Would the Death Star Need Artificial Gravity? (Because Science w/ Kyle Hill)

[@Nerdist]

Hey so I think I screwed this up a bit not being clear on my design. Many of you are right: the gravities that I calculated really only work at a right angle to the axis of rotation, so most of the workable gravity in my design would be at and around the equator -- imagine large, concentric "ring worlds" on the interior -- with gravity getting smaller and weirder (not pointing to your "down") as you move towards the top and bottom. I know there wouldn't be gravity at the center axis, and I think that's fine overall. My design also doesn't work great with windows and shafts and docking bays as shown in the canon, unless you shift those around to face "upwards." I really hate when I don't think of stuff that you all do: I'm just one guy writing these by myself, so that's on me as an educator, and I apologize if my theory doesn't check out with y'all. Thanks for keeping me honest. -- KH

[@fluffycandy5939]

Im imagining the death-star spinning around while trying to shoot at a planet, and the laser hitting everything but its target.

[@abbychavarria6205]

It's all good dude, we just have fun theorizing with you. I think it's so awesome you own up when you find something is off with your theories. Keep up the awesome work. Much love from a random person on the internet. :)

[@thechumpsbeendumped.7797]

Zorua Adventures it would work if the laser was at one of the poles.

[@lazergurka-smerlin6561]

It's ok, because without anything to test on (else than your own mind) it is hard to know exactly what would happen. And that's why i started a kickstarter to fund a deathstar (not really).

[@NinjaBearFilms]

I had always assumed it was levels curving around the center in increasingly tight rings. But what if the energy core of the death star had a proto-type of the energy core for Star Killer Base? Where in it could harvest energy/mass from a star and condense it into a super dense form. How much of a star's mass would it need to suck in for the power core to give it the artificial gravity of a planet and how much usable space do you get from that? Or to put it another way... How much would the gravity of Star Killer Base planet increase after absorbing the entire mass of a sun into it's core? Are we looking at miniblack hole creation?

[@Nerdist]

I'm thinking about posting behind-the-scenes stuff and extra rants on Instagram for y'all each week, would you be interested in that? -- KH

[@brandoagusti7288]

I'd love to see all of it in the video it self.

[@connormulholland572]

Just saying centrifugal force is really not a force, it's just inertia resisting motion, don't want people to have any misconceptions.

[@EthosO76]

definitely, yep.

[@PepinsSpot]

They might've not being evil but they were collaborators so... they had it coming.

[@pieceofgosa]

Heck yeah

[@mattcox8121]

Surprise lightsaber was surprising.

[@gi70st]

The problem is, we see docking bays and windows on the death star, so we know the orientation of the floors

[@dj1NM3]

I think Mr Nerdist has made a grave error: in that if it was a ring of 140km in diameter spun at 0.1rpm would it work as stated. If it was a hollow sphere: then as you move towards the poles (the points on the surface at the axis of rotation) and away from the equator, the artificial gravity effect would reduce to zero. This is a reason why "hard" SciFi (eg: "2001", "The Martian" and "Elysium") use ring-shaped spinning artificial gravity sections, not spherical ones. Nice try, though. . Also: If synthetic gravity technology existed like in Star Wars or Star Trek, it is shown as attraction to the floor and not to the centre of the space station or starship. That's why the presumed internal structure of the Death Star and all of the starships is a series of stacked flat floors and not concentric hollow spheres.

[@georgezwick3717]

My thoughts exactly

[@MattStryker]

I have to agree with that. I was thinking the same thing. Rings make sense, but spheres do not. Has to be a floor-by-floor generator.

[@garyboyd9704]

That's too easy to overcome though. Section off the area where the centrifical force reduces to where the gravity is unusable, Like lets say off the top of my head, the top and bottom 25% of the ship. Use one side for your fresh water and the other side for your waste water. Because you have to have a place to keep those things and they don't need gravity.

[@imnotreal1813]

Jeff NME and Gary Boyd you both made vary good points.

[@Jhakaro]

I think the best form of design would be something like a rough spherical shape with the middle the widest and getting progressively thinner in diameter at both the top and bottom except the design would consist of multiple rings so the top and bottom would be flat or even hollow for the most part and the sides or in this case equivalent of the floors, being the same width all across as opposed to in a spherical shape that would result in top and bottom section floors curving inward toward the centre of the poles. Each one would spin at a different rate, the smaller ones faster and the larger ones slower to all simulate an exact 1g of force. Even the smallest sections should be big enough by a long shot to simulate this without any adverse coriolis effect. I think you only need about a radius of .5 of a km to get away with a slow enough speed to avoid such a problem, therefore a diameter of 1km. If the Death Star is 140 km wide at it's widest point, the top and bottom sections should still be well over a km wide. Ships would have to enter through the floor essentially or if there was enough room between overlap of each section, then they could come in from on top or bottom to land. If they arrived from the side i.e the floor of the station, the whole plasma window shield thing would have to be replaced by solid mechanical doors and once closed, the ship picked up by robotic arms and placed off to the side for storage so that when the door opens they aren't just launched straight out of the station, though they'd need the plasma window thing or whatever it's meant to be to hold in the atmosphere of course. Each ring then would consist of many many separate rooms and corridors, hangar bays etc. spanning the entire circumference. Biggest problem would be getting from one ring to another if they're all moving different speeds but I believe it'd be pretty easily doable as they wouldn't be spinning all too fast due to their size. It'd be hardly noticeable for such a big structure. Still a bit weird to conceive though, imaging two spinning rings and docking tubes somehow lining up for you to jump through to where you want to go.

[@Nerdist]

Haha in my "solution" the stormtroopers on the outside of the Death Star would be flung the hell out into space. That's what you get for not taking marksmanship courses private! -- KH

[@ValGOPLock]

genius...

[@gandalftheantlion]

Well do you think its possible, to create artificial gravity generators? Or will that take a new kind of material/property to be able to simulate such a force?

[@iancook1237]

In this case, wouldnt a cylinder make more sense, there would be major gravity direction change, people would get dragged along the floor, falling towards the outside of the rotation.

[@matta8508]

Worth mentioning that beside the nº of sphere shells with workable gravities, you also would have a range of those sphere's where the circular motion would help (let's assume around the equator) and once you start move up you'd experience less and less artificial gravity until you fell.

[@joshualarson8431]

Wait, but the hangers disprove this idea, if the hangers let you go into it and it lands directly down at a 90 degree angel, not flipping 180 to be upside down on the outer shell...

[@thirdbase314159]

In the second Death Star the Emperor's Chamber has windows on the wall. If they using rotational forces to generate G's the windows would be on the floor. Also the Millennium Falcon perpendicular to the outer hull of the Death Star and lands without changing orientation. It leaves the same way. Besides artificial gravity is common in the Star Wars Universe. The ships all have it. The bases all have it. The Death Stars have it. You don't use one artificial gravity generator, you use small ones on each floor, or every couple floors, whatever the math works out to be.

