As I am currently in a war zone, I don't have many options for cabling.

I found this clothesline (steel core plastic wire rope) that appears to be one mm of diameter (steel core diameter.) 13 meters of it measured 7 ohms resistance.
Edit: It is 3.8 Ω and not 7. The first multimeter test lead probes had 2-4 resistance when shorted. A slightly better multimeter had 0.5 Ω when shorted. Both multimeters gave 3.8 after subtracting multimeters own resistances and scratching the wire ends.

enter image description here

  1. Can it carry AC 120 or 240 volts, and if so, for what distance?
  2. How many of it (doubling it) can carry DC 18V and 15 A from a solar panel arrays 10 to 15 meters away from the inverter (charge controller)? (20 W panels with open circuit voltage of 21 V).

This is just temporary solution and I hope only for few days or weeks. Air strikes blew up some transformers and high voltage lines and our concrete homes are not designed to be habitable without AC power.

Its now connected to a c32 breaker (the smallest I could find) and a breaker mounting brackets cut from a laptop battery cover. The wire is inserted in plastic bottle caps as wire wall clips and a tow heads plug is inserted to the other end (to be upgraded to three heads) because now polarity is important I think.
enter image description here enter image description here

I will connect it to a manual changeover switch when adding the solar oart after finishing the battery but that is another longer story (Lithium battery without BMS) but I may be able to reuse old laptop batteries BMS (I have more than 20 batteries): enter image description here

The solar system is only for a medium 100 W Samsung fridge and the mains are for the fridge plus two ceiling fan, one swamp cooler and three LED lights.

  • 5
    \$\begingroup\$ I see a few issues, with waste energy due to high resistance, with possible temperature rise, which will eventually affect the safety level: isolation material and isolation level aren't guaranteed. McGyver would employ it successfully to power a nuclear power plant, real life urges us to be cautionary, and use the proper type of electrical cable. \$\endgroup\$
    – LuC
    Commented Apr 27, 2023 at 8:05
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    \$\begingroup\$ @SomeoneSomewhereSupportsMonica Yes. we already done that but not over do it because we hope the power will be restored. \$\endgroup\$
    – USER249
    Commented Apr 27, 2023 at 9:35
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    \$\begingroup\$ (No offence intended: In for how long?, do you want a duration or a distance?) \$\endgroup\$
    – greybeard
    Commented Apr 27, 2023 at 17:40
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    \$\begingroup\$ @greybeard Good question! I was thinking distance "how far"(my English is not so good). Your concentration level is better than my English level. Please answer based on that. \$\endgroup\$
    – USER249
    Commented Apr 27, 2023 at 18:15
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    \$\begingroup\$ As FYI, this page is also being discussed at Hacker News, in case any relevant answers or comments show up there: news.ycombinator.com/item?id=35730074 \$\endgroup\$ Commented Apr 27, 2023 at 18:28

7 Answers 7


Steel, having just around 10 times higher resistivity than copper means it will take ten times the conductor area to match copper. If you measured 13 meter of it to 3.8 Ω, the cross sectional area would be 2 mm^2, assuming 5.95*10^-7 Ωm of resistivity for "high alloy steel" (this varies greatly unfortunately so assume +100% -50% uncertainty for all values given).

To answer your questions:

  1. How long = Time: probably many years. How long = Distance: Most devices will run happily with 10 % voltage drop. Anything universal input (100-240 V) could handle significant voltage drop due to the cable at which point it's the thermal capability of the cable which sets the limit as you don't want it to melt.

    With 3.8 Ω for 13 meter, you have 0.29 Ω/m. At 1 A 230 V AC current, you can go 39.7 meter (round trip is double distance) before you have dropped 10 % of the voltage. If you halve the current, it's double the distance. Gut feeling + experience says it would get lukewarm at 2-3 A so I would not go much above it.

Could be lethal though, as the insulation is not mains voltage rated. I would be more afraid of anyone coming into contact with the end points and the cable termination than touching the outer shell of the clothesline and somehow get zapped by it as they tend to sit outside for decades without becoming brittle by the UV exposure.

As stated below by Martin McCormick, your best bet is to put the inverter as close to the panels as you can and run AC through the clothesline versus low voltage DC current through the clothesline.

