Back momentarily, to fulfil a promise I had made in the long-running but doomed debate on flute strangling. Relax, moderators, I’m only passing through!
At Jem’s request, I had asked Prof. Neville Fletcher to clarify the mathematics behind calculating the forces and pressures a thread band could apply to a tenon. He has now done that, and I have incorporated it into the article at:
I have also tidied up some other parts of the article, added a few additional observations, provided a few useful links and so on. Because the changes are distributed, I’d recommend reading it from the top again.
Interesting, thank you … I like the comparasion with kite-flying tensions, I’ve had a fishing reel collapse for the same reason … I believe the symptoms are exacerbated if the thread in question is “stretchy”, like how nylon line will stretch compared with dacron line, (maybe this factor is incorporated in the math)
I think it’s very important to look at it as “the maximum resistance to expansion” as you highlight on your page. I don’t believe for one minute that the force being exerted constantly on the tenon is as much as you suggest, or even very much at all - I just mended a cleaning stick made out of a piece of thin dowel by using suture thread (very strong thread - much stronger than cotton or other threads used for wrapping tenons) and it was all I could do exerting my whole force on the wraps to make the split close together, which I could do easily by just holding it between thumb and forefinger. However, claiming that there is that much resistance to expansion makes much more sense to me. And it does seem like commonsense that it would be ‘multiplied’ by successive layers as well as by the whole length of the wrap.
Terry, I have to wonder at your Professor’s example.
"Now imagine that you slice along the tube on both sides and remove one half of all the threads, but that magically the thread stays in place and taut. To make this happen you have to put a weight of 700 grams on each end of each of semicircular wrap. This makes a total weight of 13 x 150 x 700 grams or 1365 kg on each end of the semicircle, making 2,730 kg altogether. "
Why would anyone imagine that? Isn’t that a Straw Man Fallacy?
Probably a 700 gram weight at the ends of the top wrap only would be sufficient to hold the others in place. Heck, tie 'em all together and dangle 1 weight each side.
I think a concept of “total force” is irrelevant and misleading in this matter: physics say that the force “per unit length” perpendicular to the surface is tension of the thread divided by the radius of the tenon: T/R. This means that for each millimetre of the thread’s length wrapped around the tenon there is a force T/R with R being the radius in millimetres. So with a tension of 1N (1 Newton, or about 100g in weight force) and a radius of 10 mm, we get a force of 0.1N/mm, or 0.1N on each mm section of the thread (Terry’s example is with 700g weight tension force, or about 7N, which is a very tight tension. 1N tension would be more normal).
Now theoretically we could multiply this with the total number of millimetres in the wrapping, like 20m of thread (*) will give us 2000N (ca 200kg force weight). Or multiply the force per millimetre with the circumference: 62.83mm X 0.1N/mm = 6.283N on one thread wrap. But this is irrelevant and misleading, since the forces are distributed around the tenon, acting from all sides, and therefore their vector sum is nil. We are summing up quantities without taking directions into account. And by talking about force weights (thousands of kg weight) we imagine crushing forces working in one direction. I think it is better not to sum up the forces like that, and instead consider the pressure directly.
Note: 20m of thread on Terry’s experimental tenon is about 320 wraps. With supposing 150 x 13 = 1950 wraps on the strangled tenon we get a lot more, but were there really so many wraps? More than 120 meters of thread?
What is the pressure under a thread wrap?
The pressure exerted is the force per unit length of thread divided by the thread width. This assumes the thread is not circular in cross section, but square. So there is a small error here which we can consider later. Let us use thread width for the moment. If our thread is 0.1mm wide, we get a pressure of 0.1N/mm per 0.1mm width, which is 1N/sqmm or 10bar or about 145psi. This is the pressure from one layer of wrap using a thread with tension 1N, which I consider not tight, but not loose either. For 7N this would be 70bar or 1015psi.
How much pressure will additional layers of wrap add?
I am not sure if each additional layer adds the same amount of pressure, i.e. if we can just multiply the single layer pressure with the number of layers. I expect some of the force goes into distorting the thread wrap, squashing it. But how much?
Hans, so for a somewhat realistic situation, you come up with 145psi, and the force added by additional layers is still unknown. I doubt a few more layers would multiply those forces significantly. Terry’s professor starts with a truly fictional example and comes to 29,000psi. Maybe he can do the math, but is unable to model a real-world example. This is like lying with statistics, it is. Hmmm, we’re in danger of ending up right back where we started. Maybe the moderators should just ban MATH in any thread. This one’s all ready to lock.
Oh, and about that compressibility of the thread layers on each other. Good idea. Wrapping is not perfectly on top of itself, thread precisely over thread, the thread will distort, compress other layers, cross over other sections, some changes of direction, etc. which probably in some way distributes those forces over a larger area.
For 1N (~100g weight force) we get 145psi on one layer. This may be multiplied with the number of layers to get the total pressure. But perhaps it is a bit less.
On Terry’s test tenon experiment I calculated 6 layers of thread, wound on firmly according to Terry. So with perhaps about 4N tension force. That is 580psi for one layer, or max 3480psi with six layers, perhaps somewhat less.
Which is precisely why my request included a plea that the example be done using a measured real-world thread tension!
Also, far be it from me to query the workings of a Professor in his own field, but one thing immediately hit me about the “cut in half tenon” tension/weight thing… surely, if you hang a thread over a bar and hang a weight on each end, you halve the total weight (compared to if you tie the thread to the bar and hang a single weight on the other end of it) at each end to get the same total tension? Using the example given, for a total weight/tension in the thread of 700g, you’d hang 350g on each end, not 700??? If I’m right, the figures thereafter (in that part of the calculation) are all wrong? (NB, not the basic formula, just that notional comparison>)
Take a scale, and wrap the platform under tension. How many pounds can you get it to read? I imagine there exists some actual device that can measure how much force is attempting to compress it towards a center.
Now what we need is somebody to document how many sockets have been caused to split because of cork that is too thick and tenons that have been forced into sockets.
And of course we need more juvenile drivel, posted in the name of lame humor, by people with too much time on their hands and not enough motivation to practice fluting.
Julia, you have correctly identified my comments as juvenile drivel, posted in the name of lame humor. Thanks for reminding me. You have incorrectly assumed, however, that I am without motivation and therefore idle. In fact, I have been turning flute tenons and playing the flute all morning. Sorry that an attempt at lame humor is the best that I can come up with in reply to the content of this thread.
Jem, if you hang a thread over a fixed horizontal pole, and hang 350g on each end, the tension in the vertical hanging parts of the thread is 350g force weight (~3.5N), not 700g. Same really as hanging two threads side by side each with 350g weights. Or holding a thread between your hands and pulling with the right hand with 3.5N force, and resisting with the left, which bears the opposite 3.5N force.
Thanks for the explanation, Hans - I did wonder about that - tried to think out the logic of it…but I bet the bit of thread at the top of the bar doesn’t agree! It isn’t the bar that is bearing the strain in the thread, though it is of course resisting 700g weight. Imagine it another way - have two support bars some distance apart, lay your thread across them and tie a (for the sake of argument) 350g weight to each end. What is the tension in the thread stretched between the bars? Assuming the thread can run freely over each bar, surely most of the combined weight/tension affects the horizontal thread? And which is a nearer parallel to the situation of a thread tensioned around a cylinder?
