Re: Fine wire tensile strengths?

Current (1/2 A) is the important factor for drop, and 120V says he has lots of wiggle room for voltage drop, no?

Grant.

rote:

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.

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.

Yes. And how much lower is proportional to the current.

John

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.

Typo. "+" should have been "=". Missed the shift key again.

John

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:

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:

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)

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:

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).

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.

Obvious troll is obvious.

Please do not feed the trolls.

Tim

--=20 Deep Friar: a very philosophical monk. Website:

He's not a troll. He actually *believes* that electricity drips off long wires somehow.

John

And you, for some reason, actually believe him. We'd all appreciate it = if you'd simply let the troll alone.

Tim

--=20 Deep Friar: a very philosophical monk. Website:

"John Larkin" wrote in = message news: snipped-for-privacy@4ax.com...

you'd simply let the troll alone.

computer screen.

--- What I'd do would be to use a length of piano wire as the strength member from which to hang the airplane and then just run the wiring next to it for the electric part.

If your lighting is incandescent I doubt whether a spike would hurt it. To be safe, though, you could connect a couple of unidirectional TVSs in aseries opposition across the load.

Something like a couple of Littelfuse 1.5KE160A:

should work.

You could also use a single bidirectional part but they're harder to find.

JF

TWO HUNDRED DOLLARS for an electret mike in a birdhouse?

John

You got it right. Use a high tensile strength carrier strand and spiral dress a bifilar pair over that, which is able to pass the needed current level.

Hard mounted battery packs and LED lighting would work too. ;-)

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You should move to wifi nets and delete the wires altogether.

you'd simply let the troll alone.

computer screen.

Related to roasting? ;^)

Grant.