  1. With just one conductor, carrying 15 A via 3.8 Ω means a 57 V drop, so not possible with 18 V at all. It would also melt. To make it work at 18 V, perhaps 20 % drop (3.6 V) could be tolerated. To get down to 3.6 V drop, you would need 57/3.6 = 16 in parallel.
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    \$\begingroup\$ @AmrBerag Distance will affect your voltage drop linearly. Half the distance and you can use half as many wires in parallel for a given voltage drop. Do you have an inverter to 120/230 V AC? If yes, place that as close to the panels as you can and do the long wiring on the 120/230 V side. \$\endgroup\$
    – winny
    Commented Apr 27, 2023 at 8:45
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    \$\begingroup\$ No, 13 meter from your 7 ohm measurement. \$\endgroup\$
    – winny
    Commented Apr 27, 2023 at 8:54
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    \$\begingroup\$ Ok sorry now I understand because 7 ohms was the resistance of the 13 meters sample. \$\endgroup\$
    – USER249
    Commented Apr 27, 2023 at 8:58
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    \$\begingroup\$ @AmrBerag, be sure to consider these suggestions another user made regarding your solar array: electronics.stackexchange.com/a/664579 \$\endgroup\$
    – daveloyall
    Commented Apr 27, 2023 at 19:07
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    \$\begingroup\$ @winny - Hi, FYI the OP has recently clarified (and this has been updated in the question) that for point (1) they meant "how far" when they said "how long" (i.e. they meant distance rather than time). It's unfortunate that the change seems to affect your answer (which was written when the question said "how long") so I thought I would send you this notification, in case you wanted to update your answer to point (1). Thanks. (As you may know, we discourage changes to a question which invalidate existing answers. However as this is a genuine clarification, it seems reasonable to allow it.) \$\endgroup\$
    – SamGibson
    Commented Apr 27, 2023 at 22:21

Agree with winny to put the inverter next to the solar panels if possible and send 230V AC (standard in Sudan) over the clothesline, despite it being more dangerous.

If you need to transfer 1000W of power, it loses less to clothesline resistance when transferred at a higher voltage.

1000W =

• 55 amps @ 18 volts

• 4 amps @ 230 volts

Power lost through the clothesline is (Current)^2 * (Resistance).

So choosing the lower 4 amps current is much less loss (~200x). The power loss dissipates as heat in the clothesline, at some point, it will burn, so it's not just about efficiency--using 230V is necessary to get this to work. You can probably carry 1-2kW with 1mm core wire at 230V AC. This is a window-unit air conditioner and some light bulbs or a couple of large fans. Maybe a microwave for short periods of time. Carefully observe for smoke of the clothesline insulator and shut it off immediately. If this happens you need to run less equipment in the house at the same time or run more parallel lines.

Use your home-harvested copper wire for the 18V DC connection between the panels and the AC inverter.

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    \$\begingroup\$ A garden hose, PVC pipe or other insulating tubes may be used to provide an outer jacket of insulation, increasing general safety. Running a ground wire and/or scavenging a GFCI and using a grounded structure from a nearby home may also provide additional safety. \$\endgroup\$
    – Xunie
    Commented Apr 27, 2023 at 21:08
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    \$\begingroup\$ I now notice you have only 14 panels of 20W or about 280 Watts total. Still 230V AC over clothesline is the way to go, but 280W is probably not enough to run a wall-unit air conditioner if that is the goal. Evaporative cooling could work (if that's the unit you have). A makeshift one could be made with an air fan blowing on wet towels. \$\endgroup\$ Commented Apr 28, 2023 at 0:25
  • \$\begingroup\$ Evaporative cooler with towels: youtu.be/DJN3lwVLGxs it should work well in Sudan with the low relative humidity there. \$\endgroup\$ Commented Apr 28, 2023 at 0:34
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    \$\begingroup\$ From your update it sounds like you indeed are using an evaporative swamp cooler, so you're probably OK then on power. The Tk-63 breaker is likely useless because the current is already going to be limited by the clothesline resistance (and the solar panels) to well under 32 amps. If you get a short the clothesline may burn before the breaker pops. Don't expect to rely on it... \$\endgroup\$ Commented Apr 30, 2023 at 3:08
  1. The AC side should be OK, as others have noted.

  2. As winny notes, you'll have a problem trying to carry 15 DC amps. However, I see you're using an array of 20W panels. You said 18V and 15A, so 270 watts; I'm going to round and guess you have 14 panels in parallel. Can you connect them in series instead of parallel, giving you a high voltage and low current?

  • 14x1 configuration would give you 252V and about one amp, a 7V drop over the measured 7 ohms or 3% loss.
  • 7x2 would be ~125V at 2A, a 14V drop so still only about 11% loss.

You'll of course have to check whether your transformer can handle the input. You'll also have less shade tolerance than in the parallel configuration, but that sounds like the least of your worries.

I just used your measurement at 13 meters for resistance. 7/13 is about .54 ohms per meter. If the panels are 15 meters away from the inverter, that's a 30m round trip, so closer to 16 ohms. That's still probably fine, or one doubling gets you back to the numbers above.


I'm not an electrician, but I can help with the circuit analysis. The formula for resistance is:

$$R = \rho \frac L A$$

where \$\rho\$ is the resistivity of the steel, \$L\$ is the length of the wire, and \$A\$ is the cross-sectional area of the wire. The first question is whether your resistance measurement is correct. Measuring small resistances with a two-wire multimeter is hard due to the resistance of the leads. You also (for testing and use) need to make sure you have good contact with the wire. The surface of metals oxidizes, which can add resistance to a weak connection.

