deglen Posted July 18, 2014 Share Posted July 18, 2014 does anyone have a formula or rule for figuring out out much loss of length would result from twisting two equal size round bars set parallel (adjacent to each other) for each complete turn? I thought it might be pi * diameter for each turn, but that assumes no gap and perfect elasticity. thanks Link to comment Share on other sites More sharing options...
Jim Coke Posted July 18, 2014 Share Posted July 18, 2014 Greetings Deg, It is simple .. NONE .. I have done many test both free hand and with my twisting machines . Even twisted one 8ft with no change . Give it a try and see ... Forge on and make beautiful things Jim Link to comment Share on other sites More sharing options...
Frosty Posted July 18, 2014 Share Posted July 18, 2014 Twisting doesn't reduce the length at all. . . UNLESS you are asking about twisting two bars around each other, about that I don't know. I'd have to give it a try, if you try please let us know what you find out. I have a rope maker friend and the wife spins, they may know something I could use as a departure point in the experiments. However, neither rope nor spinning fibers act like steel so it may be a useless thing to ask them. So, in short: twist one bar = zero shrinkage. Twist multiple bars together I'd expect some but don't know how much. Please let us know if you find out. Frosty The Lucky. Link to comment Share on other sites More sharing options...
njanvilman Posted July 19, 2014 Share Posted July 19, 2014 You can always do a test run with bars of known length, then measure again when done. Record the numbers, including # of twists done. You will have your answer. Link to comment Share on other sites More sharing options...
Judson Yaggy Posted July 19, 2014 Share Posted July 19, 2014 One bar = no shrinkage as stated above, but your 2 bars are forming a helix around each other and will get shorter. The math is out there on the net but you have to dig deep. Forming hot or cold? You could also try and form a proportional scale model , measure before and after and multiply. Link to comment Share on other sites More sharing options...
ThomasPowers Posted July 19, 2014 Share Posted July 19, 2014 One of the issues will be that depending on heat and force you may get some stretch in the bar as well. I think that this is one of those things that a couple of hours in the shop will replace several days of crunching numbers and still not being sure it matches your exact case... Link to comment Share on other sites More sharing options...
deglen Posted July 19, 2014 Author Share Posted July 19, 2014 Sorry I wasn't clearer about what I was meaning. I did make a drawing and tried to upload, but evidently didn't do it correctly. My intent was using two bars twisting in a helical pattern around each other. I did do a trial using two pieces of 18" long, 1/2" round and measured the overall axial length with each rotation, then plotted it on excel and was suprised with results that were pretty linear (my r squared value was .98 for all you math types), and my formula generated showed a .41" decrease with each complete turn. I stopped at 6 turns because it still looked pretty good to the eye, but I know from twisting rope that you reach a point where the room left between twists won't allow the diameter to fit and the rope gets all wankled up. So, my little experiment provided good enough results for me to plan my larger project which was to make a helical twist with two pieces that was a set length with a specific number of revolutions. ps. I'll try to add my original drawing again. - didn't work with .png, .bmp, jpg, or gif, so I conclude I don't know how to add a drawing to the post. Link to comment Share on other sites More sharing options...
Frosty Posted July 20, 2014 Share Posted July 20, 2014 Only one more post to clarify a question is almost a record. I don't think twisting multiple strands of steel will schlub up like rope will, steel is a lot more elastic than hemp, jute, etc. Don't quote me on that though, I'm speculating and it would depend on how tight you tried twisting the steel. Oh yeah, post some pics of your project please, we love pics. You should've been able to post the .jpg, it's easy to forget a step like actually attaching it after selecting, etc. We've all done it, don't give up. Frosty The Lucky. Link to comment Share on other sites More sharing options...
Joel OF Posted July 20, 2014 Share Posted July 20, 2014 Only one more post to clarify a question is almost a record. Haha, spot on. And if the thread hasn't gone off on a tangent before you've clarified what you mean, that's a miracle :-D Link to comment Share on other sites More sharing options...
SmoothBore Posted July 20, 2014 Share Posted July 20, 2014 Worked for a "Wire & Cable" Company for about a half-dozen years, ... and can afirm that twisting single copper strands into multi-strand conductors, ( Stranding ) ... does consume some of the length of the individual strands. Depending on the rate of twist, ( tighter twist = more flexible cable ) ... from 3% to 8% "shrink" was common. As a general practice, every length of Cable began production at 110% of the desired final length. I'm not entirely sure that any of that is relevant to your situation, ... but, ... FWIW. . Link to comment Share on other sites More sharing options...
