Some back-of-the-envelope calculations about Indigo

[kearsley]

Indigo is set on a Dyson sphere with a radius of 1 AU. I was curious if the Commonwealth having explored approximately 2000 tectons (earth-sized landmasses) was high or low.

The surface area of a sphere is 4πr2. An AU is 1.495e9 km. That means that the Dyson sphere has a surface area of 2.8e17 km2. Earth has a surface area of 5.1e8 km2. That

Comments

[magicbox]

17 seconds? What about from one tecton on one side of the sphere to a tecton on the other side? At 1 AU, that should be about 16 minutes in direct line, much higher (about 25 minutes) if the transmission can't penetrate the indigo sky. This is all based on the simple standard that if the sun burnt out, we wouldn't know for 8 minutes.

Yeah, dyson spheres are messed up when you look at the physics.

[kearsley]

I misremembered 1 AU. I thought it was 8 light-seconds, when it's actually 8 light-minutes. Post edited.

[magicbox]

Looking at my reply again, it may seem like I'm a miffed physics teacher smacking you upside the head. That was not the intent or the tone I intended. I was just curious how you got a VERY different number than I got by just thinking for a sec. :)

I think your other calculations look good though. Dyson spheres still don't work well :) The ring in Halo MIGHT work, but it's on a much smaller scale since it is smaller than a moon and the centripetal force should be able to deal with it... as long as the ring doesn't fall apart due to stresses. Again, lots of hand waving has to be done.

[kearsley]

I didn't take it that way. That was the one bit of the post that I didn't double-check and typed from memory. Well, that and the shell theorem. I made a whole lot of assumptions, based on knowing that a sphere can be treated as a point mass, thus the sphere has to be a point mass from outside, then made a few hasty calculations based on slicing a hollow sphere into multiple pieces.

I went and checked the shell theorem the next day, because I don't trust my ability to do three-dimensional integral calculus in my head, but forgot to do so with the distance to the sun. In retrospect, knowing that it's ~1 light-second to the Moon, I should have noticed that I was using the wrong units.

Honestly, I prefer showing my work and having feedback if I'm wrong. It encourages me to double-check my math before I draw conclusions

[kearsley]

For those that actually care, here's a transcript of the math used. 'setq' is equivalent to 'let'.

(setq AU 149598000.0)
149598000.0

(setq area (* 4 pi AU AU))
2.8122986530274643e+17

(setq earth-area 510072000.0)
510072000.0

(setq volume-of-water 1000000000.0)
1000000000.0

(setq number-of-earths (/ area earth-area))
551353270.3280056

(setq all-water (* number-of-earths volume-of-water))
5.5135327032800557e+17

(setq percent-explored (* 100 (/ 2000 number-of-earths)))
0.00036274383551043054

(setq water-sphere-radius (expt (/ (* all-water 3) (* 4 pi)) 0.333))
502042.71332953154

(setq sun-radius 6.955e5)
695500.0

(setq percent-of-sun (* 100.0 (/ water-sphere-radius sun-radius)))
72.18443038526694

(setq ckm-in-cc 1.0e15)
1e+15

(setq density-of-lithium (/ (* 0.535 ckm-in-cc) 1000))
535000000000.0

(setq density-of-water (/ (* 1 ckm-in-cc) 1000))
1000000000000.0

(setq weight-of-sphere (* density-of-lithium area))
1.5045797793696934e+29

(setq weight-of-water (* density-of-water all-water))
5.513532703280056e+29