[@nickelpickler6532]

"Your sister....has it." Sorry, Couldn't resist ;-)

[@thirdbase314159]

As I was typing, that was my thought, which was I put it the way I did.

[@AthAthanasius]

There's the "looking out the window as Alderaan is blown up" scene in A New Hope as well. Mirrors ? Oh, but also the scene of the Millennium Falcon being tractor beamed in heavily implies the layout is the 'flat decks' as slices through the sphere.

[@nickelpickler6532]

Ath Athanasius I think that that is just a view screen.

[@grasslion5377]

and also the scene where the death star is fired, the men are standing perpendicular to the lazer blast.

[@ravenlowes1830]

I love the "SURPRISE LIGHT SABER!" Sections.

[@MrLaxinator]

They are always... surprising.

[@bryanmaynez4117]

and light sabering

[@nerfinator6]

I don't, this guy tries too hard to be funny, and it doesn't work.

[@ChristopherOkhravi]

Too funny :D

[@lokidtt]

Even though I love this theory, I'm not convinced for two reasons : - First, how does the star target and shoot if it is constantly in rotation ? Well, I guess the axe of rotation must be aligned with the laser - Second and most important : the centrifugal force only works in a circle, not in a sphere. So people walking near the axe of rotation would simply feel dizzy, but without gravity :s The Centrifugal force would work if the death star was a tube instead of a sphere. But still, love this channel, keep up the good work !

[@ashtonhoward5582]

Pierre Wittemer another thing is that it'd be impossible to land in the Death Star, assuming the space ports are parallel to the surface (which they aren't) the ships would have to accelerate to the same speed as the outside of the Death Star before they even tried to get inside

[@carlosrangel7134]

That's pretty much what we do on Earth too though. Otherwise you would burn out for the air friction.

[@notswazo]

I just realized he's writing backwards

[@twincast2005]

they flip the image

[@notswazo]

Twincast he would've been going right to left

[@usarrr]

They mirror the image, making him seem left-handed. That's why he never wears shirts with letters on them. But you can also see his button shirts have the buttons and the holes on the wrong side.

[@AzureViking]

And in older videos when his hair was shorter and parted on one side you could see his part flip flop back and forth

[@WhysoSeriousSamual]

I think he is writing backwards. There are moments he has to think when writing, like remembering how r's look backwards.

[@jjop810]

I feel smart just by watching these videos.

[@lupusrex2515]

same

[@Jared_Grillo]

jjop810 you English good

[@onlypants2191]

Go subscribe to the channel crash course. It is great just learned some eco.

[@MrQuest0]

these videos dummy

[@lupusrex2515]

+quest0 pearl?are you watching this?!

[@louisvictor3473]

What if at the center of the death star you had a fragment of a neutron star? Sure, the death star is bigger than a neutron star, but a whole neutron star would be too massive, so we need a splinter. Now, if you could take a piece of it and hold it stable, you have enough super dense material to provide real gravity at normal rates. Would also explain why a direct hit to the core makes it go boom, cause if destabilizes that piece of neutron star, the energy release would be epic. Alternatively, could we have like a failed white dwarf at the core? Normal ones are about as massive as the sun, but as small as our planet - sooo, wouldnt something that was like one, but just a few kilometres in diameter (wheter natural or manipulated to be like that) also do the job? Heck, maybe that is why they had a back up death star, they found a suitable celestial body, stripped the garbage, split it into smaller parts, trimmed the garbage from the parts and still had enough for a second one or more. Heck, given how likely such bodies would be rich in metals, some of said garbage is actually used to build the infrastructure, solving two problems in one move.

[@offtheball87]

The first problem you're gonna have with something that dense is tidal forces. Because the radius at which you get a decent g force is so small, a small change makes a big difference. Small amounts of it put the livable range pretty close to the core but narrow. I found a density of 10^17 kg/m^3, so if you have 10 m^3 of the stuff, that puts your useful range (about 0.5 - 1.5 g) between roughly 2 and 4 km. Larger amounts push the useful range out and widen it. At 100 m^3, you've got between around 7 and 11 km. With 1000 m^3, you're at 20 to 35 km. Your best bet is around 3000 m^3, with a useful range from 35 km all the way to the outer shell. Of course, by this point you've got a mass comparable to some moons in our solar system in a space a little less than 18m in diameter. It's an engineering challenge, to be sure. As for less dense materials, you can still treat them as a point mass in your calculations. You're going to be living outside of this thing no matter what, so while you could have a less dense material, it's still got to be that massive to provide the same pull, and the gravitational pull outside of it will fall off just as fast.

[@louisvictor3473]

***** Yes, for all my ideas the goal would be to put the mass at the center and have the people living outside of it. Also, keep in mind that you don't need 1g at all layers. I am happy with having 1.5-2g at the lowest levels (which will be likely smaller and not to be used by any individual for lengthy periods of time) and get to about 0.5g-1g at the outer shell (so you're less restricted at the larger top layers, which makes covering all that distance easier, and having a comfortable and well protected living area between the two extremes).

[@offtheball87]

Well, going up to 2g does extend the range about 5km towards the core. To get decent habitable ranges, you're talking about a scary amount of the stuff. The acceleration due to gravity just a kilometre from the core, at the 3000 cubic metres I talked about earlier, would be over 2000g. Of course, they blew up a planet on a whim, so who am I to say the empire can't do it.

[@louisvictor3473]

***** Hence why I also considered other things. Chances are, once you take a chunk off it would naturally expand, so it is very likely it is surrounded in some super dense liquid under enormous pressure to keep it not as dense as when it belonged to a neutron star, but dense enough to be relatively compact and still fit at the core of the death start. Then there would be the things I called failed white dwarves (earth sized white dwarfs have a density of. 1 x 10^9 kg/m^3 which is sensibly less than the mass of a neutron star but still quite compact). Basically, the empire would be merely taking hold of a thing that happens naturally-ish, they just shaped it to their needs, rather than try and build it from scratch themselves. If they're building a shell 150KM in diameter, I assume they would have the technology to build around a preexisting mass, which is probably simpler than the technology to build it from zero. From what I calculated (I used astro.unl.edu/classaction/animations/renaissance/gravcalc.html and a sphere volume calculator), it seems that with a density of 2.14x10^6, less than even a white dwarf, and a diameter of 80KM, we would have a gravity of 2.44gs right at the surface. Not too bad. Let's say the next 10km (total 90km) is for machinery maintained by droids and remote drones. The layer at 90KM would have only 1.56g, which is quite reasonable for me, and on the outer layer at a diameter of 150KM, the gravity would still be about 0.67g, with the sweet spot for 1g-ish being at a diameter of 120-130KM. Plenty of room for people and machines, if you ask me. The mass of this core would be 5.73e+20 , or almost 0.001% Earth's mass. I think we might actually be getting into something. Whatever material it would be, it could double as a natural gravity source and a massive power source. Also one likely to produce a metric ton of heat, which might need to be emptied via some sort of exhaust system :O

[@offtheball87]

I was assuming a very dense core, but if you allow a lot of space for your gravitational core, then I could see that working. The empire is clearly not lacking resources, after all. You might be able to rely on fusion for your energy source. At the sun's rate of consumption, that much mass would last about thirty thousand years. It's likely to consume energy quicker, but I don't know how to figure that out. You're unlikely to be able to use all that heat, so you'd need an exhaust port. As for blowing it up, if you could increase the radiative pressure enough to disrupt the balance with gravity, that might do the trick. A large enough explosive, maybe one with enough energy to drive heavier elements to fuse earlier than they should in the star's life, might do the trick. So, I think you're right. The core of the Death Star is, or at least should be retconned as, a very dense, very small star. After consulting with an actual astrophysicists and not someone who did one astronomy course in university.

