Fine wire tensile strengths?
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From: Ken S. Tucker on 31 Jul 2010 10:40 On Jul 31, 12:52 am, "Tim Williams" <tmoran...(a)charter.net> wrote: > "Joe" <j...(a)somewhere.org> wrote in messagenews:i30cnt$d59$1(a)news.eternal-september.org... > > Numbers in the ARRL handbook chart of breaking load for copper > > antenna wire work out to about 39000 psi for soft drawn copper > > wire and 65000 psi for hard drawn so perhaps use 2 strands of > > varnished #24 or #26 copper magnet wire. Those strength numbers > > are higher than Tim mentioned so run deadweight tests before > > testing with model. > > 65 KSI is close to the ultimate tensile strength I saw; I specified yield strength, which is a little lower of course. > > The yield of dead soft copper is astonishingly low, it's like shiny pink clay. Only that, after you push it around a bit, it hardens up, and pretty soon it's so damn strong you can't bend it by hand anymore! Copper provides a particularly dramatic illustration of work hardening. This is a Good thread Gents, good of yo'all to be interested. To Recap, The models will be hung ~5' , the spec is to use 120V 5W & 7W xmas bulbs, although I can run a 3rd line for 12V sissy lamps. (Gnd, 120V, 12V). Using this ref http://en.wikipedia.org/wiki/American_wire_gauge I find 24 AWG to be ok. I used a Yield Tensile Strength of Copper ~10,000psi, calculated area and found it to be ~4#. (24 AWG has a dia 0.02" ) Two wires gives a strength of 8# for the Gnd & 120V, and the ohm loss calculated for 10'. Regards Ken
From: John Larkin on 31 Jul 2010 11:38 On Sat, 31 Jul 2010 00:10:30 -0700 (PDT), Nunya <jack_shephard(a)cox.net> wrote: >On Jul 30, 11:48�pm, Grant <o...(a)grrr.id.au> wrote: >> On Fri, 30 Jul 2010 23:26:44 -0700 (PDT), Nunya <jack_sheph...(a)cox.net> wrote: >> >On Jul 30, 7:01�pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: >> >> I've been asked to 'light-up' aircraft models, such as, >> >> >>http://www.flickr.com/photos/35156618(a)N03/4754110575/ >> >> >> It is suspended from a rope. >> >> I need to use very fine 120V wire (like magnetic wire), >> >> the model uses 8# test monofilament right now and weighs >> >> 1/2 # , but a strong wind requires that 8# test. >> >> Is there a table that gives wire gauge & tensile strength? >> >> The current will likely be a max of 1/2 amp. >> >> Ken >> >> > �The problem with all these replies is that he wants to >> >pass 120V over a "fine wire". �That will not get one much past >> >about twenty feet before the line drop starts to show a >> >huge drop. At least that is the idea I got. Otherwise why >> >mention 120 V at all? >> >> Current (1/2 A) is the important factor for drop, and 120V says he >> has lots of wiggle room for voltage drop, no? >> >> Grant. > > How high (long) would this line be? That too is a factor. > > And no... voltage drops with line length, not current. Sorry, no. E + I * R, and R is proportional to length for a given wire size. #24 soft copper is 26 milliohms per foot. At half an amp, that will drop 13 millivolts per foot. 100 feet would lose 1.3 volts, about 1% of the available 120 volts. >The line itself is a resistor, and that means that the >other end of the line will see lower voltage. Yes. And how much lower is proportional to the current. John
From: linnix on 31 Jul 2010 14:38 On Jul 31, 8:38 am, John Larkin <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote: > On Sat, 31 Jul 2010 00:10:30 -0700 (PDT), Nunya > > > > <jack_sheph...(a)cox.net> wrote: > >On Jul 30, 11:48 pm, Grant <o...(a)grrr.id.au> wrote: > >> On Fri, 30 Jul 2010 23:26:44 -0700 (PDT), Nunya <jack_sheph...(a)cox.net> wrote: > >> >On Jul 30, 7:01 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > >> >> I've been asked to 'light-up' aircraft models, such as, > > >> >>http://www.flickr.com/photos/35156618(a)N03/4754110575/ > > >> >> It is suspended from a rope. > >> >> I need to use very fine 120V wire (like magnetic wire), > >> >> the model uses 8# test monofilament right now and weighs > >> >> 1/2 # , but a strong wind requires that 8# test. > >> >> Is there a table that gives wire gauge & tensile strength? > >> >> The current will likely be a max of 1/2 amp. > >> >> Ken > > >> > The problem with all these replies is that he wants to > >> >pass 120V over a "fine wire". That will not get one much past > >> >about twenty feet before the line drop starts to show a > >> >huge drop. At least that is the idea I got. Otherwise why > >> >mention 120 V at all? > > >> Current (1/2 A) is the important factor for drop, and 120V says he > >> has lots of wiggle room for voltage drop, no? > > >> Grant. > > > How high (long) would this line be? That too is a factor. > > > And no... voltage drops with line length, not current. > > Sorry, no. E + I * R, and R is proportional to length for a given wire > size. > > #24 soft copper is 26 milliohms per foot. At half an amp, that will > drop 13 millivolts per foot. 100 feet would lose 1.3 volts, about 1% > of the available 120 volts. Soft copper should be fine for 10 to 20 pounds. If the OP need more, I have some #24 copper nickel alloy wire, rated at 100KSI. It should hold a couple hundred pounds. It might not be as conductive as pure copper, but it's much better than steel.
