> >Spike wrote: WOW cooool, Drexler, Bradbury, spike. There are three
> names I
> > like to see mentioned in the same sentence.
> > ...
> > Today I was doing some back of the envelope calcs and I think
> > I know why the other two wanted to know.
>
> "Robert J. Bradbury" wrote: Spike, you only get *temporary* membership
> in the club unless you take those "back of the envelope" calcs, format
> them, put
> them on the net *linked* to the references and respond to
> questions that people ask about them!
OK, but how about if I post them here first so somebody can point
me to any obvious errors?
> Or perhaps they are *virtual* envelopes...
> We trust you enough that we would only occasionally suspect
> they are "imaginary" envelopes).
HA! Got you there pardner. I do my back of the envelope
calcs in an engineering notebook, which I keep indefinely. I
will be happy to show you these next time you are in town.
As it turns out, my notebook is at work and I am at home,
but I can reproduce the admittedly rough calc from memory.
The most disturbing thing I found when doing the calculations
is that it is not necessarily the interstellarlopers which are
the problem, its the hydrogen and helium. This being the
case, I am not at all confident in my upper bound for the
abundance of the stuff. Let us stay with 200 ng/km^3
for now.
Assume an interstellar craft with frontal area of 1 km^2. It
doesnt matter: the frontal area will cancel. I just used
1 km^2 to match my units of 200 nanograms/cubic km.
Assume the craft travelling at .1c. A first order estimate
of the effect of a collision between a two particles is
made by a conservation of momentum argument: a
hydrogen atom (1 AMU) strikes a water molecule (18 AMU)
and the resultant velocity is proportional to temperature
of the water. Conservation of momentum suggests that
about 4% of the kinetic energy of the hydrogen atom
is imparted to the water molecule: the rest goes off in
photons that are not absorbed. Similarly, if a helium
atom strikes a water molecule, 4/18 or about 22% goes
to raise the temperature of water, the rest lost. If the
interstellar meduim is 75% hydrodgen and 25 % helium,
then the kinetic energy of the mixture imparted to raising
the temperature of the ice is (.22)(.25) + (.75)(.056)=.097
so about a tenth of the energy in the hydrogen and helium
is retained in raising the temperature of the ice, the rest lost.
Mass flux into the shield per unit time is then
(.1c)*(3E5 km/sec)*(2E-10 kg/km^3) =6E-6 kg/km^2 sec
So assuming a 1 km^2 frontal area, the absorbed energy
into the shield is
(.097)*(.5)*(6E-6 kg/km^2 sec) * (3E4 km/sec)^2 * (1km^2)
=260 kg km^2/sec^2 = 2.6E8 newton meters.
A calorie is about 4.2 nm, so 6.3E7 calories per second.
If the ice shield starts at 20K and the collision energy
warms it to 273K at about .5 calories per gram, then
provides heat of fusion, 80 calories per gram, then at
vacuum vaporization can occur at 273K at 540 cal/g,
about 3800 calories per gram heat of dissociation when
the water molecule is torn apart. Then there is some
further energy absorption possible in ionization, but let
us ignore that for now, for conservative calcs. That means
the shield absorbs about 4600 calories per gram, or
about 1100 joules per gram.
(2.6E8 joules per second) / (1100 joules / gram) = 240 kg/sec
of the shield lost, so since we started with a square km
of shield, then
(240 kg/sec/km^2)* (m^3/900 kg) * (24*3600 sec/day)*
(km^2/1E6 M^2 )=
.023 meters per day, or 2.3 cm/day. For 45 years.
Please somebody find a major error in my calcs. I will repeat
the calcs using tungsten shields of course, but nothing would
please me more than finding this analysis is completely wrong.
Well, actually, a peaceful human-loving Singularity, transitioning
us into a world of unimaginable wealth and wellbeing would
please me more. But nullifying this calc would please me second
most. spike
This archive was generated by hypermail 2b30 : Mon May 28 2001 - 09:50:36 MDT