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*To*: starship-design@lists.uoregon.edu*Subject*: Re: starship-design: Required microwave-antenna size. 100-light-year trip.*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Mon, 28 Oct 1996 12:16:00 +0100*Reply-To*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Sender*: owner-starship-design

Rex wrote: >1. ANTENNA SIZE REQUIREMENT (A tutorial; read at your own risk.) <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

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