0
\$\begingroup\$

I have a flat aluminum 6061, and the measured resistance is 0.5mOhms. I have read that the resistive energy deposition is proportional to the action integral of the lightning current. In the end, I would like to know the amount of power deposited into the aluminum. I was able to find the formula, and it's shown below. How do I find the action integral.

enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ Substitute the equation for the lightning current waveform into \$ I \$ and perform the integral. \$\endgroup\$ – Transistor Jan 1 at 9:58
  • \$\begingroup\$ Thank you; that helped. \$\endgroup\$ – Sam Shurp Jan 9 at 0:19
1
\$\begingroup\$

How you do this sum depends on who you are doing it for, and why.

There's nothing complicated about the action integral, once the form of I(t) is known. You're not going to stand about in a lightning storm with a current meter and a Franklin kite, so you need to get estimates of I(t) from somewhere.

If it's just for yourself, for interest, or you're protecting your shed, then you'll search the net for typical values for lightning currents and waveforms, and play around with the numbers from those.

It you're installing a lightning conductor to protect a building, then you'll look at your local building regulations, which will say something like 'the protective conductor must withstand a strike current of foo kA for bar uS without melting'. You need to use your local regulations for two good reasons. Not only does lightning storm severity varies around the world, but also these are building regulations, you have to do the sums the way the local regulators who are going to stamp your papers want them done.

|improve this answer|||||
\$\endgroup\$

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.