Creating a TL 5 Battleship using the CE VDS

[atpollard]

Let's talk about the 16" Main Guns for a moment. Under the current design, the Battleship has 3 Superbays of 500 dTons each. Each Superbay mounts three (3) 16" guns. Each 16" Gun is equal to 80 Artillery Gun-TL 4 (8D6 each) firing simultaneously at the same target.

Gaining a DM +4 bonus to hit anything on the starship-scale, ground force weaponry must divide its damage by 50 before comparing it to a starship-scale target’s armor. Because a single weapon will, obviously, be unable to punch though armor it is possible for multiple weapons to all target the starship simultaneously, and the cumulative effect can inflict damage. Every additional ground weapon beyond the first can add half its damage dice to the total before dividing the total by 50 in order to calculate damage.

Let's apply this to our 80 x Artillery Gun-TL 4.
8D6 + (79 x 4D6) = 324D6
324D6 / 50 = 6.48 = 6D6 of Starship Damage

Each 16" gun is comparable to a 50 dTon Bay on a Starship and the Battleship has NINE (9) of these Starship Kinetic Kill Bay Weapons! A full broadside by the Main Guns of a Battleship is up to 9 x 6D6 Starship Weapon hits.

 

[atpollard]

The size of the ship. The Iowa-class were designed to be 45,000 ton Standard, with the Standard defined in the 1922 Washington Naval Treaty, and which used the long ton of 2240 pounds. The 45,000 tons did not include the 9000 tons of bunker fuel carried. To compute the ship size in terms of Traveller dTons, take the 45,000 tons representing the amount and volume of salt water displaced by the hull, and multiply by 2. That will give you a good approximation of the interval volume of the hull, so 90,000 tons.

As the long ton was larger than the metric ton, you might want to divide the 90,000 by 13.5 as compensation. That equates to 6,667 dTons or so.

A method of converting from ton Standard Displacement (a figure provided in wikipedia and other common resources for real ships) to Traveller Displacement Tons (of 13.5 or 14 cubic meters) would be VERY useful. So I was wondering if you might be willing to discuss this further.

In your example, the displacement in tons Standard is approximately equal to the volume of the ship below the water line in cubic meters (or kiloliters or VDS spaces) ... all are 12-14 per Traveller dTon. So the underwater volume can be calculated exactly. From this you are multiplying by two (x2) as a "fudge factor" that the volume of the "Hull" above the water is equal to the volume of the hull below the water. Is that correct?

While that would appear (visually) about right for a typical "Cargo Ship", I have to wonder if "visually speaking" other factors might be more appropriate for ships with very large superstructures above the main deck of the hull (like Battleships and Cruise ships).

What are your thoughts about a factor of x2.5 for a Battleship?
Is it plausible that half again as much ship is out of the water compared to under the water when lightly loaded?

45,000 tons / 13.5 = 3,333.3 dTons (under water)
3,333.3 x 2 = 6,667 dTons (50% underwater estimate)
3,333.3 x 2.5 = 8,333 dTons (40% underwater estimate)

 

[Reban]

Let's talk about the 16" Main Guns for a moment. Under the current design, the Battleship has 3 Superbays of 500 dTons each. Each Superbay mounts three (3) 16" guns. Each 16" Gun is equal to 80 Artillery Gun-TL 4 (8D6 each) firing simultaneously at the same target.



Let's apply this to our 80 x Artillery Gun-TL 4.
8D6 + (79 x 4D6) = 324D6
324D6 / 50 = 6.48 = 6D6 of Starship Damage

Each 16" gun is comparable to a 50 dTon Bay on a Starship and the Battleship has NINE (9) of these Starship Kinetic Kill Bay Weapons! A full broadside by the Main Guns of a Battleship is up to 9 x 6D6 Starship Weapon hits.

Thats a useful worked piece of comparison. Thank you.

And I think I've seen that movie

 

[whartung]

Each 16" gun is comparable to a 50 dTon Bay on a Starship and the Battleship has NINE (9) of these Starship Kinetic Kill Bay Weapons! A full broadside by the Main Guns of a Battleship is up to 9 x 6D6 Starship Weapon hits.

Is that a lot?

Does the Spaceship Yamamoto have a fighting chance now?

 

[atpollard]

Is that a lot?
Does the Spaceship Yamamoto have a fighting chance now?

The CE SRD is geared towards the "Small Ship" universe of the pre-CT:High Guard Traveller Imperium. Unless I missed something, there are no 100 ton bays and no spinal weapons in the rules. The Starship Design tables (hull size vs Drive/PP number) top out at 5,000 dTons.

So the range on a WW2 Battleship is short for a Starship, but its punch within that range is as heavy as any Starship Weapon available ... not bad for TL 5.

 

[atpollard]

Time to fill in the costs ...

