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Belteshazzar
2007-07-19, 08:50 PM
In mapping my world it occurred to me that I honestly have know idea how large it should be. While it won't all be clearly mapped during a single campaign (what fun is a map without the blank spaces?) I would like to have a general idea how large I want to make my counties or how long it may take to travel from place to place.

The real trouble is that my world is placed on the inner surface of a Dyson Sphere of undecided thickness. My specialty being biology I have no idea what the dimensions of my world would have to be in order to have somewhere near earth gravity at the equator (if the poles end up weightless its not my loss).

I have seen some impressive displays of number crunching here and I am not afraid to take full responsibility for any catgirls harmed in this thread.

Jack_Simth
2007-07-19, 08:56 PM
A solid Dyson sphere (which, incidentally, isn't what Dyson proposed) doesn't have internal surface gravity. At all. By a curious quirk of mathematics, a spherical shell of even density does not gravitationally affect anything inside. Doesn't matter how thick.

Zeta Kai
2007-07-19, 09:00 PM
Well, assuming that we are talking about a campaign world with a solar primary roughly equal to out Sun (Type G2), then in order to not fry the inhabitants, they need to be roughly the same distance from said primary as we are, namely 1AU (or 93 million miles / 150 million kilometers).

A Dyson Sphere of that size would have a circumference of 942,477,796.0769 kilometers, or 584,336,233.5677 miles, give or take ~1,000 km/~600 miles.

How's that?

EDIT: Jack's right, though. The gravity would pull inward, so gravity would have to be artificially generated. Otherwise, everyone on the surface would fall off into the sun. Unless you spin the damn thing for gravity, but that would only help the equator; gravity would gradually drop to nothing as one traveled from the equatorial regions to the poles.

Douglas
2007-07-19, 09:09 PM
On the inner surface of a uniform density spherical shell, there is no gravity at all from the shell. What gravity there is would come entirely from the sun and any other spatial bodies inside the sphere, and it would all pull inwards, away from the surface.

If artificial "gravity" is acceptable, you could just spin the sphere. How large it has to be then depends entirely on the mass of the sun and how fast you're willing to spin everything, along with how strong you want "gravity" to be at the equator.

Leicontis
2007-07-19, 09:09 PM
One of the major issues of a Dyson sphere is, of course, how to keep the atmosphere where you want it. The two major ways to do this are as follows:

Double Dyson Sphere - you put a transparent inner shell, leaving only solar gravity (which is comparatively slight) to affect the inhabitants.

Gravity Generators - invent some sort of device that generates synthetic gravity for the inside of the sphere (ideally only the inside, making launching/landing spacecraft easier). This option is probably better for what you want, because it wouldn't require you to rewrite all the rules for movement.

As for spinning, if you want equatorial gravity around 1G and a solar primary like Sol, you'll need to remember that there is no known or theorized material that can withstand the forces generated by its own weight at that spin rate. Floor material would thus be virtually indestructable.

There are also other options, such as ringworlds. For a more thoroough treatment, read the essay "Bigger than Worlds" by Larry Niven.

Khosan
2007-07-19, 09:09 PM
The actual radius could be anything in terms of gravity. It'd just need to spin fast enough to achieve earth-like gravity on the inside of the sphere.

What would be needed to figure out is how large it'd need to be to achieve some kind of bearable temperature. Fortunately, we have an idea of that already, 149,598,000 km (1 AU, the distance from the earth to the sun), since we know there's a bearable temperature for humans at that distance. So, you'd have 281,229,865,302,746,438 sq. km to work with.

Edit: Ninja'd. But I bet I can come up with a rotational speed! 29,787,771 m/s. Your poles would be screwed though.

Ulzgoroth
2007-07-19, 09:09 PM
On the other hand, the equatorial regions could retain an atmosphere just fine if the thing is spun, and you'd still have a nearly unimaginable amount of habitable space (though it would effectively be a ringworld, with weirder edges.)

Belteshazzar
2007-07-19, 09:10 PM
A solid Dyson sphere (which, incidentally, isn't what Dyson proposed) doesn't have internal surface gravity. At all. By a curious quirk of mathematics, a spherical shell of even density does not gravitationally affect anything inside. Doesn't matter how thick.

I had heard this mentioned once but was unsure as to the veracity of this statement. So how does one get around this. I noticed you specifically mentioned that an uniformly solid Dyson Sphere has no internal gravity. Does this imply that it could have internal gravity somehow?

P.S. The Sun is not an actual sized sun but a divinely magical superweapon created to destroy the previous universe for it's misdeeds (think Aboleths ect..) This means that that it is much smaller (only a few miles in diameter) and is unlikely to unintentionally fry the inhabitants of my world. Atmosphere is also rather thick in this universe so air is unlikely to be a problem, as whole draconic empires have thrived on the Nightshroud (the cloud of flying islands, rocks, ruined airships ect... orbiting the micro-sun that gives the inner surface some measure of night and day)

Khosan
2007-07-19, 09:17 PM
P.S. The Sun is not an actual sized sun but a divinely magical superweapon created to destroy the previous universe for it's misdeeds (think Aboleths ect..) This means that that it is much smaller (only a few miles in diameter) and is unlikely to unintentionally fry the inhabitants of my world. Atmosphere is also rather thick in this universe so air is unlikely to be a problem, as whole draconic empires have thrived on the Nightshroud (the cloud of flying islands, rocks, ruined airships ect... orbiting the micro-sun that gives the inner surface some measure of night and day)

Oh. Well then.

Feel free to apply whatever values you see fit in this case.

Douglas
2007-07-19, 09:17 PM
I had heard this mentioned once but was unsure as to the veracity of this statement.
I learned it in physics class at high school. We actually went through the math to prove it.


So how does one get around this. I noticed you specifically mentioned that an uniformly solid Dyson Sphere has no internal gravity. Does this imply that it could have internal gravity somehow?
The only way to get actual gravity on the inner surface of a solid sphere is to vary the sphere's thickness and/or density. If you do that, it will have gravity over the spots where it's thicker and denser but anywhere else things will tend to fly off into space towards the center and away from the shell.


P.S. The Sun is not an actual sized sun but a divinely magical superweapon created to destroy the previous universe for it's misdeeds (think Aboleths ect..) This means that that it is much smaller (only a few miles in diameter) and is unlikely to unintentionally fry the inhabitants of my world.
So the sun's output is arbitrary. In that case, you can just decide on whatever size you like and both the sun's luminosity and the sphere's rotation speed can be adjusted as necessary to match. This will end up with no gravity at the poles, but you said that didn't really matter.

Inyssius Tor
2007-07-19, 09:18 PM
Take the easy way out: have your sun "radiate gravity", pushing everything away from it!

Ulzgoroth
2007-07-19, 09:18 PM
It's simply impossible for a uniform spherical shell to produce a gravity field inside itself. It cancels out. The external field is obtained by just adding up all the mass inside the sphere and calculating it as if it's at the center...you could live on the outside of an appropriately sized and weighted dyson sphere, but it would essentially be a gigantic planet with no sun (the sun being trapped in the core).

A ringworld at one AU needing a surface 'gravity' of one g, if I haven't botched the math, should rotate in 776037 seconds, which is just a little under 9 days. Though I'm not sure anyone on the surface could tell.

Jayabalard
2007-07-19, 09:18 PM
my advice... unless you really like that sort of realism, ignore the laws of physics and just give it an arbitrary diameter, and artifical gravity.

If you want realism, I suggest reading sci-fi, and Ringworld (http://en.wikipedia.org/wiki/Ringworld) and Ringworld engineers would be a good start(there are more sequals but those two cover the basics). Ringworld engineers has some corrections and clarifications that cover the physics impossibilities, and was written because, among other things, there were MIT students chanting "The ringworld is unstable" at worldcon.

Ringworld is huge...

Radius: 0.95×108 miles (~1.5×108 km) (~1 AU)
Circumference 6×108 miles (~9.7×108 km)
Width 0.997×106 miles (1,600,000 km)
Height of rim walls 1,000 miles (1,600 km)
Mass 2×1027 kg (1.8×1024 short tons) (1,250,000 kg/m², e.g. 250 m thick, 5,000 kg/m³)
Surface area 6×1014 sq mi (1.6×1015 km²); 3 million times the surface area of Earth.
Surface gravity 0.992 gee (~9.69 m/s²)
Spin velocity 770 miles/second (~1,200,000 m/s)
Sun's spectral class G3 verging on G2; "barely smaller and cooler than Sol".
Day length 30 hours
Rotational time 7.5 Ringworld days (225 hours, 9.375 Earth days)

TheRiov
2007-07-19, 09:21 PM
Well hold on. If you assume a non-rotating Dyson sphere, the Escape Velocity at 1 AU is something like 42 km/s (escape velocity of the Earth is only 11.2 km/s
)

Thats something like 4x the gravitational force if people lived on the exterior of your Dyson Sphere.


On the INSIDE you could null out that force (at least along the equator) by rotating your sphere, but the poles (with little angular momentum) are still going to want to collapse--there is a reason that any form of object with significant angular momentum forms a disk.


Out of cureosity, why a dyson sphere? If you're worried about having enough space to work you have got to figure that the interior of such a sphere is something on the order of (assuming I've done my math right) 551,359,235 times the surface area of the Planet Earth.


You might consider moving to a Ringworld type setup--the physics arent QUITE as nasty --(Either way its a technological impossiblity with the physics we know today, the Strong Nuclear force doesn't even have the tensile strength necessary to sustain the forces involved. )

Jayabalard
2007-07-19, 09:22 PM
Either way its a technological impossiblity with the physics we know today, the Strong Nuclear force doesn't even have the tensile strength necessary to sustain the forces involved. )hooray for scrith...

Bosh
2007-07-19, 09:26 PM
The surface area of the inside of a Dyson sphere is sphere shaped. The forumla for finding the area of a sphere is 4pir(squared). As has been noted the earth is 93 million miles from the sun. So the surface area of the inside of a Dyson sphere is 108686516832000000 square miles. That's a rather large amount of surface area. The total surface area of the earth is 197,000,000 square miles so you've got the surface area of about 551,700,000 times the surface area of the earth. You've got a lot of space.

Ulzgoroth
2007-07-19, 09:27 PM
You might consider moving to a Ringworld type setup--the physics arent QUITE as nasty --(Either way its a technological impossiblity with the physics we know today, the Strong Nuclear force doesn't even have the tensile strength necessary to sustain the forces involved. )
Er, can you clarify that? I don't see why a ringworld has to be under any stress at all, inherently. You could just spin it at one rotation per year and it would pretty much fly in formation with no support at all, though it would have to be enclosed for atmosphere.

TheRiov
2007-07-19, 09:30 PM
That assumes you want a null-gravity ringworld. You couldn't keep atmosphere in, and it wouldn't be livable. What would be the point?

Now if you rotate it fast enough to accomplish 1G then the tension goes WAY WAY WAY up.

Jayabalard
2007-07-19, 09:42 PM
Er, can you clarify that? I don't see why a ringworld has to be under any stress at all, inherently. You could just spin it at one rotation per year and it would pretty much fly in formation with no support at all, though it would have to be enclosed for atmosphere.I don't want to come across wrong... niven is a generally a fairly hard sci-fi writer and the numbers that you see thrown around are from real calculations, not pulled out of thin air. The strain and instability were actually pointed out by other people (mit students, physicists and such), though he did have to invent a rather absurd material (scrith) to account for the need for sugh high tensile strength

The phyisics work out similar to they would if as if it were a span (a bridge with no supports except on the ends) the length of it's circumference.

it has a mass about the same as Jupiter

you can't spin it as slowly as you suggest, or you don't generate the centripetal force to have pseudo gravity. Spin speed is ~770 miles/second, and works out to ~40.5 revolutions per earth year iirc.

I'm tired or I'd dig up a more in depth and mathematical description.

other trivia:
the energy involved in bringing it up to spin would be ~1.6*10^39 joules of kinetic energy. This is equal to the entire energy output of a Sunlike star for 130,000 years.

Fax Celestis
2007-07-19, 09:45 PM
...why not merely have the Dyson sphere continuously radiate an altered reverse gravity (http://www.d20srd.org/srd/spells/reverseGravity.htm) spell, one that sucks things down instead of pushes them up?

This answer is also known as the "screw physics, magic rocks" answer.

TheRiov
2007-07-19, 09:55 PM
That figure assumes the material has no initial KE. Your construction blocks come from somewhere, you're either disassembling a few jupiter sized planets or a star. Lets assume you're taking a companion star and somehow generating your Unobtainium by some complex fusion reaction from another star (or even taking a high mass star and just pealing off portions of it--, thats a sizable chunk of KE there.

We're of course talking about a manufacturing process so far outside the realm of our known physics any meaningful discussion is probably moot.

Lets assume that any civilization who wants to undertake this multi-thousand year task, has solved the power problems, is somehow holding everything in place with force fields instead of any a physical surface, why is this being built in the first place?

If your society can create matter, generate near limitless power, etc. why risk chaining them to one star? We're talking about a society that would want to survive beyond the next few billion years. Branch out, theres millions of stars in this galaxy alone..

TheRiov
2007-07-19, 09:57 PM
...why not merely have the Dyson sphere continuously radiate an altered reverse gravity (http://www.d20srd.org/srd/spells/reverseGravity.htm) spell, one that sucks things down instead of pushes them up?

This answer is also known as the "screw physics, magic rocks" answer.

if you're running a sci-fi campaign you aught to at LEAST pay lip service to Science. You can get away with a LITTLE more by switching to space opera but you better not be throwing around words like Dyson Sphere in space opera.

TheRiov
2007-07-19, 10:01 PM
I should retract an earlier statement. There is ONE reason to built such a surface if not for habitable living space: Power.

If you wanted to capture the entire output of a star, you COULD basically build a series if rotating rings (rotating to null out their own gravity so they dont fall into the sun) of low mass and then beam that energy somewhere.