[@keenanmeyer5511]

The deathstar is like an ogre, it has layers...

[@Neonext331]

well i think its layed out horizontally, because when the laser is fired we can see workers inside who are standing horizontally , because the lazer isnt traveling upwards from where they're standing, but left to right

[@cebenify]

wouldn't the death Star have aiming problems while spinning?

[@twincast2005]

Negligibly so.

[@cebenify]

ic

[@DissociatedWomenIncorporated]

Twincast But even if you spinned it with the primary weapon used as one pole of an axis, reorienting it to aim would be a massive pain, not just in terms of manoeuvrability but also in terms of the internal "gravity" suddenly lurching in a different direction for a few minutes. Artificial gravity does seem to make more sense. Including using it to _negate_ centrifugal forces whenever it _does_ spin around to aim.

[@mikemarx9360]

It can't splin and shoot.

[@KindredBrujah]

Based on Kyle's calculation it would perform one full rotation every 10 minutes. So it could be problematic depending on how long the main gun needs to be on-target for. Alderaan seemed to be destroyed pretty quickly, so I don't think that's a big consideration.

[@H4N2]

Could you have a extremely dense artificial core that creates the gravity?

[@skintstudios]

not true, not all parts of a sphere can rotate at the same velocity, so at the 'equator' you would have strongest 'weight' and at the 'poles' there would be no 'weight' so yeh... this wouldn't work.

[@fordas8949]

shameless plug

[@bascep]

OldFordas exactly what i was thinking

[@sebastianahrens2385]

It also doesn't get rid of the gravity gradient toward the center either :/

[@fordas8949]

he did explain that though starting at 5:30

[@thechumpsbeendumped.7797]

OldFordas I came here to say the same, his solution would only work with a space ship with similar shape to the one that Arthur C Clark designed in the 2001 film.

[@damiantremmor7825]

According to the Wookipedia: It was over 120 km (74.56 miles) in diameter and had 357 internal levels. Due to its enormous size, it had a heavy gravitational pull which forced TIE fighter pilots to adjust their ships' thrusters for the kind of takeoff more common within a planetary atmosphere than on a space station.

[@bentomio]

You forgot that the death star isn't circular, it's a sphere, which means only people along the spinning line would experience that gravity.

[@x-iso]

if each circle-floor would rotate on it's own, then yes. Though i would imagine a nightmare to workaround transition between these moving halos.

[@reedcoffman3210]

Yeah the further the the poles of the death star you got the less gravity you would experience

[@RadeUeMasq]

Just one thing you left out. If the whole structure is spinning to create out gravity. How would the death star fire it's weapon with out haulting the rotation & stopping the gravity? Simple! Make the equatorial ring spin while the northern & southern hemispheres stay stationary. Allowing the death star to fire it's weapon with out putting personal at risk.

[@GibbonStreet]

What about other ships which appear to have gravity, such as the Millennium Falcon. These much smaller ships travel through space with seemingly normal gravity applied and no rotational force. It seems more plausible to me that within the star wars universe, gravitational generators have been created in order to fabricate a gravitational field which would pull objects within the effect with the same force as gravity. This could likely also be the technology behind tractor beams, creating a force of pull to counteract the mass of the designated object.

[@FrostWolfPack]

You got it ships and space stasons juse a gravuty plates that work like weak tractor beams to make shure that sentients dont start to float around.

[@PepeSi1via]

I think it would be a cost issue - given the Death Star has to rotate to aim its weapon, it would just be easier to simply apply the rotational gravity instead of purchasing the gravitational fields on such a large scale.

[@FrostWolfPack]

We dont know how much the gravity plates cost and we have seen them at wide juse in sw universe allso emperor cinda was all cost to the wind when it came to death star.

[@barghestblue731]

except that it would have to keep rotating, they couldn't stop to aim it (like they do in the movies) and if they missed they would have to wait a minute for the station to rotate again for the target to actually be in position for the weapon try to hit it again, if they used gravity plates and not rotation they could actually aim at their targets.

[@FrostWolfPack]

Allso doking ships would be a nightmare.

[@ptonpc]

Could the spheres be spun at different speeds? That would allow 1g throughout the station. You could use the lifts to get from sphere to the other.

[@Nerdist]

That would be super complicated, but awesome. -- KH

[@PcGameGold]

Why not tether two death stars and spin them both to create gravity?

[@dumpeeplarfunny]

That theory sounds great, unless you've seen any of the Star Wars movies. None of the windows in the Death Star are in the floors, and nothing outside of it, viewed from inside, appears to be moving, which means it does not use centrifugal force to simulate gravity. The ships clearly don't either, so that shouldn't be a surprise.

[@keepironman14]

ikr i was thinking the whole time, "um dude the deathstar doesn't work that way. it's clearly horizontal floors. even just the logic of how they move about along with the fact that the hangers aren't in the floor they are in the wall. and afterall, if the station has the tractor beam strong enough to catch han solo flying so far out they are still having a hard time seeing that it's a station, i think a full gravity system should be cake work

[@dumpeeplarfunny]

Exactly.

[@iamprodigyable]

Stephen Walker Or maybe they're just screens and they're watching live footage of whatever they're shooting at

[@dumpeeplarfunny]

The pull of gravity is parallel to the outer walls of the Death Star in every scene, some of which show ships entering and leaving, so those are definitely not video screens.

[@PuddingXXL]

well they can walk on an building sized ship so yea I doubt they use this kind of centripetalic design.

[@jonathanmacgregor765]

Hate to be a nit-picker because I'm such a huge fan of this show. But I was working through the calculations for centrifugal acceleration and noticed that the values of the radius were inconsistent regarding the exponents, showing both a=(v^2)/r and 9.8 (m/s^2)= (v^2)/(70km^2). Fortunately, the math discussed was 100% correct. When I replaced (r^2) with (r) as shown in the initial onscreen formula, the values matched those in the video. Just wanted to post in case anyone else noticed the same discrepancy. Also, writing backwards is hard, Kyle. You make it look so incredibly easy. Keep up the good work!

