[TowerTalk] NEC (modeling, not code) question

[jimlux]

On 4/8/20 11:18 PM, Máximo EA1DDO_HK1H wrote:
> Hi Jim,
> 
> I have just tried that, first on one specific wire, then on all wires.
> To be honest, I expected a error message but didn't.
> I can only guess second overrides first.
> 
> Then I've tried the same command twice, but different materials, copper 
> first then aluminium, and look what the wire info shows;
> 
> 
> **
> 
> **
> 
> <http://foro.ea1ddo.es/>
> 
> **Not sure how it behaves. New alloy?
> 
> 
> 73, Maximo.
> 
> ------------------------------------------------------------------------
> *De:* TowerTalk <towertalk-bounces at contesting.com> en nombre de jimlux 
> <jimlux at earthlink.net>
> *Enviado:* jueves, 9 de abril de 2020 0:13
> *Para:* towertalk <towertalk at contesting.com>
> *Asunto:* [TowerTalk] NEC (modeling, not code) question
> If one does a LD type 5 (wire conductivity) on the same wire twice, does
> the second one override the first, or does it essentially sum it?
> 
> Example
> 
> LD 5 800 0 0 2.5E7              makes wire 800 out of aluminum
> LD 5 800 0 0 2.5E7              does it again
> 
> So, is the resistance of wire 800 now twice what it was?
> _______________________________________________
> 
> 

I tried some experiments, trying to infer what was going on from 
measurements of feedpoint impedance, but ultimately, I just dug into the 
source code:

The loads are summed.
What happens is that for each segment being loaded, NEC calculates the 
equivalent resistance using the conductivity and permeability (mu).  It 
then adds that to the resistance for that segment.

So, if you have a scenario where some of your structure is aluminum, and 
some is copper, the way to model it is to figure out what the 
conductivity of a material that summed with copper will come out right.

For instance, aluminum is 2.5E7, copper is 5.2E7
We're working in resistivity, so you want to solve an equation that 
looks like this:

1/3.9 + 1/x = 1/2.5  - that is, the resistance of copper plus special 
material = resistance of aluminum.

Rcu + Rx = Ral

It's actually a bit trickier, because of skin effect - the resistance 
per unit length goes as  skindepth * resistivity

resistivity is 1/conductivity
but skindepth goes as sqrt(resisitvity)...

So the resistance of a given segment goes as 
sqrt(resistivity)*resistivity or
resistivity ^1.5

so we have

1/(5.2^1.5) + 1/(x^1.5) = 1/(2.5^1.5)

Grind through this with some rearrangement
and you get
X = (1/Al^1.5 -1/Cu^1.5)^(-1/1.5)

which works out to 4.04E7 for Material X.

So, if you do a

LD 5 0 0 0 3.9E7	loads the entire structure as copper
LD 5 900 0 0 4.04E7	loads tag 900 as Aluminum

You'd wind up with the right thing.