[email removed] writes:
> >From: Mark F. Cook [email removed]
> >Subject: (2075) Black Globes redux
>
> >B) Any solid striking a black globe will demonstrate all the properties
> > of a perfect inelastic collision, with all the kenetic energy tranferred
> > to the globe. The only thing the occupants of the globe would notice
> > is an energy raise in their capacitors.
>
> This would make black globes next to useless against incoming missiles.
> <Assume a 50 kg missile has been accelerating for 20 minutes at 6G.>
>
> for a time of 20*60=1200 seconds.
>
> v = v0 + a.t = 0 + 9.8 * 1200 = 70600m/s
0 + 9.8*6*1200 = 70560 m/s
> 2 11
> ke = 0.5 m v = 0.5 * 50 * 4984360000 = 124609000000 = 1.246 * 10 J
>
> Is that a big number or what! Splat goes one space ship!
Two points.
First, that assumes that the target ship isn't moving. If the target takes
any evasive action, some of that acceleration is going to be used up following
the target. Second, if the target is moving towards the missile at the time of
impact, the target's velocity gets added to the missile's velocity; if the
target is moving away from the missile, the target's velocity gets subtracted.
However, these figures will probably do for an order of magnitude analysis.
The MT Referee's Manual says that 1 kilolitre of capacitors can absorb 650 MW
of energy. Which just shows what a heap of junk the MT Referee's manual is -
MW is power, not energy. If 1 kilolitre of capacitors can absorb energy equal
to 650 MW coming in for one full turn, i.e.
6 11
650 * 10 * 1200 = 7.8 * 10 J
then that kinetic energy will hardly be noticed. It all depends on how long
that 1 kilolitre of capacitors can absorb 650 MW.
- - --
"Keyboard? How quaint!" - M. Scott
Adrian Hurt | JANET: [email removed]
UUCP: ..!ukc!cs.hw.ac.uk!adrian | ARPA: [email removed]