[@alexkaen1701]

In the Death Star Technical Manual, yes they made a book for it, this issue is actually addressed. The Death Star is arranged with its floors rising vertically from one pole to the other. There is even a memo from a Moff describing how inefficient this design is for the artificial gravity, which you explain perfectly in this episode. However, the Emperor wants to have this layout anyway, basically, just because he can. He built the Death Star as inefficiently as possible just to show off what the Empire could do.

[@edlippincott6205]

if the death star was spinning to make artificial gravity wouldn't that mess up the lasers ability to aim

[@garavonhoiwkenzoiber]

Not if it's spinning with the laser as the axis!

[@dommorris8163]

and the centre would be incredibly awkward to stand on

[@jmonseur5322]

Not if the lazer was one of the axis

[@Kzorith]

It wouldn't have to be. You'd just have to align the death star so that when it was in firing range the laser was oriented towards your target. It would be easy enough to shift the rotation of it of the station with thrusters and unless it was close enough to another gravitational source, likely the people would barely notice the shift.

[@GamesFromSpace]

Why would the center be awkward to stand on? The death star in this scenario rotates once every ten minutes. You wouldn't even really notice.

[@thecountrybrickman2709]

So how would the gravitational field on the second death star work then? Because the outer rings were never completed then there wouldn't be gravitational field right?

[@TheLublume]

The second death star appears to be much larger and actually inside a planet/moon of some sort, therefore it's a different logic

[@JasonTatseos]

I think you're confusing the second death star (Return of the Jedi) with the star killer base (force awakens)

[@dommorris8163]

TheLublume really? that sounds like the thing in the force awakens.

[@Samael1113]

The simplest answer is that all the Death Stars effectively uses a small star or incredibly dense molten core to generate the energy needed, which would have the side effect of providing gravity to the base.

[@DennisXiloj]

Samael 11 still you have to have onion-layered floors, but it doesn't look like that in the structure of the second death star

[@SyntheticReign]

But that spinning happens in one dimension right? The Death Star is a sphere, meaning it couldn't possibly be rotating on every single axis simultaneously, which means that not only would there be different gravitational forces in each layer, but each location of each layer. Because science.

[@Zurenza]

I'm fairly sure that's one of the reasons we can only use this form of artificial gravity if it's a ring shaped ship.

[@Rainfall7]

yeah, that's what I thought

[@colinfleming9376]

Correct spin a sphere and everything will collapse towards the gravitational plane, this is how solar systems, galaxies, and planatery rings form.

[@SyntheticReign]

Colin Fleming All of which exist on a single plane

[@Nerdist]

Yeah, flaw in the design is that you pretty much have to stick to in and around the equators of each sphere; should have mentioned that. -- KH

[@bubbykins4864]

"Is the Death Star big enough to generate it's own gravity?" Since it has mass, yes. And we're talking having people in it, so that's not the problem anyway.

[@WrongParadox]

alternatively the core of the Death Star might be very massive with a much greater density than steel however this doesn't help the "sidewise" gravity effect shown in the film at the spacedock - the dock's exit isn't up but instead sideways ...the people are walking around with a wall facing (not a ceiling facing) pointed to space. basically neither a spinning nor a very dense core solves the issue.

[@reecevince2166]

Can you do a because science episode on how and assassin from assassins Creed would survive leaping off massive buildings without dying

[@Vegeta8300]

The British Guild of Assassins There is already a Game Theorists episode about that.

[@lordbubax3929]

The British Guild of Assassins matpat already did that

[@PepeSi1via]

I think you'd need a bigger bail of straw than what is depicted in the game...a lot more.

[@Herdatec]

in short: it is not the depth that is killing you it is the rapid deacceleration within the 1m high heap of straw.

[@joemucci3895]

he did that already

[@lemonwedge4640]

You and Clerks have stumbled upon the "oh, man! most of those guys were just doing their jobs!"

[@sweetjaysus4507]

ha! i knew i wasn't the only one who thought of that scene!

[@markoproloscic4492]

Lemon Wedge film theorist also had an awesome episode where he explained that destroying the Death Star would also throw the entire galaxy into a recesion.

[@pieceofgosa]

Marko Prološčić lol #LegalizeSpice it would invigorate the galactic economy and create thousands of new jobs.

[@lemonwedge4640]

Marko Prološčić I saw that video! Fascinating.

[@s980845]

do can a city like stargates Atlantis really float in real life?

[@vsGoliath96]

Bradley Hayes Float? Probably. I want to know what it would take to make it fly!

[@lasarith2]

vsGoliath that's Easy a ZPM , 😉 Since it's 5KM across , though I'd guess the weight would be a few million tonnes, just need to know mass ~Energy to be able to lift it , I'm guessing that would be a Kinetic energy calculator once -know the mass of Atlantis.

[@WeAreGRID]

i mean, do boats float? giant ass tankers float, as long as it has big enough surface displacement, anything can float.

[@zeonthefox]

also the arms/petals of the city could be pressurised containing enough air to creat buoyancy

[@theordinarytime]

I don't think it's all that nice to pop up a city that's 5km across from an ocean with that speed... The resulting tsunami would have hit the mainland areas pretty hard me thinks. Would probably serve them right if the tsunami came back around and hit them..

[@Gaarafan007]

A notable flaw to this theory is the orientation of the hangar bay the Millennium Falcon is tractor beamed into. It is oriented in a North=up/South=down fashion. And as a side note, yes, spinnging would be cheaper, but in a space station meant to inspire fear, the price of the thing is just another factor in that, like "OMG, the empire has so much wealth that thy built a moon sized space station with so many artificial gravity generators to allow a North=up/South=down orientation!"

[@hewhoSees1]

Minor issue. The /centripetal/ acceleration (or the catching force) applied by the floors of the station to the feet of the Imperial menace would be at right angles to the axis of rotation. The axis of rotation is always pointing in the same direction, meaning "gravity" also always points the same direction, i.e. horizontally inwards. So the floors couldn't be spherical shells. They'd have to be cylindrical shells. Otherwise, towards the poles, the floors would become increasingly steep with respect to the "gravity", which would also get progressively weaker. So if you used spheres, only the portion near the equator would be a reasonable approximation of level-ground type gravity.

[@its_A_me_Njobe]

Battle Stations Operations. hahaha I died

[@Nerdist]

I've been doing Palpatine in the office constantly.

[@Telsion]

+Nerdist pssst, you forgot the KH part XD

[@ronpetersen2317]

at Nerdist Some guy named Archer told me to give you a message. "Phrasing!!!"

[@MrOuchiez]

+1, Mr Petersen

[@grieverlion]

Your face at 2:24 should be a new meme jajajaja

[@RuneForumwalker]

Kylo Hill: Asgardian Lord of the Sith.

[@argumentgenerator1.937]

Muuro Kylo is a disgrace to the Galactic Empire, he did jack shit in the movie.

[@RuneForumwalker]

That doesn't matter.

[@fancyblackpantz]

What empire bro? they're gone.

