Another strangled flute

[Denny]

In the strangled flute example, the tenon was 15mm long, and about 20mm in diameter. So the area would be PI x r2 x L = 3.1412 x 1 x 1.5 = 4.7 sq cm

I'm don't remember where I grabbed the 14cm number, copied the wrong number.

But you solved for volume not for lateral area which would be 4.7cm3 not cm2

Punch in 1.5cm height and 1.0cm for radius here -
http://www.calculatorsoup.com/calculato ... linder.php

It's 9.42cm2

Terry just halved his diameter for the radius and lost the 2 in the 2nd step, it should be
PI x r2 x L = 3.1412 x ( 1 x 2 ) x 1.5 = 9.42

[George]

In the strangled flute example, the tenon was 15mm long, and about 20mm in diameter. So the area would be PI x r2 x L = 3.1412 x 1 x 1.5 = 4.7 sq cm

I'm don't remember where I grabbed the 14cm number, copied the wrong number.

But you solved for volume not for lateral area which would be 4.7cm3 not cm2

Punch in 1.5cm height and 1.0cm for radius here -
http://www.calculatorsoup.com/calculato ... linder.php

It's 9.42cm2

Terry just halved his diameter for the radius and lost the 2 in the 2nd step, it should be
PI x r2 x L = 3.1412 x ( 1 x 2 ) x 1.5 = 9.42

He used the formula for volume and came up with the volume 4.7cm3

From the website -

Calculate volume of a cylinder:
V = π r2 h
Calculate the lateral surface area of a cylinder (just the curved outside):
L = 2 π r h

[Terry McGee]

Terry, I know I am still confused by this - any chance of asking your physics guru to look at this thread and contribute (first hand) a full, exampled explanation/working, rather than just the bald formula? I sympathise with George to the extent that I am not convinced anyone has yet completely, demonstrably correctly, fully and convincingly interpreted and used it, at least, not so as us Maths/Physics dunces can properly and clearly understand.

I'll see what I can do, Jem. I don't like bothering Neville more than I have to, for obvious reasons.

But, BUT, PLEASE, can the professional exemplar and exposition be done with a realistic estimate of working thread tension, NOT with the breaking strain?

Then we argue what is a realistic estimate! The good thing about the breaking strain is that it is definable. It also has relevance in my 'immoveable object' theory. But, it doesn't really matter. Since the force applied on the tenon will be directly proportional to the tension in the thread, I can start with breaking strain, and anyone can scale it down to what they think is reasonable.

If he could also be bothered to examine each hitherto offered interpretation/calculation and explain why each is right or wrong, that would be a bonus, but maybe a big ask. At present we just have contrary accusations that opposing parties are "wrong" (and implications of wrong-headedness), unelucidated by simply telling each other there is error/misconception without sufficient or final explanation/demonstration of that alleged error; and no referee! Simply referring back to the formula, over which there is clearly a conflict of conceptualisation (which I at least cannot reach a conclusion about) does not achieve a resolution. That is just like the bad, old-fashioned Maths & Science teaching in schools that used to (maybe still does) put so many otherwise intelligent folk off the subjects!

That would definitely be too big of an ask. Although Neville has well passed retirement age, he maintains an office at the ANU and continues his work in a number of areas of physics. But hopefully, if I can clarify what he means, perhaps with a fully worked example, that will suffice.

Terry

[Terry McGee]

Curiouser and curiouser, cried Alice ...

I've just logged the results of the 2nd bout of humidifying. Trends are the same. But I've noticed something else. Not only is the bore at the middle of the tenon continuing to compress, but the bore and outside diameter at the ends of the tenon are growing. We expect them to swell during a humid cycle, but they don't seem to be fully recovering during a drying cycle. As I mentioned previously, this might just be because I'm not giving them a long enough drying cycle - I'll test for that in the drying cycle we've just entered (I'll extend it and take enough measurements to ensure we reach equilibrium). But it definitely looks like a trend.

Now what makes that interesting is that the three tenons on the strangled flute, now all without wrapping, are all too big at the tip to enter their respective sockets without rubbing. This puzzled me before - why would the sockets have got smaller or the tips of the tenons bigger? I don't normally see that in old English flutes (this is Australia, after all, and even in a La Nina weather pattern, it isn't anywhere near as damp as London!). But fascinating that the In Extremis tenon is showing the same distortion pattern.

Curiouser and curiouser!

Terry

[George]

You don't need to make a tenon to see that there's not massive pressure being created.

Take 5min and wrap some thread around your finger.



According to Terry's method of adding up force this is putting 185lb of total force with 32.74psi

I can tell you first hand that's not the case, it's more like a light squeeze.

Diameter ~ 2.25in
Radius ~ 1.125in
Length ~ 0.8in

Surface Area ~ 5.65in2

140 Loops
~100g

600g*140 = 84,000g
185lb
32.74psi

As to the number of wraps not stacked multiplying the force.

Look, push like this -



Then like this with the same force -

Same total force just over a larger area, it's not multiplying

[Terry McGee]

Um, increasing the area of the pinch will keep the same force but reduce the pressure. To emulate what I'm talking about, you're going to need to find 149 other people to pinch you at the same time. And, indeed, 13 lots of them.