Jayabalard
2007-07-19, 10:02 PM
if you're running a sci-fi campaign you aught to at LEAST pay lip service to Science. You can get away with a LITTLE more by switching to space opera but you better not be throwing around words like Dyson Sphere in space opera.They use the term "Dyson Sphere" in star trek, which definitely has some space opera qualities.

and the correct term is "scrith" which is a specific type of "unobtanium" :smallbiggrin:

Really, Fax's suggestion is the easiest by far... don't explain very much and wave your hands alot when you do explain anything.

TheRiov
2007-07-19, 10:12 PM
and the correct term is "scrith" which is a specific type of "unobtanium"


Scrith is just Niven's word for the specific material used to construct his ringworld. unobtainium is a generic term for any theoretical material (or material you cannot obtain)

Fax Celestis
2007-07-19, 10:59 PM
if you're running a sci-fi campaign you aught to at LEAST pay lip service to Science. You can get away with a LITTLE more by switching to space opera but you better not be throwing around words like Dyson Sphere in space opera.

When the original poster says "The Sun is not an actual sized sun but a divinely magical superweapon created to destroy the previous universe for it's misdeeds," it's a safe assumption that magic is involved. Why not just solve the problems with magic and go about your business, so that some smartass player doesn't get up on their high math-horse about how "ZOMG your physics are wrong?!?!oneoneeleventyone"

Belteshazzar
2007-07-19, 11:00 PM
It would appear that a perfect sphere causes more gravity problems than I would have expected (as I said biology is my specialty not physics)However, I wonder if some sort of ellipsoid wouldn't be more feasible? Perhaps a football or seed shaped world would have enough in common to fulfill both my needs (fully encloses it's sun, large enough to not be obviously curved.) Perhaps something shaped like the Chulips from Nadesico with the ridges/ mountain ranges to provide backbone.

Fax Celestis
2007-07-19, 11:00 PM
if you're running a sci-fi campaign you aught to at LEAST pay lip service to Science. You can get away with a LITTLE more by switching to space opera but you better not be throwing around words like Dyson Sphere in space opera.

When the original poster says "The Sun is not an actual sized sun but a divinely magical superweapon created to destroy the previous universe for it's misdeeds," it's a safe assumption that magic is involved. Why not just solve the problems with magic and go about your business, so that some smartass player doesn't get up on their high math-horse about how "ZOMG your physics are wrong?!?!oneoneeleventyone"

Belteshazzar
2007-07-19, 11:18 PM
An over-reliance on magic goes against the flavor of my world. This sun also happen to have faint antimagical properties (I implement possible spell failure for all spells.) My world is supposed to serve as the incubator for all the remaining life in the universe because everything outside the world is a cold starless expanse filled with dead, ruined worlds and the misbegotten descendants of those who survived the destruction ( and other darker things).

PS. Thanks for all of the prompt responses I have gotten to my request. I probably should have posted more details in my original post but I honestly didn't expect such an enthusiastic response. If anyone else wants to post other ideas they have had for oddly arranged worlds (magical or otherwise) feel free to post them.

Khosan
2007-07-19, 11:36 PM
It would appear that a perfect sphere causes more gravity problems than I would have expected (as I said biology is my specialty not physics)However, I wonder if some sort of ellipsoid wouldn't be more feasible? Perhaps a football or seed shaped world would have enough in common to fulfill both my needs (fully encloses it's sun, large enough to not be obviously curved.) Perhaps something shaped like the Chulips from Nadesico with the ridges/ mountain ranges to provide backbone.

Nah, it works as best as you can expect. With a Reverse Gravity, everything works perfectly well. In the real world, a Dyson's Sphere is silly for the primary reason of creating gravity. The only thing you can do is spin it fast enough so people are forced to stick to the outside.

The best shape would probably be a cylinder. Just have the sides perpendicular to the axis of rotation and not have people living on them. Gravity might be a bit strange near the ends, as gravity from the star gets weaker and more toward the inside of the sphere (it'd feel a bit like walking uphill, despite it being perfectly flat).

Dervag
2007-07-20, 12:19 AM
As for spinning, if you want equatorial gravity around 1G and a solar primary like Sol, you'll need to remember that there is no known or theorized material that can withstand the forces generated by its own weight at that spin rate. Floor material would thus be virtually indestructable.Yeah, that's a serious problem; you could set off a nuke against Dyson Sphere floor material and barely scuff the finish.


The actual radius could be anything in terms of gravity. It'd just need to spin fast enough to achieve earth-like gravity on the inside of the sphere.

What would be needed to figure out is how large it'd need to be to achieve some kind of bearable temperature. Fortunately, we have an idea of that already, 149,598,000 km (1 AU, the distance from the earth to the sun), since we know there's a bearable temperature for humans at that distance. So, you'd have 281,229,865,302,746,438 sq. km to work with.

Edit: Ninja'd. But I bet I can come up with a rotational speed! 29,787,771 m/s. Your poles would be screwed though.Your speed is off by several orders of magnitude; a Dyson sphere of that radius spinning that fast would have a centripetal acceleration ('surface gravity' from the perspective of someone standing on the inside) of roughly six hundred gees. Nothing remotely human could even stay alive at that acceleration.

The formula is a = (v^2)/R, where a is the surface acceleration (which we want equal to 9.8 m/s^2, the acceleration due to gravity at the Earth's surface), v is the speed of the spin at the equator, and R is the radius of the sphere.

I'm not sure where your math slipped, but when I redid the calculation I got 1.21*10^6 m/s, or 1210 kilometers per second. And yes, that is per second, not per hour.

For reference, that is about 250 times faster than the speed of an orbiting satellite, and about 16 times faster than the fastest possible meteor that could strike the Earth.


I had heard this mentioned once but was unsure as to the veracity of this statement.One uses really really stiff floorboards for the Dyson sphere. There's no other way. Since this is D&D, the first suggestion that springs to mind is some variant of a Wall of Force. The wall of force is invulnerable to physical attack, so no amount of pressure or tension can damage or stress it, and it could support an unlimited load. Since an uber-god put it there, you don't have to worry about it being dispelled by accident (unless you really want to... in which case the world blows apart and goes flying into space).


So how does one get around this. I noticed you specifically mentioned that an uniformly solid Dyson Sphere has no internal gravity. Does this imply that it could have internal gravity somehow?The only way to give a solid spherical shell gravity on the inside pulling towards the outside is to make the mass of the shell lopsided. You have to put more mass on one side than on the other, or the pulls of the two sides cancel each other out. And even then you only get gravity pointing outward in the area directly over the lopsided bit of mass.

Here's my suggestion. The shield around this divine superweapon is made of a wall of force, coated with a layer of rock on the inside. This shield and rock layer do not have noticeable gravity; the gravity of the shield cancels itself out everywhere inside the shield.

The shield is not spinning, or is spinning only at a speed sufficient to keep it orbiting the 'Sun' (which produces negligible spin gravity).

However, there is one bit of the shield where, for some reason, the Creator saw fit to add an extra 'pancake' of heavy matter (such as rock) that is hundreds or thousands of miles thick, creating a sort of plateau on the inside of the disk.

As long as you stay near the center of the plateau, there will be substantial gravity pointing outward (away from the 'Sun') due to the disk. As you move towards the edge of the plateau you find yourself climbing 'uphill', because the plateau's gravity attracts you to its center of mass. And the gravity will get weaker, because you're moving farther away from the center. If the plateau is, say, several tens or a few hundreds of thousands of miles across, there will be a roughly Earth-sized land area in the center where the gravity appears to point just about straight down with nearly constant strength.

By manipulating the area and thickness of the plateau I can make this region with Earth-normal gravity as big or small as you like.

Of course, all the water will tend to flow 'downhill' to the center of this area, so you'll end up with a ring of continents that get 'higher' as you move away from the central ocean. Only major terrain features will produce drops in elevation great enough to cancel out the effect of getting farther away from the center of the disk.

The farther away from the center that you move, the more the gravity slopes relative to the surface of the plateau and the weaker it gets. So you find yourself climbing a steep slope in weak gravity. By the time you reach the edge you're climbing almost straight up (even though the 'Sun' is still directly over your head), but the gravity is so weak that you could probably bench-press a hippo in it. I could give you equations for the angle of the slope of gravity and the strength of gravity, but that would require me to make a skill check (so to speak).

Now, if you make it over the lip of the plateau, you find yourself standing on what appears to be level ground, but the 'Sun' is now off to one side rather than being directly overhead. You can walk around the lip of the 'cliffs' at the side of the plateau and feel no adverse effects, but the gravity is very weak, so it will be like moonwalking only more so. Then, if you step 'down' onto the surface of the shield, you're climbing up a cliff again (away from the mass of that big plateau, remember?). But the gravity is getting even weaker, until eventually you only notice it if you make no effort to secure yourself against it and just drift with the current. Finally, it gets so weak that you're effectively weightless and can move around like an astronaut in outer space.

However, taking this marvelous journey will require you to travel millions of miles, or at least hundreds of thousands, so it's very unlikely if not impossible for anyone to actually do it. As far as most people (except the cosmologists) are concerned, the world is a shallow bowl ringed by impossibly high mountains, with the Sun sitting directly overhead all the time. Incidentally, this also explains where the air goes; you can keep a 'pool' of air on the center of the plateau to cover the Earthlike region.


P.S. The Sun is not an actual sized sun but a divinely magical superweapon... a few miles in diameterOK, if we're going to do this calculation we need to know how heavy it is, and how much light it gives off compared to the Sun we know and love.

If it gives off less light the Dyson sphere has to be smaller to preserve a comfortable temperature, but a smaller sphere has to spin faster if you want to use spin to generate gravity. If it gives off the same amount of light and is as far away from your world as the Sun is from the Earth, then its much smaller disk would make looking directly at it unbearable; it would burn holes in the backs of your eyes. On the other hand, if the space between you and the 'Sun' is full of air, then the air will diffuse the light enough to make it tolerable... I think.


Atmosphere is also rather thick in this universe so air is unlikely to be a problem, as whole draconic empires have thrived on the Nightshroud (the cloud of flying islands, rocks, ruined airships ect... orbiting the micro-sun that gives the inner surface some measure of night and day)Unless the flying islands are very close to the Sun, or many many thousands of miles wide, they can't create a night-day cycle. And for anything as bright as the Sun, no floating island could survive less than a few tens of millions of miles away without some kind of magical shield.

So for that to work, your 'Sun' has to be much weaker as a light source than the Sun we know and love, and the sphere has to be much closer to it... in which case I can't imagine any way to generate gravity for it without using a magic gravity generator; my plateau idea won't really cut it.


Take the easy way out: have your sun "radiate gravity", pushing everything away from it!That's actually probably a good solution. If this thing can destroy universes, maybe it does so by being an incredibly potent source of anti-gravity that can scatter the universe into a huge cloud of widely separated atoms. In which case it's currently turned off and running at 'standby power levels', generating just enough antigravity to permit life to evolve on the inner surface of its containment vessel.

Trauco
2007-07-20, 03:50 AM
My world is supposed to serve as the incubator for all the remaining life in the universe because everything outside the world is a cold starless expanse filled with dead, ruined worlds and the misbegotten descendants of those who survived the destruction ( and other darker things).

Should have read that before i drew this.
http://farm2.static.flickr.com/1281/857851988_6301ebd74c_o.jpg

star A is inside the sphere, star b is outside the sphere, if the sphere does not rotate you got gravity on one hemisphere, if it does rotate you could have a eternal wave inside the sphere.

PD: I made some other drawings:


However, there is one bit of the shield where, for some reason, the Creator saw fit to add an extra 'pancake' of heavy matter (such as rock) that is hundreds or thousands of miles thick, creating a sort of plateau on the inside of the disk.

http://farm2.static.flickr.com/1219/857926742_a36a95b842_o.jpg

The verson with the wave

http://farm2.static.flickr.com/1003/857097607_d59ccaf47f_o.jpg

Ulzgoroth
2007-07-20, 04:38 AM
The turning world with the external gravity source could be incredible. You have to be nomadic, or at least move occasionally...otherwise the land tilts from under you and the air pulls away. But everything that goes around eventually comes back. And while towers and such wouldn't survive being turned upside down very well (at least with traditional architecture), underground dungeon-type things might well, so after sufficient time the front edge of the life zone will be returning ancient constructions to the world...

It would have to turn pretty slowly though, or not even completely nomadic populations could stay ahead of it.

TheRiov
2007-07-20, 07:10 AM
An over-reliance on magic goes against the flavor of my world. This sun also happen to have faint antimagical properties (I implement possible spell failure for all spells.) My world is supposed to serve as the incubator for all the remaining life in the universe because everything outside the world is a cold starless expanse filled with dead, ruined worlds and the misbegotten descendants of those who survived the destruction ( and other darker things).

PS. Thanks for all of the prompt responses I have gotten to my request. I probably should have posted more details in my original post but I honestly didn't expect such an enthusiastic response. If anyone else wants to post other ideas they have had for oddly arranged worlds (magical or otherwise) feel free to post them.


Well.. lesse here. If you're making the rest of your universe a near void hows this:


your sphere creators have the technology to breach space/time and tap an extradimensional power source. This breach does NOT allow travel, but simply lets them tap de Sitter space for energy. They use this to power their artificial sun, but without generating gravity. Since they knew the rest of the universe was on the way out (or at least near space) they gathered what material they could (remains of an trinary star system, whatever) and build a sphere around their power source. Since we've solved the power problem (unlimited energy) this world has near infinite lifespan.
Lets also postulate that since the sphere builders can tap other power sources they've mastered artificial gravity. The interior surface of the sphere doesn't need to spin to maintain gravity or keep itself from falling anywhere --it only has to support its own mass, and that is held in check by your grav field.

