[BC] need a NON-technical explanation

[Harold Hallikainen]

> Although it is widely used, "mV/m/kW" for the IDF at some distance is not
> scientifically accurate, IMO.  It implies that the IDF is directly
> proportional to power.
>
> If the IDF was "306 mV/m/kW," this terminology might lead one to believe
> that with 10 kW, the IDF would be 3,060 mV/m.
>
> But IDF is not related directly to the change in power -- it is related to
> the square root of that change.  With 10 kW, the IDF in this example would
> be 306 * SQRT(10) = 967.7 mV/m.
>
> A more accurate form would be:  mV/m at 1 km for X kW of applied power.
>
> RF

Should the unit be mV sqrt(kW)/m ? After all, noise is often measured in
nV/sqrt(Hz). Or, if we go back to the basic units, it would be V*sqrt(W)/m
.  Time to make up a new unit? Perhaps the Fry? Or should this
"efficiency" just be an antenna gain in dBi?

Harold