Describe a beamed power solution. ( Digest Number 135)

here is a list of Criswells papers I ended up reading one of Criswells papers somewhere on the net while following links that were provided earlier. I hit one statement that caused me to scratch my head. It said that his scheme was predicated on using the antenna near field. I have been doing airborne radar simulations for decades for flight training simulators. I have never had to consider the effects of the near field in any of those simulations some of which have been very intense. So I went back to a primer on radar and found a page discussing the near field in relationship to reflections from rain.
If I got the math correct it will require a square antenna 66.4 Kmeters per side (square is better than round for these equations for some reason). This has to be a phased array antenna which is totally mind boggling considering the cost and difficulties in creating an earth based 20 ft phased array antenna.
For those who like technical details I provide the following:
"The electromagnetic field in the Fresnel (near field) region is quite different from that in the Fraunhofer region although there is a smooth transition from one to the other. The most significant feature is that the radiation is largely concentrated within a cylinder whose cross section is the aperture area until the distance from the aperture is approximately that designated as the transition range. This behavior is also indicated by the on-axis power density which at large distance varies inversely as the square of the distance but in the Fresnel region oscillates about a constant value. It may also be noted that at the transition range k*r0 a cone with solid angle "theta" would intercept an area of (k**2)Ae ......"
So the transistion range is defined as k*r0. The rest of the task is trying to find what these letters stand for.
r0 = Ae / wavelength
Ae = (0.4 to 0.7 for different types of antennas) * Area of square antenna
Area of square antenna = Length of side **2
wavelength = speed of light / 2.4 Ghz
speed of light = 2.99*10**8
K = (1.0 to 2/3 depending on antenna shape, square = 1)
Dl = Distance from the earth to the moon = 384,401,000 Km
Solve for Length of side.
Also the radius of the moon is 1737Km which will also affect such a large antenna.
Mitchell James
www.InnerTransit.org (Homebase for collaborative engineering)