I wanted to get a good comparison of the Death Stars to the Earth and Moon, so I did. Then, I did some calculations regarding the inside of the Death Stars, and was very surprised with want I found. Assuming I didn't put a decimal in the wrong place or make a major error, this should be right. Let me know what you think. (Planning to use this in my Geometry class, didn't just make it for fun! BTW)
Awesome. Wow. That scene in clerks discussing all the poor engineers and electricians is really hitting home right now.
I don't know what I appreciate more, this accurate scale or the fact that you capitalized Moon when referring to our satellite #NerdAstronomer
I love what you did but I just do not believe the data are accurate. There are conflicting reports regarding how big the 2 Death Stars are, so I just choose to believe my own eyes that while the 2nd Death Star is bigger, it isn't 5 times bigger than the original. But don't get me wrong.... I wouldn't change a thing. I think your graphics and calculations are really interesting.
The RoTJ novelization called the DS2 "nearly twice as big (as the DS1)" And the newest book about the Death Stars (Death Star Owner's Workshop Manual/Death Star Owner's Technical Manual) goes back to the old figures of 120km and 160 km respectively, after lobbying by the authors. Part of the motivation may have been that the Forest Moon's size had previously been established by books like The Essential Atlas and its predecessors as quite small (4900km diameter) - which does not work with a 900km DS2. Regarding habitable volume - it's generally portrayed in the books as having people live and work mostly in the crust - with the interiors of the Death Stars having very few people working there.
That's great. Well done. If you have a spare minute can you please work our how big a Kenner DS would've had to have been for my 4 inch figures to reenact ANH to the correct scale?
I like calculations- working out how dense a 4900 km Forest Moon would need to be to have half-decent gravity, and so forth.
Of course, even restricting yourself to the crust, still allows a huge amount of space. http://www.irregularwebcomic.net/109.html
Yeah, I just choose the biggest sizes I could finds, and listed them in the graphic. Since it isn't stated in the movie, there is no canon size. I do agree that at the 900km size isn't very plausible, even in the Star Wars universe. Oh, and with a diameter of 900km, it wouldn't be "5 times bigger." It would be a little over 5 times wider, but the bigger one would be 177 times the volume. I do like the idea of using the crust and having spherical floor-levels, but that's not they way it's portrayed in the movie. What would you think the "thickness" of the habitual crust layer to be? How many levels could we estimate?
Ok. I went back to the movie to try to get a canon size for this thing. The shots of the Death Star in space don't work well, they don't show the whole planet so it's hard to do a size comparison. The shot of them coming out of hyperspace could work, but the Death Star is closer to them than the moon, so again its hard to do a comparison. However, that shot does pretty much match the hologram, so lets assume its to scale and use it. Setting the Endor moon at 4900km would make the Death Star in this image 383km. The problem with this size is gravity. While it's not impossible for a small moon to have Earth-like gravity, it does mean the moon needs to be very dense. That's not impossible. If it had a relatively larger metallic core than the Earth does, like Mercury, then it could be more dense. But, according to my calculations, if it were 4900km wide, it would need to have a density of 14.3g/cm3 to have Earth's gravity. The density of iron is only 7.8g/cm3. Lead is just 11.4! So, even if the entire interior were made of the abundant heavy metals we find in Earth (iron, nickel and cobalt), it still wouldn't be dense enough. (For comparison, the density of Earth is 5.5g/cm3, and it's the densest planet or moon in the solar system). So, unless I've made a mistake, the 4900km Endor moon is implausible based on it having similar gravity to Earth. If we only have what we see in the movies to go by, the simplest assumption is that since Endor has about Earth's gravity, it is about the size and density of the Earth. Well, if we put Endor at a diameter of 12000km (about the size of Venus, which has 90% Earth gravity), the Death Star in this image comes out to 938km in diameter. That works out kinda nicely.
This is the GFFA we're talking about here. If characters can have normal gravity walking around a ship in the middle of space, then I think we can ignore how much gravitational pull the Death Star should have on surrounding bodies. Heck, the gravity of the Death Star itself isn't even oriented towards the center of the structure, so let's just throw everything we know about the laws of gravity out the window.