[@adamcarroll9613]

Actually, he captured a high powered force sensative, and then fought two force sensatives after getting hit square in the chest by what is effectively a .50 BMG round. So... overall, he did pretty well. Oh, also [PURGED BY MANDATE OF THE ORDO HERETICUS] legendary general Han Solo.

[@RuneForumwalker]

Jerald Lutney Two force sensitives? There weren't three force sensitives in the movie. Well i guess there was since Luke is in the movie along with Emo Ren and M'Rey Sue.

[@imofage3947]

Yes, we get it. The Battle of Yavin was essentially like 9/11 for Storm Troopers. Except for ONE KEY DIFFERENCE. Very few if any civilians were present on the Death Star when it was destroyed, making it a legitimate military target. Serving aboard the Death Star required security clearance because, at least in Legends, it's location was considered classified. So all 1+ Million people aboard the Death Star worked for the Imperial Navy. I'm sure nearly everyone aboard considered it's destruction to be impossible just like the HMS Titanic was considered unsinkable. But the rational skeptics know that the universe isn't required to conform to our expectations. Was the destruction of the Death Star an act of terrorism? Perhaps, from a certain point of view. But it was also a desperate act of self defense. The Empire had demonstrated it's willingness to destroy a planet and billions of civilians with no connection to the Rebel Alliance with the destruction of Alderaan.

[@masterSageHarpuia]

to be accurate, the death star would be the equivalent of the pentagon, or a whole carrier group. Doesn't really feel wrong to blow it up when it just nuked all of France into a radioactive waste land killing everyone.

[@feelthepony]

hiroshima and nagasaki?

[@tyleroneill8312]

I used fans to make artificial gravity so it has floors. This is because fans on the floor would make a vacuum causing air to be pulled from the air filter. This air rushing into the room will push you to the ground. This would mean the floors are grated

[@Benny_Blue]

I haven't watched a New Hope in a while, but if I remember correctly, the X-Wings that exploded against the surface of the first Death Star didn't damage the surface a bit. Doesn't this mean that the Death Star needs to be made of something with a strength much greater than that of X-Wings? I don't think there's enough room on the scale of metal strength for space-faring X-Wings to be that much weaker than the surface of the Death Star, and yet for the Death Star to be made of a steel-like metal.

[@Jessie_Helms]

Benny Blue the Death Star was made of Beskar, I believe, which is Mandalorian iron. Iron, yes, but much stronger than dura steel (basically steel). But, steel is the best analogy IRL. He should have made it, say, 25% denser than actual steel for the calculations

[@GamesFromSpace]

I always assumed it was arranged as concentric spheres, since even "magical" antigravity generators probably behave like fields.

[@GamesFromSpace]

Though if it's using centrifugal force, spheres would not be ideal. You'd instead want it arranged as concentric cylinders, since gravity would only be at a right angle to the sphere's surface directly on the equator. This reduces the available volume by (a third?) without careful design.

[@kinggoten]

I don't think spining would be to great I mean yea to target a planet it would be easy enough to target a planet but in rotj they fire the super laser at capital ships which would be moving(slow but still moveing) the rate of the spining would have to be increased decreased of possibly even reversed which would really screw with everyone's gravity

[@GamesFromSpace]

Zero Cool Maybe they simply screw with the gravity when needed. They could have artificial gravity generators in place for battle, and use rotation to save power when just floating. Besides, they need some sort of inertia canceling magic technology to fly around at a decent speed.

[@DanteYewToob]

Joshua Pearce What is the outer shell acts like that BB8 toy, and the outtermost shell can rotate independent of the internal layers? Just a thought... Although the cannon is way too "deep" for that right? It's generally shown to pass about 1/3-1/2 way through the Deathstar?... idk just an idea.

[@Binyamin.Tsadik]

Each floor would need to rotate in a different direction to stop the entire structure from counterspinning.

[@interstel]

Good catch I forgot to cover that one in my original rebuttal.

[@sonicpsycho13]

Since it's a sphere, you'd experience less acceleration as you approached the poles of the axis of rotation.

[@Nerdist]

Yep. -- KH

[@hwurst38]

you are totally right and because of this i came to the idea that it has to be designed as a cylinder with the giant Barrel thing on top and rocket-motor thing at the bottom

[@IIStaffyII]

Couldn't u solve that by having one of the inside spheres rotate around a separate axis? So for all the floors below the first gravity is simulated more correctly? Or am I missing something?

[@hwurst38]

centrifugal forces do always point at a right angle away from the rotating axis. if you are standing on or in a sphere halfway between the horizon and a pole, the centrifugal forces would drag on you at an angel of 45° measured from the floor. rotating one solid sphere in multiple axis can not compensate this, it would just result in a weird distribution of these angles. you would have to move every single point on the sphere at the same speed and this isn't possible. imagine a rubix cube spinning all its middle-layers "horizon-layers" at once, it would rip apart.

[@jonathanwine4235]

You're forgetting one thing, the axis on which it is spinning. Also, when the millennium falcon lands in the Death Star hangar from the tractor beam, it lands perpendicularly compared to the outer ring around the center of the Death Star. It is shown many times that there are hallways on board the Death Star that make turns and have stairs and ramps (anyone who has played Star Wars battlefront 2 knows what I'm talking about), not spheres within spheres. This makes the axis of rotation very important to the g-forces felt on the Death Star. Just wanted to point that out.

[@dan_fantastic9353]

In the star wars canon the Death Star’s outer layers work exactly the way you said they should but the layers closer to the center are just a bunch of floors stacked on top of one another like most of us assumed.

[@obi-wankenobi8406]

Or have you ever heard of interdictior ships in star wars thus it's easy to just use those generators

[@supremeleadersmeagol6345]

Yeah, I thought of the same thing watching this video. I wish these people who do these kind of videos would look more closely into the existing in universe explanations. Pet peeve of mine.

[@mkd2839]

That's complete bullshit. This is "Because *Science*", not "Because BS Made Up Fantasy Magic"

[@crucial1303]

The point of these videos are to try and use scientific reasoning not science fiction reasoning

[@supremeleadersmeagol6345]

That doesn't mean they should disregard existing sources completely. The canon Death Star plans for instance clearly show it's floors to be horizontal, not spherical. Also just looking at one of the hangers outside of the Death Star or the level layout on Death Star 2 proves that isn't true. Now, I have no problem with the kinds of videos that say "this is what would actually happen in real life due to science" but is this one, it seems to me Kyle tries to sell this theory as what is actually is happening in the movie.

[@obi-wankenobi8406]

StardustNotHötzendorf it creates artificial gravity

[@WulfricUlfang001]

At the end, it is Clerks all over again.

[@AGHybrid]

What if..... The Deathstar is built around a small white dwarf? A neutron star would be better for its size being that to a small city. I know there will be tonnes of radiation. But, you know. Alien, future space magic shielding right?

[@Gergo049]

too much gravity then. a neutron star has a lot of mass.

[@einname9986]

A Micro Black hole at the bottom would be better as it doesn't emit radiation and enables the design of several floors that are not necessary to be circular. I think it could also be used to convert mass more or less directly into energy via an accretion disk.

