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Re: [TowerTalk] NEC (modeling, not code) question

To: Máximo EA1DDO_HK1H <ea1ddo@hotmail.com>, towertalk <towertalk@contesting.com>
Subject: Re: [TowerTalk] NEC (modeling, not code) question
From: jimlux <jimlux@earthlink.net>
Date: Thu, 9 Apr 2020 07:16:38 -0700
List-post: <mailto:towertalk@contesting.com>
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@contesting.com> en nombre de jimlux <jimlux@earthlink.net>
*Enviado:* jueves, 9 de abril de 2020 0:13
*Para:* towertalk <towertalk@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.

_______________________________________________



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