(ooh ooh.. you could get funky and use a variant of what they suggested--their 'sun' is indeed an anti-grav generator, so instead of being 'flung' outward by rotation to simulate gravity, they're actually PUSHED by your 'sun' down. That means to travel to the sun, the closer you get the harder it gets to approach it. (stronger near the source)


The anti grav force would also act as a limited defense mechanism for the dyson sphere, forcing rogue asteroids, etc and deflecting them away.



I'm breaking about 50 laws of physics here at least BUT you can kinda get away with it because we're suggesting a power source that taps another dimension and therefor you can attribute all sorts of wierdness to the mixing of one universes physical laws with our own.

Citizen Joe
2007-07-20, 07:36 AM
Use cartoon physics/Wile E. Coyote

You could simply NOT explain the physics and then when one of the players says something about it, you say "You're right... you float off into the sky and die in the sun. Any more questions?" Suddenly, nobody questions physics.

Iku Rex
2007-07-20, 08:06 AM
Why not have people live on the outside of the sphere? Build it close enough to the star that the total volume inside makes for a "star" large enough to have 1G gravity on its surface. The inside would be covered in a superconducting energy transfer material. Light on the outside could be provided by another star, huge moving mirrors in orbit reflecting power transferred from the inside of the star or even flying magic stars.

TheRiov
2007-07-20, 08:24 AM
Why not have people live on the outside of the sphere? Build it close enough to the star that the total area inside makes for a "star" large enough to have 1G gravity on it's surface. The inside would be covered in a superconducting energy transfer material. Light on the outside could be provided by another star, huge moving mirrors in orbit reflecting power transferred from the inside of the star or even flying magic stars.

that basically makes the sphere with 4x the radius, and 16 times the surface area (and 16x the total volume of material needed to construct it-- but 1/16th the received energy per unit area.
(The Sun's pull is fantastic, --4x what we're pulled by the Earth. we dont notice it because we're in orbit around the Sun, in the same way that earth still pulls on objects in orbit, they just dont notice it relative to their own point of view)

Bender
2007-07-20, 08:28 AM
My suggestion:

Do you want the area of the world to be bigger than earth? Since earth already has an extremely large surface compared to most campaign worlds, lets assume the same size is ok. Then the inside of the sphere would have to be the same size as the outside of our earth.
A spin to generate earth like gravity would have to be about 8000 m/s which is about 25 mach. A manageable speed, and you won’t notice a thing on the inside of the sphere.
For a material to hold everything together: I’m not going to bore you with the calculations, the result is: tension=2*g*r*rho
g=9,81; r=6000000 m, rho is density of material
suppose a material with the density of water, the tension becomes approximately: 2*10*6*10^6*1000 or 10^11 N/m^2
steel can have a tensile strength of 20^12 N/m^2, even with a density of 7000, there is still some reserve. If you want more reserve, take industrial plastics like Kevlar or PEEK and you are safe.

Suppose these cables are the main substance of your sphere, with the reserve you can add all the soil and rock you want. Putting most of the cables in a central mountain chain is a nice touch, and probably better, but not necessary.

The gravity field would be cylindrical, however, so walking away from the equator would also be like walking uphill. Once you are about 250 km from the equator, you would start to notice I think, but you can go a lot further without really being hindered, I’d say several 1000 km would be comfortably inhabitable, over a 40000 km ring. Most water would also be around the equator.
The gravity would decrease with the cosinus of the angle (slowly in the beginning, more and more rapidly when you reach the poles).

Now the sun: to make it appear as the same size as we see the sun, it has to be about 60 km in diameter in the centre of the sphere. Some large clouds or pieces of debris might can be stationary around it, as they are not affected by the rotation of the sphere.

Some nice touches: the atmosphere would also be thickest at the equator, almost none at the poles.
You could have the gravity outside be comparable to earth’s gravity. On the equator, you would fly of, but at the poles, you could easily walk on the outside, there might be tunnels through the shell, between the cables connecting in and out.
On a clear day, you can see the other side.

Downside: the poles would probably collapse due to the gravitational pull. If your players are geeky enough to point that out, you can always say that they are very light as they don’t need the reinforcement cables anyway.

That’s all, folks

Citizen Joe
2007-07-20, 08:47 AM
Actually, just put holes at the poles. This thing IS a weapon, so it needs to be able to fire through the sphere somehow, the poles are the best location. It would look like a pulsar.

Keld Denar
2007-07-20, 08:53 AM
or how long it may take to travel from place to place.


This one is easy. According to V as proven in http://www.giantitp.com/comics/oots0145.html
The distance from point A to point B is always 1 random encounter.

And now you know, and knowing is half the battle!
GI JOE!!!!!

JEntropy
2007-07-20, 08:58 AM
I really liked Dervag's idea:


Here's my suggestion. The shield around this divine superweapon is made of a wall of force, coated with a layer of rock on the inside. This shield and rock layer do not have noticeable gravity; the gravity of the shield cancels itself out everywhere inside the shield.

The shield is not spinning, or is spinning only at a speed sufficient to keep it orbiting the 'Sun' (which produces negligible spin gravity).

However, there is one bit of the shield where, for some reason, the Creator saw fit to add an extra 'pancake' of heavy matter (such as rock) that is hundreds or thousands of miles thick, creating a sort of plateau on the inside of the disk.

As long as you stay near the center of the plateau, there will be substantial gravity pointing outward (away from the 'Sun') due to the disk. As you move towards the edge of the plateau you find yourself climbing 'uphill', because the plateau's gravity attracts you to its center of mass. And the gravity will get weaker, because you're moving farther away from the center. If the plateau is, say, several tens or a few hundreds of thousands of miles across, there will be a roughly Earth-sized land area in the center where the gravity appears to point just about straight down with nearly constant strength.

By manipulating the area and thickness of the plateau I can make this region with Earth-normal gravity as big or small as you like.

Of course, all the water will tend to flow 'downhill' to the center of this area, so you'll end up with a ring of continents that get 'higher' as you move away from the central ocean. Only major terrain features will produce drops in elevation great enough to cancel out the effect of getting farther away from the center of the disk.

The farther away from the center that you move, the more the gravity slopes relative to the surface of the plateau and the weaker it gets. So you find yourself climbing a steep slope in weak gravity. By the time you reach the edge you're climbing almost straight up (even though the 'Sun' is still directly over your head), but the gravity is so weak that you could probably bench-press a hippo in it. I could give you equations for the angle of the slope of gravity and the strength of gravity, but that would require me to make a skill check (so to speak).

Now, if you make it over the lip of the plateau, you find yourself standing on what appears to be level ground, but the 'Sun' is now off to one side rather than being directly overhead. You can walk around the lip of the 'cliffs' at the side of the plateau and feel no adverse effects, but the gravity is very weak, so it will be like moonwalking only more so. Then, if you step 'down' onto the surface of the shield, you're climbing up a cliff again (away from the mass of that big plateau, remember?). But the gravity is getting even weaker, until eventually you only notice it if you make no effort to secure yourself against it and just drift with the current. Finally, it gets so weak that you're effectively weightless and can move around like an astronaut in outer space.

Now, I'm no physicist (I had to look up Dyson sphere to figure out what it was) but I know a plot device when I see one.

We all know that, pre-columbus, most Europeans believed that if you sailed long enough in one direction, you'd fall off the face of the Earth. You could do an interesting twist on this with your world. Instead of sailing, though, it involves rock-climbing.

Mention, off-hand, to your players that people believe if you climb up over the mountains you will just sort of float away. This won't make a lot of sense to them, since (if I understand Dervag right), as you walk further from the center of the plateau, you'll feel like gravity is weighing heavier on you, not lighter. Mention that expeditions into the mountains have never returned, or something flashy like that.

They will probably approach it with the post-Columbus skepticism we all know and love. If you ever let them travel into the mountains, it will certainly make for an interesting gaming session, as once they make it over the mountain, they reach a flat plain; if they keep going, they'll start feeling themselves float away...

I think you could dampen the effects of lower gravity in the mountains (the benchpressing a hippo part) by hinting that they feel much stronger the higher they climb; they may assume they're moving closer to some powerful, ancient magical source, and they probably won't have a hippo anyway :smallwink:

Sorry I don't have any great physics for the OP, but maybe you'll get two cents out of that.

factotum
2007-07-20, 09:24 AM
Actually, just put holes at the poles.

Wouldn't that make it a really, really big version of Sigil? :smallwink:

Iku Rex
2007-07-20, 10:11 AM
that basically makes the sphere with 4x the radius, and 16 times the surface area (and 16x the total volume of material needed to construct it-- but 1/16th the received energy per unit area.
(The Sun's pull is fantastic, --4x what we're pulled by the Earth. we dont notice it because we're in orbit around the Sun, in the same way that earth still pulls on objects in orbit, they just dont notice it relative to their own point of view)You're not making a lot of sense here.

Where is the number 4 coming from? :smallconfused:

TheRiov
2007-07-20, 10:56 AM
the force of the sun's gravity at the distance of one AU is approx. 4x that on Earth's surface.

To half the force of gravity you have to double the distance to the sun. (2 AU)

To Half that force (get you down to ~1g) you'd have to move to 4 AU.

At 4 AU you're 4x further from the sun, but you're receiving 1/16th the sunlight per unit of surface area (4 squared)

Lapak
2007-07-20, 12:26 PM
The 'antigravity sun' idea seems to be the most sensible to me, as well. Especially if you're already making it an anti-magic sun.

Perhaps it's a anti-everything sun. The superweapon converts matter to antimatter, gravity to antigravity, positive energy to negative energy, negative to positive... it's a giant conversion bomb with a perpetually renewed area of a effect. The life-giving heat and light, and the beneficial gravitational side effects, are simply byproducts of the incomprehensible destruction within the actual area of effect, waste energy that happens to allow the survival of a blessed few.

Khosan
2007-07-20, 12:29 PM
Your speed is off by several orders of magnitude; a Dyson sphere of that radius spinning that fast would have a centripetal acceleration ('surface gravity' from the perspective of someone standing on the inside) of roughly six hundred gees. Nothing remotely human could even stay alive at that acceleration.

The formula is a = (v^2)/R, where a is the surface acceleration (which we want equal to 9.8 m/s^2, the acceleration due to gravity at the Earth's surface), v is the speed of the spin at the equator, and R is the radius of the sphere.

I'm not sure where your math slipped, but when I redid the calculation I got 1.21*10^6 m/s, or 1210 kilometers per second. And yes, that is per second, not per hour.

For reference, that is about 250 times faster than the speed of an orbiting satellite, and about 16 times faster than the fastest possible meteor that could strike the Earth.

*snip*

Yea, I wouldn't doubt I mucked it up somewhere. It was kind of late, and I started getting ideas for other things. I think it might've been when I converted everything to meters. And it's been around...Six months since I did anything relating to gravity.

Also, I came up with a mini-theory that the 'best' (in terms of uniform gravity) Dyson's Sphere would actually be some kind of hyperbola/parabola rotated around the star. I haven't done any math to prove it, but it seems like something I could waste more of my Summer on.

Edit: Hang on, I spotted it. I was taking into account an object with the mass of our sun in the center of the sphere. Yours don't. So I'm right if this thing were floating around Sol; your equation would be correct if it were hanging around an empty void.

Iku Rex
2007-07-20, 12:41 PM
Where are you getting your numbers TheRiov?

I'll try some math. These calculations do not include the mass of the sphere/shell, only that of the Sun. Don't get hung up on the exact figures used - the rounding is fairly random.


These proportionalities may be expressed by the formula g = m/r^2, where g is the surface gravity of an object, expressed as a multiple of the Earth's, m is its mass, expressed as a multiple of the Earth's mass 5.976·10^24 kg and r its radius, expressed as a multiple of the Earth's (mean) radius (6371 km). -- http://en.wikipedia.org/wiki/Surface_gravity

Right now the Sun (http://en.wikipedia.org/wiki/Sun) has a mass of 332 946 Earths and a radius of 6.955×10^5 km. 695 500 / 6371 km = 109 Earth's radius. That makes 332 946/109^2 = 28 g . Yupp, that's about right. (Wikipedia says 27,94 g.)

In order to get 1 g the radius of the sphere (in Earth radiuses) would have to be the square root of the mass (in Earth masses). That's 577 earth radiuses. (332 946/577^2 = 1)

6371 km (Earth's radius) * 577 = 3 676 067 km radius for the sphere.
(5,3 times the current radius.)

1 AU (mean distance between the Earth and the Sun) = 149 597 871 km
Mean distance Sun - Mercury (http://hyperphysics.phy-astr.gsu.edu/hbase/solar/soldata2.html#c1): 57 900 000 km
Mean distance Sun- Venus (http://hyperphysics.phy-astr.gsu.edu/hbase/solar/soldata2.html#c2): 108 200 000 km.

So unless I've made some mistake above [quite possible] the sphere would be relatively close to the Sun. Based on some more quick calculations the sphere would have the surface area of roughly 332 500 Earths (??). 332K Earths should be enough for anybody. :smallwink:

Lorthain
2007-07-20, 01:47 PM
the force of the sun's gravity at the distance of one AU is approx. 4x that on Earth's surface.

To half the force of gravity you have to double the distance to the sun. (2 AU)

To Half that force (get you down to ~1g) you'd have to move to 4 AU.

At 4 AU you're 4x further from the sun, but you're receiving 1/16th the sunlight per unit of surface area (4 squared)


Forgive me to disagreeing, but the quoted materiel looks wrong. By my quick calculation, the sun's gravity at 1 AU is ~0.06% that of Earth's surface gravity (which is why we don't fly off the planet during the daytime). Doubling the distance from the sun would reduce its gravity to a quarter (just like solar radiation); maybe you are using the wrong equation or misunderstanding the escape velocity mentioned in an earlier post?


Anyway, for the Dyson Sphere I would suggest making use of that new element called Gravitunium. In addition to gravitons, Gravitunium also produces gravituns; gravitun particles behave exactly like gravitons except that they disappear from existence a few miles away from their origin (slipping into a slightly different dimension).