Terry

[Terry McGee]

It's getting harder to talk about the tenon in extremis test without an image. So, I've created a short web page where the data will be displayed as it arrives. I haven't linked that to my index page yet, so the only way of getting there is via:

http://www.mcgee-flutes.com/Effects_of_ ... apping.htm

You'll need to go there for a full explanation of what these curves tell us, but maybe that's enough for some:



Terry

[George]

Um, increasing the area of the pinch will keep the same force but reduce the pressure.

Yes that's the point!

Why do you maintain that adding loops to the length of a thread wrap is any different than increasing the area of the pinch?

They're both applying a constant force by tension over an increasing area, in your mind one multiplies and the other doesn't.

To emulate what I'm talking about, you're going to need to find 149 other people to pinch you at the same time. And, indeed, 13 lots of them.

It doesn't matter how many people are contributing if the force is the same, each person's fingers would just be increasing the area over which that same force is being applied.

[Terry McGee]

It doesn't matter how many people are contributing if the force is the same, each person's fingers would just be increasing the area over which that same force is being applied.

My point is that it would hurt more!

Terry

[highwood]

Why do you maintain that adding loops to the length of a thread wrap is any different than increasing the area of the pinch?

Because it is.

[Casey Burns]

So Terry, will you try this also with a corked tenon inside a socket? Also, do you have a test piece without thread wrap made from the same boxwood, to see if the wood is moving on its own?

Also, why not test with a thicker piece. Maybe what we see in the historical instruments with this type of compression is simply a factor of wall thickness of the tenon. I for one would never make my tenons as thin as your test piece. It would be interesting to see what thickness is the threshold for this.

Casey

[George]

My point is that it would hurt more!

No your point was that it'd have less pressure, the opposite of hurting more. Look you just said it -

Um, increasing the area of the pinch will keep the same force but reduce the pressure.

*Same force* with a higher pressure hurts more, with a lower pressure hurts less.

(I have taught high school physics...)

you just do not understand.
... and now

Because it is.

Instead of telling me I don't understand or "because I said so" answers, explain it to me like you would to one of your high school students. I dunno like you actually want me to understand what you're saying instead of telling me how little I know.

You & Terry are saying -

1. As a thread wrap increases it's area while maintaining the same force, it's actually increasing it's total force and pressure?
2. But that's not the case with the pinch example or the tube clamp example and goes against the P=F/A

[Terry McGee]

So Terry, will you try this also with a corked tenon inside a socket?

No plans to. I haven't seen any sign of bore compression with corked flutes. But if I were to find such a sign, I'd probably want to investigate it. I still have in mind that odd phenomenon of some old corked flutes that have cork too thick to enter the sockets safely. But I can't see that this relates to this issue. Indeed, I'm not sure what to make of that one!

Also, do you have a test piece without thread wrap made from the same boxwood, to see if the wood is moving on its own?

That would be a good thing to include in a more careful test. The "in extremis" test is a quick test intended to find out if there is an issue, before putting any real effort into designing serious tests.

Also, why not test with a thicker piece. Maybe what we see in the historical instruments with this type of compression is simply a factor of wall thickness of the tenon. I for one would never make my tenons as thin as your test piece. It would be interesting to see what thickness is the threshold for this. Casey

Certainly, it would be worth proceeding with more serious tests, and with real-life tenon thicknesses. But it would be quite a task to do properly. You'd want a range of tenon thicknesses, to explore if there is a threshold thickness one should never go below. And in different woods. And I guess you'd want a range of wrapping threads and tensions. So there would be a matrix of samples. But the reaction time of thicker samples will be much slower, so the experiment is going to take a very great time longer to conduct and might yet be unclear in its outcome. Heeding Julia Delany's grim forecast of lengthening waiting lists, I think I'd better leave this one to someone else!

Terry

[Terry McGee]

OK, a bit more data, and some further explanation, all at:

http://www.mcgee-flutes.com/Effects_of_ ... apping.htm

Terry

[Arch_Angel]

In the strangled flute example, the tenon was 15mm long, and about 20mm in diameter. So the area would be PI x r2 x L = 3.1412 x 1 x 1.5 = 4.7 sq cm

I'm don't remember where I grabbed the 14cm number, copied the wrong number.

But you solved for volume not for lateral area which would be 4.7cm3 not cm2

Punch in 1.5cm height and 1.0cm for radius here -
http://www.calculatorsoup.com/calculato ... linder.php

It's 9.42cm2

I believe your on the right track. It doesn't matter if the 150 wraps (w/13 layers) covers 9.42cm^2 or 1 wrap (w/13 layers) covers .0628cm^2. The only real question is how to properly determine if the layers add, multiply, or even only add a factor of force to the inner most PSI.

I'm not at all surprised on the moist/dry experiments. Taunt tread is going to resist moisture (like squeezing a towel dry) creating a unmovable force - perhaps minor shrinkage (as PSI is lost). While the wood would act like a rolled up blanket, shrinking while wet but re-expanding once dried. I would expect the wood to become more oval while drying due to friction against the tread while expanding.

Mob example (dis)proof:
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX13 layers of 150 wide tread mob
------------------- Barricade tenon pressure
XXXXXXXXXXXXXXX Wood riot police (layers unknown)

Not-
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX 150 x 13
________-________vs.
________X________1

Now we are back to the question of how much PSI a taunt thread produces over the .0628 cm^2 area of the (round[ish]) tenon.