Lightning strikes have been known to cause massive amounts of damage. The stats on a lightning bolt are:

current levels sometimes in excess of 400 kA, temperatures to 50,000 degrees F., and speeds approaching one third the speed of light

These are massive numbers, but lightning protection systems are designed to draw the lightning away from the building or structure that they are protecting. Lightning protection systems can be simply thought of as lightning rods connected to the ground via cabling (downconductor).

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The NOAA specification for lightning protection requires that lightning rods be at least 0.5in (13mm) in diameter. The downconductor is a similar size copper cable (4/0 AWG or 12mm). The allowable amperage for this type of wire is only about 250A for constant current. I realize that this is more of a heat limit rather than a instantaneous current capacity limit.

From this paper on lightning protection (page 28):

Positive feedback on the operation of a lightning protection system is seldom documented and most often not even noticed. Only in some rare cases can it be documented that a lightning protection system has been struck if it works properly and there is no damage. There is sometimes evidence at the strike termination point which can be noted during a careful inspection, but it is seldom cost effective for the owner of a lightning protection system to obtain the expertise necessary to conduct such a careful inspection.

How can a seemingly small 0.5in (13mm) piece of metal handle a lightning strike with little or no visible damage much less without being completely destroyed?

  • $\begingroup$ A solid half inch rod can withstand a lot of current and heat! $\endgroup$
    – Paul
    Feb 22, 2015 at 19:49

2 Answers 2


The current limit specification for a wire is limited by the heat that current will produce, and how much heat the wire can dissipate before getting too hot. "Too hot" depends on the circumstances. You will see higher current ratings for the same type of wire in chassis wiring applications than the electrical code allows for home installations, for example. This is mostly due to how hot too hot is. The ultimate limit for extreme applications is that the conductor not melt. Temperatures anywhere near that would be unsafe running along wooden supports inside a wall in a house.

As you say, the wire is rated for 250 A continuous. Lightning is decidedly not continuous. 1 ms is "long" for the time of the main lightning stroke. There may be several strokes in one event, but the total time is still short, and the other non-main strokes will have significantly less current.

Do the math. You say the wire is 12 mm in diameter, so has a cross-sectional area of 113 mm² = 113x10-6m². The resistivity of copper at 20°C is 1.68x10-8 Ωm. A 1 meter length of this wire therefore has a resistance of

(1.68x10-8 Ωm)(1 m)/(113x10-6 m²) = 149 µΩ

The power with 400 kA thru this resistance is then:

(400 kA)²(149 µΩ) = 23.8 MW

Times the 1 ms time the current is applied yields the energy:

(23.8 MW)(1 ms) = 23.8 kJ

The density of copper is 8.93 g/cm³, and our 1 m length has a volume of 113x10-6 m³, which is 113 cm³.

(113 cm³)(8.93 g/cm³) = 1010 g total mass of copper

The specific heat of copper is 0.386 J/g°C.

(23.8 kJ) / (0.386 J/g°C)(1010 g) = 61°C

This means putting 400 kA through a 12 mm diameter copper wire for 1 ms will cause a temperature rise of 61°C. That's a rather extreme value for a lightning strike. The main stroke is usually substantially shorter than 1 ms, and the other strokes have substantially less current. However, even with these numbers it shows that while the wire will certainly get toasty for a while, it is well within the wire's ability to handle without any structural failure.

  • $\begingroup$ Worth noting that the limiting factor on the current capacity of the cable is usually the failure of the insulation - not the conductor. That said, cable insulation is typically rated at 75 or 90 degrees Celsius continuous. The copper will heat up effectively instantaneously and the heat will then dissipate through the insulation. $\endgroup$
    – Dale M
    Feb 23, 2015 at 4:18
  • 2
    $\begingroup$ @DaleM, are the down connectors usually insulated? Here in the UK they're often uninsulated flat copper strips say 25×6 mm, in which case insulation temperature rating isn't an issue. $\endgroup$
    – Chris H
    Feb 23, 2015 at 12:21
  • $\begingroup$ @ChrisH Yes, that is one method. For high-rise buildings you can also use an attractor with a single down conductor that is insulated, at least in Australia. $\endgroup$
    – Dale M
    Feb 23, 2015 at 22:40
  • $\begingroup$ @DaleM fair enough - I'm not an expert and I wasn't thinking of high rise buildings anyway. $\endgroup$
    – Chris H
    Feb 24, 2015 at 8:42

[ This started as an answer. But the calculation ended up 3 orders of magnitude short.
So, this is more of a comment, I guess. ]

A single lightning bolt has a finite amount of energy 1. The ballpark number is $5 \cdot 10^9 \text{J}$, which is equivalent to 145 L of petrol2.

Let's assume that all of the energy of the lightning bolt is dissipated in the rod. Let's say that out lightning rod in 12mm diameter and 10m long. Density of copper is 8960 kg/m3,
so the mass of our rod is $10.13 \text{ kg}$.
The specific heat of copper is $\text{0.385} \frac{\text{kJ}}{\text{kg °C}}$. How much energy does it take to warm up the copper rod from 25°C to the melting point of copper, which is $1083 \text{°C}$? It's $4.05 \cdot 10^5 \text{J}$.
The heat of fusion for copper is $\text{213} \frac{\text{kJ}}{\text{kg}}$. How much energy does it take to melt our rod? It's $2.13 \cdot 10^6 \text{J}$.

I have made an assumption that all of the lightning's energy is dissipated in the rod. But, I don't know of that's a sane assumption.

1 Even though it has very high peak power and current.
2 Source
3 Who were Preece and Onderdonk? The article about early (1880, 1928) equations describing currents that cause wire to melt.

  • 5
    $\begingroup$ I think only a very small fraction of a lightning strike's total energy gets dissipated in the rod. The current is the same everywhere along the lightning, so the energy dissipated at each location will be proportional to the resistance of the conductor. A few miles of ionized air should have significantly more resistance than a few 10s of meters of copper wire. $\endgroup$ Feb 22, 2015 at 21:45
  • 4
    $\begingroup$ The bulk of the energy is discharged in the ground. $\endgroup$
    – Dale M
    Feb 23, 2015 at 22:51

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