Lace your Dyson Sphere in Gravitonium and you can get standard gravity on the inside, no spin or people falling off the poles required. The Nightshroud idea should still work since it is orbiting beyond the influence of the Dyson Sphere Gravitunium, but still having it's own Gravitunium.

TheRiov
2007-07-20, 02:20 PM
oh you're actually going to make me do math arent you... ok

Loth, your reasoning is faulty--no matter which side of the earth you're on you're still being accelerated to the sun at all times. Orbiting (the sun) nothing more than falling at the sun and missing... continously.

The slight changes in gravitational force from one side of the earth to the other are simply a function of radius(the night side is farther away) those tidal differences are overcome by earths gravity certainly.


(goes to find a real calculator)

Belteshazzar
2007-07-20, 04:13 PM
I am glad I had this idea previewed before I started nailing facts down into my world. I had a feeling something felt a little off with a perfect sphere.
I think I will modify it by stretching the world out into a kind of sloped cylinder or football (the American kind) with rounded ends and reiforced ribs (mountain ranges with enormous mineral veins) running the length of my world from tip to tip. I will try and keep the inner surface area comparable to that of the earth. The inside will still be lit by the micro-sun (holy weapon) with the Nightshroud and various small moons providing a sporadic and sudden night cycle. Gravity will be simulated by centripetal force with null gravity areas at both poles. To Top off the strange topography one pole will be blistering hot while the other remains icy cold because the entire construction (I feel this cannot be called a planet any longer in the strictest sense) is propelled though the universe by what amounts to an enormous continual flamestrike. Forget orbiting some dead gas giant or star this world is ready to race Diskworld any day of the week.
Sound Good?

Benejeseret
2007-07-20, 07:52 PM
Hey

I am also a biologist so my solution bypasses physics (ie. forget screwy gravity stuff)

Think of swinging a bucket around and centrifical force holds the water in. Same idea but for your sphere world.

Have the whole thing being swung around something outside of the sphere...maybe some giant super-nano-tube connecting directly to the center of the galaxy (maybe this was what the superweapon energy would travel down to the center of all creation to destroy it).

Your world is then spun around as the center of the galaxy rotates via this super-nano-tube, and the seas, land, and atmosphere stays all to the one side just like water in a bucket. The superweapon star is fixed in place or attached to the inner side (to be connected to the super-nano-fibre to the center of the galaxy to blast down along)

Your dragon races have plenty of 'sides' to climp up along and create cliffface type dwellings and empires along. The sea is one big roundish water mass in the very center opposite the super-nano-fibre attachment.

The "great forests" could also be massive vine-like plants that crawl up the sides and towards the sun-weapon on the opposite side.

Bene

Citizen Joe
2007-07-20, 08:22 PM
There's some major problems with centripetal force artificial gravity...

YES it works well for solid structures.
It does NOT work so well for sloshing liquids (need baffles)
It works HORRIBLE for gasses

So the primary problem is that the water would evaporate and thus leave its containment on the rotating surface. With no gravity, it would all drift into the sun. It certainly wouldn't rain down.

goat
2007-07-20, 09:02 PM
You don't want a Dyson sphere, that's expensive and complex.

It's much easier to build a Buuthandi.

Douglas
2007-07-20, 10:08 PM
You might be right about easier, I really don't know, but I doubt it would be cheaper to build something quite literally named for how expensive it is (http://www.schlockmercenary.com/d/20020309.html).

Ulzgoroth
2007-07-20, 10:14 PM
You don't want a Dyson sphere, that's expensive and complex.

It's much easier to build a Buuthandi.
Um...A Buuthandi is a Dyson sphere, I'm pretty sure. What he's building isn't either, because he's really trying to build inside-out land. (Yes, it certainly is cheaper than building a rigid construct of much the same size...)

I second the 0.06% g from the sun at 1 AU calculation, BTW.

technomancer
2007-07-20, 10:18 PM
That wasn't nice. At least you can provide a link: http://www.schlockmercenary.com/d/20020309.html

Jack_Simth
2007-07-20, 10:29 PM
You might be right about easier, I really don't know, but I doubt it would be cheaper to build something quite literally named for how expensive it is (http://www.schlockmercenary.com/d/20020309.html).
Considering the sheer amount of material required for a solid shell, and the amount of stress that material is required to take? Oh yeah, it'll be pricier.

Dervag
2007-07-21, 12:47 AM
I am glad I had this idea previewed before I started nailing facts down into my world. I had a feeling something felt a little off with a perfect sphere.

I think I will modify it by stretching the world out into a kind of sloped cylinder or football (the American kind) with rounded ends and reiforced ribs (mountain ranges with enormous mineral veins) running the length of my world from tip to tip. I will try and keep the inner surface area comparable to that of the earth. The inside will still be lit by the micro-sun (holy weapon) with the Nightshroud and various small moons providing a sporadic and sudden night cycle. Gravity will be simulated by centripetal force with null gravity areas at both poles. To Top off the strange topography one pole will be blistering hot while the other remains icy cold because the entire construction (I feel this cannot be called a planet any longer in the strictest sense) is propelled though the universe by what amounts to an enormous continual flamestrike. Forget orbiting some dead gas giant or star this world is ready to race Diskworld any day of the week.
Sound Good?That will work. If you want me to do any more calculations for you, PM me. If the sphere is roughly the size of the Earth, then the hollow shell can be spun for gravity at manageable speeds (though I still think you'd be best off using a wall of force on the outside, just in case your superweapon hits a piece of space debris).

Be warned that anywhere but the equator, the gravity will appear to point 'downhill' towards the equator at some angle (because gravity points along the equatorial plane everywhere along the shell), and that it will start to get weaker than normal very rapidly as you leave the equator. So the habitable world will actually be only a narrow band around the equator, unless the shell is elongated like a cigar along its axis of rotation.

By the way, there are no seasons on this world, or at least no safe way to create them that I can think of. Every part of the world will have the same climate year round, and that climate will get hotter at the equators and colder near the poles (with the exception of the super-heat-source at one pole).


Well.. lesse here. If you're making the rest of your universe a near void hows this:

your sphere creators have the technology to breach space/time and tap an extradimensional power source...You misunderstand how this thing came to be. The creator of the sphere also created the weapon that turned the previous universe into a near-void. The world that the PCs are moving around in is actually the inside of the superweapon's containment vessel.


Use cartoon physics/Wile E. Coyote

You could simply NOT explain the physics and then when one of the players says something about it, you say "You're right... you float off into the sky and die in the sun. Any more questions?" Suddenly, nobody questions physics.That's kind of jerky to the players.

Besides, it's more fun to create a consistent explanation for how the world works and then deal with and explore the consequences than to assert that the world works a certain way and slap down anyone who points out the inconsistencies.


Why not have people live on the outside of the sphere? Build it close enough to the star that the total volume inside makes for a "star" large enough to have 1G gravity on its surface. The inside would be covered in a superconducting energy transfer material. Light on the outside could be provided by another star, huge moving mirrors in orbit reflecting power transferred from the inside of the star or even flying magic stars.But there isn't supposed to be anything on the outside of the sphere. Also, I gather that anything that is outside the sphere is nasty and monstrous and horrible, so that you'd be a lot happier if there were a Dyson sphere between it and you.


that basically makes the sphere with 4x the radius, and 16 times the surface area (and 16x the total volume of material needed to construct it-- but 1/16th the received energy per unit area.
(The Sun's pull is fantastic, --4x what we're pulled by the Earth. we dont notice it because we're in orbit around the Sun, in the same way that earth still pulls on objects in orbit, they just dont notice it relative to their own point of view)Actually, the effect of the Sun's gravity at the radius of the Earth's orbit is really tiny. You can calculate this using the formula for centripetal acceleration, a = (v^2)/R: the Earth orbits the sun at roughly 30000 m/s, with a radius of roughly 150 000 000 000 meters. That gives an inward acceleration of 0.006 meters per second squared, or less than one tenth of one percent of the gravity we feel due to the Earth's pull at its surface.

If the Sun's gravity were as strong as you say, then the Earth would have to orbit the Sun at over 2000 kilometers per second to have a stable orbit. In which case the year would be only four and a half days long.


My suggestion:

Do you want the area of the world to be bigger than earth? Since earth already has an extremely large surface compared to most campaign worlds, lets assume the same size is ok. Then the inside of the sphere would have to be the same size as the outside of our earth.That works if the source of light for this hollow spherical world is faint and small enough that you can survive being only a few thousand kilometers away from it. Which is OK, but I was under the impression that the original poster was designing a hollow sphere the size of the Earth's orbit, not of the Earth proper.


Mention, off-hand, to your players that people believe if you climb up over the mountains you will just sort of float away. This won't make a lot of sense to them, since (if I understand Dervag right), as you walk further from the center of the plateau, you'll feel like gravity is weighing heavier on you, not lighter.No, that is not how it works.

The pull of gravity that you feel due to the plateau pulls roughly toward the geometric center of the plateau (remember, the center is hundreds or thousands of miles underground). As you move towards the edge, gravity will in fact get weaker, because you get farther from the center and the gravity of a distant object is weaker than that of a closer object with the same mass.

But as you move away from the middle region where gravity behaves normally, the center is no longer directly under your feet. So instead of gravity pointing perpendicular to the surface, it points at an angle. In effect, even though the plateau appears to be flat to the eye and is flat geometrically, you still feel as if you're climbing uphill to move away from the center. The slope is very shallow near the center and very steep near the edge. However, as you move away from the center the gravity gets weaker, until you could climb the 'slope' with an elephant strapped to your back by the time you reach the rim (assuming, of course, that the plateau is large enough to have a large region of Earth-normal gravity at the middle).

The problem is that the plateau must be hundreds of thousands of miles wide for that to work. So maybe the reason nobody's ever reached the rim is that you'd have to climb for hundreds of thousands of miles, uphill all the way, to reach it. Even though the gravity would be weak by the time you neared the rim, that's still a long way to go wrong.


They will probably approach it with the post-Columbus skepticism we all know and love. If you ever let them travel into the mountains, it will certainly make for an interesting gaming session, as once they make it over the mountain, they reach a flat plain; if they keep going, they'll start feeling themselves float away...Well, they never 'float away', but the gravity holding them on the ground gets weaker and weaker. So if they actually reach the rim and try to jump off the edge (which will appear to them as the top of a near-vertical cliff by this point), they'll find themselves carried for miles and miles out into the air beyond the rim, but they'll still eventually fall back onto the 'side' of the plateau. Which will seem like more or less level ground to them.

It's a good idea for an adventure, but I'd really have to do strange things to the mass distribution of the disk to make it work. I think I could make it work, but it would not be easy and it would be virtually impossible for the PCs to make it to the rim and back in their lifetime.


the force of the sun's gravity at the distance of one AU is approx. 4x that on Earth's surface.

To half the force of gravity you have to double the distance to the sun. (2 AU)

To Half that force (get you down to ~1g) you'd have to move to 4 AU.

At 4 AU you're 4x further from the sun, but you're receiving 1/16th the sunlight per unit of surface area (4 squared)Even if your calculation of the Sun's gravity at the radius of the Earth's orbit were correct, that is not how gravity works. Gravity follows an inverse square law. Doubling the distance reduces the gravity by a factor of four, to one quarter the strength, not by a factor of two to one half the strength.

At 4 AU you're four times farther from the sun, receiving 1/16th the sunlight per unit of surface area, and getting 1/16th of the gravity.

Which, since the Sun's gravity is not as strong as you think, would be a mere 0.000375 meters per second squared, or about one three-hundred-thousandth of the gravity we feel at the Earth's surface.


Edit: Hang on, I spotted it. I was taking into account an object with the mass of our sun in the center of the sphere. Yours don't. So I'm right if this thing were floating around Sol; your equation would be correct if it were hanging around an empty void.The Sun's gravity is negligible for our purposes. You can use the formula for centripetal acceleration to calculate the gravity we feel due to the Sun at the distance of the Earth from the Sun, and it is measured in millimeters per second squared. So that extra component doesn't change the required speed of the Dyson sphere appreciably; we can spin our Dyson sphere as if solar gravity did not exist and we'd never notice the difference due to solar gravity except on sensitive scales.

Corolinth
2007-07-21, 03:10 AM
I haven't read the rundown of all of the posts in the thread, so some of what I post will be redundant. Deal.

In order to understand the responses to your question, you have to first understand what gravity is. Gravity is an attractive force based on the distance between two objects, and their respective masses. All objects exert gravity! The greater the mass of the two objects in question, the greater the force of gravity between them. The greater the distance between the two objects' gravitational centers, the less gravitational force they exert on each other. This is why we do not go flying towards the sun, despite it being larger than the earth by a factor of 100,000.

If you dig a hole to the center of the Earth, the force of gravity acting on your body decreases. The deeper you dig, the more Earth mass is on the other side of you. If you dig a hole a mile deep, you will have a mile's worth of Earth's mass pulling you in the opposite direction. Once you reach the center, you are in a zero gravity environment, because there is equal mass exerting gravity on you in all directions.

The problem you encounter with gravity inside of a Dyson's sphere is that the mass of the sphere is spread out over a vast distance. There is a very small amount of sphere mass exerting gravity on you beneath your feet, and the rest of the sphere is so far away that its gravity is negligible. Likewise, a Dyson's sphere with a sufficiently large radius could make the gravity of the sun negligible. Spinning the Dyson's sphere would create the illusion of gravity along a band of the sphere perpendicular to the axis of rotation. You have a velocity tangent to the rotation of the sphere, and a normal force being exerted by the wall of the sphere that pushes you inward towards the center. Your inertia holds you against the inner wall of the sphere. Were you to jump, however, you would have no forces acting against you, and would continue to travel "up".

This is the part of the lecture where I point out that centrifugal force does not exist. Centrifugal force is no more than an objects own inertia.