From: John Larkin on 31 Jul 2010 14:55 On Sat, 31 Jul 2010 08:38:34 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Sat, 31 Jul 2010 00:10:30 -0700 (PDT), Nunya ><jack_shephard(a)cox.net> wrote: > >>On Jul 30, 11:48�pm, Grant <o...(a)grrr.id.au> wrote: >>> On Fri, 30 Jul 2010 23:26:44 -0700 (PDT), Nunya <jack_sheph...(a)cox.net> wrote: >>> >On Jul 30, 7:01�pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: >>> >> I've been asked to 'light-up' aircraft models, such as, >>> >>> >>http://www.flickr.com/photos/35156618(a)N03/4754110575/ >>> >>> >> It is suspended from a rope. >>> >> I need to use very fine 120V wire (like magnetic wire), >>> >> the model uses 8# test monofilament right now and weighs >>> >> 1/2 # , but a strong wind requires that 8# test. >>> >> Is there a table that gives wire gauge & tensile strength? >>> >> The current will likely be a max of 1/2 amp. >>> >> Ken >>> >>> > �The problem with all these replies is that he wants to >>> >pass 120V over a "fine wire". �That will not get one much past >>> >about twenty feet before the line drop starts to show a >>> >huge drop. At least that is the idea I got. Otherwise why >>> >mention 120 V at all? >>> >>> Current (1/2 A) is the important factor for drop, and 120V says he >>> has lots of wiggle room for voltage drop, no? >>> >>> Grant. >> >> How high (long) would this line be? That too is a factor. >> >> And no... voltage drops with line length, not current. > >Sorry, no. E + I * R, and R is proportional to length for a given wire >size. Typo. "+" should have been "=". Missed the shift key again. John
From: John Fields on 31 Jul 2010 17:14
On Sat, 31 Jul 2010 00:10:30 -0700 (PDT), Nunya <jack_shephard(a)cox.net> wrote: >On Jul 30, 11:48�pm, Grant <o...(a)grrr.id.au> wrote: >> On Fri, 30 Jul 2010 23:26:44 -0700 (PDT), Nunya <jack_sheph...(a)cox.net> wrote: >> >On Jul 30, 7:01�pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: >> >> I've been asked to 'light-up' aircraft models, such as, >> >> >>http://www.flickr.com/photos/35156618(a)N03/4754110575/ >> >> >> It is suspended from a rope. >> >> I need to use very fine 120V wire (like magnetic wire), >> >> the model uses 8# test monofilament right now and weighs >> >> 1/2 # , but a strong wind requires that 8# test. >> >> Is there a table that gives wire gauge & tensile strength? >> >> The current will likely be a max of 1/2 amp. >> >> Ken >> >> > �The problem with all these replies is that he wants to >> >pass 120V over a "fine wire". �That will not get one much past >> >about twenty feet before the line drop starts to show a >> >huge drop. At least that is the idea I got. Otherwise why >> >mention 120 V at all? >> >> Current (1/2 A) is the important factor for drop, and 120V says he >> has lots of wiggle room for voltage drop, no? >> >> Grant. > > How high (long) would this line be? That too is a factor. > > And no... voltage drops with line length, not current. >The line itself is a resistor, and that means that the >other end of the line will see lower voltage. The load will see lower voltage _and_ current if the wire resistance increases. Assuming that the load draws 1/2 ampere when there's 120V across it, then the load will look like: E 120V R = --- = ------- = 240 ohms I 0.5A From the table at: http://www.dossert.com/technicalinfo/barecopperwire.htm It appears that multiplying the cross-sectional area of the wire, in square inches, by 38.5e3, yields the maximum breaking strength of annealed or soft-drawn copper wire. Then, since the OP needs an 8 pound breaking strength, the minimum cross-sectional area would need to be: 8lb S = -------- = 0.00021 square inches 3.85e3 Consulting a copper wire table, we find that 26AWG has a cross-sectional area of 0.0001996 square inches, and 25AWG 0.0002517. Choosing the #25 would yield a breaking strength of about 9.7 pounds, and it would have a diameter of 0.02 inches, or about a half a millimeter for you metric freaks out there. ;) To hang an airplane 20 feet from from a ceiling, say, would require 40 feet of it, and with a resistance of 39.37 ohms per thousand feet would wind up having a total resistance of 1.3 ohms. So, we originally had: (view in Courier) 120V | [24R] | GND with a current in the load of half an amp and the load dissipating: P = IE = 0.5A * 120V = 60 watts. Now with the wire connected we'll have: 120V | +---E1 | [1.3R] R1 | +---E2 | [240R] R2 | GND The current in the circuit will be: E1 120V I = --------- = -------- 0.497 ampere, R1 + R2 241.3R and the voltage across the load will be: E2 = I R2 = 0.497A * 240R = 119.28 volts, so you can see that when resistance is added to the circuit the voltage across as well as the current through the load will decrease. In this case, hardly enough to worry about, since instead of 60 watts the load will be dissipating about 59.3 which will be undetectable to the eye if the load is lighting. The wire will be dissipating about 0.646 watts, and for a 40 foot length of it, I doubt whether its increase in temperature over ambient will even be noticeable. |