Hello fellow Travellers,
at first a short question on the ship designs I send to the list. Till now
I added the calculation tables to them. Is anybody interested in this tables
or should I skip them in further postings? They have one advantage: It is
easy to modify the design if you don't like it.
Okay, now some comments and questions for Brilliant Lances:
Questions:
1. Fuel Cells and chemical Power plants: Is the oxygen needed for this
power plants contained in the figures mentioned in the Technical Booklet ?
2. Sorry Loren, but once again the question: Where can I find the crew
condition table mentioned in the Rules Book?
3. How do I calculate the MV value of hulls with other displacement than
those named in the Hull Size Table? Is there a closed formula to
calculate this values? (I'm a bit lasy, so I havn't tried to find
it out myself)
4. To the rules: If the base difficulty to hit a ship is Formidable
because of range, can a MFD cut this difficulty down?
Example:
The Long Range of a TL-14 150Mj Laser Turret is 8. Assume a TL-14 MFD
is installed.
Target is a ship 9 hexes away. The base difficulty is Formidable. The
diff mod because of target size is -1. The absoulte range give a diff
mod of +3. The TL-14 MFD ignores 5 diff mods.
What is the end difficulty?
a) Formidable, because only the +2 diff mod is ignored
b) Easy, because the weapons range difficulty is ignored too
I personally tend to a), but I would like to be sure.
Comments:
- - I wonder if I got the fuel consumptions of Jump drives correct:
Assuming a 100 kl hull we get the following figure:
Volume of fuel per Jump
Jumpnumber Volume of drive 1 2 3 4 5 6
1 2 kl 10 - - - - -
2 3 kl 7.5 15 - - - -
3 4 kl 6.7 13.4 20 - - -
4 5 kl 6.25 12.5 18.75 25 - -
5 6 kl 6.0 12.0 18.00 24 30 -
6 7 kl 5.83 11.67 17.5 23.33 29.17 35
where "Jumpnumber" is the maximum Jump performace of the drive.
Interesting enough the higher the possible Jump number of the drive is,
the less fuel is consumed for a lower Jump. A one parsec jump using
a Jump-6 drive consumes only 58% of the fuel a Jump-1 drive would need.
- - To the Technical Booklet page 15, Step 7:
"Each ship requires 0.5 MW of power output per displacement ton (...) to
achive an acceleration of 1 G."
Okay, assuming a 10 tons ship, this is 5 MW per g-rating. I only need
1g, so I assign 5 MW as Thrust power.
The book continues:
"To determine the acceleration rating of the ship, multiply its power
devoted to thrust (...) by 10 and divide the product by its hull number,
rounding fractions down."
Okay, using this formula on the above mentioned drive give:
5 MW * 10 / 10 tons = 5
So using this formula the 5 MW thrust power result in a 5 possible
maneuver g's.
Because most of the ships (I havn't checked all) in the Technical Booklet
use the first figure, you should translate the sentence above to:
"To determine the acceleration rating of the ship, multiply its power
devoted to thrust (...) by 2 and divide the product by its hull number,
rounding fractions down."
- - some lines later the books says:
"Each Thrust MW consumes 0.25 cubic meter of reaction mass per hour in
addition to the normal fuel cost of the power plant."
Please note that one combat round is 1/2 hour, so a g-turn use
0.125kl of fuel per MW.
Any comments are strongly recommended,
Juergen
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Juergen Kirsch
Institut fuer Informatik, Universitaet Bonn
Germany
[email removed]