[ExI] If you follow the developments with Tabby's star . .
spike66 at att.net
Sun Sep 18 15:51:20 UTC 2016
From: extropy-chat [mailto:extropy-chat-bounces at lists.extropy.org] On Behalf
Of Keith Henson
Sent: Sunday, September 18, 2016 8:08 AM
To: ExI chat list <extropy-chat at lists.extropy.org>
Subject: Re: [ExI] If you follow the developments with Tabby's star . .
On Sat, Sep 17, 2016 at 12:49 PM, Anders <anders at aleph.se> wrote:
>>... Let's see if I get the basic argument: you have a shell of radius R.
> The luminosity L is absorbed, and in the standard model assumed to all...
>> ...extra IR now is radiated all over the place. So this gives L/(32 pi^2
> R^4 )extra input of heating per square meter. That doesn't *seem* too
>...If it is really a shell, then radiation to the inside will be in net
equilibrium. Only the outside will radiate the energy from the star...So if
we continue to find no excess IR from this star, it's supportive of certain
classes of space industrial objects that radiate heat directionally...Keith
Anders and Keith, there is an approach I have been struggling with, which
uses Bessel functions, but I need some adult supervision if someone here can
offer it, or knows someone who can, specifically someone with access to idle
graduate students armed with Matlab and such.
Assume a sunlike star and assume away all planets and debris (I am not a
mathematician, but I sometimes act like one when it is time to assume away
planets and debris.)
OK now assume a 1 square meter reference plane perpendicular to a line from
the center of the square meter thru the center of the star. Out at 1 AU,
there are nearly half a mole such meter squares, so it shouldn't be hard to
imagine picking one.
OK now imagine a kind of truncated square based pyramid (frustum) opening
outward from there, such that the included angle formed by lines to the
center of the star remains constant. Imagine the pyramid going out 5 AU so
that the frustum base is 5 meters on a side.
We now need only calculate the heat load on that small end square meter base
(which is about 1400 W) and the heat emission at the 25 m^2 big end out into
3K space. The heat load thru the sides of the frustum is irrelevant, since
the heat going in vs the heat going out is identical along the entire face
always and forever amen. That simplifies the model to heat in at the small
end, and heat load out the big end, ja? Are we ready to Bessel?
OK, with that model, and some clever Matlab coding (I no longer have access
to Matlab, oy) we should be able to create thermal distributions along the
length of that frustum.
I did that using uniform distributions and discovered that my Bessel
functions predict we overheat inboard if we extract too much energy from
that mass distribution (of MBrain nodes) within the Frustum.
If the mass distribution of MBrain nodes within the frustum is sufficiently
low, most of the energy passes thru, and the temperature distribution stays
good. But if the mass distribution is high, the inboard part overheats. At
some point, there is a uniform mass distribution along the length of the
frustum in which the peak temperature is a nice balmy 300K. I propose we
call this mass distribution the Bradbury density. Or we could call it the
Bradbury300 density, so we can calculate a new density for Bradbury350 and
The Bradbury density assumes no directional reflection, so all the energy
has to come in the 1 meter square reference plane at 1 AU and be emitted
from the 5 meter square 5 AU plane.
We could of course use other numbers. I propose a name for peak
temperatures designated as inner diameter in AU, dash, outer diameter in AU,
Bradbury, peak temperature in Kelvin.
The above thought experiment assumes no low-entropy reflection and the peak
temperature would be called:
I propose this name because had Robert lived, he would have embraced this
notion, assuming I invested several hours arguing with him over it (he
didn't do MBrain thermal models much and didn't cotton to them, but he liked
my doing them.)
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