Morning PDT AnotherDilbert,
AnotherDilbert said:
snrdg121408 said:
A hull's volume is based on its interior space not the exterior base. TNE FF&S states that basic hull volume is based on the dimensions of a sphere. The other configuration options modify the volume.
No. The hull size it the total volume of the hull. The configuration is the geometric shape of the hull which modifies surface area and hence how much hull material (armour) we need to enclose the hull.
A 1000 dT streamlined ship is has 1000 dT available completely unmodified by the configuration. The hull and any armour is 20% more expensive.
A 1000 dT Planetoid ship has 800 dT available and 200 dT not usable, it is still a 1000 dT hull. All calculations based on hull, like hull cost or drive sizes, uses 1000 dT as base value. Look at the asteroid ship on p128, the hull size is 300 dT of which 60 dT is not usable and 240 dT available.
Example: A 1000 dT hull is 14000 m3. It can either be a sphere with radius of 14.95 m or a cube with a side of 24.1 m, either way the volume is 14000 m3. The surface area is not the same so hull and armour costs are not the same.
Dang-it another time my memory was out to lunch and proof I should be looking at the source material more. Of course I admit that when I was replying I was not able to find my copy TNE FF&S Mk I Mod 1. This morning PDT I was able to find the book without a problem. I'm blaming gremlins for hiding the book.
A 300 d-ton hull using standard, streamlined, sphere, close structure, or dispersed structure configuration is able to use 300 d-tons for installation of the components. To fit the same 300 d-tons in a planetoid hull configuration the planetoid's total volume needs to be 375 d-tons.
In my example the hull's total volume of 1,000 d-tons was to fit as close to 1,000 d-tons of components as possible. A 1,000 d-ton planetoid hull, as you pointed out, looses 200 d-tons leaving only 800 d-tons of space. To achieve the same 1,000 d-tons of standard, streamlined, sphere, close structure, or dispersed structure volume for components a planetoid hull needs to be 1,250 d-tons.
I agree that turning a rock into a hull configuration of planetoid with a total volume of 1,000 d-tons would leave 800 d-tons or one with a total volume of 300 d-tons would leave 240 d-tons in which to install components. In both cases the volume to install components is smaller than a hull built with a configuration of standard or, streamlined or, sphere or, close structure, or dispersed structure.
To have the same volume for installation of components as standard or, streamlined or, sphere or, close structure, or dispersed structure hull configurations planetoid and buffered planetoid hull configurations have to start with a larger total volume.
snrdg121408 said:
My understanding of the above material is that a hamster cage is an alternate, like a double hull, means of generating an 1G field in areas that are continuously occupied by crew and passengers instead of the entire hull. The cost of Cr50,000 per d-ton based on the information in the Non-gravity Hulls section includes the installation of grav plates.
Using both grav plates and a hamster cage does not make sense to me.
Quite, you do not need it, it is not generally done, but it could be done for special purposes, e.g. something like the Lab Ship. I just wanted to show the cost calculation since it is about the most complicated case for hulls.
Thank you for clarifying that the example does not need the hamster cage and that I appear to have a basic understanding of the hamster cage rule.