What you would need to generate Earth-gravity is a Dyson's sphere with a wall of a width roughly equal to the diameter of the Earth, and a density roughly equal to that of the Earth. Actually, it would be more accurate to say that you would need walls with roughly the same mass as that of the Earth, because it is mass that causes gravity. Thickness and density are largely moot, it's the mass that's important. The sphere would have to have a very large radius to make the gravitational force exerted by the mass of the other half of the sphere negligible. Since you are on the inside of a sphere, the ground immediately surrounding you has most of it's gravitational force canceled out by the mass immediately on the other side of your body, leaving only the component which acts "down" contributing to your overall net gravity. Because more mass is generating "down" gravity than just that which is directly beneath you, the walls need not be exactly Earth-thickness at Earth-density. However, the smaller the radius of the sphere, the more gravity will be exerted by the sun, so the mass of the walls will need to be adjusted to compensate. Likewise, the less mass the sun has, the smaller the radius of the Sphere needs to be.

Norsesmithy
2007-07-21, 04:57 AM
http://img49.imageshack.us/img49/2665/centifugalbondrh3.png

Attilargh
2007-07-21, 05:13 AM
You're my new hero.

Norsesmithy
2007-07-21, 05:17 AM
Go here:

xkcd.com (http://xkcd.com/c123.html)

Bender
2007-07-21, 06:17 AM
This is the part of the lecture where I point out that centrifugal force does not exist. Centrifugal force is no more than an objects own inertia.

As pointed out in the cartoon: there would be a centrifugal force for those living in the sphere.
(disclaimer: if you don't know anything about virtual forces and the implications of rotating coordinate systems: ignore previous statement and certainly never use it in a discussion :smalltongue:)



What you would need to generate Earth-gravity is a Dyson's sphere with a wall of a width roughly equal to the diameter of the Earth, and a density roughly equal to that of the Earth. Actually, it would be more accurate to say that you would need walls with roughly the same mass as that of the Earth, because it is mass that causes gravity. Thickness and density are largely moot, it's the mass that's important. The sphere would have to have a very large radius to make the gravitational force exerted by the mass of the other half of the sphere negligible. Since you are on the inside of a sphere, the ground immediately surrounding you has most of it's gravitational force canceled out by the mass immediately on the other side of your body, leaving only the component which acts "down" contributing to your overall net gravity. Because more mass is generating "down" gravity than just that which is directly beneath you, the walls need not be exactly Earth-thickness at Earth-density. However, the smaller the radius of the sphere, the more gravity will be exerted by the sun, so the mass of the walls will need to be adjusted to compensate. Likewise, the less mass the sun has, the smaller the radius of the Sphere needs to be.

If the sphere was the same thickness everywhere, as pointed out before, its gravity would exert no resulting force on you. None at all. You would float around. I can prove that with math, but I'm lazy now. Just know that as the the mass of the sphere above you might seem neglectable because of the distance, it is also much larger than the mass underneath you. Very large multiplied with almost neglectable in this case results in 1, not 0.


There's some major problems with centripetal force artificial gravity...

YES it works well for solid structures.
It does NOT work so well for sloshing liquids (need baffles)
It works HORRIBLE for gasses

So the primary problem is that the water would evaporate and thus leave its containment on the rotating surface. With no gravity, it would all drift into the sun. It certainly wouldn't rain down.

Due to viscous forces, the water and air would be dragged along with the sphere, to eventually rotate at the same speed as the shell. This would result in the almost same centripetal (or centrifugal if you live there) forces as on the sphere itself. They would be slightly smaller, since the atmosphere is slightly closer to the centre.

kjones
2007-07-21, 07:54 AM
the force of the sun's gravity at the distance of one AU is approx. 4x that on Earth's surface.

To half the force of gravity you have to double the distance to the sun. (2 AU)

To Half that force (get you down to ~1g) you'd have to move to 4 AU.

At 4 AU you're 4x further from the sun, but you're receiving 1/16th the sunlight per unit of surface area (4 squared)

False. Remember, F = (M*m*G)/(R^2). By doubling the distance to the sun, you decrease the force of gravity by a factor of 4.

TheRiov
2007-07-21, 11:06 AM
I haven't read the rundown of all of the posts in the thread, so some of what I post will be redundant. Deal.

In order to understand the responses to your question, you have to first understand what gravity is. Gravity is an attractive force based on the distance between two objects, and their respective masses. All objects exert gravity! The greater the mass of the two objects in question, the greater the force of gravity between them. The greater the distance between the two objects' gravitational centers, the less gravitational force they exert on each other. This is why we do not go flying towards the sun, despite it being larger than the earth by a factor of 100,000.

Right answer wrong reason. Gravitational attractive force is a function of mass true, but accelleration is due to a force is inversely proportional to the mass of the object. (ie, the heavier something is the harder it is to get moving) --thus the mass of the object we're calculating the acceleration (your person, the earth, etc) cancels out of the equation and only the other object's mass matters. (ie. accelleration due to gravity from the sun applies the same to a human, a planet, a paperclip) A random human in the same orbit as the Earth, with no other gravitational forces acting on it, will maintain the same orbit as the Earth.





If you dig a hole to the center of the Earth, the force of gravity acting on your body decreases. The deeper you dig, the more Earth mass is on the other side of you. If you dig a hole a mile deep, you will have a mile's worth of Earth's mass pulling you in the opposite direction. Once you reach the center, you are in a zero gravity environment, because there is equal mass exerting gravity on you in all directions.

The problem you encounter with gravity inside of a Dyson's sphere is that the mass of the sphere is spread out over a vast distance. There is a very small amount of sphere mass exerting gravity on you beneath your feet, and the rest of the sphere is so far away that its gravity is negligible. Likewise, a Dyson's sphere with a sufficiently large radius could make the gravity of the sun negligible. Spinning the Dyson's sphere would create the illusion of gravity along a band of the sphere perpendicular to the axis of rotation. You have a velocity tangent to the rotation of the sphere, and a normal force being exerted by the wall of the sphere that pushes you inward towards the center. Your inertia holds you against the inner wall of the sphere. Were you to jump, however, you would have no forces acting against you, and would continue to travel "up".

This is false.

While you would have no FORCES acting on you, You still have your own momentum that would keep you pointing in a straight line from the moment you left the surface. That momentum would keep you moving in the same direction and you'd just impact the surface of your sphere again

http://img.photobucket.com/albums/v662/theriov/physics.gif
note the red arrows are the force/momentum you added by jumping--but you didn't cancel out your linear momenum. with the rotational speeds we're talking about you're not going to do that.




What you would need to generate Earth-gravity is a Dyson's sphere with a wall of a width roughly equal to the diameter of the Earth, and a density roughly equal to that of the Earth. Actually, it would be more accurate to say that you would need walls with roughly the same mass as that of the Earth, because it is mass that causes gravity. Thickness and density are largely moot, it's the mass that's important. The sphere would have to have a very large radius to make the gravitational force exerted by the mass of the other half of the sphere negligible. Since you are on the inside of a sphere, the ground immediately surrounding you has most of it's gravitational force canceled out by the mass immediately on the other side of your body, leaving only the component which acts "down" contributing to your overall net gravity. Because more mass is generating "down" gravity than just that which is directly beneath you, the walls need not be exactly Earth-thickness at Earth-density. However, the smaller the radius of the sphere, the more gravity will be exerted by the sun, so the mass of the walls will need to be adjusted to compensate. Likewise, the less mass the sun has, the smaller the radius of the Sphere needs to be.

Patently false. The sum of the mass on the far side of the sphere is greater but farther away. In a uniform sphere or shell only the mass INSIDE your radius affects you.
http://img.photobucket.com/albums/v662/theriov/physics2.gif

no matter where you are inside the sphere the mass closer to you is cancelled out by the mass on the far side.

Citizen Joe
2007-07-21, 12:35 PM
I got an idea. Make a Dyson sphere around a star... The sphere itself has a net zero effect on gravity within it, but I'm sure there is a huge pull on anything outside. Now inside the sphere, floating around the star, put a planet... with normal gravity.

OK now pretend the Dyson Sphere doesn't exist:smallamused:

TheRiov
2007-07-21, 01:27 PM
that works from a physics point of view. If you're going to do this though it might make more sense to just build a giant 'balloon' to absorb all outgoing radiation from the star to power your civilization. The balloon would be 'inflated' by radiation pressure from the star, The mass requirements are far less, and you would basically have a sun to power all your energy.

(in fact some variants of this dont require a sphere on any kind, just a series of satellites--while the engineering required to do this is difficult (we have no materials that are light enough yet) its not outside the realm of physics.


In fact, it may be possible to actually move stars around using a variant of this (albiet very very very very slowly) but you could actually deflect the path of a star by constructing a gigantic mirror--I dont explain it very well but there's an interesting wikipedia article on the subject:

http://en.wikipedia.org/wiki/Stellar_engine

Lapak
2007-07-21, 01:50 PM
I got an idea. Make a Dyson sphere around a star... The sphere itself has a net zero effect on gravity within it, but I'm sure there is a huge pull on anything outside. Now inside the sphere, floating around the star, put a planet... with normal gravity.You realize that at this point you've reinvented Spelljammer.
OK now pretend the Dyson Sphere doesn't exist:smallamused:Aw, now we're back to normal.

Khosan
2007-07-21, 02:09 PM
The Sun's gravity is negligible for our purposes. You can use the formula for centripetal acceleration to calculate the gravity we feel due to the Sun at the distance of the Earth from the Sun, and it is measured in millimeters per second squared. So that extra component doesn't change the required speed of the Dyson sphere appreciably; we can spin our Dyson sphere as if solar gravity did not exist and we'd never notice the difference due to solar gravity except on sensitive scales.

Yea, spotted my error again. This time it was becaus I forgot it was GM/r^2 (possibly because the way I was originally calculating the acceleration was by shortening the equation to GM/r=v^2, remembering that I needed to have 9.8 m/s^2 outwards, and plugged it in as 9.8=v^2/r - GM/r or something). That does have a rather noticeable effect.

You have trounced my physics quite thoroughly, my good man. I take my non-existent hat off to you.

Dervag
2007-07-21, 02:18 PM
I haven't read the rundown of all of the posts in the thread, so some of what I post will be redundant. Deal.Now, that kind of prima donnage is often well warranted...


The problem you encounter with gravity inside of a Dyson's sphere is that the mass of the sphere is spread out over a vast distance. There is a very small amount of sphere mass exerting gravity on you beneath your feet, and the rest of the sphere is so far away that its gravity is negligible.But not fully warranted in this case.

What's actually going on is that the net gravitational force due to a uniform hollow spherical shell is zero everywhere in the shell, as others have pointed out here. Anywhere in the Dyson sphere, the gravitational force due to the sphere itself will be zero.

You can prove this with calculus, and the only reason it's true is that the gravity due to the 'rest of the sphere' is not negligible. In every place inside the sphere, the gravity due to the nearby part of the sphere (which is smaller but closer) and that due to the faraway part of the sphere (which is bigger but farther off and therefore exerts less gravitational force).

If the gravity of the faraway part were in fact negligible, then you would feel a net gravitational force towards the surface you were standing on, with strength determined by the thickness of the spherical shell.

The reason the net gravity force at the center of the Earth is zero is that from the point of view of something at the center of the Earth, the Earth can be imagined as a bunch of nested hollow spherical shells, much like the layers of an onion. Each shell exerts zero net force on objects inside it, and you're inside all the shells when you're at the center of the Earth.


Likewise, a Dyson's sphere with a sufficiently large radius could make the gravity of the sun negligible. Spinning the Dyson's sphere would create the illusion of gravity along a band of the sphere perpendicular to the axis of rotation. You have a velocity tangent to the rotation of the sphere, and a normal force being exerted by the wall of the sphere that pushes you inward towards the center. Your inertia holds you against the inner wall of the sphere. Were you to jump, however, you would have no forces acting against you, and would continue to travel "up".No, you wouldn't; eventually you'd hit the ground again. If you could jump at hundreds or thousands of meters per second the trajectory would not resemble the parabolic flight path that a jumping person follows on Earth, but you would hit the ground again.

Your body would follow a straight line relative to the sun, or to whatever is at the center of the Dyson sphere. But that straight line would eventually intersect the sphere at another point. So no, you would not keep floating forever; you would instead go up and down again from the point of view of somebody standing in the place you just jumped off of.

Now, the path you follow as a jumper will not look quite the same as it would on Earth, but the effect is small for the speeds of ordinary human jumpers.


This is the part of the lecture where I point out that centrifugal force does not exist. Centrifugal force is no more than an objects own inertia.People love to say that, but it doesn't mean much. Because while centrifugal force appears fictitious, there is a force pushing on any object that is rotating or spinning. If there weren't it would not be spinning, it would be going in a straight line. For instance, if you are standing on the surface of a spinning Dyson sphere, you feel a force pushing you against the ground which you call centrifugal force. And there is a force, but it's pushing the ground against you and not the other way around. Pressure sensors (like the ones in your skin that tell you when you're feeling a force) can't tell the difference, and the difference is arguably academic. A force pushing the ground against you and a force pushing you against the ground are two sides of the same coin because of Newton's Third Law (for every action there is an equal and opposite reaction).


What you would need to generate Earth-gravity is a Dyson's sphere with a wall of a width roughly equal to the diameter of the Earth, and a density roughly equal to that of the Earth. Actually, it would be more accurate to say that you would need walls with roughly the same mass as that of the Earth, because it is mass that causes gravity.No, that will not work. If you're walking around inside a hollow spherical shell, it does not matter how thick the walls of the sphere are. They could be as thin and light as the skin of a balloon, or they could be as thick and heavy as a million miles of lead. It doesn't matter. For a perfect spherical shell, where the outside and inside surfaces of the shell are both spherical and both have the same center, the net gravitational force everywhere inside the sphere is zero. Once again, you can prove this.

Gravity doesn't have the effect you assert it does.


I got an idea. Make a Dyson sphere around a star... The sphere itself has a net zero effect on gravity within it, but I'm sure there is a huge pull on anything outside. Now inside the sphere, floating around the star, put a planet... with normal gravity.

