Death Star Spaceship

[FeiLin]

I'm sure this has been done before, and it's not really a serious exercise, but I tried to model the Death Star from Ep IV with the rules from the Spaceships series (RAW). Obviously, it would be bigger than SM15, so I had to improvise a bit. Also, I just glanced a bit at the normal Wiki for some basic stats.

It's more than 100 km, which I translate to SM31 (sphere shaped). I don't know about TL, but let's assume it's 10 (it's not _lower_ than that, at least). I have no idea how many J the main battery is, but I assume it's A LOT, so let's say it's a spinal battery. Let's also just have some fun and assume it's a large system, so it takes up not 3 but 9 slots (I think that's allowed according to RAW). I included the cheapest armor in each section, and filled the two cores with power plants for 2*4 power points that power the main gun now at 9 (the trick is to find the weak spot to bypass the armor in order to kill it, although I don't know if there's a rule that would let a fighter-size missile do the trick, unless you do some kind critical hit or special rule). Even so, it still needs another power plant, but let's make it half a system, so it has 10 in total.

There's a control room with 20 000 stations, Complexity and Comm/Sensor level both at 19 (arbitrarily extrapolated). To go with that, there's an array system with level 21, and just because I had a small slot over an ECM.

I struggled a bit with guns, since you could only get 30 per tertiary battery, but I'd say the DS probably has them in the thousands at least. It _would've_ worked if you could simply say a small system be "recursively small", so you could get the tertiary batteries down to SM10-15 (there'd definitely be enough then... but I didn't find that possible).

There's a small fusion torch reaction engine (I think, but I'm a bit at a loss with all the techno-babble for the engines, and I haven't looked more into the SW lore, so maybe there's better suited ones). Also, I slapped on a stardrive-1.

There's of course a hangar ("small" one, though, at SM30), with a 3*10^12 ton capacity and a launch rate of 500 million (again, arbitrarily extrapolated, but I'd say at least 5000 for SM15, so still way more per minute than I recall from the movie...). There's a cargo area for 5*10^12 tons. There's 6*10^11 habitats, which gives room 1.2*10^12 people in regular rooms, so definitely room for the 1.7 million that suppose to work there, and some 20 million open areas to have some fun with.

Then, finally, having that many people, you need something for them to do, so let's build them a factory for them to work (that's perhaps what the Empire does best).



The stat line (but columnised for simplicity) is as follows:
TL 10
Spacecraft
dST/HP 500000
Hnd/SR -10/5
HT 14
Move 0.5G/25 kps
LWt 3*10^14 ton
Load (didn't calculate)
SM 31
Occ 1.2*10^12 ASV
dDR 30000
Range 1x
Cost $ 5.122*10^19



This is the system configuration (3 in the same slot means small systems, and 2 are half systems):
Front
Front 1 Armor (steel)
Front 2 Spinal
Front 3 Spinal
Front 4 Spinal
Front 5 Hangar Cargo Factory
Front 6 Fuel Power Plant (antimatter)
Front Core Power Plant (antimatter)
Central
Central 1 Armor (steel)
Central 2 Spinal
Central 3 Spinal
Central 4 Spinal
Central 5 Tertiary Battery (tractor) Tertiary Battery (?) Power Plant (antimatter)
Central 6 Stardrive
Central Core Power Plant (antimatter)
Rear
Rear 1 Armor (steel)
Rear 2 Spinal
Rear 3 Spinal
Rear 4 Spinal
Rear 5 Control Room Multipurpose Array Open Areas
Rear 6 Fission Engine Habitat ECM

[Varyon]

An oversized Spinal Battery should take up 10 systems, not 9 (Spaceships follows a 1-3-10 progression). As a normal Spinal Battery requires at least one Core system, so should this, and it wouldn't be out of the question to require both Core systems be used up by the Spinal Battery. Looking at the Wookiepedia entry, the death star actually had a diameter of 100 miles (160 km), putting it at SM+32 as a sphere. An SM+33 Spinal Battery would be 3 ZJ. As a laser (as it's described) or particle beam (most tend to treat SW blasters as particle beams), it would deal 3dx1,000,000 dDam; as a plasma weapon (which SW blasters resemble more closely) it would deal 6dx1,000,000 dDam. Unfortunately, both of these fall far short of what would really be needed to blow up Alderaan - ZJ is 10^21 J, while according to Atomic Rockets' Boom Table blowing up Earth requires around 6x10^31 J, and the Death Star's maximum output was actually around 10^38 J. SI doesn't even have prefixes that go that high! Even using the Cosmic Power switch doesn't help here, as that only multiplies output by 1000. Without Cosmic Power, the SS Death Star is on the same level as the last time Yellowstone erupted. With it, the SS Death Star is on the same level as the Wilkes Land Crater. Earth has tanked both of those just fine.

