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Scale of FT (longish and contains physics......)

From: Stephen Pugh <mafb90@d...>
Date: Tue, 19 Nov 1996 18:59:57 -0500
Subject: Scale of FT (longish and contains physics......)

There is actually enough information in More Thrust to calculate a 
time and a distance scale for Full Thrust. The results are somewhat 

On page 13 of More Thrust we have the rules for planetary orbits. If 
we follow the sugestions there and take an Earth-type planet to have 
a radius of 6" then we can get to work.

The radius of the Earth if 6,378 km. straight away we have our distance 
scale 1" = 1063 km. We can round it down to 1000km, but won't for 
reasons that will be explained later.

Time for a bit of physics. One outcome of Newton's Laws of motion and
gravity is that for a circular orbit T^2 = (4 * Pi^2 * r^3) / G * M
where T = the period of revolution
      r = radius of the orbit
      G = Gravitational constant 
      M = mass of object being orbited (ie the planet)
In this case r = 12" = 12,756,000 m and M =  6 x 10^24 kg
Hence, T = 14309 s

Now a ship in this orbit moves 6" a turn around the circumference of 
a 12" radius circle.
circumference = 2 * pi * 12 = 75.398"
At 6" per turn 1 orbit = 12.566 turns.
12.566 turns = 14309 s
And so 1 turn = 1138.7 s = 18.98 minutes

So we have our time and distance scales. Now we can calculate some 
derived quantities. How much acceleration does 1 point of thrust 
represent? If we assume thrust is applied over the entire turn then 
this formula applies-
acceleration = change in velocity / time
1 thrust = 1" / 1 turn squared
	 = 1,063,000 / 1138.7 ^2
	 = 0.8198 m/s^2
So what? This is nearly exactly one twelth of 1g!
If we had made 1" = 1000km then it would have thrown this neat 
coincidence out. If we make the time 19 minutes (1140s) then it
makes it even closer.

There is a problem. Say a ship starts from rest and accelerates at 
this velocity. After 1 turn it is moving at 1"/turn. By the rules 
it should have moved 1". By physics it should have moved 1/2". 
The only way the rules can be correct physically is if a _very_ 
large acceleration was maintained for a _very_ short period of time. 
However, if we apply one of the golden rules of Full Thrust, "always 
round things up" we can achieve a solution. A thrust of 8 must move 
the ship by just over 7".

This comes out with an acceleration applied for 1/4 of a turn and 1
thrust now equals 1/3g.
A thrust 8 escort can put out a thrust of 8/3g which is fairly good
for what a human crew could stand for extended periods and still 

Summary of what we've got so far:
1"	  =  1063 km
1 Turn	  =  19 minutes
1 Thrust  =  1/3g applied for 4.75 minutes

The 36" range of pasive sensors and A batteries is equivalent to 13% 
of a light second, which when coupled with the length of the turn  
rather spoils the thread about targetting. There's enough time to
perform over 2000 ping-analyse-shoot cycles in one turn. Is that 
enough to give you a 50% chance of hitting?

To move at 1% of the speed of light a ship would need a velocity of
over 3000"/turn!

At this scale our sun would have a radius of 655"!

But wait, I hear you muttering, surely the rules in More Thrust are 
optional, what about Jim Webster's idea of defining one table edge as
the planet's atmosphere?

Well, this makes all the ranges much smaller. If we assume that the 
short edge of a table 4' = 48" is the diameter of a planet. Then a
new scale emerges 1" = 265 km or 1/4 of what we had before.
How we scale time and thrust in this case is trickier as we have no
data on orbits. But if we assume that this scale implies a rather low
level of technology then gravity control is unlikely and our vessels 
will have to operate at thrusts that won't cripple the crew. So we keep
the thrust at 1/3g. This gives us 1 turn = 9.5 minutes or half of our
previous value. If the planet is wider than the table then the scale
is further reduced.

(NB For constant acceleration the time will scale as the square root 
of the scaling on the distance.)

That's quite enough for now.

|			Stephen Richard Pugh			     |
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