OK now pretend the Dyson Sphere doesn't exist:smallamused:That actually works. Of course, the pull of the Dyson sphere will actually be quite small unless the Dyson sphere is many, many times the mass of the star, but that works.


Yea, spotted my error again. This time it was becaus I forgot it was GM/r^2 (possibly because the way I was originally calculating the acceleration was by shortening the equation to GM/r=v^2, remembering that I needed to have 9.8 m/s^2 outwards, and plugged it in as 9.8=v^2/r - GM/r or something). That does have a rather noticeable effect.

You have trounced my physics quite thoroughly, my good man. I take my non-existent hat off to you.My last job was tutoring freshman physics students, and my next job will be grading freshman physics students' exams. It becomes second nature at some point.

Citizen Joe
2007-07-21, 03:54 PM
I think its fairly important to remember that the point of a Dyson Sphere is to capture all the radiation from a star, it was never intended as a habitat. So all of our problems with gravity and maintaining an atmosphere stems from those cross purposes.

So, if you are dead set on living on the sphere itself (and not on a planet within the sphere) then you just need to find a depth into the sphere shell where the gravity is what you desire. So as you move outward, the gravity increases until you reach the outer surface. At that point, the gravity is probably crushingly strong. The environment would be completely artificial, but at least you don't have to worry about falling off the surface.

Ulzgoroth
2007-07-21, 05:44 PM
The surface gravity of the (1 AU) sphere only reaches one G if it masses around 1500 times as much as the sun. If you build it as thin as you really want a proper dyson sphere to be, it should weigh quite a lot less than that. Of course, if you're planning to live outside the sphere you don't need it so large...and probably want it a bit thicker so you don't accidentally put a foot through it.

Citizen Joe
2007-07-21, 06:47 PM
OK, advance math question (or maybe not)

Lets say that the center object (star/device/whatever) is sufficiently large and sufficiently close that it generates 0.5g acceleration from the inner surface of the Dyson sphere. Standing on the outer surface of the Dyson Sphere you feel 4g's (combined star/dyson sphere).

1. How big is the star?
2. How far to the inner surface of the sphere?
3. How thick would the sphere be?
4. How heavy would the sphere be?
5. How far through the sphere crust would you experience 1 g?

ZebulonCrispi
2007-07-21, 07:08 PM
Another interesting problem of a Dyson sphere is that the gravitational forces on the sphere by the sun cancel each other out in all directions, so there's nothing to stop it the sphere and the sun from drifting in slightly different directions until they collide. Unless you want to outfit the entire dyson sphere with massive correcting thrusters...

Valairn
2007-07-21, 07:20 PM
For an interesting idea, you could move the campaign to the outside of the sphere, and have small suns powered by what's going on inside come out of the sphere and go back in like solar arcs.

For a larger scope, gravity is functional again because you are on the outside of the sphere, so the there is a general pull towards the "center". You should be able to have lots of fun with exploring the inside of the sphere.

Douglas
2007-07-21, 07:23 PM
1. How big is the star?
2. How far to the inner surface of the sphere?
3. How thick would the sphere be?
4. How heavy would the sphere be?
5. How far through the sphere crust would you experience 1 g?
Insufficient information.

Valairn
2007-07-21, 07:24 PM
Another interesting problem of a Dyson sphere is that the gravitational forces on the sphere by the sun cancel each other out in all directions, so there's nothing to stop it the sphere and the sun from drifting in slightly different directions until they collide. Unless you want to outfit the entire dyson sphere with massive correcting thrusters...

To respond to this directly, the gravity of the star effects the sphere equally in all directions, and the reason the sphere doesn't fall back into the star is because its structure resists that pull. The whole thing in essence becomes a large star of its own. If the star moves closer to one side of the sphere, the gravity shifts and one side of the whole sphere becomes heavier, pulling the far side towards it. Assuming a steady mass across the whole of the sphere, it should stay in relative equilibrium.

Jack_Simth
2007-07-21, 07:24 PM
Another interesting problem of a Dyson sphere is that the gravitational forces on the sphere by the sun cancel each other out in all directions, so there's nothing to stop it the sphere and the sun from drifting in slightly different directions until they collide. Unless you want to outfit the entire dyson sphere with massive correcting thrusters...
Variable reflective layer (aka, your solar panels flip over and become mirrors). You use light pressure to move the thing. Doesn't move very quickly, mind, but it moves. So long as you don't have to deal with anything massive (and you shouldn't) you can deal with drift.

Citizen Joe
2007-07-21, 07:29 PM
Another interesting problem of a Dyson sphere is that the gravitational forces on the sphere by the sun cancel each other out in all directions, so there's nothing to stop it the sphere and the sun from drifting in slightly different directions until they collide. Unless you want to outfit the entire dyson sphere with massive correcting thrusters...

Actually with the solar winds and photonic pressure, it would tend to equalize or be easily countered. Its more likely that they orbit each other, but from sphere's perspective, it just looks like the star is sort of wobbling.

Khosan
2007-07-21, 07:31 PM
OK, advance math question (or maybe not)

Lets say that the center object (star/device/whatever) is sufficiently large and sufficiently close that it generates 0.5g acceleration from the inner surface of the Dyson sphere. Standing on the outer surface of the Dyson Sphere you feel 4g's (combined star/dyson sphere).

1. How big is the star?

Any size, really. It could be very small, super-dense star or a ridiculously huge, but not too dense at all.


2. How far to the inner surface of the sphere?

Again, anything. For the hell of it and the sake of my math, I made it 1 AU with no spin. This means the star would come in at 1.643*10^33 kilograms; about 800 times more massive than our sun.


3. How thick would the sphere be?

Anything. Same case as the star. I'll just assume it's 1 AU thick.


4. How heavy would the sphere be?

It'd change depending on the thickness and the distance the outside and inside are from the star. In this case, 2.64946*10^33 kg. Also known as ridiculous expenses.


5. How far through the sphere crust would you experience 1 g?

Answering this would require complex integrals and calculus that I really don't want to do. Since you're actually inside the shell, it makes the work a bit tricky.

Douglas
2007-07-21, 08:28 PM
Answering this would require complex integrals and calculus that I really don't want to do. Since you're actually inside the shell, it makes the work a bit tricky.
Plenty of physicists have done all those complex integrals for you already, though. All you need to know is that it all comes out to be exactly the same as if any part of the sphere outside your position doesn't exist.

Citizen Joe
2007-07-21, 08:29 PM
OK, it seems you chose poorly. :smallamused:

How about this. A white dwarf with 1 solar mass. How far would the inner surface of the dyson sphere to generate 5 m/s/s acceleration (about .5g)?


A = 5 m/s/s = GM/(r^2)
r^2 = GM/5 m/s/s
G = 6.67 × 10^−11 N m^2 kg^−2
M = 1 solar mass = 1.9891 x 10^30 kg
GM = 6.67 x 1.9891 x 10^19 = 1.3267 x 10^20
r^2 = 2.6545 x 10^19
r =5.1512 x 10^9 m = 5.1512 x 10^6 km = 0.03434 AU

That's pretty close, but we're using a white dwarf so it isn't that close.

Assuming an average density of 2000 kg/cubic meter, how thick would the sphere have to be to have 40 m/s/s acceleration (when combined with the sun's) on the surface?


Inner radius = r
inner volume = 4 X pi X r^3
thickness = t
Outer Radius = r+t
outer volume = 4 X pi X (r+t)^3
shell volume = outer volume - inner volume = 4/3 X pi X ((r+t)^3 - r^3)
Md = Mass of dyson sphere = shell volume X 2000 kg/cu. m.
A = 40 m/s/s = G(Md + Msun)/(r+t)^2
At this point the math gets ugly so I'll hand it off...

Corolinth
2007-07-21, 10:08 PM
Were you to jump, however, you would have no forces acting against you, and would continue to travel "up".I was in error in posting this, because I forgot to account for the fact that the body in question is already in motion. You would have no forces acting against you, but you are currently moving in a line tangent to the rotation of the Dyson's sphere at the point in which you jumped.

Corolinth
2007-07-21, 11:19 PM
no matter where you are inside the sphere the mass closer to you is cancelled out by the mass on the far side.According to the math, you are correct, because math says you calculate gravity from the object's center of mass. This is where physics and math separate. The force of gravity that a body of mass exerts is the net value of the sum of the gravitational force exerted by all of its component particles. We are talking about a hollow sphere, however, and not a solid body.

In the center of the sphere, what you have posted is correct. If the sphere is large enough, all of the mass of the sphere is far enough away from you that its gravity is negligible. If the mass of the sphere is closer, you are in a "zero gravity" environment because the sum total of all gravitational forces acting on you is zero. We are not talking about an object in the center of the sphere, we're talking about an object on the inner walls that is in close proximity to part of the mass.

Every subatomic particle that makes up the walls of the sphere has mass, and therefore exerts gravity. The particles on the opposite side of the sphere are at a distance of 2AU (assuming we have a sphere with a radius equal to the distance at which the Earth orbits the sun). At that distance, the sum total of the gravitational force exerted by the total number of particles on the opposite side of the sphere is negligble. By comparison, the sum total of the particles composing the wall of the sphere directly beneath you is at a distance ranging from your point of contact with the ground out to the total width of the sphere's walls. The total force of gravity acting on your body would be the net total of both sums.

You can not calculate gravitational force from the mass center of the sphere, because there is no mass there. We don't have to do this for the Earth because the Earth is a solid body.

If we make a Dyson's sphere with three inch thick walls, you would be correct because the gravity exerted by the mass of the walls would be negligible. If we make it three thousand miles thick, now the gravity is no longer negligible. Now if we make the radius of the sphere equal to 5AU, the gravitational force generated by the opposing mass is too far away for us to care. Experimental data supports your argument, but experimental data has not been taken with a Dyson's sphere that has such a large radius.

According to your argument, if we were to create a three dimensional elliptical solid by wrapping aluminum around the Earth and the sun to close it off and create a Dyson's football, we would suddenly be in a zero gravity environment. Actually, according to your hypothesis, since the majority of the mass of the system is at the sun, we would all go flying towards the sun when the last aluminum plate was placed to close off the system. However, a center of gravity already exists between the Earth and the sun. Meanwhile, we're all stuck to the surface of the Earth because the sun, massive though it is, is too far away from us to overpower the gravity of the Earth.

That the inside of a Dyson's sphere should have no gravity is a quirk of mathematics, but it is not a quirk of physics. Matter has mass. Mass exerts gravity. The fact that you are inside of the sphere does not change this. The proton beneath your feet exerts a greater force of gravity than the corresponding proton on the opposite side of the sphere because it's closer. When you find the individual gravitational force vectors for you and every single subatomic particle in the wall of the sphere, and add up all of those force vectors, you will find that there is indeed a gravitational force acting on your body, and that it is acting away from the center of the sphere towards the part of the wall you happen to be standing on.

For the same reason, there will be a vacuum in the center of the sphere. All of the gas particles will gravitate towards the walls of the sphere where the mass is.

Roog
2007-07-21, 11:38 PM
According to the math, you are correct, because math says you calculate gravity from the object's center of mass.
Thats not the math that it used to prove this.


This is where physics and math separate.
I'm sure that the implications of this have escaped you.


The force of gravity that a body of mass exerts is the net value of the sum of the gravitational force exerted by all of its component particles. We are talking about a hollow sphere, however, and not a solid body.
Yes. So you do full calculations. And the gravitational forces of all the component particles of the sphere cancel out at every point within the sphere.

That does not mean (as you seem to think) that there is no gravity inside the Dyson sphere. It simply means that if the sphere is perfect (or an acceptable approximation thereof), then you can ignore the sphere for all gravitational calculations you make regarding it's internal space.


According to your argument, if we were to create a three dimensional elliptical solid by wrapping aluminum around the Earth and the sun to close it off and create a Dyson's football, we would suddenly be in a zero gravity environment.
No. It means that we could ignore the gravitational effects of the aluminum sphere, when calculating gravitational effects on earth.

Dervag
2007-07-22, 12:28 AM
I think its fairly important to remember that the point of a Dyson Sphere is to capture all the radiation from a star, it was never intended as a habitat. So all of our problems with gravity and maintaining an atmosphere stems from those cross purposes.The catch is that this 'Dyson Sphere' is not a giant solar power array around a star. It is some other massive-scale construction around something that, while it may glow like a star, is not a star.


So, if you are dead set on living on the sphere itself (and not on a planet within the sphere) then you just need to find a depth into the sphere shell where the gravity is what you desire. So as you move outward, the gravity increases until you reach the outer surface. At that point, the gravity is probably crushingly strong. The environment would be completely artificial, but at least you don't have to worry about falling off the surface.Huh?

You can have gravity on the outside of the sphere. How much gravity depends entirely on the thickness of the sphere and on its size. A thick sphere will have lots of gravity, a thin one very little.

Unfortunately, if we were to use a central mass (a 'star') equal to that of the Sun, and a sphere with the same radius as the Earth's orbit, we need to make the Dyson sphere have several hundred times the mass of the Sun in order to get acceptable gravity on the outer surface.

With a smaller sphere, we can go for a (much) lower mass. The required mass will decline proportionate to the inverse square; making it half as big across lets us make it one quarter as heavy, one third the diameter gives us one ninth the mass, and so on. Of course, the actual thickness of the sphere will remain constant because the area of the sphere also drops off proportionate to the inverse square, so your shell will still have to be X kilometers thick regardless of how big it is.

The catch is that living on the outside of a Dyson sphere sucks, because there's no light. There's plenty of heat; in fact you may have to install radiator fins on the outside of the shell to make it cool enough to walk on depending on how big it is. But there's no light except starlight (which, in the original poster's universe, doesn't even exist).

Not good.

The inside is a happier place to live because there's sunlight, but gravity is a problem. Spinning something the size of the Earth's orbit for gravity is effectively impossible; it's just too big and you'd need invincible building materials to hold it together on the outside. In a D&D universe a Wall of Force will do, but that's a contrivance. Not good.