To get the Death Star's performance at its size, you'd need some new weapon. To blow up an Earth-sized planet to the same extent as Alderaan, you'd need to deal around 6dx1,000,000,000 (6dx2B for plasma) dDam. To match the maximum output of the Death Star, you'd need to deal around 2dx500,000,000,000 (2dx1T for plasma) dDam. That calls for either x2000 damage or x1,000,000 damage. GURPS rules are a bit more lenient - with Earth having 200,000,000 dHP (and ignoring its DR) based on its mass, Total Destruction can be obtained with "only" 2,200,000,000 dDam, which on average requires a bit over 6dx100,000,000 dDam (30,000 YJ for a laser, 3,000 YJ for a plasma weapon). That's achievable, if we use Cosmic Power (x1000 ouput, so 3 ZJ becomes 3 YJ) and scale your Death Star up to SM+40, for a diameter of around 2,500 miles. That's around the size of our own moon (2,169 mile diameter).

As a side note, I think setting it as an oversized spinal battery is probably correct. The Wookiepedia entry notes that all the weapons (other than the superlaser), habitats, etc are located in a relatively thin outer shell, with the innards taken up by the superlaser, reactor, sublight engines, and hyperdrive.

[Agemegos]

Originally Posted by FeiLin

I have no idea how many J the main battery is, but I assume it's A LOT…

For what it's worth, it is possible to estimate the energy of the main armament on the Death Star from the fact that it blows up Alderaan (an Earth-like planet) and scatters the debris at a measurable speed. 10^31 J are required to overcome the planet's gravitational binding energy, but a minimal-energy disintegration involves the debris leaving at escape velocity, which is only about 10 km/s, and that would result in the cloud of debris taking eleven minutes to expand to double the size of the planet. Alderaan explodes to double size in a few seconds, which implies a lot of kinetic energy in in excess of the gravitational binding energy. I haven't done the calculations myself, but those who have tell me that the answer is 10^38 J. That's one hundred trillion yottajoules.

The authors of Star Wars ignored physics, and perhaps you should too.

[Anders]

I read somewhere that the Death Star firing requires turning a mass equivalent of Himalaya into energy.

[Agemegos]

Originally Posted by Anders

I read somewhere that the Death Star firing requires turning a mass equivalent of Himalaya into energy.

Let's see.

E = mc²

m = E/c²

m = 10³⁸ J / (3 × 10⁸ m/s)²

m = 10³⁸ / (9 × 10¹⁶) kg

m = 1.1 × 10²¹ kg

The Himalayas are roughly 2,400 km long, 200 km wide, and say 5 km high. 2.4 million cubic kilometres. 2.4 × 10¹⁵ cubic metres. They are mostly sedimentary and metamorphic rock of marine origin, so say about 2750 kg/m³. That's 6.6 × 10¹⁸ kg.

That's not enough. We would need about 170 Himalayases converted into kinetic energy to destroy Alderaan as violently as we saw on the screen.

<postscript>

What is the mass of the Death Star itself?

[AlexanderHowl]

I generally use a volumetric analysis, with the density of liquid hydrogen as the baseline, for massive installations). If we assume Mimas as an example of a small moon, we have a volume of 32 million cubic kilometers. A comparable Death Star could be 1 quadrillion tonnes to 3 quadrillion tonnes, depending on density.

If we go with 3 quadrillion tonnes, the result is a more modest SM+27. The spinal battery would have a 3 YJ release (assuming cosmic energy), which ends up being 6d×100,000 d-damage, which is about 1% the damage required to destroy a planet. You would need around a million times the energy.

[Hide]

So, from what you say. Are you implying that the energy sources in the book cannot match the DS power as seen in the movies?

[Agemegos]

Originally Posted by Hide

So, from what you say. Are you implying that the energy sources in the book cannot match the DS power as seen in the movies?

Originally Posted by Agemegos

The authors of Star Wars ignored physics, and perhaps you should too.

Looks like a "yes".

[AlexanderHowl]

Even the Cosmic Power lens is insufficient. A 10^33 J surge would be required (given the fact that a significant amount of energy will be wasted), and the waste heat alone would vaporize the station. You would need something on the order of 6×10^19 grams of matter and antimatter (30 trillion metric tons of each). It would actually make more sense to drop the antimatter on the planet and not even worry about a laser, as it would be less dangerous to the station.

[Varyon]

Originally Posted by Hide

So, from what you say. Are you implying that the energy sources in the book cannot match the DS power as seen in the movies?

Neither the weapons nor the power plants can, at least not at the size of the Death Star. As I noted, using Cosmic Power, a 10-system (50% of the spacecraft's mass) Spinal Laser requires a space station that's a bit larger than Luna (Earth's moon). Such a weapon requires 10 Cosmic Power Points, which can be provided by 2 Cosmic Total Conversion systems (which seem appropriate for Star Wars' hypermatter reactors). So long as the Death Star's various turbolasers and force shield (if it even has one; I don't recall it being mentioned for DS1, and DS2 used a shield located on Endor) don't require Cosmic Power, that should be enough (the stardrive and possibly thrusters might require Cosmic Power, but those aren't used at the same time as the superlaser so we're fine there). That leaves 8 systems for the stardrive, thrusters, armor, defenses, hangar, and living areas.

Originally Posted by cptbutton

I would assume the DS turbolaser actually somehow converted a large portion of Alderaan's core into antimatter via [magic physics].

Yeah, that's the interpretation I often go with.