Re: (1297) I'm SOOO confused

<from Peter Berghold>
> I was once again trying to design a ship under the new MT design
> rules.  This time, (I thought) something simple.  A 20T (OT)
> fighter.
>
> Under High Guard, this ship had a power plant with a rating of 12
> and needed 12% of the ship's volume for fuel, or 2.4T.  This
> roughly equates to 32.4 KL under the MT rules.  The fighter I am
> designing is to have a 6G manueverability rating.  When I tried to
> calculate the size power plant to accomplish this deed I came up
> with the following figures:
>
>           Item                Units           Volume            Wt
>        Manuever 6              3.4             45.9             119
>        Power Plant            80                7.2             160

I don't know where you got this figure ------------^ it should be 80.

>
> I arrived at the power plant figure by determining that the
> manuever drive would need 238 power units as input power for their
> survival.  I doubled that figure in the interest of maintaining
> excess power for agility.  SO far so good, but now comes the fuel
> calculations:
>
> 80 power plant units use 0.009 KL of fuel per hour.  I calculate a
> 30 day supply as 30 X 24 X 0.009 X 80 as 518.4 KL.  Not much, but
> the whole vessel only has a volume of 270KL!
>
> What am I doing wrong?   What kinds of guidelines do other folks
> use in this instance?
>

What are you doing wrong? First you are not reading the power plant
chart correctly, and second, you are not designing the ship
for its intended use.

I'll address the second one first. The fighter, without staterooms
(as the OT fighters, and any 20T fighter would be designed), cannot
support a pliot for 24 hours, never mind 30 days. Only use the fuel
storage that is required. I typically use 3-10 days for fighters.
(actually I typically put small staterooms in fighters (50T+) and
have them stay on station for several weeks).

The first part: I deduce from your design that the fighter is Tech 15
(this is the only TL case in high guard where the power plant rating,
and PP volume are  identical).
You indicate you want 2x238 for the Power plant (to include agility),
noting at this point a laser (250Mw) cannot be installed - Missiles only.
(This doubling is a good starting estimate, but will have to be
revised after the ship is sketched out, as agility is a function of mass).

So 476 Mw of power, do not require 80Kl (units) of TL 15 power plant. You
forgot to account for scale efficiencies. Any PP over 14Kl, gets triple
effieincy. ie. 18 Mw per "unit" of power plant. Thus you plant can be
26.44Kl (round to 27, which gives 10Mw to power life support, computers, etc.)

(This is the problem with A-M power plants, as they don't get scale
efficiency, fusion power plants are quite often better at the higher techs)

Thus your fuel for 30days is .009*27*24*30= 175Kl within your volume
requirements, if you only go for 10 days this becomes just 59 Kl.

					-Dan Corrin

P.S. I am still	working on a C starship design package. (With Paul Dale, and
Jo Jaquinta) We are (essentailly) out of the design phase, and have
started coding.