35

The Death Star is large enough to have its own gravitational field. One would expect that the preferred "down" direction points toward its center, instead of running from pole to pole. However, the pictures of the incomplete station show a structure that favors two perpendicular directions. How come?

  • 4
    Suppose gravity pointed toward the center. Would it get weaker as you moved "down"? If so, it would be strange for people to feel their weight change as they move between levels. If not, there will be some sort of singularity near the center, and people on the inner levels will feel strange lateral forces. It would seem the most comfortable for gravity to have the same direction and magnitude everywhere, especially for personnel who are used to living on large planets. And as James Sheridan's answer suggests, this is attainable with period technology. – Nate Eldredge Oct 2 '14 at 6:48
  • 7
    (And now I'm imagining a Death Star on which everyone suffers from chronic vertigo, a la Lucille Two.) – Nate Eldredge Oct 2 '14 at 6:53
  • 13
    Whether it is "large enough" is irrelevant, the requirement is that it's massive enough. – O. R. Mapper Oct 2 '14 at 7:10
  • 5
    @RobertF because a cube could never be mistaken for a moon, thus depriving society of one of the most quoted lines from the SWU. – mwotton Oct 3 '14 at 6:16
  • 8
    @RobertF: Because that allows for a maximum of internal volume at a minimum of external surface area maybe? – O. R. Mapper Oct 3 '14 at 8:47
49

A radial structure would give the floors and ceilings a noticable curvature, which would get more pronounced closer to the interior. Flat objects, like regular furniture, wouldn't be flush with the ground: all of their weight would be carried by their centre, with the edges sticking out above the floor (or, more accurately, the floor would stick out under the edges).

Not only would this require expensive custom furniture and equipment, but the curvature on inner levels would be greater than the outer levels, so equipment would only sit properly within a particular range of levels.

"Local flatness", eg. per room, wouldn't work either:

  • Items nearer the walls would roll towards the centre, which is technically "down hill"
  • The layout of rooms on each level would have to depend on those above and below it, to avoid floors and ceilings "pinching" together.
  • Alternatively, some extra space could be allocated to allow for pinching (eg. with adjustable floor/ceiling tiles on pistons), but this reduces the number of levels that can fit in the structure.
  • Floor plans couldn't be revised by simply moving walls around: the floors would need to be altered and walls from one place might not fit in another.
  • There would be abrupt changes between the sections, eg. at doors and along corridors.
  • Large spaces like hangars would need to be split into several areas with different floor orientations.
  • 7
    It is good to see that they spend wisely their money when building a death star. – BlueTrin Oct 2 '14 at 14:21
  • 4
    Actually, this is a remarkably good answer. – ThePopMachine Oct 2 '14 at 14:39
  • 16
    On the edge of a 150 km-diameter sphere, this effect would be so small as to be unnoticeable: assuming a room 10 meters wide, built to follow the curvature, the center would be a sixth of a millimeter higher than the edge. Of course the effect gets larger as you move toward the center. But wasn't the middle of the Death Star hollowed out to serve as a power station anyway? – David Z Oct 2 '14 at 16:50
  • 3
    @BlueTrin I attended the University of Sheffield, and according to Wikipedia its Arts Tower was originally designed to be cylindrical. They changed it to a cuboid when they worked out the cost of bespoke furniture to account for the curved walls. I figured a spherical "building" would be even worse! – Warbo Oct 2 '14 at 20:09
  • 6
    @David Z's answer is a beautiful example of demolishing an entire academic argument with one simple sentence. – Contango Oct 3 '14 at 12:49
81

The Death Star's natural gravitational field would be exceedingly weak.

The Death Star was described as a "small moon" in Star Wars; Wikipedia lists the first Death Star's diameter as 140 to 160 kilometers. A large moon like ours, 3500km of solid rock, only produces a gravitation field 17% of Earth's at its surface. The Death Star, besides being much smaller than our Moon, is also mostly hollow, the reason being that you don't build something that big unless you really need the interior space. The Death Star at its surface would not have a gravitational field of even a thousandth that of Earth, and the field would decrease as you moved into the interior of the station. Given the weak field, there's nothing to be gained by orienting the floors relative to it.

A radial orientation means all windows are skylights.

If you're lucky enough to have a window office in the Death Star, it would be much nicer if you could sit in your executive chair and look out at the stars without craning your neck. But if the floor is always pointed toward the center of the station, then the windows out are necessarily on the ceiling. The Emperor seemed to prefer normal windows.

emperor, vader, luke look out window at the stars

  • 25
    It's good to see that the engineering decisions of the Death Star depend mostly on interior design considerations (windows on the walls, please!) :-) – David Richerby Oct 2 '14 at 9:05
  • 30
    If you're supreme evil emperor of the universe, I think you automatically reserve the right to make quirky design choices – Jon Story Oct 2 '14 at 10:15
  • 7
    @DavidRicherby: Well, I mean, there's this little hole. It was kind of an aesthetic choice by the architect. And if you shoot a laser into this whole, the station blows up. – James Sheridan Oct 2 '14 at 11:50
  • 6
    @Wlerin, as you move closer to the center, you leave some of the mass behind, which now pulls the other way, so it starts to cancel out. As you get to the center, the mass fully surrounds you from every direction and the gravity force fully cancels out. Amazingly, you can easily prove using Gauss law that anywhere inside the void of an ideal hollow sphere, no matter how thick or thin, the gravity force from the shell cancels out completely and it has to be exactly zero (Newton proved this first, the hard way, with calculus). – Euro Micelli Oct 2 '14 at 12:24
  • 5
    Note that the Emperor's throne room is at the top of a tower on the north "pole" of the Death Star, so presumably it would function the same either way independently of the Death Star's gravity orientation. – Milo P Oct 2 '14 at 17:39
20

The design you're describing is very similar to that of Centrepoint Station, a large space station in the Corellian system. Its construction pre-dates the invention of artificial gravity without spin in the Star Wars universe. By the time the Death Star is constructed, however, artificial gravity is commonplace. It makes sense, therefore, for the Death Star to use the pre-existing and commonly used artificial gravity convention used in all other Imperial construction, rather than pursuing a radically different design philosophy, no matter how much sense it might make from a physics background.

9

All the above answers are good, but I'd also like to add that with a radial layout, only one layer would be on the surface of the station: a single shell surrounding the entire sphere. Obviously you could divide that "surface level" into different zones for different operations, but it would quickly become unwieldy to have, for example, a docking station where ships come down from the roof and then you have to go down a series of floors for different levels of operation.

Much better to have a sphere with flat levels equivalent to latitude, so that the re-supply docks can be at a certain latitude while the military docks can be elsewhere, etc. With a flat level structure every floor has a ring of surface contact around the edge, allowing for more efficient transfer of objects and ships between the interior of the station and the outside, and easier localization of different departments and facilities along a single floor of the facility.

7

Given fixed density, gravitational acceleration at the surface of a spherical body is proportional to the radius of the body (mass goes up as the cube of radius, but gravity falls off with the square of distance from the center of gravity).

Thus, the gravity of a solid rock Death Star of the same density as Earth's moon would be on the order of 0.007 Earth gravities at the outer edge, and decreasing as you go deeper. The Death Star isn't solid rock; a large fraction of its volume filled with nitrogen and oxygen gas, hundreds of times less dense.

This means the artificial gravity generators still have to do more than 99.3% of the work even if you orient floors center-ward. The Empire was obviously not concerned with engineering efficiencies of that order.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.