Fine wire tensile strengths?
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From: John Fields on 31 Jul 2010 17:58 On Sat, 31 Jul 2010 16:14:03 -0500, John Fields <jfields(a)austininstruments.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. >>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] <-- oops... 240R | 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.
From: Nunya on 31 Jul 2010 19:22 On Jul 31, 2:14 pm, John Fields <jfie...(a)austininstruments.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. > >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. Well any stretching will also cause the diameter to reduce, but the other effect that such stretching has is micro-structure fracturing. That measn the resistance will be even higher than the same diameter wire that has not been stretched. The other thing to worry about is that one does not twist the wire pair together as any of the aforementioned stretching will also cause a fracturing of the enamel on the wire as well, and a shorting event is far more likely. I would use three strands of a smaller wire, twisted together and use two lengths of that, one for each run,source and return. The three smaller strands will have a better "stretch" characteristic than any single strand will. So, if a single #26 or #24 was chosen, then I would use like 3 #28 segments twisted together to yield the same current capacity, but far better tensile numbers. But sure, for such a short run, it will be negligible. Use Silver and get better electrical and mechanical numbers (not plated... solid Silver).
From: Grant on 31 Jul 2010 22:20 On Sat, 31 Jul 2010 11:55:00 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >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. Most people see a typo and look at the keyboard to see what it might be :) I didn't even notice the typo, no doubt others will jump on it! Get Larkin for a lark! :o) Grant. > >John >
From: Tim Williams on 1 Aug 2010 01:36 "Nunya" <jack_shephard(a)cox.net> wrote in message news:c64626d4-e940-41b0-a93f-69930ad66775(a)z15g2000prn.googlegroups.com... > And no... voltage drops with line length, not current. Obvious troll is obvious. Please do not feed the trolls. Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms
From: John Larkin on 1 Aug 2010 02:17
On Sun, 1 Aug 2010 00:36:55 -0500, "Tim Williams" <tmoranwms(a)charter.net> wrote: >"Nunya" <jack_shephard(a)cox.net> wrote in message news:c64626d4-e940-41b0-a93f-69930ad66775(a)z15g2000prn.googlegroups.com... >> And no... voltage drops with line length, not current. > >Obvious troll is obvious. > >Please do not feed the trolls. He's not a troll. He actually *believes* that electricity drips off long wires somehow. John |