[ExI] On vacation until July 2 Re: Tabby's star

[Keith Henson]

It's been a very long time since the problems of computing out in
space were discussed here.  We may be seeing a K2.  It looks like they
are using a huge structure to hold the power and radiating equipment.
This way you get a fixed speed of light delay on the computing
elements.


The diameter of Tabby's star is ~1099206 km

Crossing time from center to center (the dip) is around a day.

D/24*3600 about 12.7 km/s.  In the solar system, the orbital speed for
Jupiter is 13 km.

M is 6.6743 × 10+11 * 1,43
G is 1.98847)×10-30  18.97845280903 10+19 v is 12700 m/s

r = GM/v**2  0.11766664274927149854299708599417  10+13, ~7.8 Au

Jupiter is 5.2 AU

luminosity is 4.68, so at 1 au 6388 W/m^2, this power is divided by
7.8^2 tp get 105 W/m^2,  ~0.1 GW/square km.

The area of Tabby's star is D/2 squared x pi .  22% blocked would be
208771274655 square km, 409 times the area of the earth.  As a square,
456914. km on one side..

The input power is over 1220 times what the human race uses

To radiate 105 W at 65K (measured), the radiator surface will need to
be about 50 m^2 for every square meter of light input.  This looks
like it is optimized for computation.

Because the radiators can't view either other, the whole thing might
be implemented as a deep V-shaped wedge with the sunlight going down
the middle and the radiators on the outside

Comments welcome.  If you want to check the math, please do.

Keith




>
>