Chassis = MCr 300 x 4 / 25 = MCr 48
Armor = MCr 48 x 5% x 4 points of armor = MCr 9.6
Power Plant = MCr 168 x 0.006 x 4 PP = MCr 4.032
Screw Propeller = MCr 84 x 0.25 x 0.1 x 4 PP = MCr 8.4
Agility = MCr 48 x 0.5 = MCr 24
Fuel = Cr 830 x 6,048 spaces = Cr 5,019,840
Controls = Cr 2,000
Communications = Cr 2,000
Control Cabin = (1 Std x Cr 20,000) + (132 Ext x Cr 5,000) = Cr 680,000
Staterooms = 270 x Cr 500,000 = MCr 135
Military 2-bunk Barracks = 1215 x Cr 100,000 = MCr 1.215
Primary Armament Turrets = 3 x 6000 x Cr 16,000 = MCr 288
Artillery Gun-TL 4 (x720) = Cr 160,000 x 720 = MCr 115.2
Secondary Armament Turrets = 10 x 60 x Cr 16,000 = MCr 9.6
Artillery Gun-TL 4 (x20) = Cr 160,000 x 20 = MCr 3.2
Quad-40mm Large Turrets = 20 x 63 x Cr 16,000 = MCr 21.16
Rocket Artillery-TL 5 (x80) = Cr 6000 x 80 = Cr 480,000
Large Turrets (x49) = 49 x 6 x Cr 16,000 = MCr 4.704
Heavy Machine Gun-TL 5 (x49) = Cr 6000 x 1.5 x 49 = Cr 441,000
Artillery Gun Ammo (8640 spaces) = 8640 x Cr 4000 = MCr 34.56
Artillery Gun Ammo (240 spaces) = 240 x Cr 4000 = Cr 960,000
Rocket Artillery Ammo (2400 spaces) = 2400 x Cr 5000 = MCr 12
Heavy Machine Gun Ammo (15 spaces) = 15 x Cr 5000 = Cr 75,000



TL 5 Battleship (Iowa)

Chassis

• Base = Cr 48,000,000
• Configuration = Cr 0
• Iron Armor = Cr 9,600,000

Power Plant

• Internal Combustion = Cr 4,032,000

Propulsion

• Screw Propeller = Cr 8,400,000
• Increased Agility = Cr 24,000,000

Fuel

• Hydrocarbon = Cr 5,019,840

Controls

• Basic = Cr 0
• Autopilot = Cr 2,000

Communications

• Class III = Cr 2,000

Sensors

• None = Cr 0

Computer

• None = Cr 0

Accommodations

• Control Cabin = Cr 680,000
• Standard Staterooms (x270) = Cr 135,000,000
• Military 2-bunk Barracks (x1215) = Cr 1,215,000

Additional Components

• None = Cr 0

Armaments

• Three 500 ton Barbettes [Primary Armament]
• Large Turrets (x3) = Cr 288,000,000
• Artillery Gun-TL 4 (x720) = Cr 115,200,000
• Bonus gunners & loaders (x 237) = Cr 0

• Ten 5 ton Barbettes [Secondary Armament]
• Large Turrets (x10) = Cr 9,600,000
• Artillery Gun-TL 4 (x20) = Cr 3,200,000
• Bonus gunners & loaders (x 30) = Cr 0

• 20 Quad-40mm "Large Turret" Mounts [Anti-Air Armament]
• Large Turrets (x20) = Cr 21,160,000
• Rocket Artillery-TL 5 (x80) = Cr 480,000

• 49 Singled-20mm "Large Turret" Mounts [Anti-Air Armament]
• Large Turrets (x49) = Cr 4,704,000
• Heavy Machine Gun-TL 5 (x49) = Cr 441,000

Ammunition (Thirty minutes [300 x 6 second combat rounds] worth of ammo for our Armaments)

• Artillery Gun-TL 4 (x720) = Cr 34,560,000 [Main Armament]
• Artillery Gun-TL 4 (x20) = Cr 960,000 [Secondary Armament]
• Rocket Artillery-TL 5 (x80) = Cr 12,000,000
• Heavy Machine Gun-TL 5 (x49) = Cr 75,000

TOTAL COST = Cr 726,330,840

 

[Timerover51]

Recalculating Fuel Requirements:

Hydrocarbons require more space than Hydrogen Fuel (x3) and each Power Plant W requires 42 tons of Hydrogen per 2 weeks. So my 4 Power Plant W would need 168 tons of Hydrogen for 2 weeks, or 504 tons (168x3) of Hydrocarbons.
[/B]

"Hydrocarbons require more space than Hydrogen Fuel (x3)"

One final comment on this exercise.

Since when do hydrocarbon fuels weigh LESS than Liquid Hydrogen?

A metric ton of Liquid Hydrogen occupies 14.114326 cubic meters or 498.44 Cubic Feet. A long ton of WW2 Bunker fuel, larger than a metric ton, occupied about 39 cubic feet. A gallon of WW2 or current motor gasoline weighs 6 pounds per gallon, 44.88 pounds per cubic foot, or 1585.02 pounds per cubic meter, which comes out to 1.39 cubic meters per metric ton. Gasoline is about the lightest hydrocarbon used for liquid fuel. Ten metric tons of it occupies slightly less space than one metric ton of Liquid Hydrogen. Twelve and three-quarter long tons of WW2 Bunker fuel, or just under 13 metric tons, would occupy about the same amount of space as one metric ton of Liquid Hydrogen.

Edit Note: My value for the weight of bunker oil may be low, as another source gives the weight only slightly less than water. That would push the weight of 14 cubic meters of bunker oil very close to 14 metric tons.