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Originally Posted by Alan Wheeler  I should qualify that . Because the lightning has just traveled several miles through a virtual insulator (Air) the last few meters via the mast means nothing to the strike... |
Wheels, just want to correct a minor point here, as I understand it, the lightning doesn't really travel through the air, rather it travels along a path of ionized gas (ionized air in this instance). The ionized air has a much much lower resistance than air. The air is ionized when a sufficent voltage exists to cause the ionization procress, this occurs whenever any spark (big or small) is produced in air.
Off the topic, but have you ever noticed that you can't create a spark from a 12 volt battery by shorting out the terminals as the conductor is brought up to the battery post. The spark only occurs after the conductor is brought in contact with the other terminal and then removed. It is the back EMF generated by the collasping of the current flow in the conductor that raises the voltage high enough to ionize the air grap and allow current to continue to flow (as a spark). I don't remember the exact figure but IIRC, one needs more than 10,000 volts to ionize air at sea level. Of course we all know that shorting out a 12 volt battery is very dangerous and should never be done at home
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Gord,
Thanks for the information but my question remains is how does an "Air Terminal" bolted to the masthead of an Al mast really help. I figure the masthead itself would effectively be an "air terrminal" with the mast being the conductor. Of course it maybe not as perfect as a purpose built one.
If one considers the sacrificial aspects of a say a 3/8 copper rod as the air terminal, then I thinking the said copper air terminal could disappear very fast during a strike leaving the mast head becoming the "new" air terminal while there is still a considerable charge to be dissipated.
But perhaps I have the wrong end of the cat here and missing something basic.