From: Robert J. Bradbury (bradbury@www.aeiveos.com)
Date: Fri Sep 17 1999 - 19:31:11 MDT
On Fri, 17 Sep 1999, Robin Hanson wrote:
> You are suggesting that an optimum is m1 = 0 and m2 = M, putting all
> the metal at one star so as to get every last photon and reradiate at near
> 3K, while completely leaving the other star alone. This requires that
> P''(m) > 0 on average, with *increasing*, not diminishing, returns.
No. I'm suggesting that there are always decreasing *local* optima.
If there were *no* costs to transfering material (or information) from
1 to 2, then doing that would make sense. But if the costs of the
material/information transfer *exceed* the cost of local manufacture
then it makes sense to reject remote information/material (if you
have to pay for it).
>
> Instead I expect any plausible model of metal-limited Dyson sphere to
> show decreasing returns: The first few tons gives lots of power while
> the last few tons gives a lot lot less. This implies that to get the
> most power one shoul spread the metal evenly across the two stars:
>
> m1 = m2 = M/2 .
>
I would agree with this *in principal* -- [i.e. we are subject to
significant diminishing returns with metal usage]. Even a single
solar cell harvesting layer is likely to get you 20-30% of the
power of the star. All successive layers only get a cut of what
is left over. But *the* most efficient use of the local energy
will harvest it at the highest temperature possible (the melting
point of the harvesting materials) and radiate waste heat at the lowest
possible temperature (the background radiation, unless you are too
close to some other "hot" object).
So in your presentation if star m1 is rich and star m2 is poor
it would make sense to balance the situation. However this ignores
the costs of transfering m from s1 to s2. It might make more
sense to manufacture m at s2 and keep them there.
If you can make a case that I harvest 20-30% of the energy
in a system using a minimal amount of available mass and then
transfer the remaining available mass to a "mass-short"
system, and you can make the case that this is optimal
in terms of computational throughput then I will accept
this as a reasonable strategy.
Robert
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