Economics of shipyard

[AnotherDilbert]

What if a ship is a hollow sphere? Lets say it is 100 meters in radius, ...

I would call it a 300 kT hull with some empty space unallocated.

 

[EldritchFire]

What if a ship is a hollow sphere? Lets say it is 100 meters in radius, but the decking is from say the Hull to 10 meters in and inside is nothing but empty space which the spaceship encloses. It could be filled with an atmosphere but no structural components? So how do you calculate the volume of such a ship for construction purposes?

Empty space is cargo space. Just give it loads of cargo and you'll be good.

 

[AnotherDilbert]

Empty space is cargo space. Just give it loads of cargo and you'll be good.

Effectively yes, but technically cargo space have to be allocated (p22). We can leave space unallocated.

 

[Tom Kalbfus]

What if a ship is a hollow sphere? Lets say it is 100 meters in radius, ...

I would call it a 300 kT hull with some empty space unallocated.

Lets say we used this map for the ship's deck plan. Lets use the scale of 18 meters per hex, and lets suppose this map constitutes one deck which wraps around an empty space, the space from floor to ceiling is 3 meters. By counting the number of hexes across I can determine it is 35 hexes in circumference or 630 meters. Dividing by pi gives us a diameter of approximately 200.5 meters.
Volume of a sphere is

Volume = 4*pi*(100.26^3)/3
Volume = 4222509.67 cubic meters = 301,607.83 dtons
That is one answer, it is the answer you would use for determining the size of the jump drive you would need for this ship.
However for construction costs you might want to use this formula
Volume = number of hexes * distance from floor to ceiling * 6 equilateral triangles in a hex * 9 meter hex radius * cos(60 degrees) * 9 meters / 2.
Volume = 500 * 3 * 6 * 9 * cos(60 degrees) * 9/2
Volume = 182,250 cubic meters = 13,017.86 dtons.

I think the construction cost should be based on the 13,017.86 dton figure instead of the 301,607.83 dton figure. I think you will need a jump drive for a 301,607.83 ton hull if you want a starship, but the construction cost should be based on a hull that is 13,017.86 dtons, as it doesn't cost anything to built the empty space in the center. Does anyone disagree? I think the ship construction rules assume spaceships full of decks rather than a large empty hollow space. If you took this sphere and rotated it 3 times a minute, the equator area would experience a g-force of 1 Earth gravity.

 

[AnotherDilbert]

Easier:
Outer hull: ⁴/₃π × 100³ ≈ 4 186 667 m³
Inner hull: ⁴/₃π × 90³ ≈ 3 052 080 m³
Difference: 1 134 586 m³ ≈ 81 000 dT

The outer hull is just as big as a 300 kT hull, so the hull and armour would be based on 300 kT.
The drives would be based on 300 kT.
The bridge and crew might be based on the smaller size, if the DM is feeling friendly?

 

[Condottiere]

If it's empty inside, or mostly, you're just constructing the hull, as probably modules that are just welded together.