Re: starship-design: Required microwave-antenna size. 100-light-year trip.

Rex wrote:

<Tutorial removed>

>If we set the Fraunhofer-diffraction angular width equal to the
>angle subtended by the sail diameter (Ds) at the range R, we can
>get an expression for the diameter of the emitting antenna (De)
>that is required to put "84 percent" of the emitted power within
>the area of the sail:
>     De = 2.44 * lambda * R/Ds   .
>
>For microwaves with lambda = 1 cm and a sail diameter of 100 km,
>say, the required diameter of the emitting antenna grows 2.31
>million km for each light-year of distance to the sail, to keep
>the main lobe just within the sail diameter.  This is for an
>antenna whose figure (shape) is correct to a fraction of a wave-
>length.
>
>For an emitting-antenna diameter of 2.31 million km (and wave-
>length of 1 cm) for a 1-lt-yr distance to the sail, the boundary
>between near and far fields is at a range of about 23 thousand
>lt-yr, far enough away to say we're operating in the near field.
>[Note: at 1 lt-yr distance, the spot size is
>(1 lt-yr/23,000 lt-yr) * 2.31 x 10^6 km = 100 km, the diameter of
>the sail, as desired.]
>
>Are we prepared to build an expanding antenna this large?  It
>looks as if the M(aser)ARS should be a L(aser)ARS.  For any wave-
>length, however, figure control is going to remain a headache.

What if
lambda = 500 nm = 5E-7 m, (Blueish green)
R = 10 ly = 9.46E16 m
Ds = 300 km = 3E5 m
Then
De = 2.44 * lambda * R/Ds = 2.44 * 5E-7 * 9.46E16 / 3E5 = 3.85E5 m = 385 km

range > De^2/(2.44 lambda)  ---> 9.46E16<1.215E17

So we are just around the border between the near and far field.

300 and 385 kilometer diameters don't seem like a headache...

Timothy