[@matthewsmith6057]

One relatively simple alternate explanation is that the Death Star is constructed around something super dense, with a planet-sized mass, but a much smaller volume. Maybe a tiny black hole or a bit of dwarf star matter. You'd have to reinforce the heck out of the inner levels, but it would create the proper gravity, and probably serve as a pretty handy power source.

[@ashtonhoward5582]

To land on the Death Star, you'd have to match the speed it is spinning, though that would only work assuming the space ports were parallel to the surface, which we know is not true.

[@pieceofgosa]

nope, anyone working on the deathstar knew who they were working for and took that risk upon themselves (this has actually descended into a Kevin Smith movie)

[@davemartino4953]

John Alexander well... do you think every German wanted to be drafted and fight for Hitler?

[@pieceofgosa]

Dave Martino it was a joke man, haven't you seen Clerks ?

[@pieceofgosa]

Dave Martino trust the internet to invoke Hitler within 2.8 seconds

[@davemartino4953]

John Alexander nope, good movie?

[@pieceofgosa]

Dave Martino I love Kevin Smith but dammit it if he didn't peak with his first movie :) It's awesome, I would highly recommend it.

[@JEilonwyn]

Hey Kyle, Great episode. But I have a question about the Flash: If the Flash runs at relativistic speeds, why does the age of everything not increase while he is running? Its like time dilation is both working and not at the same time.

[@NinjabeeRedtricity]

J_Eilonwy it might have to do with the speed force n such, if it didn't exist, he would run so fast that he'd create a nuclur explosion(by running into atoms and breaking them) but hey, not a bad suggestion though L:

[@JEilonwyn]

Ryan Applegate Yes... but its not that Flash ages slowly (that I could get behind). Its that at Relativistic Speeds time should be SLOWER for Flash than everything else: thus Flash should age slowly but all others SHOULD age rapidly due to time dilation.

[@JEilonwyn]

Ryan Applegate Right... so unless time moves faster for flash than any other thing at relativistic speed; why does no one age relative to to his speed (especially when he runs at or near light speed)?

[@JEilonwyn]

Ryan Applegate To put it in more specific terms... If I am stationary and flash runs at 80% lightspeed around the city saving people and it takes him lets say even 30 minutes to do that mean that for me almost an hour has passed. Now that doesn't sound like much but how many times does Flash do that in a day... shouldn't everything age relative to his speed??? Or am I missing something?

[@JEilonwyn]

Ryan Applegate OK... Lets try a different example then. Lets say I'm in a spaceship traveling at 80% lightspeed doing donuts around the earth for what I perceive as six months. Does not special relativity mean that one year has passed for my family and friends who stayed on earth? If that is true (And that is really the point of the question I suppose) then would not Flash undergo a similar effect?

[@TheDutchWolfx]

Thats no moon, thats a massive metal space onion.

[@Nerdist]

Sounds better, doesn't it. -- KH

[@JBdiGriz]

Two problems: 1) Remember the landing scene in Episode IV, where the Falcon lands in the Death Star? The Falcon approached the DS in a 90° angle and landed on a platform WITHOUT changing its heading. So spheres are not plausible. 2) Even if you rotate the Death Star in order to ‚produce‘ some sort of artificial gravity, 2.a) that gravity would only work on the ‚equator‘ of the rotation, and (much more important), so the rest of the station is without or lesser gravity. A sphere shaped battle station would be a massive waste of payload space and material. 2.b) it makes landing on and especially targeting with the main weapon much more complicated, unless the weapon is at one of the rotational poles.

[@noahm.9091]

Except that scene of Aldderan being destroyed shows the death star's beam being sent through a giant corridor past a person, it would have to be a horizontal corridor, so the centrifugal force theory doesn't work

[@franciscor755]

Well the Death Star needs to aim its laser. It couldn't do that very well if it's constantly spinning. And stopping to aim wouldn't be very good for the crew or the equipment.

[@TheLublume]

They can calculate the rotation to match their aim. Not that difficult for aerospacial engineers.

[@Ponsvoid]

that might be why there is so many people required to fire the laser, because they have to line up the shot perfectly in order to fire it at a planet. OR... the other shell doesn't spin, just the inner layers.

[@ThePheonixOfThe6]

that would be a fair enough explanation for the shot to match up with how it was presented in the movie. another thing that bothers me about this multi-spherical structure theory is that the famous "fatal flaw" of the death star would be an impossibility, at least the way im seeing it. if perhaps the the spherical shells spun on an axis that held the core of the station then sure this still makes sense

[@jakobzamorski8688]

What about all the laser components inside the shell, where would they go?

[@ThePheonixOfThe6]

Jakob Zamorski just looked at an image of the design for the laser, at least according to some light googling... and yeah the laser would not be feasible anymore with an interior spherical structure

[@nachoolo]

But, we have a problem. They we see spaceships docking on the Death Star. If your assumtions were correct, then the exit would be located up from the landing platform but we saw it being on the side of the landing platform. So your theory of how the Death Star generates gravity doesn't stand up in the Star Wars cannon.

[@Zurenza]

Honestly it's not a theory it's more of a correction, the creator of star wars likely didn't know at the time (just like the rest of the world) how creating a sphere shaped space ship would work. He most likely designed it as a floor based system with floors that run perpendicular to the equator. We didn't actually figure out how to create artificial gravity until about 20 year's ago and star wars is much older than that, infact most of our knowledge of space exploration and physics in space come from the last 20 years. So really Kyle's isn't a theory just a correction on how the death star SHOULD work not how it actually did in the show.

[@unvergebeneid]

Ha, my thought exactly!

[@rmsgrey]

What? Just taking an example from the video, 2001 was released in 1968, almost a decade before Star Wars, and they knew about centrifugal "gravity" when making that film. The idea that gravity and acceleration are indistinguishable is one of the central ideas of general relativity, which is over a hundred years old - and the observation that acceleration and gravity feel the same goes back much further than that. Centrifuges have been around for well over a century (and the idea goes back at least a century before that). If George Lucas didn't know about spinning objects in space to create fake gravity, it's not because the idea wasn't there; it's because he didn't do any research...

[@Zurenza]

rmsgrey Let me rephrase so you'll understand, the idea of creating artificial gravity was not until very recently, yes the idea of Centrifuges was there but no one had put two and two together to discover it could be used to create artificial gravity until the space age. In the film A Space Odyssey they created a space craft that did create artificial gravity through rotating a ring with Centrifuges, BUT this concept was not confirmed to actually be plausible until 30 year's later, which is the most likely reason George Lucas didn't use it himself when creating the Death Star. Now unless he ever comes out and says "Hey this is how the Death Star's built" then we will never know.

[@joemucci3895]

* canon

[@oriwan3786]

Hi Kyle, are you writing mirrored? I always wondered how the writings were done, or is it all editing?

[@LevityMire]

Yup, mirrored and I was so impressed at his ability to write backwards.

