A couple of weeks ago we had a big storm roll through the Chicago area. I watched as the city's skyscrapers were struck multiple times by lightning throughout the night and told my coworker about it the next morning.

That got us both thinking, is there a maximum load that a lightning rod/lightning dissipation system can handle? Is it possible to overload or burn out a rod on a building like the Sears Tower? How massive/long-lasting of a lightning strike would you need for that to happen?

(This is also posted on the Earth Sciences SE where someone suggested I post it here)

  • 1
    $\begingroup$ I don't think it would so much as overload as it would melt. But I'm interested to see what everyone else says $\endgroup$
    – Jim
    Jul 10, 2014 at 19:54
  • $\begingroup$ It definitely is possible. Lightning is just an electric discharge so you could drive a large current which paired with the resistance of the rod would cause a large amount of heat to accumulate in the rod. Like @Jim said this could cause the rod to melt. If you don't get an answer by tonight I'll get you specifics when I have some extra time. $\endgroup$
    – jkeuhlen
    Jul 10, 2014 at 20:08
  • $\begingroup$ The weakest link (highest resistance and/or impedance) section of the lightning rod and its grounding system is going to heat up the most and is most likely to melt through (fail). In a skyscraper, generally there's good enough conduction through riveted steel beam connections to let the frame act as a path to ground, although copper straps may be added. For a reinforced concrete structure, there would probably be an explicit lightning grounding system. $\endgroup$
    – Phil Perry
    Jul 11, 2014 at 16:03
  • $\begingroup$ Lightning rods can suffer blast damage and erosion from repeated strikes, and need repair or replacement. Metal statues atop monuments and buildings act as lightning rods, and after a while get strongly pitted and need repair. $\endgroup$
    – Phil Perry
    Jul 11, 2014 at 16:05

1 Answer 1


Yes, it's possible. Though, I'm sure the engineers of the Sears Tower took that into account.

Catastrophic failure of the rod is pretty straightforward. Lighting is a ton of electrical current, and the rod has some resistance. Current through a resistance makes heat by Joule heating, which says that the power is proportional to the resistance times the square of current.

A big strike can have currents up to 120 kA. The resistivity of copper is about $1.68 \cdot 10^{−8} \Omega \mathrm m$, so the resistance of a typical ground rod used in US residential construction (5/8 in in diameter, 8 feet long) is something like:

$$ \frac{1.68 \cdot 10^{−8} \Omega \mathrm m \cdot 2.5 \mathrm m}{2\cdot 10^{-4} \mathrm m^2} = 0.21 \mathrm m \Omega $$

Now, lighting is a pretty fast pulse, so a lot of the energy will be high frequency, so our copper rod will be subject to skin effect, which will effectively increase the resistance by some factor dependent on frequency. To keep our estimate brief let's skip the full Fourier analysis on lighting and just say the resistance increases by an order of magnitude. So the resistance of a ground rod is something like $ 2 \mathrm m \Omega $.

Now according to Joule, the power in the rod is:

$$ 2\mathrm m \Omega (120 \mathrm{kA})^2 = 28800 \mathrm{kW}$$

Wow! But, lighting strikes are also really brief. A big strike transfers maybe 350 coulombs of charge. Working backwards, if the current were a constant 120kA, then to transfer that much charge would take:

$$ \frac{350 \mathrm C}{120 \mathrm{kA}} \approx 3 \mathrm {ms} $$

So all that power for that 3mS means a total energy of:

$$ 28800 \mathrm{kW} \cdot 3 \mathrm {ms} = 84 \mathrm{kJ} $$

Wolfram Alpha puts that in perspective as about the energy released by burning two grams of gasoline.

Now, you would go on to calculate the heat capacity of our grounding rod, and figure out how much hotter this energy would make the rod, and determine if that's enough to melt it. But, just through intuition and experience, I can tell you that with two grams of gasoline you can make a grounding rod pretty hot, but not melt it.

Of course, there are all kinds of effects for which we are not accounting. Lightning strikes are so fast that we must consider the inductance and thermal resistance of everything. The parts of the grounding system that carry most of the current (the surface of the rod) will get the hottest fastest. But also, the inductance of the entire grounding system helps to limit the current by spreading it over a longer time (storing the energy temporarily in the field of the inductance). And since power is proportional to the square of the current, moving the same amount of charge over a longer period of time means significantly less Joule heating in the conductors.

Also, Sears tower doesn't have just one little ground rod. It has huge steel piles that go way deeper than 8 feet and are many orders of magnitude more massive.

As far as multiple strikes accumulating to cause damage, I doubt it. Even in a big storm, the time between strikes is very long relative to the duration of the individual strikes. In this entire time, anything that got hot has time to transfer its heat to less hot things nearby. The bits in the ground have the whole Earth as a heat sink. The bits above ground are very massive.

This might seem counter-intuitive, since after all, lightning is really powerful, right? I mean, it's super bright and loud. But consider, most of the energy in the strike goes into the flash of light, and heating the air, making thunder. Though a lightning strike does indeed release a ton of energy, most of it is expended working on the air and to electromagnetic radiation. All we need to do is provide a path for the current, and it takes much less energy to move current through a metal rod than it does through air. So, the metal rod needn't absorb most of the energy.

  • 3
    $\begingroup$ In short, YES it is possible, but lightning rods and, more importantly, lightning strike management systems are engineered to withstand the strikes. $\endgroup$
    – jkeuhlen
    Jul 10, 2014 at 20:54
  • $\begingroup$ Why do your calculations based on the thick, rigid ground rod hammered into the ground, and not on the (relatively) flexible 1/4 inch copper wire or 3/8 inch aluminum wire connecting the lightning rod to the ground? $\endgroup$
    – David Cary
    Jul 11, 2014 at 0:34
  • $\begingroup$ @DavidCary the lightning protection system of a skyscraper is made from materials that are way, way beefier than even the puny ground rod in the calculations. Like, steel beams and stuff. In applications where there isn't a handy skyscraper (radio broadcast towers, for example), the connections are made with beefy copper grounding strap, usually 0.022" thick and 2" wide. The electrical wiring (to the outlets and what not) in a properly designed system doesn't take any of the strike current. $\endgroup$
    – Phil Frost
    Jul 11, 2014 at 2:11
  • $\begingroup$ Ah, right. However, several 2-story buildings in my area have a wire from the lightning rods on the roof to a ground rod pounded into the ground. I hear "Lightning Cable must have at least 59,450 circular mils of conductivity." (which is approximately 1/4" diameter). A 0.022"x2" strap is about 56,000 circular mils -- less beefy, right? P. 5 of "How Does a Lightning Rod Work" mentions "The US lightning code suggest approximately 1/4 in. diameter copper wire". $\endgroup$
    – David Cary
    Jul 11, 2014 at 3:13
  • $\begingroup$ I think you meant $\text{ms}$ as your time unit? (i.e. lowercase) $\endgroup$
    – David Z
    Jul 11, 2014 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.