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Re: [TowerTalk] Grounding base slab and pier tower bases

To: Roger D Johnson <>,
Subject: Re: [TowerTalk] Grounding base slab and pier tower bases
From: Jim Lux <>
Date: Tue, 04 Jul 2006 13:36:41 -0700
List-post: <>
At 11:31 AM 7/4/2006, Roger D Johnson wrote:
>David Robbins K1TTT wrote:
> > 20' isn't all that much rebar.  There's probably that much in a 3'x3'x4'
> > block for rohn 25/45 type base when its done right.  And I would be
> > surprised if anyone could put such a limit on 'absorption' by a ufer ground
> > like that, sounds like something pulled out of thin air to me.  The truth
> > is, concrete is a pretty darn good conductor in its own right, 
> especially in
> > ground where it stays relatively moist.  The whole reason for concrete
> > encased ground rods is to improve the conductivity and spread the current
> > out more so that the current density is lower when it gets into the
> > surrounding soil that is a worse conductor.
>I really need to find Ufers paper on the subject. As I said originally,
>my understanding is that the capacitance to grounds important to absorb
>the charge which then leaks off through the concrete. You seem to be
>treating it as a pure resistance. I have my doubts whether the average
>tower base will carry 10-20 Kamps and, even if it does, it has to dump
>that charge into the surrounding earth.

The reference is on my lightning info web page:
It's an obscure conference paper, so if you find a source, please let 
everyone know where you found it.  I'm perfectly willing to pay copyright 
copy fees for it.

But, to return to the Capacitance issue, and whether the C holds the entire 
charge, or if it's just a component in the circuit... we can ballpark this 
a bit, to see if it's reasonable.  Two ways to consider it:

First, let's consider it as a sphere in free space (usually a lower bound 
on C), for which C = about 110pF/m radius.  Let's say that it's a 3 meter 
radius sphere. The C is then about 330 pF.  The charge transfer in a 
typical lightning stroke is around a Coulomb, so we can figure what voltage 
this would be: Q=CV; V=Q/C; V=1/0.33E-9; V=3E9 volts...  Nope, way, way too 
big to be even quasi reasonable.

Let's try another strategy.. Let's say the slab is a big plate on a 
capacitor, separated by 1mm from earth.  Let's further say that it's 3x3 
meters (10x10 ft, or thereabouts).  C=A/d*epsilon0 = 8.84E-12 *A/d = 
8.84E-12 * 9/1E-3 = 80E-9.  Run the same voltage calculation as above: 
V=1/80E-9 = 12.5 MV... still awfully big.

Now, let's look at the reactance of that C, at, say, 100 kHz (a standard 
lightning impulse is 2 microsecond rise and 50 microsecond fall, so let's 
say that it's like 100 kHz). X=1/(2*pi*100E5*80E-9) = 1/(2*pi*8000E-4) = 
1/(6*0.8) = 1/5 -> 0.2 ohms....  That's pretty low.. MUCH lower than the 
resistive impedance to the earth from a ground rod (5-20 ohms).  With that 
20 kA impulse, you're looking at a 4 kV voltage drop, which isn't all that 
huge.  Given that the rise time is actually much faster, and the reactance 
lower,  the voltage will be less.

It would appear then, that at least as far as ballparking goes, the Ufer 
ground looks like a capacitor to ground for the lightning impulse.  (And, 
the existence, or not, of an insulating vapor barrier is immaterial).



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