In this video here, almost 1000A is being run through a 10mm steel bar, and after a minute or so, it melts.


So in terms of charge, 60,000C applied for 1 minute destroys the conductor.

However, it would take a lot more current for shorter times. A good example being that if lightning struck such a bar, it would survive since lightning carries only 15C per flash on average.

Lightning averages 30,000A for a brief moment. So its carrying 30C per millisecond and discharges 15C then the flash lasts 0.5ms.

So what would be the smallest gauge cable that could sustain 30kA for 0.5ms? Is it possible to work this out mathematically, from Ω⋅m and the melting point? Do engineers do this or do they just buy a product that has been rated/tested?


Actually, it suddenly occurred to me that fuses must have been computed which led me to find the Preece equation and Onderdonk equation and this calculator.


I'll post the question anyway since it clearly has an answer.

  • 1
    \$\begingroup\$ I suspect that architects know how thin iron features can be made without risking them going molten on a lightning strike. It's not of much interest to engineers because we'd use copper, which has less than 1/5 the resistivity. Iron is practically a heating element by comparison. Fusing current is probably unnecessarily pessimistic because they're going for steady-state heating in air rather than taking Russell's mentioned factors into consideration. Copper wires from lightning arrestors are not huge \$\endgroup\$ Nov 12, 2014 at 13:24
  • \$\begingroup\$ @SpehroPefhany Info only - not a comment on your input per se - My answer was aimed at establishing a 'laws of physics' upper bound based on some already rather tenuous data - and at possibly addressing a few possible misconceptions, possibly :-). It started out as a quick comment as I rushed off to bed with aching tooth in order to get not enough sleep before an 8am Dentists's appointment. I can see that the attempt at being brain filter expanding if not especially useful in real life was very largely a waste of my effort and time. I'll try to remember that. .... \$\endgroup\$
    – Russell McMahon
    Nov 13, 2014 at 6:46
  • \$\begingroup\$ .... I'll address Dave's lack of analytical skills & failure to reach his master's level of ability in rudeness separately. Perhaps. [Yes. Tooth that is not there hurts more than before it was extracted. Does not help my level of tolerance for rude stupidity or stupid rudeness. (No, no, not you Spehro :-) ). \$\endgroup\$
    – Russell McMahon
    Nov 13, 2014 at 6:48
  • \$\begingroup\$ @Luke Given the sort of effort and information put into your question (eg quantising lightning charge levels, current flows, time constants, ...) I'd expect that you'd not have too much trouble wading through my hastily written response below. What is interesting is that I arrived at a bar size about the same as your example which, at 300A may survive indefinitely, and at 3000A will melt in a moment - and your example current is in the middle of that range. Maybe my answer is closer to correct than I expected :-) \$\endgroup\$
    – Russell McMahon
    Nov 13, 2014 at 6:54
  • \$\begingroup\$ Thanks Russell, I'll have a look. A ton of work landed on me after I posted but I left my browser tabs open and will try and model it in Excel. \$\endgroup\$ Nov 14, 2014 at 8:42

2 Answers 2


After doing some research, here's my own answer.

The table here contains two fuse time values for Onderdonk's equation for each guage of wire http://en.wikipedia.org/wiki/American_wire_gauge

Here's the math for calculating the current to fuse a 4AWG cable within 1 second.

Amps = Area*SQRT(LOG(((Tmelt-Tambient)/(234-Tambient))+1)/(Time*33))

Tmelt:    1084.62
Tambient:   10
Time:        1
Area:    41700
Ifuse:  = 6341.733   Amps

So taking a conservative 1ms for a lightning strike, 4AWG cable will fuse if the bolt is 200,000A

According to that math, a copper cable of 7700cmils will fuse after 1ms of 37,000A and an average lightning bolt is 20-30,000A in the UK, meaning that 11AWG cable 2.3mm in diameter might survive! [?]

  • \$\begingroup\$ +1. I was using iron or steel which was what was started with and is somewhat a moving target as reported characteristics vary wildy. You could plug values for Copper into my "mess" and see how it compares. I think for safety you could use thermal energy needed to bring to melting temperature but no latent heat of fusion (as "half melted" is not usually what is meant by "survives ...A". After that you need to start looking at radiation from the surface and thermal conduction for longer periods. I doubt that convection has much effect at ms timespans. \$\endgroup\$
    – Russell McMahon
    Nov 15, 2014 at 23:09
  • \$\begingroup\$ I'm going cut out the 'live blog' stuff from my answer and reduce it down to the useful information and mark my own as the answer. With the benefit of hindsight, my question was, "What's the math to work out the fusing time for a conductor?" and Onderdonk's equation above does just that. \$\endgroup\$ Nov 17, 2014 at 12:27
  • \$\begingroup\$ @RussellMcMahon I wanted to say thanks for your input but also that I understand why your answer got down-voted. I'm a computer programmer and have spent lots of time on StackOverflow, for years. SO was 'invented' because programmers/brainy (like you) people love to discuss and philosophize problems. But at the expense of productivity. I still get annoyed by the 'rules' of SO, but the hardline (often rude) approach does work, keeps answers concise, clear, high-quality and has changed our industry - I don't know what we'd have done without it. \$\endgroup\$ Nov 17, 2014 at 12:44
  • \$\begingroup\$ Luke - you started out talking about a steel bat but now seem more focused on Copper. The question did actually indicate that was your preference as you started with "the bar" and ended up asking about cables. So, if you took Onderdonk's equation and applied it to steel, how would it fare? Resistance is far higher and as you specify 30 kA, dissipation per length at a given cross section will scale with resistivity. Then for short periods I'd expect most energy would stay in the bar and contribute to heating. Onderdonk does not account for either latent heat , or latent heat of fusion. .... \$\endgroup\$
    – Russell McMahon
    Nov 17, 2014 at 13:02
  • \$\begingroup\$ .... He does account for melting point. So it seems likely that the currents used may need to be scaled by the inverse square root of the resistivities, and probably needs to account for heating to MP and melting energy differences. So overall, while Onderdonk may do well enough for Copper, I suggest in general terms that his results could be far enough out for other materials to be dangerous to rely on. No? \$\endgroup\$
    – Russell McMahon
    Nov 17, 2014 at 13:05

Rushing past ... .

