Dyson Spheres revisited

[aramis]

Justin:

the far bits would have FAR less pull than the near bits. Inverse Square Law and all. But the near bits, on even an 0.1 AU diameter ring would be, for all local intents, nearly a plane. But you are right that even a DU layer a meter thick would be negligible gravity. (It would be *just* enough, on such a large structure, to pull down some dust, and mostly to the center of the band.)

 

[rancke]

the far bits would have FAR less pull than the near bits. Inverse Square Law and all. But the near bits, on even an 0.1 AU diameter ring would be, for all local intents, nearly a plane. But you are right that even a DU layer a meter thick would be negligible gravity. (It would be *just* enough, on such a large structure, to pull down some dust, and mostly to the center of the band.)

You forget that when handwavium is heated, it produces gravitons.


Hans

 

[HG_B]

Without artificial grav the inside of a Dyson sphere would be 0 G no matter how thick.

http://up-ship.com/blog/?p=1774

 

[JustinInOz]

HG_B is right

I was thinking about it on the way into work.

HG_B has hit the nail on the head. Nice diagram by the way.

I remember doing the calculations in 1st year physics at uni. The thought experiment was: what would happen if a hole was bored directly through the earth and you jumped in it.

The answer is, assuming the hole did not collapse from the liquid core and that you did not burn up from the heat etc, is that you would bob up and down like a pendulum. You experience only accelleration from the mass of the earth inside the the radius that you were currently at. When you are at the centre of the earth you are experiencing no accelleration due to gravity from the earths mass. You are however moving at peak velocity at this point. You then experience increasing accelleration until you come to a stop at the point where you jumped but at the opposite end of the earth. The journey then begins again but starting from the opposite side.

Aramis, you are right too that the inverse square realtion causes the "far bits" to have much less influence per mass. There are equally more of them however. So the whole thing does cancell out. You experience no g's from the mass of the sphere when you are "inside" it. You would experience very little g's when outside of it too. This is because when outside you are very distant from its centre of gravity. It would be only a small fraction of the mass of the star. Its centre of mass would be exactly the star. The only significant g's you would be experienceing would be that of the mass of the star plus the small change of the dyson sphere. Even using the heavy duty Dyson-Niven sphere it would be less than 1/3 of 1 % of the mass of the Sun.

This would only leave artificial gravity or spin to give you a reason to stick to the inside of the sphere.

 

[aramis]

Without artificial grav the inside of a Dyson sphere would be 0 G no matter how thick.

http://up-ship.com/blog/?p=1774

http://up-ship.com/blog/wp-content/uploads/2009/01/dyson11.jpg

Doesn't actually show the math, nor prove, anything, HGB, other than he's confounded the calculations for a tunnel through a solid object with the calculations for a thin object.

Further, it's well accepted that, given a flat plane, one winds up with a uniform gravity near surface, dropping off and angling further only near the edges.

Given a system of units, each with a pull of 1 at distance 1, and a radius of 1000 from the center of the system, when I run it out, I get a net gravity for an object at 0,999, a pull of 0.9999874963.

Rerunning it at 1° increments, I get +1.0000873153 net ∆V towards the rim...

At 0.25°, 1.02535784309999∆V towards the rim. (Math is being done at 10 place precision. No vector is 0 save those at cardinal.)

At 0.05°, 2.35703288309999.

The more tightly I run it, the more the local gravity will be (because I'm not dividing total mass out. The thing is, that gravity will fall off quickly as I leave the ring, and will be higher on the outside.

So a quick run of the math, and I've established for myself that whomever you quoted never actually DID the calculations for a RING rather than a solid sphere.

It's a useful approximation for solid object... but not for a ring where the net effect is well past the drop-off.

Just for fun, I'm plugging in different distances in 1 unit every 0.05°:
At 0,999: Ay=2.35550448449998
at 0,998: Ay=1.1183615257
At 0,995: Ay=0.452423306699999
At 0,1001, Ay=-2.37303040020001
At 0,950, Ay=0.0420742615999998

 

[HG_B]

Doesn't actually show the math, nor prove, anything, HGB,

Gauss theorem - vector calc

 

[aramis]

Gauss theorem - vector calc

So, you're arguing that connecting the objects means their grvitation suddenly and magically transforms the local gravitational effect to a null?

