Sand casters

Jerry Han wrote:
>Somebody else can figure out the MASS and mechanics of the lauchers;
>I have to get back to work here.  (8-)

I posted the following numbers the last time this thread came up, a
little
over a year ago.

At the most commonly used scale (1" = ~1000km), the density of a sand
cloud
is not very cost effective compared to the amount of mass required; if
you
evenly dispersed a quantity of sand equal to the mass of a destroyer
into a
1" cubic volume, the density would be so low that a capital ship might
fly
through the resulting cloud without hitting a single grain.

Jeff

-----

At 05:00 PM 8/19/1998 -0500, Jeff Lyon posted this:
>I was afraid someone would ask that...okay, here are some rough
numbers:
>
>Let's assume a theoretical substance called "sand" with uniform 1 mm
>cubical crystals and a density of 2.5 g/cm^3 (silicon dioxide is 2.27;
I
>like round numbers).
>
>A one centimeter cube would have 1000 grains of "sand" in it.
>
>A one meter cube would have 1 million cubic centimeters and 1 billion
>grains of "sand" and would have a mass of 2500 kilos or 2.5 metric
tonnes.
>
>One standard unit of mass in the "official" FT background is 100 metric
>tonnes or the equivalent of 40 cubic meters of "sand".
>
>A cubical cargo hold 10 meters on a side would hold 1000 cubic meters
of
>"sand" which would have a mass of 2500 metric tonnes or 25 mass units.
>
>If dispersed uniformly throughout a 1" volume of space (assuming that
the
>most commonly accepted scale is ~1000 km per inch) then your 10-meter
wide
>cube of sand would expand into a 1 million-meter wide cube.  This is a
>dispersion ratio of 1:100,000.
>
>Therefore you would have an average separation of 100 meters between
each 1
>mm grain of "sand".  A cubic kilometer of space would have a cubic
>centimeter of "sand" (1000 one-mm grains) dispersed uniformly
throughout it.
>
>Obviously, higher densities would require proportionally higher mass.