[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: starship-design: Into detail

In a message dated 7/3/97 10:53:10 PM, TLG.van.der.Linden@tip.nl (Timothy van
der Linden) wrote:

>Regarding your questions:
>>Actually The largest scope I mentioned was Hundreds of thousands of miles
>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

Not bad over interstellar distences.  If you see anything really interesting
you could probably get funding to scale up the scope array.  A 4 order of
magnitude improvement should get you 1 meter res.  8 order centimeter rest.
 Enough to study life forms (if any).

Expensive, but not compared to the starship project.

>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)

The scope aray or post processors could compensate for the rotation.

>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

Ah, I make that 2.5e7 meters per side.  How'd you get 14,500?


>P.S. If anyone has more hard data, I may be able to give a closer estimate.