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*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: Into detail*From*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Date*: Thu, 03 Jul 1997 21:23:12 +0100*Reply-To*: TLG.van.der.Linden@tip.nl (Timothy van der Linden)*Sender*: owner-starship-design

Kelly, Regarding your questions: >Actually The largest scope I mentioned was Hundreds of thousands of miles >across. Ah, I must have read hundreds OR thousants. >Thou the calculations are a good Idea. To a degree you could scale >up the scopes to any size nessisary. How big would it need to be? Well a factor 2 bigger 300,000 miles instead of 3,000 miles will resolve objects of: d=1.22 * 9.45E16 * 5.3E-7 / 5.556E8 = 110 metres Besides Rayleigh's criteria, there is the point of brightness. To create a visible image on each separate telescope I need to know the minimal brightness (in Watt/m^2) that those CCD cameras can "see". I don't know this "minimal brightness", but assume that one needs at least 1 photon per pixel. Assuming that the planet we look at reflects 400 Watt per square meter (in the visible range), than at Sol we have only 400/(2 pi r^2) = 400/(2 pi 9.45E16^2) = 7.1E-33 Watt/m^2 (I use 2*pi*r^2 which is the surface of half a sphere over which the light is reflected) Luckely we didn't want 1 meter detail, but110 meter detail, so we have 110^2 more Watts here at Sol: (110^2)*7.1E-33=8.6E-29 Watt/m^2 A single photon has an energy of h*c/lambda=(1E-34)*(3E8)/(5.3E-7)=5.7E-20 Joule So that means about 1 photon per second per 6.6E8 square meter. Say we need 1E6 photons for a photograph (theoretical 1000x1000 pixels). We make a 1 second photograph. (Can't do longer, otherwise the planet has turned much more than 110 meters) Therfore we need an aperture surface of 1E6*6.6E8=6.6E14 square meters. Hmmm, that means a aperture radius of 14.5 kilometers. Most of my estimates have been quite optimistic, so it could well be that one needs a few orders bigger. Timothy P.S. If anyone has more hard data, I may be able to give a closer estimate.

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