Much better is to use a Dyson sphere that spins slowly or not at all (probably with some built-in attitude jets to make sure it doesn't go unstable and fall into the star), and to build a big heavy pancake of matter under the part of the disk that you want to have gravity. With proper shaping of the pancake you can make a large habitable region; you can even build walls around the top of the pancake to hold the air in. You can even put multiple pancakes on the same sphere; in theory there's almost no upper limit.

This allows you to create numerous large habitable 'worlds' on the inside of your Dyson sphere and to build the Dyson sphere itself out of relatively lightweight material.


OK, advance math question (or maybe not)

Lets say that the center object (star/device/whatever) is sufficiently large and sufficiently close that it generates 0.5g acceleration from the inner surface of the Dyson sphere. Standing on the outer surface of the Dyson Sphere you feel 4g's (combined star/dyson sphere).

1. How big is the star?
2. How far to the inner surface of the sphere?
3. How thick would the sphere be?
4. How heavy would the sphere be?
5. How far through the sphere crust would you experience 1 g?Either the star is so huge that it's completely unstable, or the sphere is mindblowingly tiny (so tiny that the star's light would heat up the sphere until it glowed cherry red).

Remember, as has been shown earlier on this thread, the acceleration due to the gravity of the Sun at the position of the Earth's orbit is less than 0.01 g. To get something like 0.5 g you would have to go way inside the orbit of Mercury. Or you would have to use a star dozens of times larger than the Sun, in which case anything as far away from it as the Earth is from the Sun would be just about as hot as Mercury.

Not good.

We assume that the Sun must have massive irresistible gravity because it's big and because it hold the Earth in its orbit. That isn't true. You could bench-press a bull African elephant, possibly even a whale, against gravity with the strength of the Sun's gravity at the position of the Earth. We don't even notice the Sun's gravity compared to the Earth's gravity.

The Sun may be a million times heavier, but it is also almost a million times farther away. And distance is more important to the strength of gravity than mass is.


Another interesting problem of a Dyson sphere is that the gravitational forces on the sphere by the sun cancel each other out in all directions, so there's nothing to stop it the sphere and the sun from drifting in slightly different directions until they collide. Unless you want to outfit the entire dyson sphere with massive correcting thrusters...This problem was first discovered by physics and engineering students looking at Larry Niven's Ringworld concept, giving rise to the following jingle:

"Oh, the Ringworld is unstable, the Ringworld is unstable;
he did the best that he was able and it's good enough for me!"
(sung to the tune of "The Farmer in the Dell)


According to the math, you are correct, because math says you calculate gravity from the object's center of mass. This is where physics and math separate. The force of gravity that a body of mass exerts is the net value of the sum of the gravitational force exerted by all of its component particles. We are talking about a hollow sphere, however, and not a solid body.I'm not sure if you grasp the concept here. I know we are talking about a hollow sphere. The gravity from the distant parts of the sphere is not negligible. You literally can not ignore it if you want your result to be even vaguely accurate.

And when you do include the gravity due to the other side of the sphere, rather than pretending that the only part of the sphere which exists is the part directly beneath your feet, a quick double integral (really, it is quick) will reveal that the gravity of the sphere cancels out at any point inside the sphere. Again, this is not a difficult proof.

I can prove this, as can hundreds of thousands, if not millions, of other people who possess the requisite calculus skills. It is not difficult; in fact the proof is done routinely in many intermediate level courses on physics and even in some freshman level courses.


In the center of the sphere, what you have posted is correct.Yes, but it's also true everywhere else inside the sphere, even if you're standing on its inner surface.


If the sphere is large enough, all of the mass of the sphere is far enough away from you that its gravity is negligible.That is not true. The mass does not simply go away because you get farther away from it. You can't justify calling it negligible without first doing the calculation to show that it is negligible, and I have every faith that when you do the calculation you will see that it is not negligible.


Every subatomic particle that makes up the walls of the sphere has mass, and therefore exerts gravity. The particles on the opposite side of the sphere are at a distance of 2AU (assuming we have a sphere with a radius equal to the distance at which the Earth orbits the sun). At that distance, the sum total of the gravitational force exerted by the total number of particles on the opposite side of the sphere is negligble.Again with the 'negligible'. Don't be so cavalier with 'negligible'. Dismissing a physical quantity as 'negligible' when talking about a physical system is very dangerous. For instance, the engineers who built this (http://www.youtube.com/watch?v=HxTZ446tbzE) bridge were firmly convinced that the effects of wind on the bridge's center span were 'negligible'. They proved horribly wrong; the bridge resonated with the wind blowing down the river and tore itself to pieces.


You can not calculate gravitational force from the mass center of the sphere, because there is no mass there. We don't have to do this for the Earth because the Earth is a solid body.In a word, duh.

I did no such thing. If I had, I would never have asserted that there is no net gravity inside the sphere. I would instead have asserted that you would feel a gravitational force pulling towards the center, just as you would on Earth. And I would be wrong.

I know this is a counterintuitive result. It freaked me out the first time I saw it. But the proof is very straightforward (well, if you consider calculus straightforward; it's as much a learned skill as carpentry). And this is not a case of math producing an 'unreal' result. Instead, it is a case of the laws of physics not living up to your idea of how they should behave.

If you do not believe me and do not feel up to the task of doing the calculation yourself, you are welcome, indeed encouraged, to contact any person with a bachelor's degree in physics, ask them to do the calculation, and to give you the result. They will tell you the same thing. For that matter, even if you do believe me and/or do feel up to doing the calculation yourself, I invite you to do so. Do not take my word for it; get a second opinion from a qualified person. They will give you the same answer.


If we make it three thousand miles thick, now the gravity is no longer negligible. Now if we make the radius of the sphere equal to 5AU, the gravitational force generated by the opposing mass is too far away for us to care.Do you have any idea how much a sphere 5 AU in radius and three thousand miles thick will weigh? We're talking about a mass many times greater than that of the Sun. It will in fact have very considerable and powerful gravity, even at a distance of 5 or 10 AU. You can't pretend that it's negligible just because "ohmigod that's a long way away!"


Experimental data supports your argument, but experimental data has not been taken with a Dyson's sphere that has such a large radius.The laws of physics don't work that way. We know that the inverse square law of gravity is correct. This has in fact been tested by observing objects the size of a Dyson's sphere (such as our own solar system, and for that matter entire galaxies, things transcendantly larger than any Dyson's sphere ever imagined). It is a trivial application of the law of gravity to prove that the net gravitational force inside a hollow spherical shell due to the shell is zero. You can check this in the textbooks. You can do the calculus yourself, and if you do it right you will get the same answer that a chain of people running from Sir Isaac Newton to the present day got: the net gravitational force inside a hollow spherical shell due to the shell is zero.


According to your argument, if we were to create a three dimensional elliptical solid by wrapping aluminum around the Earth and the sun to close it off and create a Dyson's football, we would suddenly be in a zero gravity environment. Actually, according to your hypothesis, since the majority of the mass of the system is at the sun, we would all go flying towards the sun when the last aluminum plate was placed to close off the system.OK, I call bull. This indicates a total misunderstanding of my statement, apparently due to the inability to distinguish between the gravity due to the shell and the gravity due to objects inside the shell.

If we built a Dyson sphere around the solar system outside the Earth's orbit, it would not make the slightest difference to anything inside the shell. The shell would exert no net force on anything inside it; the Earth would continue in its orbit just the same as if the shell did not exist. The gravity between the Earth and the sun would not be changed in the slightest. We would not "all go flying towards the sun" any more than we are flying towards the sun right now. We would continue to spin around it in big circles.


That the inside of a Dyson's sphere should have no gravity is a quirk of mathematics, but it is not a quirk of physics.It is only a quirk if you choose to call it a quirk. It is a direct and straightforward application of the law of gravity derived by Newton and used successfully ever since the mid-1600s. There is no magical line at which physics ceases to follow mathematical laws that you can use to say that the math describing a Dyson sphere is somehow wrong.


Matter has mass. Mass exerts gravity. The fact that you are inside of the sphere does not change this. The proton beneath your feet exerts a greater force of gravity than the corresponding proton on the opposite side of the sphere because it's closer.But there are a hell of a lot more protons on the opposite side of the sphere than there are directly beneath your feet.

There's more to physics than reciting a bunch of clichés and putting together a syllogism from them.


When you find the individual gravitational force vectors for you and every single subatomic particle in the wall of the sphere, and add up all of those force vectors, you will find that there is indeed a gravitational force acting on your body, and that it is acting away from the center of the sphere towards the part of the wall you happen to be standing on.You will find no such gravitational force, and the fact that you claim this indicates that you have never actually added up the force vectors.

I have, using the techniques of calculus, as have dozens of people I know personally and hundreds of thousands of people around the world over a period of several centuries. And once we checked our math we all got the same answer. Again, you are welcome, and in fact invited, to check this with other people who can do the math, and with the physics textbooks. Don't take my word for it, but you will get the same answer from them that you do from me.

I am not making this up, bizarre as it may sound.

Demented
2007-07-22, 01:47 AM
In case you need a link:
http://en.wikipedia.org/wiki/Shell_theorem

If only somebody could find something detailing other shapes...

Citizen Joe
2007-07-22, 08:22 AM
OK, using the shell theory and adjusting for these assumptions:
1) Central mass is a 1 solar mass white dwarf
2) Inner surface of shell is positioned so that the acceleration towards the dwarf is 5 m/s/s (about 0.5 g)
3) Gravity at the outside of the shell is 1.5 g
4) Average shell density is about twice that of water to account for strong materials and open spaces.

The results are:
Inner sphere radius: 0.03434 AU = 5.1512 Million km
Shell thickness: 18,000 km
Habitable zone 9-11 m/s/s gravity: 8100 km to 9900 km from inner surface.
Surface area at inner sphere: 111 trillion sq. km.
Except for the need of an artificial light source, this world could be very similar to Earth. The day night cycle could even be simulated with periodic heat purges appearing as bright fountains of light.

Lorthain
2007-07-23, 02:23 AM
Something cool that I didn't noticed mentioned is that, for a spinning sphere, if a person moves against the spin at a matching speed they will loose gravity (and gain extra gravity if the move with the spin). A mean way to kill somebody is to hurl them out of a catapult against the spin (making sure to match speed, plus a little extra to get through the atmosphere) and letting them float up into the sun.



The laws of physics don't work that way. We know that the inverse square law of gravity is correct. This has in fact been tested by observing objects the size of a Dyson's sphere (such as our own solar system, and for that matter entire galaxies, things transcendantly larger than any Dyson's sphere ever imagined).


It is a direct and straightforward application of the law of gravity derived by Newton and used successfully ever since the mid-1600s. There is no magical line at which physics ceases to follow mathematical laws that you can use to say that the math describing a Dyson sphere is somehow wrong.



I have, using the techniques of calculus, as have dozens of people I know personally and hundreds of thousands of people around the world over a period of several centuries. And once we checked our math we all got the same answer.


I don't mean to defend all of the odd stuff Corolinth said, but he does vaguely hit on one good point; math and physics are wrong (in a sense) in the way they treat hollow spheres. The calculus applied by basic physics textbooks to spheres fundamentally assumes a uniform mass distribution; by contrast real objects are made up of individual particles/waves (atoms and their constituents). Of course, the textbooks purposefully apply the 'wrong' math to obtain a cleaner result. This is ok as long as it is made clear that the textbooks are using a less than accurate model and that a more detailed approach could potentially give different results.

I worked through a quick particle-by-particle scenario just to see what sort of results it would give. If the sphere is made up of six particles (a rather extreme, but noteworthy case), there will be drastic gravitational fluctuations internally. Bumping it up to eight particles also has gravitational fluctuations, though not quite as drastic. Being lazy, I stopped there. As noted, the infinite particles scenario yields no gravitational fields. I figure a real Dyson Sphere exists somewhere in between the extremes, with any internal gravitational fields (due to sphere itself, of course) being 'negligible' on a human scale. Who knows until someone builds it (though I'd also settle for a better model).

The gravity quotes are included above because they lead to something interesting: Newton's Law of Gravitation gives terrible results on a galactic/universal scale. A fudge factor (dark matter) was introduced as a fix, but Newton's version of gravitation is certainly in doubt and has a number of people trying to modify/replace it.

Roog
2007-07-23, 02:30 AM
The gravity quotes are included above because they lead to something interesting: Newton's Law of Gravitation gives terrible results on a galactic/universal scale. A fudge factor (dark matter) was introduced as a fix, but Newton's version of gravitation is certainly in doubt and has a number of people trying to modify/replace it.

Yes, I hear that some people are suggesting a theory known colloquially as Relativity.:smallwink:

Dervag
2007-07-23, 03:25 AM
Something cool that I didn't noticed mentioned is that, for a spinning sphere, if a person moves against the spin at a matching speed they will loose gravity (and gain extra gravity if the move with the spin).Well, if the sphere is a megascale object (large on the scale of a solar system), then the speed you'd have to reach to lose a significant amount of your spin-generated artificial gravity is mindboggling because the sphere is spinning mindbogglingly fast.


A mean way to kill somebody is to hurl them out of a catapult against the spin (making sure to match speed, plus a little extra to get through the atmosphere) and letting them float up into the sun.It would be much easier to fling them into space using the same catapult on an Earth-like planet; the speed of the spinning sphere is much higher than the Earth's escape velocity.


I don't mean to defend all of the odd stuff Corolinth said, but he does vaguely hit on one good point; math and physics are wrong (in a sense) in the way they treat hollow spheres. The calculus applied by basic physics textbooks to spheres fundamentally assumes a uniform mass distribution;Well, when the object are more than a hundred million kilometers across and the individual wavicles making up the sphere are less than a nanometer across (which they are), then the approximation is justified.

