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Old May 22nd, 2019, 03:52 PM
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Begin at the beginning ... Vehicle Design Checklist

1. Choose a Vehicle Chassis
a. Determine chassis configuration
b. Determine open or closed chassis
c. Install armor (optional)

Since we need "Displacement" as the first step in designing the Chassis ... how large is an Iowa Class Battleship?
Start with the box that would surround the ship (length x width x height).
I use round numbers because small fractions are silly at this level of precision, so call it 270 meters long x 33 meters wide (straight from wikipedia).
Eyeballing a plan and side view of the ship shows that it is roughly as tall as it is wide, so I will use 33 meters for height as well.

That makes the box around the ship 270m x 33m x 33m = 294,030 cubic meters [about 21,000 dTons for anyone curious].
So we know the ship has to be smaller than 21,000 displacement tons (dTons).
But how much smaller?

First the HARD way ... I traced a plan of the ship and divided the area of the length x width by the area of the deck (using a drawing program) and came up with about 0.75, then I did the same for the side view and came up with about 0.75 as well. When we multiply 0.75 x 0.75 we get 0.56, meaning that the volume of the ship is about 56% of the volume of the box.

294,030 cubic meters x 0.56 = 164,657 cubic meters [11,761 dTons for anyone curious].

That required a fancy drawing program and a lot of time and work, so here is the EASY way ... for almost all relatively normal ships (may not apply to catamarans or other exotic hulls) if you ignore any sails or masts, then the ratio of BOX to SHIP almost always comes out somewhere around 0.6 (or 60%). So let's just use the easy way ...

270m x 33m x 33m = 294,030 cu.m. BOX
294,030 cu.m. x 0.6 = 176,781 cu.m. SHIP

We need dTons and 1 dTon = 14 cu.m., so ...

176,781 cu.m. SHIP / 14 = 12,601 dTons

So the Iowa Class Battleship is between 11,800 dTons [Hard Calculation] and 12,600 dTons [Easy Calculation], but we only have 2 digit accuracy anyway, so let's use 12,000 Displacement Tons.

Using the TL 9 Destroyer as a guide:
  • it has an 800-ton hull [the Battleship is a 12,000-ton hull]
  • it has a "closed" configuration [the Battleship will not be a convertible either.]
  • it has Titanium Steel (Armor 6) [the Battleship will need a TL 5 armor and this is a design variable that we may need to play with to balance the design for the rules.]

so we have 12,000 tons (dTons) at 12 vehicle "spaces" per dTon = 144,000 spaces.
Construction time appears to be 9 hours per ton, so the 12,000 ton ship has a base construction time of 108,000 hours (12 years) ... I guess that we will need to use the Starship Construction time for large watercraft [good to know].
We only have Iron Armor (TL 4) available on the list (Titanium appears at TL 7). Since Iron Armor is installed in 2 point per 5% space increments up to a maximum of 10 points of armor, it is easy to calculate a quick reference for all available options for TL 4-6 Armor
  • Iron Armor 1 = 2.5% of vehicle = 3600 spaces
  • Iron Armor 2 = 5% of vehicle = 7200 spaces
  • Iron Armor 3 = 7.5% of vehicle = 10,800 spaces
  • Iron Armor 4 = 10% of vehicle = 14,400 spaces
  • Iron Armor 5 = 12.5% of vehicle = 18,0600 spaces
  • Iron Armor 6 = 15% of vehicle = 21,600 spaces
  • Iron Armor 7 = 17.5% of vehicle = 25,200 spaces
  • Iron Armor 8 = 20% of vehicle = 28,800 spaces
  • Iron Armor 9 = 22.5% of vehicle = 32,400 spaces
  • Iron Armor 10 = 25% of vehicle = 36,000 spaces

The Destroyer used 10% of the space for armor, so we will use that as our first guess.
Now we have a Chassis

... Base = 144,000
... Configuration = 0 [closed]
... Armor = - 14,400 [armor 4]
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