A bottom survival limit is set by the thermal capacity of the material and melting is determined by the phase change energy. Most people don't need to know the latter for steel :-). BUT cooling, thermal insulation, what's beyond the section in question may all add to survivability by taking heat away. As lightning delivers the energy effectively instantaneously as far as thermal issues are concerned, I'd expect lightning to need fewer C (Coulombs) to melt a bar than you'd need over a a longer period. Thermally its Watts per volume that matter.

Iron / Steel:

Latent heat 420 J/kg/k
LH of fusion: 96-140, 272 kJ/kg different sources

Melt 2100-2800 F-> Say 1400C

Ambient to melting temperature ~= 1400 x 420 = ~~ 600 kJ/kg
Melt - Say 300 kJ/kg to get it nicely splattered.

Heat + melt = 600 + 300 = Say 1000 kJ

30,000 A will dissipate I^R.t J
Use 20,000 A for 1 mS

E = I^2.R.t
R = E/t/I^2 = 1000 kJ /.001 / 30,000^2 =~~ E3 x E3 / E9 ~= 0.001 Ohms.
Higher than I would have expected.

To not melt from sheer heat absorption alone a 1 kG sli=iug of steel has to have a face to face R of
R <~~~= 1 milli Ohm

SG of steel is ~~~= 8
Resistivity of steel ~~~= 1 to 2 x 10^07 Ohm.meter.

1kg = 1/8 SG x 1/1000 tonne = 1/8000 m^3 = 0.000125 m^3

Using square wire because it's late :-)

t = thickness of slice in m R= t/A Rho = t^2/V.Rho
so t = sqrt(R . V . E^7)
t = sqrt(0.001 x 0.000125 x E^7)
= sqrt(1.25) =~ 1.1m
Which has to be wrong as it makes cross section (1/8000)/1.1 m^2
or about 11 mm square iron bar to resist 30,000 A for 1 mS.
Hard to believe.

However - the same energy to achieve full melting of the same kg of steel would equally well be delivered by 3000 A in 100 mS.
Or 300A in 10 s IF no heat energy was lost.
So if the answer is correct (unlikely) you may just about not melt a 11mm square steel bar in open air at 300A.

Note that whereas for water the energy for vaporisation is > energy for heating from ambient. because steel has such a long way to go from ambient, the heating energy is about 2 x actual melting energy (if I read the tables right).

Anyone is welcome to revisit both the assumptions, constants, theory and figurings above.
The odds of there not being an error are close to zero.
I'm off to sleep - dentist's appointment for sore tooth 6.5E0 hours from now.

Open tabs. Relevance tbd:






http://goo.gl/1S4cm :-)

  • 2
    \$\begingroup\$ Ugh, another Russell McMahon brain-spew. It's like a drive-by shooting, where the bystanders are left to clean up the mess and retrieve anything useful out of it. Please come back and turn this into some sort of coherent argument. \$\endgroup\$
    – Dave Tweed
    Nov 12, 2014 at 14:49
  • \$\begingroup\$ I see somebody thinks theanswer is not useful. So be it. Sad for them though. Luke seems to be able to handle the material well emough. I may well have come back (sans a tooth and in more pain now than before it was extracted :-) ) and formated the answer better. But, after @DaveTweed's crass rudeness [although he falls disappointingly short of his master's abilities :-( ] plus complete lack of interest by anyone other than Luke, the game's not worth the candle. .... \$\endgroup\$
    – Russell McMahon
    Nov 15, 2014 at 23:01
  • \$\begingroup\$ .... It's interesting to note that Onderdonk takes no account of either thermal capacity or melting energy of the material concerned. He may be assuming Copper? | Onderdonk posits: I = A*(log(1 + (Tm-Ta)/(234+Ta))/33*s).5 Where I = current in Amps, A = cross sectional area in circular mils, Tm = melting temperature of the material in oC, Ta = ambient temperature, also in o C, and s = time in seconds. | He has melting temperature there so not fully assuming Cu, but .... .... \$\endgroup\$
    – Russell McMahon
    Nov 15, 2014 at 23:01
  • \$\begingroup\$ .... What could have turned into an interesting discussion amongst informed minds (and surrounding parts) instead not quite dies aborning. I guess @DaveTweed has achieved his aims then. \$\endgroup\$
    – Russell McMahon
    Nov 15, 2014 at 23:05
  • 1
    \$\begingroup\$ I think you've been here long enough to know that "an interesting discussion" is not what this site is about. It's about clear answers to technical questions, and wading through your stream of consciousness (complete with mistakes and irrelevant references) is not likely to be helpful to future visitors. I wish that when you get "inspired" by a topic like this, you'd write the stuff above into a private file on your own machine, then come back after a good night's sleep to create a proper answer for the site that we could all be proud of. \$\endgroup\$
    – Dave Tweed
    Nov 16, 2014 at 0:47

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