Timerover51
SOC-14 5K
I have been going back over the 1977 edition of the LBBs, specifically the ship design area. The requirement for the Power Plant is that its letter code matches the letter code of the Maneuver Drive, and it does not have to match the Jump Drive at all. This does lead to some excess space in the Engineering Section.
The standard design Subsidized Merchant-Type M is a 600 dTon ship with 80 dTons of volume set aside for Engineering. The ship is equipped with Jump Drive-J, 50 Ton mass for 90 Million Credits. It also has Power Plant-D and Maneuver Drive-D, massing and costing 13 Tons/32 Million Credits and 7 Tons/16 Million Credits respectively. The total for the drives is 70 tons, leaving 10 ton unused. The cost savings of the smaller Power Plant is 40 Million Credits, not an inconsiderable sum for an approximately 220 Million Credit ship. Note also that if the Power Plant matched the Jump Drive, there would be only 2 tons remaining for a maneuver drive, which would not move the ship. I would say that is why the Engineering Section increased to 85 dTons in the 1981 edition so as to allow for the 7 tons of maneuver drive.
If you go to an 800 dTon Hull, it has 165 dTons set aside for drives. In designing a commercial ship, I put in Jump Drive-M, giving me Jump-3, and Power Plant and Maneuver Drives-F. Tonnage and cost respectively for the drives and power plant are jump drive 65 tons/120 Million Credits, maneuver drive 11 tons/24 Million Credits, and power plant 19 tons/48 Million Credits. Now, I have a ship capable of Jump-3 with 1-G Maneuvering capability. The smaller power plant saved 48 Million Credits. However, the total mass or dTon Volume of my drive plant is only 95 dtons. I have 70 dTons of space left over. If I had gone with an F Jump Drive, I would have 100 dTon of unused space. If I had gone with Type M for everything, I would still have 40 dTons of unused volume left over.
Therefore, my question is, what can I do will all of that unused space?
The standard design Subsidized Merchant-Type M is a 600 dTon ship with 80 dTons of volume set aside for Engineering. The ship is equipped with Jump Drive-J, 50 Ton mass for 90 Million Credits. It also has Power Plant-D and Maneuver Drive-D, massing and costing 13 Tons/32 Million Credits and 7 Tons/16 Million Credits respectively. The total for the drives is 70 tons, leaving 10 ton unused. The cost savings of the smaller Power Plant is 40 Million Credits, not an inconsiderable sum for an approximately 220 Million Credit ship. Note also that if the Power Plant matched the Jump Drive, there would be only 2 tons remaining for a maneuver drive, which would not move the ship. I would say that is why the Engineering Section increased to 85 dTons in the 1981 edition so as to allow for the 7 tons of maneuver drive.
If you go to an 800 dTon Hull, it has 165 dTons set aside for drives. In designing a commercial ship, I put in Jump Drive-M, giving me Jump-3, and Power Plant and Maneuver Drives-F. Tonnage and cost respectively for the drives and power plant are jump drive 65 tons/120 Million Credits, maneuver drive 11 tons/24 Million Credits, and power plant 19 tons/48 Million Credits. Now, I have a ship capable of Jump-3 with 1-G Maneuvering capability. The smaller power plant saved 48 Million Credits. However, the total mass or dTon Volume of my drive plant is only 95 dtons. I have 70 dTons of space left over. If I had gone with an F Jump Drive, I would have 100 dTon of unused space. If I had gone with Type M for everything, I would still have 40 dTons of unused volume left over.
Therefore, my question is, what can I do will all of that unused space?