The math is correct. The physics that underlies the math makes a simplifying assumption that is true on the scale of the sphere. Now, if you look at the sphere with an electron microscope you will find that the sphere is in fact made up of separate particle/waves. But that doesn't matter. The graininess of the sphere's structure is so fine that it might as well not exist for purposes of calculating the gravitational force inside the sphere, and the fact that the sphere is made of the same material everywhere and of the same thickness everywhere means that, to any sane level of approximation (say, one part in a trillion trillion), the gravity does cancel out.


by contrast real objects are made up of individual particles/waves (atoms and their constituents). Of course, the textbooks purposefully apply the 'wrong' math to obtain a cleaner result. This is ok as long as it is made clear that the textbooks are using a less than accurate model and that a more detailed approach could potentially give different results.There is no more detailed approach. You can not treat the sphere as being a collection of particles, because the computation can't be done by any computer that now exists in any reasonable time frame for the actual number of particles that would make up a Dyson sphere. This can be proven by doing a quick estimate of the sphere's thickness and diameter, which gives its volume, and then assuming that it has some reasonable density. Even if it's as non-dense as a fog bank, it will still contain trillions of trillions of trillions of particles. No computer on Earth can do enough calculations to find the effect of every particle in any sane human time scale. And if a computer could, then it would find that the gravitational effects were so close to zero that the difference from zero could never be detected, even in theory. Other things would swamp any possibility of detecting the effect.

This is why calculus was invented. It gives 'approximations' that are so accurate that instruments built by mortal hands can't tell them from the perfectly true reality. The localized gravitational fields created by the fact that the atoms in the sphere are not perfectly evenly distributed don't even affect the other atoms.


I worked through a quick particle-by-particle scenario just to see what sort of results it would give. If the sphere is made up of six particles (a rather extreme, but noteworthy case), there will be drastic gravitational fluctuations internally. Bumping it up to eight particles also has gravitational fluctuations, though not quite as drastic. Being lazy, I stopped there. As noted, the infinite particles scenario yields no gravitational fields. I figure a real Dyson Sphere exists somewhere in between the extremes, with any internal gravitational fields (due to sphere itself, of course) being 'negligible' on a human scale. Who knows until someone builds it (though I'd also settle for a better model).The actual Dyson sphere exists so close to the infinite particle extreme that there is no detectable difference; dwelling on the undetectable difference is not useful.


The gravity quotes are included above because they lead to something interesting: Newton's Law of Gravitation gives terrible results on a galactic/universal scale. A fudge factor (dark matter) was introduced as a fix, but Newton's version of gravitation is certainly in doubt and has a number of people trying to modify/replace it.Well, in and of itself there is no a priori reason why all the matter in the universe should be readily visible to us.

If the only matter that exists is matter we can see, then Newton's Law of Gravitation would break down on the intergalactic scale. But since Newton's Law of Gravitation shows every sign of working on the stellar and interstellar scales, it makes more sense that we're seeing the effect of masses we can't percieve directly than that the law should be chucked and replaced by something else. Especially since the 'something else' would have to act just like Newton's Law in all the places where we see Newton's Law working with perfect accuracy.

By analogy, the planets Uranus and Neptune were first discovered not by people looking in telescopes but by astronomers who calculated that the outer planets weren't following the orbits Newton's Law predicted for them. They concluded that there must be an unknown mass disturbing the orbits of those planets, distorting them from what Newton's Law would predict due to the known masses in the solar system alone.

Now, this was as much of a 'fudge factor' as dark matter, no more and no less. But that fudge proved correct and the extra, unknown planets were found. That doesn't prove that dark matter will be found, but it's a good reason to give a well tested, extensively verified, and highly reliable theory the benefit of the doubt until you know the pieces aren't fitting together any more.

Bender
2007-07-23, 03:39 AM
I want to second everything Dervag said. I'm one of those millions of people that can easily prove it, but I don't have to, because the link to wikipedia is already given.

saying a dyson sphere is in between 6 particles and infinity is somewhat odd. Earth has about 10^50 particles, a dyson sphere would have even more. In most mundane calculations 1000 is already a good approximation of infinity. I don't know a lot of calculations that uses numbers larger than 10^10, and don't make some kind of abstraction of it. There is a reason they use "mol" in chemistry and not the actual number of particles. Another example is pressure, you don't calculate the force every single particle in a gas exerts, even though the pressure distribution of a container with only 6 particles would be very heterogeneous.

What I want to say is: there is hardly any difference between considering every single particle in a sphere or assuming the sphere is perfectly homogeneous.

There might be a difference if there are big fluctuations in the kind or density of particles in the sphere, for example large lead veins, or big "mountains". They still would have to be few and heterogeneously spread over the surface. It's like the idea of the pancakes, if I understood that correctly.

EDIT: and what Dervag said just now, while I was typing this, darn you :smalltongue:

Golthur
2007-07-23, 09:44 AM
Yes, I hear that some people are suggesting a theory known colloquially as Relativity.:smallwink:

This made me laugh. Thank you. :biggrin:

Yes, I also am one of the people who could prove "no gravity inside a hollow sphere", but, as Bender says, it's really not necessary; Wikipedia's got it covered.

As Dervag says, you can't calculate the gravitational effects of a Dyson sphere by enumerating individual particles - it just wouldn't work in a pragmatic sense.

With 2 particles, you've only got one interaction to consider; with 3, you've got 3; with 4, you've got 6; with 5, you've got 10; with 20, you've got 190; 100, you've got 4950; with 1000, 499500; and it just gets uncountably huge once you start entering into the realms of, you know, visible chunks of matter. No computer today could run that simulation in anything close to a human lifespan.

Bender
2007-07-23, 10:41 AM
Implications of relativity:

If you rotate a sphere with 1 AU radius fast enough for 1g gravity, it's diameter would shrink a little bit, for every km it would be about 5 mm shorter. This means if you calculate Pi on it, the result would be 3.14157520 (instead of 3.14159265). Of course, this would not be the case for every circle you draw there, but Pi would be direction dependant.
Every day would last about 0.5 s longer.
Your mass would be slightly more, but most scales wouldn't be able to measure that.
I have to admit I don't know the exact implications on the gravity issue, but I think the shell-prove would still be valid.

Of course, the mass of the dyson sphere would slow down time as well, but much less (time for gps satellites goes only 45 µs/day faster than on earth due to this effect)

In an earth-sized sphere, the effects are much smaller, in fact negligible for most mundane purpuses

Dervag
2007-07-23, 12:34 PM
EDIT: and what Dervag said just now, while I was typing this, darn you :smalltongue:Consider me darned; the hole in my toe is gone now. Thank you.

Lorthain
2007-07-23, 01:58 PM
I am sorry if I have offended anyone. What I meant to accomplish by pointing out the difference between distributed mass and real particles was to defend Corolinth, who was getting ragged on pretty bad. A lot of his ideas were wrong, sure, but his idea that the math and physics were potentially being misapplied was not entirely without merit and may have originated from valid concerns.

Of course I am aware calculus is applied in this situation for practicality reasons, and I think I mentioned that in my post. The 'better model' I suggested would not involve modeling every particle of a Dyson Sphere, but simply enough particles to show the internal gravitational forces becoming negligible on a human scale. (By the way, the calculations for these particles would scale up on a 1:1 basis, since the only concern is the gravitational influence of a given particle on an internal point.) I would do the model for myself if I could think of an easy way to evenly distribute the particles around the sphere.


The graininess of the sphere's structure is so fine that it might as well not exist for purposes of calculating the gravitational force inside the sphere, and the fact that the sphere is made of the same material everywhere and of the same thickness everywhere means that, to any sane level of approximation (say, one part in a trillion trillion), the gravity does cancel out.

I provide this quote because it leads into what specifically bothers me: using calculus implies that the differences between particles and distributed mass are 'negligible.' Corolinth figured that the gravity from the opposite side of the sphere is 'negligible.' There was no argument provided for why Corolinth's assumption is wrong and calculus' assumption is right, he was just told to do the calculus for himself. So, I didn't mean to offend anybody, but I felt the double standard was undeserved.



Well, if the sphere is a megascale object (large on the scale of a solar system), then the speed you'd have to reach to lose a significant amount of your spin-generated artificial gravity is mindboggling because the sphere is spinning mindbogglingly fast.

You are correct, but I thought I read that the sun in this situation would be much smaller than our real sun, implying the Dyson Sphere would be much, much less than 1 AU in radius and the spin speed (at the internal surface) could be small enough to kill people the way I suggested. I meant it as an interesting possibility.



By analogy, the planets Uranus and Neptune were first discovered not by people looking in telescopes but by astronomers who calculated that the outer planets weren't following the orbits Newton's Law predicted for them. They concluded that there must be an unknown mass disturbing the orbits of those planets, distorting them from what Newton's Law would predict due to the known masses in the solar system alone.

Now, this was as much of a 'fudge factor' as dark matter, no more and no less. But that fudge proved correct and the extra, unknown planets were found. That doesn't prove that dark matter will be found, but it's a good reason to give a well tested, extensively verified, and highly reliable theory the benefit of the doubt until you know the pieces aren't fitting together any more.

You bring up an interesting analogy; the obvious counter-analogy would be Newtonian Mechanics (NM) and relativity. As I'm sure you're aware, NM worked great for a long time until people started making observations near the (relativistic) speed of light and NM ceased giving accurate answers. The equations ended up being rewritten, though this didn't happen without a good deal of debate. Newton's gravitation could repeat the past, appearing correct over small (solar system) distances, failing at great (inter-galactic) distances, and being rewritten after some debate.

For now, the model (Newton's gravitation) doesn't fit the (inter-galactic) observations, implying either the model or the observations are wrong. Neither assumption is necessarily more correct than the other, though changing the model has more opposition due to it already being written in all the textbooks (among other reasons). I am looking forward to seeing how it all turns out.

Citizen Joe
2007-07-23, 09:49 PM
You could make the Dyson sphere for its collecting abilities and then hang a bunch of rotating tube environments from the inner surface. That way you don't have to spin the whole sphere and you can contain the atmosphere easier. Additionally, each tube can be a different environment and gravity.

sikyon
2007-07-23, 11:38 PM
The thing in the middle doesn't have to be a star, right?

If not, let's just make it a mass of exotic matter with negative gravitational mass (possible but not yet discovered). Thus, in this sphere, gravity is a push, not a pull!

Also, for calculus is pretty exact. Finite analysis of particles indicates that atoms are described as waves. In fact, these waves are solved using calculus. ACTUALLY it is impossible (at least currently) to find exact solutions using calculus to these wave functions. They are solved numerically by approximation. At very high compuational power. So asking to calculate precise gravit with regards to "granularity" is impossibvle

Bender
2007-07-24, 01:22 AM
I am sorry if I have offended anyone. What I meant to accomplish by pointing out the difference between distributed mass and real particles was to defend Corolinth, who was getting ragged on pretty bad. A lot of his ideas were wrong, sure, but his idea that the math and physics were potentially being misapplied was not entirely without merit and may have originated from valid concerns.

I too hope I didn't offend anyone.
As a mechanical engineer I know the concerns of theory versus application (engineering = applied science). I'm not hoping to discourage Corolinth, on the contrary. There are a lot of misconceptions about physics, even simple things like gravity. People tend to believe their own intuition more than mathematics (why else would anyone spend a lot of money on lottery tickets). In most conflicts between human intuition and physical theory, theory is right and human intuition is wrong.
For every physical calculation you make you are working with an idealised model, and it is very important to know the limitations of a certain model and when it has to be improved. A very popular idealisation is consider matter as homogeneous. Because the guarguantuan amount of particles involved, this is a very good and very convenient model. The bigger your objects, the better the approximation in fact.


I provide this quote because it leads into what specifically bothers me: using calculus implies that the differences between particles and distributed mass are 'negligible.' Corolinth figured that the gravity from the opposite side of the sphere is 'negligible.' There was no argument provided for why Corolinth's assumption is wrong and calculus' assumption is right, he was just told to do the calculus for himself. So, I didn't mean to offend anybody, but I felt the double standard was undeserved.

I see how this can be confusing. The link provided (http://en.wikipedia.org/wiki/Shell_theorem) shows why the mass at the opposite side is not negligible.
Proving that the error of a model with distributed mass is negligible would require a lot of calculations (not difficult ones, but an awful lot). Dervag and I both provided solid arguments why it would be though.


You are correct, but I thought I read that the sun in this situation would be much smaller than our real sun, implying the Dyson Sphere would be much, much less than 1 AU in radius and the spin speed (at the internal surface) could be small enough to kill people the way I suggested. I meant it as an interesting possibility.

With a radius the size of earth it might work, and is an interesting possibility. I think the speed of 8000 m/s would still be a bit high, but you can make a reasonable argument.


You bring up an interesting analogy; the obvious counter-analogy would be Newtonian Mechanics (NM) and relativity. As I'm sure you're aware, NM worked great for a long time until people started making observations near the (relativistic) speed of light and NM ceased giving accurate answers. The equations ended up being rewritten, though this didn't happen without a good deal of debate. Newton's gravitation could repeat the past, appearing correct over small (solar system) distances, failing at great (inter-galactic) distances, and being rewritten after some debate.

For now, the model (Newton's gravitation) doesn't fit the (inter-galactic) observations, implying either the model or the observations are wrong. Neither assumption is necessarily more correct than the other, though changing the model has more opposition due to it already being written in all the textbooks (among other reasons). I am looking forward to seeing how it all turns out.

Oh, but Newton's mechanics doesn't even explain the movements in the solar system perfectly. It is just a very good approximation, because at mundane speeds, sizes and masses, the formulas for relativity simplify to Newton's mechanics. As I said, I don't know the exact implications of general relativity concerning gravity, but I don't think it would interfer in this discussion. Any possible error caused by it, no human would be able to detect with his own senses, and it certainly can't creaty enough gravity inside the dyson sphere.

Again, I don't wand to offend anyone by providing counterarguments. I just like to explain physics and do something about misconceptions.