[@LevatekGaming]

LevityMire I think it is mirrored, but it is flipped. So he's writing normally, and in editing the video is just flipped.

[@raetglias3274]

LevityMire No he write normally and the video is flipped, you can tell if you look at the shirt he's wearing and how the designs flip sides when he writes

[@ryanm7832]

+Nerdist I have one possible example for the interior design of the Death Star that doesn't add up with this idea: the hangars. We are shown craft entering/exiting, as well as personnel standing just inside the force-field-blast-door, and the layout of these scenes would suggest that it is layered like a cake, instead of an onion. I understand Kyle's explanation completely; generating gravity through spinning is fairly common knowledge. I just can't figure out how to make this explanation stick on the Death Star, based on the various hangar scenes. I don't poke holes out of spite, I do it out of genuine curiosity, and to learn others' viewpoints; I love an intelligent discussion or debate. I think the likely answer is just that Lucas didn't delve too far into the scientific accuracy of a giant, spherical space station. I'd love to hear what y'all think about this.

[@jarosbodytko6462]

That last part reminds me of Clerks where they were talking about the contractors that the Empire hired to build the 2nd Death Star and most likely were present at it's destruction. All the construction workers that were just doing their jobs were killed when the 2nd Death Star was destroyed.

[@MinybjuMC]

How does that explain the hangers where the ships land. When a ship lands on the death star its horizontal.

[@Nerdist]

In my design the hangars would have to be facing upwards, parallel to the axis of rotation. -- KH

[@georgeconnett7812]

Wouldn't you just experience the artificial gravity at a right angle from the centre of the axis of rotation though?

[@georgeconnett7812]

Or at least get the desired affect there

[@henryfolds5713]

George Connett that's what I was thinking!!

[@Painteagle]

Yeah. What would happen at the poles?

[@georgeconnett7812]

Squirefig you'd just float like in free space

[@Painteagle]

George Connett That's what my assumption was. I don't think his Death Star design works.

[@PtolemysEye]

Am I the only person who rewinds the "SURPRISE LIGHT SABER" like 3 times?

[@Nerdist]

I do too, is that weird? -- KH

[@PtolemysEye]

...Yes

[@JonathanDaniel1986]

There's so many comments so this one will never get noticed...BUT, the problem with spherical floors (versus a more optimal cylindrical structure) is that you only get the optimal acceleration on one small arc of the station. The farther you move closer to the axis of rotation (on any floor or "shell") the more your "g" will decrease until you will slowly start to experience the weak natural gravity of the station and sink to the "ceiling". It wouldn't eliminate livable volume, but would certainly make livable efficiency a factor. It would however be fun to have long-jump semi-orbit contests around weak g points :)

[@paulwatrobski8277]

I actually just had the same realization during this video, remembering the scene where Luke looks out the window at his "friends" before his "battle". How could they be standing in that orientation and looking out the window at the same time?

[@ChaseTheTech]

But if the floors were spherical and you generated artificial gravity via centrifugal force, then then you wouldn't be able to get close to the axis

[@Nerdist]

You could if you wanted to float like a badass. -- KH

[@ChaseTheTech]

Nerdist Fair enough, I think that'd make an ideal place for prison cells. I imagine that just throwing criminals in a room left to float would be pretty punishing. P.S. OH MY GOD YOU REPLIED, I LOVE YOU.

[@daman7387]

He's like Cisco from The Flash in real life.

[@thelittlepasty8360]

da man Except less annoying and not an excuse for cheap writing

[@tlotloselei8085]

I told you I was a genius Kyle...I did my own research before watching this....I was a bit off but STILL

[@tlotloselei8085]

P.S I saw the question on Twitter #FollowKyleOnTwitter

[@Nerdist]

I agree. -- KH

[@gabrielgonzalez1993]

I almost got died when you impersonated emperor Palpatine "other 40% for battle station operations"... keep the videos coming bro

[@jackurquhart7994]

the death star was understaffed at the time of its destruction and a large amount of the people on the station were hired contractors and their families that were staying there at the time, so this makes the rebels even worse than before!

[@alexdavidson7498]

There is no such thing as looking to closely!

[@richardfarrer5616]

Of course, that means that there is no such thing as "looking to closely" so alex was right after all :-)

[@bubbykins4864]

That's because Closely isn't a real place!

[@jamesyeoman794]

CENTRIPETAL NOT CENTRIFUGAL!!! HOW DARE YOU MIX THOSE TWO UP! YOU SHOULD KNOW WHAT YOURE DOING! I TRUSTED YOU AND YOU TOOK THAT TRUST AND CRUSHED IT! why... Why do this to me

[@archdukeferdinandofthe3rdc9]

Centrifugal is correct. The spinning gives you outward momentum, but since you're on the inside of the sphere (even though this only works for circles and would fail miserably if applied to a sphere), your momentum pushes you against the floor, simulating gravity. While centripetal force is a real thing, you seem to have confused the two.

[@Nerdist]

It is centrifugal. I checked. -- KH

[@jamesyeoman794]

I am studying A-Level Physics. I have just done the Circular motion module and have recently endured the lecture on how Centrifugal Force is a misconception and the force people are referring to is Centripetal Force

[@jamesyeoman794]

***** www.regentsprep.org/regents/physics/phys06/bcentrif/default.htm hyperphysics.phy-astr.gsu.edu/HBASe/cf.html Centripetal force is an actual force whereas Centrifugal force is a fictitious force as seen here: en.wikipedia.org/wiki/Centrifugal_force I'm not arrogant but I did have to sit through a 1/2 hour lecture that was done so that people didn't end up using the wrong term.

[@jamesyeoman794]

***** btw I love vids like this one where sci-fi stuff is applied to our understanding of physics. Now, if you'll excuse me, I'm off to think about why there isn't an Iron Man in real life yet even though there are many billionaires

[@leofisher5391]

Are you left handed

[@Nerdist]

Maybe -- KH

[@Hmc260]

I believe that the image is as a mirror, otherwise what he writes would be backwards to the spectators, that way it seems that he is left handed... because logic.

[@matthewbaugh2898]

The thing about that quote that gets me is that the Death Star is not a station at all. The word "station" implies the thing is more-or-less stationary. Modern space stations are relatively stable in their relationship to the earth. Even if they're not in geosynchronous orbit, they are always in earth orbit. But the Death Star travels from place to place. It's a ship, not a station.

[@thulean.uruk-hai]

@6:20 - I agree with you taht it's the most logial design choice, and that the curvature wouldn't be noticed, etc, but you forgot something: viewports (windows). There are scenes where they are looking out through viewports such as when Luke is watching the attack fleet getting ambushed by an actually functional Death Star. Also, you can see the layout of the hangar bay when landing and that it is congruous with the interior as they go through the hatches. Your design is logical, but not the one they actually used.

[@NimhLabs]

That "different layers of cores" is actually reflected by the scenes in "Return of The Jedi" remake. Though the docking bay doors point directly out into space on their walls, not their roofs or floors.

