Describe a beamed power solution. ( Digest Number 135)

>If using a reflector, then we can eliminate the RF to DC and DC to RF
>conversion in the middle...
>
>That puts us at 50.8% efficiency... instead of 38.5% for the relay...
>
>Tom M.
>TomM@...

You need to add in a loss due to angle of the reflector. When the
reflector is not perfectly perpendicular to the incomeing wavefront
there is a power loss due to phase shift of the wave as each peice
of the wave front gets to the reflector. Power =K (sin(pi(L/wavelength)sin(angle))/(pi(L/wavelength)sin(angle)))squared.
Where L is the length of the reflector. If you think of the reflector
as rotating around one end of the reflector. If the distance that
the other end makes from perpendicular is equal to the wavelength
there is complete cancelation of the signal.

A reflector is going to act like a solar sail. Station keeping due
to momentum transfer will have to be calculated. The relay should
have the same problem. I don't know if the momentum transfer would
be signficant.

Mitchell James
mejames@...
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