Prev: Re: [FT] SML-AF Was Re: More weapon... Next: Re: [SG] Vehicle design

[SG] Vehicle design

From: Jaime Tiampo <fugu@s...>
Date: Thu, 21 Dec 2000 21:26:17 -0800
Subject: [SG] Vehicle design

I'm working on some of my Gear designs for SG and I'm trying to find
some info on weapons and systems capacity points cost. Specifically I'm
trying to build an artilery Gear and I can't find info on how big an
artilery piece should be. And EW equipment, does it take up any space in
the vehicle?

Jaime
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From: "Laserlight" <laserlight@quixnet.net>
To: <gzg-l@csua.berkeley.edu>
References:
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Subject: Re: System Thrust was Re: Rules you use...
Date: Thu, 21 Dec 2000 21:42:43 -0800
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> >Remember that d = 1/2 at^2, so (without working it out to verify)
you
> >should be able to use 10^2 FT turns or about 1 day (25 hours).
>
>   Hi Laser,
>
> Could you expound on this one for me. I'm not getting it off the top
> of my head...
>
> Schoon

The formula for constant linear acceleration is d = 1/2 at^2 (d =
distance, a = acceleration, t = time)

In FT scale, d = 1000km, t=.25hour, and a= about 2.47m/sec^2
(if you used 1 gee and 7.5 minutes you'd get close to 1000km which is
why I keep harping on it)

You're saying use d = 10,000,000km (increase from FT scale by 10^4).
However, the acceleration is multiplied by t *squared*.  So increase
the time by 10^2 and you should work out right.  Plugging it into the
spreadsheet, I get a=2.47m/sec^2, t=90,000sec, and d does indeed turn
out to be 10,000,000km (actually, if you use 2.47m/sec^2 you get
10,003,500km).

90,000sec divided by 3600sec/hour = 25 hours.  So a FT ship under
constant acceleration covers 1 "system MU" in just over a day.

The problem is that, if you accelerate for 2 days, you should move
4MU, not 2.

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