Re: [FTFB] Not ships, exactally. . .

On Sat, 13 Feb 1999 DracSpy@aol.com wrote:

> Given: that 1 Manuver Unit (MU, 1" or 1cm) = 1000 km.
> Given: that 1 turn = 15 minutes (to fit neatly with DS2)
> Then: 1 Thrust over 1 turn = 1/15th of 1 g (1 standard Terran gravity,
set to
> 10m/s/s), or 0.0167 g, or 66.67 cm/s^2
> 
> now if my schooling serves me that would (keeping the turn at 15
minutes) that
> would make MU =11,00km, I think.

Starting from rest and from position zero, the distance travelled under
constant acceleration is:

s = 1/2 a t^2  

where

s = distance in meters
a = acceleration in m/s^2
t = time in secs

0.01 g is (approximately) 0.1 m/s^2, 15 minutes is 900 sec.  So the
distance is for this case:

s (meters) = 1/2 (0.1 m/s^2) (900 sec)^2 = 40,500 meters = 40.5 km

Using 1/15 g = 0.6667 m/s^2 (I don't quite see where the 0.0167 g came
from, but maybe I'm missing something obvious - at any rate we get the
same thing for m/s^2!)	this is:

s (meters) = 1/2 (0.6667 m/s^2) (900 sec)^2 = 270,000 meters = 270 km

To get 1000 km/MU you need (forgive me for showing all the work, I'm
working this out on the screen - always dangerous <g>):

1,000,000 meters = 1/2 g (900 sec)^2

=> g = 2 (1,000,000 meters) / (900 sec)^2 = 2.47 m/s^2 = 0.25 g

which is a number I've seen tossed around here, I think.  But keep in
mind
that in order for the Full Thrust engines to include enough fuel for one
ten-turn battle, 0.01 g thrust is the best you can do with current
technology (for example, chemical rockets do a lot better than this, but
you need so much fuel that the rocket is -mostly- fuel, which just
doesn't
fit in the Full Thrust model).

Someone please check my math, though, I should know better than to not
do
it on paper first!

Keith