Hmm, I wasn't talking about the gravity of the Death Star. I was referring about the gravity of the Endor moon, and explained how a world only 4900km wide size couldn't plausibly have Earth gravity, unless they give some sort of explanation for it. Here's one they could have used: "The empire has activated a gravity field in a 100km radius around the shield generator, so you'll be experiencing standard gravity." But, they didn't give an explanation for the standard gravity, so the simplest assumption is the Endor moon is of similar size to Earth. That's an easier assumption than they have different laws of gravity. Based on the hologram of the Endor moon and DS2, it comes out to be around the 900km figure if you assume the Endor moon to be of similar size to Earth. It's pretty standard in science fiction to have artificial gravity on ships and space stations, which we assume they are using on the Death Star. I'm sorry, but I really like this kinda stuff. Closest I can get to the experience of a detective. I think my next goal is going to be to derive the size of the original Death Star using the height of the hanger they brought the Falcon in, using that to get the width of the Equatoral Trench, and from there the radius of the Death Star.
Science fiction has artificial gravity, such as the rotation of the space station in 2001 creating a centifugal force which simulates gravity. Science fantasy, however, doesn't bother with explanations in the slightest, like how Luke and Obi-Wan can just walk around the Falcon like normal even though it is in the middle of space. (And I don't really care what the nonsense that is the EU says by way of explanation)
I tend to think of it the way Star Trek makes it out to be: artificial gravity is created by grav plating in each of the decks, which emit a gravitonic energy equivalent to 1 standard G. I don't know if it's ever explained how it works in Star Wars, so this is the assumption I will go by until it's stated otherwise.
According to Saxton: http://www.theforce.net/swtc/orbs_habitable.html The Earth's escape velocity is 11.2 km/s. Venus also retains a significant atmosphere and has an escape velocity of 10.4 km/s. Mars has an escape velocity of only 5.02 km/s and it does not retain a very dense atmosphere. However the problem with Mars' atmosphere is probably more to do with the lack of reprocessing of carbonate rocks in tectonic subduction zones. A planet with a 5 km/s escape velocity might keep a hospitable atmosphere if it were tectonically active. That depends on a complicated details of geology and composition. http://www.theforce.net/swtc/orbs.html#endor The Endorian gravity is unlikely be greater than 8m/s² (versus Earth's 9.8m/s²) even if the moon is wildly enriched with heavy chemical elements. Light gravity may also be necessary to justify other evidence in Return of the Jedi, particularly the ability of rebel heroes to fall many metres from an ewok trap without breaking any bones, and the ability of an ewok combat glider to carry not only an adult ewok but also a load of rocks heavy enough to annoy an AT-ST. The enormous size of the Endorian trees and dangerous mega-fauna is directly attributed to the low gravity [The Illustrated Star Wars Universe]. Forest Moon's diameter is 4900 km (The Essential Atlas) Earth's diameter is 12756 km (equatorial) Forest Moon's diameter (and radius) is 0.384 x that of Earth. (so Earth's is 2.603 x the Forest Moon's) In order to have a gravity of 9.8m/s², the Forest Moon's density would need to be 2.603 x that of Earth. In order to have a gravity of 8m/s², the Forest Moon's density would need to be 2.125 x that of Earth. This is much less dense than the densest metals. And if the gravity is lower than 8m/s², the Forest Moon can get away with being even less dense than this. A Forest Moon with a large iridium or osmium core - could have enough gravity to hold on to an atmosphere.
Which still makes no sense because gravity doesn't exactly work in one direction. If you're on a deck in the middle of the ship, the grav plating in the deck above you would be 1G, the grav plating in the deck below you would be 1G, so you wouldn't exactly be able to walk around like normal anyway. Well you can assume whatever you'd like, or you can just realize that it is a fantasy universe and worrying about such things is pointless. If I can accept a laser that can destroy an entire planet by making it explode to smithereens, I can accept that there is Earth-like gravity everywhere the characters go. The fact that such things break about 1,000 laws of physics doesn't bother me because I suspend belief for the sake of storytelling.