Your measurement of 7 ohms for 13 meters give a resistivity of:

$$\rho = \frac {R A}{L} = \frac {7\Omega \cdot \pi \cdot (0.5 \mathrm{mm})^2} {13 \mathrm m} = 4.23 \times 10^{-7} \Omega \cdot \mathrm m$$

which seems to be in the right range according to this and other sources, so your measurement seems reasonable. But you didn't say how you're doing the measurement.

Resistance is linear with distance -- halve the length of the wire and you halve its resistance. Double the length of the wire and you double its resistance. Two wires in parallel will have half the resistance of one, three wires will have a third of the resistance of one, and so forth. Your wire seems to be \$0.5385\ \Omega/\mathrm m\$. If you want to make a cable with around 1 ohm of resistance, you could use:

  • 1.85 meters of 1 wire
  • 3.7 meters of 2 wires in parallel
  • 7.4 meters of 4 wires in parallel
  • etc.

If you want to carry your large 15 amp current, you'll need to go lower. Maybe a 0.1 ohm cable can work:

  • 1 meter of 6 wires in parallel
  • 2 meters of 12 wires in parallel
  • 4 meters of 24 wires in parallel
  • etc.

For comparison, copper's resistivity is around \$1.7 \times 10^{-8} \Omega\cdot\mathrm m\$. Your steel's resistivity is around 25 times higher.

Aluminum is almost as good as copper at \$2.7 \times 10^{-8} \Omega\cdot\mathrm m\$, so if you find any aluminum wire (or long strips of aluminum if you get desperate) that would be much better than steel. Be sure to insulate -- high currents can be just as dangerous as high voltages if they get shorted out or the connection suddenly breaks.

Good luck! I hope your situation improves soon.

  • 1
    \$\begingroup\$ Thank you. We have prepaid electricity and water billing system and the system is down and government and the electricity company said in social media that we can do whatever it takes to restore power. Sometimes there are unused cables hanging from poles at the end of lines or neare buildings that switched from overhead wires to under ground cables. These very Thick wires appear to be 25mm^2 or more insulated aluminum. \$\endgroup\$
    – USER249
    Commented May 2, 2023 at 21:37
  • \$\begingroup\$ That would be roughly 0.001 ohms per meter, which is great! \$\endgroup\$
    – Adam Haun
    Commented May 3, 2023 at 6:18

Source: https://news.ycombinator.com/item?id=35730074

Credit: https://news.ycombinator.com/user?id=ltbarcly3

TLDR: connections must be protected by some kind of anti oxidation coating, if you have nothing else use grease but something designed for electrical connections is better. If you have nothing else, melt some lead and dip the exposed part of the wire in that to coat it. Lead should be readily available in a war zone? Long term the wire WILL melt at some random point along the wire so it is much better if this wire is kept away from anything flammable.

The risks of using steel cable for power are going to be corrosion/rust. The primary way this will cause problems is around connections, but long term it is also an issue away from connections. At connections, corrosion can cause the wire to become loose, something like ox-gard can be used to delay this significantly. It isn't designed for steel but some kind of protection must be done at connections or arcing will be a major risk. Longer term the wire itself will rust, and at some point the conductive cross section of the wire will be compromised to the point it overheats and melts. This might be a year and it might be 100 years depending on factors which are hard to predict and control.


If you have access to lead or Tin.

  1. Strip cloth line to get the steel core (Blade) (If burned, scrub so the outer oxidation is removed).
  2. Vasiline or try other available items to be used as a flux. Cover steel wire with the hack-flux.
  3. Run Dip the wire in lead , you can also pull the wire through the molten lead.
  4. Once lead is cooled, do the same with enamel or use carefully keeping distance between +/-

The outer lead will do the conduction, inner steel provides the mechanical support.

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    \$\begingroup\$ Note the OP's circumstances: War zone. \$\endgroup\$
    – DKNguyen
    Commented Apr 27, 2023 at 17:48
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    \$\begingroup\$ @DKNguyen: Future readers might have different situations, or who knows, maybe a lot of lead (spent bullets?) could be lying around. (Although somewhere to heat this much lead could be hard to find). This is creative, and alternatives to the enamel step include maybe using a garden hose as housing / insulation, as other commenters suggested, with separate hoses for each conductor. (Although something to protect the metal from the air would be good, since I assume there'd be parts of the steel that didn't get perfectly coated.) \$\endgroup\$ Commented Apr 28, 2023 at 15:34
  • \$\begingroup\$ Do you mean the entire wire? What abot insulation? \$\endgroup\$
    – USER249
    Commented Apr 29, 2023 at 20:34

In a war situation you use everything that is available. You should just be cautionous for:

  • voltage drop because of the high resistance
  • bad contacts because of oxidation
  • overheating because of the high resistance
  • failure of the insulation because of overheating
  • dangerous situations because the plastic is not really designed as isolation and may not be able to withstand the actual voltages

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