/or thick-walled and deeply troughed) tenon that will use more than 6-7m of this thread, if that much. If I am starting with a full length hank of my reduced thread, I would expect to have several winds around my palm left from an average-fit upper body upper tenon, the largest one.

Now, here's a conundrum for our maths/physics bods: I am using a triple thread (we'll ignore that each of those is 2-ply!). I apply tension to it as a single thread, but that tension will be roughly equal in each of the three strands. If I understand things correctly thus far, it doesn't make any difference to the number-crunching because the numbers representing the three component strands' tension, area covered etc. are equivalent to those for the triple, composite thread - the tension figure does not multiply by three - the strands have the same tension as the thread they compose but cover 1/3 of the area. Right?