====================SKIN_EFFECT=====================================

skin effect         high frequencies as follows:
                          With ro    = radius of the round wire (meters)
                               del   = skin depth (meters)
                                     = 1/(pi*f*mu*sigma)^.5
                               f     = frequency
                               mu    = material permeability (henrys/meter)
                                       (4*pi*10^-7 for good conductors)
                               sigma = material conductivity (mhos/meter)
                                       (5.8*10^7 for copper, for example)
                               Ro    = 1/(pi*sigma*ro^2)     (DC resistance)
                               if      ro/del < 2  (low frequencies)
                                       R/Ro = 1 + (1/48)*(ro/del)^4
                                       (approx. 6 0.000000e +00rror at ro/del = 2)
                               if      ro/del > 2  (high frequencies)
                                       R/Ro = ro/(2*del)
current               | Iz/Io | = e^[-(ro-r)/del]
                      where r is distance from center of to the edge,
                      the current has decreased to e^(-2) of value at edge.)
                     
                      > piece of wire due to the skin effect at high 
                      frequencies. I have two
For copper wire.      divide 66 by the square root of the frequency in Hertz.  
                      That will give you the skin depth in milimeters.
                      For 100 KHz, that works out to about 0.2 mm.
                      Since wire is already smaller than skin depth,
                      resistance will change little if at all.
virtually all current exists within two skin depths.
                      For a round copper wire, no plating,
                      R=(1/(2a))*sqrt(rho*f*mu/pi)  Ohms /meter length
                      where a =radius of wire in meters
                      rho= copper resistivity Ohm-meter = 1.74*10^-8
                      f =freq Hz
                          mu=permeability of copper 4*pi*10^-7 Henries per 
                      meter
                      R = 4.17*10^-6 *1/a * sqrt(f)   for copper
                      Condition: radius a >> skin depth

resist_skin_effect    The formula is Skin Depth) = 1/sqrt(Pi*f*mu*sigma)
                      mu = pemeabillity of the metal
                      sigma = conductivity of the metal
                      All you have to remember is two facts. One, 
                       resistance of 40 AWG is about 1 ohm
                      per foot. Two, resistance doubles (or halves) 
                      every three gauges. So,
                       resistance of 37 gauge is 0.5 ohms/foot, 
                      34 gauge is 0.25 ohms/foot....
                      40 AWG is really 1.046 ohms/foot, 
                      but for a quick and dirty, nothing beats it.
                      Ohms per thousand feet = 10^(0.1 * Gauge Number - 1)
                      ->>     where Gauge Number is American Wire Gauge.
                      Actually the correct formula is:
                      ohms/ft = 2^((AWG-40)/3)     or:
                      ohms/1000 ft = 0.001 * 2^((AWG-40)/3)
                      Funny, looking at it this way, 
                      where you can find the thermal resistivity of copper
                      as 401W/m.K at 300K
                      A 1 ounce per square foot copper trace is 
                      34 microns thick, so a typical
                      12 thou (305 micron) wide trace will have 
                      a conductivity of 4.2uW/m.K