[@AlmightyYogs]

as someone who is the biggest nerd ever when it comes to star wars, I own the Death Star blueprints/specifications. yes it would require artificial gravity because the outer layer is made up of two levels of floor that are oriented as if it were a planet, while the internal part of the space station is levels/floors oriented with the latitude of the space station. so the "north" pole of the station would be the top floor, and you can catch turbo lifts to all the other levels, right down to the "south" pole. The technology for this artificial gravity has existed within the star wars universe for millennia. This fact is corroborated by several lines in the Correllian Trilogy books, by author Roger MacBride Allen. Specifically the third book, Showdown at centerpoint, where Han, Leia, and their children go for some forgotten reason (it has been many years since I read it) to Centerpoint, which is a planet sized space station that is largely uninhabited. It was created around the year 100,000 BBY, so it is the oldest artifact in the galaxy that survives to the time after the Empire. it has no artificial gravity, and it rotates to great its hold on it's surface.

[@johannestruppel7639]

not just a little obvious, but also explains the countdown and why they need nearly a moon full of scientists and computers to calculate operations

[@GogiRegion]

The problem is that the spinning theory is that that would have inconsistent gravity. You can’t have a sphere have that kind of gravity. The answer magnetic floors and boots. Your body is pulled down. The reason that everyone had that gravity is because every ship is legally required by the empire to have the same strength of magnetic flooring and boots. This makes every ship have identical feeling gravity.

[@mattpharois9719]

In the movies, the 'energy beams' travel outwards. This would mean that the beams would appear to flow downwards if you were observing them from close range, or at least mostly downwards, give that we use centrifugal force. In the movies, however, we see that an operator is standing sideways to the beam, so centrifugal gravity does not work in this instance.

[@TheDemigans]

We can see in the movies that it cant be turning the deathstar. First, it would make hitting enemies harder as you would have to wait for the Death Star to turn towards the target each time. Second, the Emperor sits in a heightened tower on the surface and falls to the Death Stars core, as does the super star destroyer when its bridge is taken out.

[@Bambuzzsprosse]

Oh damn dude, that Imperator voice was sooooo perfect :D

[@BrainTitusK]

I've gotten used to the idea of "surprise lightsaber" whenever I see him making a starwars movie, that I anticipate every notion he explains with a feeling of expectation that its coming during a transitioning. :3

[@stydras3380]

You could spin the spheres at different rates ... Thus the "gravity" would be same on every sphere. Funfact: if you spin a sphere theres always an axis it spins around. ==> if you would walk in the direction of the point where this axis meets the sphere the "gravity" you experience would decrease. In fact ON the point where they intersect you would spin around your own axis and wouldn't experience any "gravity". Altough great video, SURPRISE LIGHTSABER and out

[@cutthroatsanddespots1956]

OK people, it works like this; The whole thing has an equitorial spin, and as for perpultion...The engine core and hyperdrive are embedded pole to pole, (In other words, it moves north) filling all the space on board that has bad gravity with vitle machinery . The hanger bays at equator are equipped with artificial gravity tecnology (tractor beams)

[@gundamfan2020]

I've watched all the videos and it's just hit me and realised kyle writes back to front for us to read. that must be so hard to do ontop of working equations and dishing alien drawings. that's insane.

[@FlagCutie]

I still remember the day in 9th grade when I learned the difference between centripetal and centrifugal force. It shook me to the core.

[@Terriss0421]

This is pretty much how I pictured the Death Star working. Either that or fancy, SAI-FYE gravity generators.

[@HellaVonFreyja]

I have an issue with a rotating death star for gravity, but only from a few scenes. If you look carefully during New Hope (When the Falcon lands/takes off, and I think one scene with Han chasing some storm troopers) and Return of the Jedi (When the Emperor arrives, and when Luke leaves. I can't remember any in Empire strikes back), you'll notice that the landing bays open directly into space. I think there was some comment about semi-permeable force fields resisting gas escaping but allowing ships to pass or something, but it still shows that the station cannot be rotating. The deck plates are perpendicular to space, not floors with gaping holes that you could jump down to get into space, after all.

[@Johnrich395]

My Solution would be that the Death Star is powered by a black hole, or series of black holes, massing approximately 1 earth. This sits in the core, and people walk on the outsides of the shells. This also solves 2 issues. 1. if you are spinning, it's more difficult to hit a specific target. 2. if you are spinning you don't get apparent gravity at the poles.

[@kentbarrett6249]

This channel teaches me more practical math than school

[@Seeriosa]

They could also potentially have the floors rotating in comparison to one another, to increase or decrease the centrifugal force. This would also explain why A; the shot to destroy it was theoretically impossible (it would have to have aligned perfectly) and B; why it takes so long for the death star to charge when it uses kyber crystals instead of conventional power sources. It's not charging per se, but it's turning to the correct configuration to be able to fire.

[@user-oy9zy4ds9m]

My guess is the Death Star interior is laid out upright but they have special fictional gravity producing machines similar to how land speeders can float in midair and not using centrifugal force

[@jeffbyrd6003]

I think another thing people forget is that even a small amount of simulated gravity can be helpful, things as low as .1g help lot of different day to day activites in space compared to what they are with almost no simulated gravity

[@jollygoodfellow3957]

What would happen if the Death Star fired at an actual star? From white dwarf to red giant and so on, and a Neutron star.

[@finalfreak56]

funnily enough this is the structure that seems most natural to me from the beginning. i've always just kind of assumed it was laid out like this.

[@atomicskull6405]

a lot of Star Wars fans already assumed the floors were set up like that internally already. Horizontal floors inside a 140km sphere just don't make sense even when you don't consider gravity. Even since I was a kid and saw the original movie I always assume it was spherical layers surrounding the giant reactor in the center.

[@Brain_Sync]

People were standing according to the vertical axis of the station on the docking bay, where the Millennium Falcon is tractor-beam-landed, so I assume they must have had a mystery gravity system that follows the same axis all throughout the space station. My wild guess is, a metal core with enough mass and density, floating in the center, spinning at incredible speed in random directions, could generate a strong enough gravitational field, similar to that of a planet or a moon. But that would imply people would be standing outwards to the center, with concentric levels.

[@user-roninwolf1981]

As a veteran of the DoD, in response to the final segment about the relativity of evilness with Luke killing the personnel in the Death Star; the personnel working in the Death Star were Imperial Armed Forces personnel, and that makes them Lawful Combatants, not to mention that the Death Star itself was classified a Legal Target. I would say that George Lucas has some awareness of the Law of Armed Conflict (LOAC). The only ways Luke would be in the wrong in killing the personnel in the Death Star would be if the Death Star was declared a medical station, a religious temple station or a station full of refugees.

[@Frostieyy]

the interior design of the deathstar cant be multiple circles because in episode 4, the falcon enters the space station from the side and when they entered, the floor was horizontal. If the deathstar had a circular interior, the floor would be vertical.