Re: Starship Fuel

> Date: Wed, 23 Jan 91 21:17:04 PST
> From: Mark F. Cook [email removed]
>
> 2.  Fuel duration & multiple jumps by a starship.
>
>   I have a design for a modified Beowulf-class Free Trader that has
>   jump-2 and maneuver-2.  It burns up 405 Kl. of fuel per jump and
>   has a total fuel capacity of 859 Kl.  It also has a duration of
>   30/90 days (whateverthehell that means) according to the design
>   rules in the Refs. Manual.  If I understand the rules correctly,
>   that means that it can make one jump (jump-1 or jump-2, it doesn't
>   matter 'cause the both burn 405 kl. of fuel, thanks to those
>   mysteriously vanishing 'jump governors').  On the remaining 454
>   Kl. of fuel, the ship can stay powered for 90 days in some sort
>   of 'low-power mode (or '8-hour work days' as the Refs. Manual calls
>   them).  This implies that the ship burns 5.0444 Kl. of fuel per day
>   in that mode, right?
>

I don't recall what the rulebook says about multiple jumps etc and the
govenor problem, but if you notice the fuel required for a jump is
proportional to the numer of jump units required for the drive. (I think
62.5klxnumber of jump units). But examine the following table (from memory):

	Jump #		1  2  3  4  5  6
	100T Jump Units 2  3  4  5  6  7

You'll note that it doesn't take twice as much fuel to go jump 2 as it does
to go jump 1 (unlike the old traveller rules). This is the main reason
why a jump-2 ship can't go J-1 and J-1. In my campaign I allow the players
to use their fuel in Jump Fuel Units (JFU). I rate their tanks in JFU, and
then they can subtract the amount they use for a jump [That is J-n requires
n+1 JFU]. Thus your free trader with 810 kl of tanks would have 6 JFU, which
is enough for 2xJ-2 or 3xJ-1. I presume that the designers didn't want to
confuse the players with this sort of partial fuel useage.

>  [...]
>   I realize this completely ignores the other factors such as in-system
>   transit time, etc.  It also doesn't address whether or not this low
>   power consumption figure assumes the M-drives are going or not.
>   However, I don't think the rules explicitly forbid this sort of
>   action, do they?  All of this hinges on my understanding of the 30/90
>   duration figure, which is zip.  So somebody enlightening me, okay?
>

The 30/90 duration has no deep understanding. The power plant is computed
for the requirements of the ship, and you provide fuel for n days of
continuous output at 100% power. The 3xn duration assumes that the ship
is not being manned around the clock, but only 8 hours per day.

There are two points to be made here, first the only states that the
power plant can have (at least those mentioned in the rules) is 100%, idle
and off. The presence of the idle state (mentioned with vehicle plants, I
belive in the skills section) is so that you don't have to cold start the
engine every time. However I would think that if the plant can run at idle
(ie. keep the fusion reaction, and magnetic bottle or whatever) it can run
at any intermediate state.

The longer duration assumes 8 hours of 100% output followed by 16 hours of
idle. This would mean no life support, sensors, computer and especially
gravity. However during those 8 hours of activity everyhting is not running
so things can average out. I for one am not satisifed with this type of
explanation however.

> One possible interpretation of the 30/90 duration (mine :-)).
>
>   The 90 day duration is max. duration if the M-drives are NEVER run.
>   This gives use a life-support (et.al.) fuel usage of 5.0444 Kl./day.
>   The 30 duration is max. duration if the M-drives are run CONSTANTLY.
>   Since we know life support burns 5.0444 Kl./day, we know that life-
>   support burns 151.3333 Kl. of fuel in 30 days.  This means that the
>   remainder of the 454 Kl. (which is 302.6667 Kl.) is 30 days of con-
>   sumption by the M-drives, which works out to 10.0889 Kl./day.  Now
>   that we've got these 2 figures:
>         M-drives ------- 10.0889 Kl./day
>         life support ---  5.0444 Kl./day
>   We can derive per hour consumptions and determine right to the
>   credit how much it will cost to refuel at that next starport! :-)
>

This is similar to the way I would implement it. However you are
still assuming that the power plant can be set to variable output. One
way to avoid making this assumption is to install multiple power plants
as long as each is > 14kl (252Mw at tech 15?) there is no degridation
of power output. (This of course flaunts the only one ship system at
a time in operation rule).

Consider the power consuming devices:
	Maneuver (high)
	Agility (very high if present)
	avionics/communications/sensors (low)
	Weapons (very high)
	Screens (very high)
	Gravity/Compensators (high)
	life support (low)
	Computers (low)
	misc (berths, purifier) (low)
If you were to collect the correct things together and assign them their
own power plant then the fuel could last much longer by only using the
appropriate power plants at the correct time.
ie. Electonics, life sppt, misc, and Gravity on one plant, maneuver
on another, and weapons, screens and agility on a third (assuming they
meet the 14kl requirement). Then the fuel could be really conserved
by only running the life sppt et al, somewhat conserved when maneuvering
as well, and with the combat power plant you probably won't be concerned
with your fuel consumption. Note that it takes some time to bring a plant
up from idle, and a military ship that is trying to power up while under
attack would be to say the least under a disadvantage.

Consider the following fusion plants at TL 15. one 24kl, two 12kl,
four 6kl and eight 3kl plants. Each set would draw the same amount of
fuel from the tanks, but consider the power output:
The 24kl  generates 1*24*6*3   = 432Mw
The 2x12kl generate 2*12*6*2   = 288Mw
the 4x6kl  generate 4* 6*6*1.5 = 216Mw
the 8x3kl  generate 8* 3*6*1   = 144Mw

Thus it probably won't save fuel substituting the smaller plants for a
single large one. I don't know why the same fuel would generate such a
conflicting power output, but perhaps the larger plant if run at 50%
output would require 75% of the fuel or so. Ie the fuel to power output
would not be linear, which is why they don't mention power level settings
for the plants.

					-Dan

Dan Corrin, System Manager, Mechanical Engineering, UWO, London, Ontario
TML/CZ FTP site coordinator: [email removed] ...!watmath!julian!engrg!dan