Other people have already tried similar things - for both Death Stars: http://www.st-v-sw.net/STSWdeathstarsizes.html http://www.st-v-sw.net/STSWdeathstarsizes-2.html However- these estimates are often fiercely criticised.
Checked these calculations and it looks like they pretty much match what I found. 2.6 x Earths' density (5.5g/cm3) gives the 14.3g/cm3 that I calculated. If you want to go a bit lower, that's fine. The lower at 2.125 x Earth's density gives you 11.7g/cm3, still way higher than the common metals that make up the rocky planets in our solar system. I specifically said it wasn't impossible for it to work out, but prefer to obey Occam's Razor and go with the theory with the simpliest assumptions. Based only what is in the movie, it is simpler to assume the moon is of similar composition as other terrestrial worlds, than to assume it is special and composed almost entirely of some of the rarest metals elements found in the planets we know. If you want to go by the books and other sources, that's fine. I'm not saying they are wrong. Implausible doesn't mean impossible. I even offered a way to explain having Earth gravity on a smaller world in my previous post. What was this whole thing about? Oh yeah, that the Death Star 2 could have been 900km diameter. The only reason I used that number was because when I googled "death star size", I found this: "There is a broader range of figures for the second Death Star's diameter, ranging from 160 to 900 kilometers." I choose the larger size. If someone can show me something from the movie that would make one measurement more likely than another, then I'll change my mind. I didn't make the graphic to start a debate or arguments, I made it for my classroom and thought it would be neat to share. Thanks for taking an interest! But this will be my last post in this thread. Didn't mean to get so caught up in it.
The above links include lots of photo comparisons - beginning with the size of Imperial Shuttle, extrapolating size of launch bays from that - then extrapolating size of equatorial trench, then the smooth band in which the trench lies, then the whole Death Star. And also try to demonstrate that many of movie shots are not consistent with a Forest Moon to Death Star 2 size ratio of 11:1. No problem - I enjoy debates in general.
Well, after some time off I just couldn't resist coming back for more. This time it wasn't for my classroom, just for my own stupid amusement. Also, I have to admit I must be a little bias in wanting the Death Star to be as huge as it can possibly be. I love large-scale engineering in sci-fi. The novel spin-off of the Dyson Sphere Star Trek episode is one of my favorite books. A television adaptation of Ringworld has been talked about again recently, and I really hope it makes it this time. Lets start with the Super Star Destroyer. The Executor dwarfs the other Star Destroyers, you know, like the one that flew over the camera in the first scene of ANH, to give us a sense of how big the ships are. Yeah: THOSE ARE TINY NOW. Then, we see that Super Star Destroyer CRASH into the FRIGGIN Death Star. Which was amazing in itself. But, we also get another moment of "bigness", which to me was more powerful than the ANH fly-over. As big as the Executor was, the spherical Death Star looked nearly completely flat when the ship crashed into it. Now, we are never given any sizes of any ships in the movies. So, I looked it up and found a reported length of 19km for the Executor. Compare that to the two reported Death Star II sizes of 160km and 900km. The 900km clearly is a better match for what we see in the movie, if you accept the 19km size for the Executor. If you don't, and want to set the Death Star at 160km, then that shrinks the Super Star Destroyer, and thus the normal Star Destroyers, to a size much too small. Take that shrinkage all the way back to ANH, and the Tantive IV might only be about 30 meters long (didn't do the math on that, though) Well, this might not change anyone's mind, but at least I can add "argue with the internet" as a reason why it's important to be able to solve proportions!
It's certainly true that the Death Star 2's apparent size is pretty inconsistent from shot to shot. No matter what size you use - some will end up being iffy. This article on the subject http://www.facebook.com/notes/star-...f-the-size-of-the-death-stars/641724579210986 from CeiranHarmony, one of the better known posters from the Fleet Junkie thread, is pretty interesting.