Nobody Special Posted July 21, 2014 Share Posted July 21, 2014 The correct answer isn't deliberately twist too long and cut to desired length? Sigh....... I've been doing it all wrong, I guess. Link to comment Share on other sites More sharing options...
yahoo2 Posted July 21, 2014 Share Posted July 21, 2014 For a perfect 45 degree helix I think the magic number is 1.414 (that's the % cross section diameter of the rod at a 45 degree angle) So each full twist height will be 1.41 times the rod diameter and the starting length will be 1.41 times longer than the finished twist. Or the twist will be 0.707 of the starting length if you prefer to look at it that way. I cant back it up with a mathematical equation, it just feels right when I roll it around in my head and compare it to calculating pitch. it will be something like Sin(twist angle) X length = twisted length.......I think Link to comment Share on other sites More sharing options...
yahoo2 Posted July 21, 2014 Share Posted July 21, 2014 Here is where I admit a was wrong ..again :( . My rough calculation was for half a turn or a single helix full turn and it still comes up slightly short if you double it. this link should get you closer if you can guesstimate the height you want for a complete turn. Calculating the length of two twisted copper wires http://mathforum.org/library/drmath/view/53155.html Link to comment Share on other sites More sharing options...
bigfootnampa Posted July 21, 2014 Share Posted July 21, 2014 Truly this question is a bit INTERESTING! There exists no way of calculating that will suffice because the variables are VAST and flexible!! It is quite UNTRUE that a single bar twisted will be unchanged in length!!! This is easily proved!!! Length change is dependent on MANY factors though! The shape of the bar... a flat bar will easily shrink a LOT with a tight twist... a square bar will usually shrink more than a round bar. The HEAT... a very hot bar is pretty stretchy... a cooler bar of the same shape and size might shrink much more even with the SAME TWIST! Bars with texture can behave quite differently from smooth bars... a fluted bar might behave more like several separate bars twisted together! ALL very interesting... but INCALCULABLE!!! Link to comment Share on other sites More sharing options...
Grant Posted July 21, 2014 Share Posted July 21, 2014 Good morning all In the vain of beating a dead horse, i think we are confusing twisting and wrapping. If we twist around an axis, that axis does not change, does not change if it is the center of a 1" square bar or the center of four 1/4" square bars. Now wrapping is a hole different deal. I expect we have all built some fence and the angle the wires cross determine the number of wraps for a given length of wires and this is clearly a case for an empirical study. -grant Link to comment Share on other sites More sharing options...
yahoo2 Posted July 22, 2014 Share Posted July 22, 2014 I think that is a good distinction between the two. The point where it stops being a wrap (or helix) and starts being a twist should be dictated by how much free space there is for metal to move into. four square bars has none so it is a twist right from the start. three square bars would wrap with only minor deformation until the space is gone, then extra energy would be needed to stretch the metal to start a twist. two round bars will have quite a lot of free space around the total diameter to move into. I have done far to much fencing, it is quite obvious with soft tie wire when you get to that point and everything tightens up and becomes hard to turn. Link to comment Share on other sites More sharing options...
Jerel Posted January 31, 2020 Share Posted January 31, 2020 From 1998: http://mathforum.org/library/drmath/view/53155.html *** Date: 11/16/98 at 12:19:55 From: Doctor Peterson Subject: Re: Calculating the length of two twisted copper wires Hi, Roland. If we picture one of your wires making a helix of radius R, with one turn taking H inches, it will look like this: |<--R-->| | | ******|****** | ** ** | ******o****** |---- | o| | ^ | o | | | |. o | | | ooo. | | | | . | | | | .| | | | |. | | | | . | | | | . | H | | .| | | | | | | | | | | | o | | | o | | | o| | | | o | | | ******|****o* | | ** | o ** v ******o******------ The length of the wire can be found by unwrapping it from the cylinder to form a right triangle: o o | o | o | o | o | o | o | o | L o | H o | o | o | o | o | o | o | o | o | ********************************************************* 2 pi R We'll get: L = sqrt(H^2 + (2 pi R)^2) as the length of one turn of wire, so the ratio of the length of the twisted pair to the length of the wire will be: L / H = sqrt(1 + 4 pi^2 (R/H)^2) The radius of the cylinder about which the center of the wire will spiral will be about the same as the radius of the wire itself: ******** oooooooo ******** *** ooo *ooo *** ** oo ** ** oo ** * o * * o * * o * R o * * o *----------o * * o * o * * o * * o * * o * * o * **** oooo* **oooo **** ******** oooooooo ******** So in your example, with R = .020 in and H = 1 in, we get: L / H = sqrt(1 + 4 pi^2 .020^2) = 1.0079 and for a 100 ft pair, each wire is 1.0079 * 100 ft = 100.79 ft. That doesn't sound like much difference. It will increase if the wires don't touch tightly all the way around. You may have to adjust the radius to correspond to reality! *** Link to comment Share on other sites More sharing options...
ThomasPowers Posted January 31, 2020 Share Posted January 31, 2020 Are they assuming no stretching of the material? Link to comment Share on other sites More sharing options...
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