Absurd.

A solid massive object's local effect can be approximated accurately by subdividing the object into subunits, and measuring the effects of the subunits on the test spot.

It's orbit and effect on objects significantly away from itself will be essentially point source, but it still will warp spacetime locally based upon local mass, not uniform point source.

And NASA uses that effect in measuring density by observing orbital deviations of satellites.

 

[HG_B]

So, you're arguing

I'm not arguing at all. Have fun with the calcs.

 

[JustinInOz]

For those who are interested here is the maths:
http://www.physicsforums.com/showthread.php?t=121120
http://www.merlyn.demon.co.uk/gravity1.htm#FiSSh

Aramis, as to your recourse to the assumption physics of a plane, there is a slight problem with your reasoning. The plane assumption works as long as it is practically a plane. No distance inverse relationship kicks in until the plane is far enough away for it to start to resemble a point rather than a plane. So as long as there are insignificant "edge effects", it still "looks like a plane, quacks like a plane".

That being the case, you are ignoring the equally large plane "above". If it is a plane in one direction, it is a plane in the other. We are then back to where we started, zero g. But this line of argument is ignoring the integration of forces due to gravity of a hollow shell in all 3 dimensions mentioned in the first e-mail I mentioned above.

My apologies for distracting people the example of the thought experiment from uni. I meant to illustrate the matter not muddy the waters.

Either way we are arguing about the native g forces being experienced inside a Dyson-Niven Sphere as being either exactly zero or practically zero. Not much point in arguing about it.

 

[aramis]

Either way we are arguing about the native g forces being experienced inside a Dyson-Niven Sphere as being either exactly zero or practically zero. Not much point in arguing about it.

No, sorry, but wrong. I suspect were talking cross purposes.

I'm not treating the ring as a plane; I'm saying that, to the person standing upon it, it's close enough to a plane for their local reference, at least when not near the mountains at edge.

make the ring thick enough and dense enough, and local effect of gravity is sufficient to hold things down without spin grav. And the far side won't be strong enough at the distances involved to make much of a difference. the trick is finding enough mass, and making it spin at orbital speed, so that it doesn't fall in upon itself. It's the only viable solution to the structural issue... Say, SG 18 (DU), 20,000km wide, 2000km thick along the full circumference of the ring, and then build your scrith walls up from it's edges... and you get something that holds an atmosphere on both sides.... and it's going to have a surface gravity close to earth's. ( 1/2 the thickness, 3x the density, essentially planar to those upon it.)

 

[HG_B]

Not much point in arguing about it.

Right, it's been proven for longer than I've been alive.

But, the subject of a "Dyson sphere" is great fodder for a sc-fi game.

 

[JustinInOz]

No, sorry, but wrong.

Seriously Aramis, have a look at the maths, set up the integral and do it.

Don't listen to my arguments. Read the reference from the physics forums.

Inside a hollow sphere, the mass of the sphere cancells itself out.

I am not trying to jerk your chain here. It is simply a question of geometry and phisics. The answer is an unequivocal zero.

I have got no more to say on the matter.

 

[Blue Ghost]

Right, it's been proven for longer than I've been alive.

But, the subject of a "Dyson sphere" is great fodder for a sc-fi game.

Which was kind of the purpose of my thread. When I think of one of these babies, I think of the classic "ST;TNG" Relic episode, which may or may not have been inspired by the 1970's Trek novel "The Starless World" which I read in middle school. The cover art shows our beloved original series Constitution class starcruiser flying into a circular opening of one. We glimpse a view of green pastures, lakes, hills, clouds and other terrain hugging the spherical surface inside.

How does the engineering work? It's never explained. But then again that story told of an intelligent star, discovered by Spock mind melding with the thing. How real is that? Not very. Okay, psionics or para-neuro-"science" isn't the same as structural engineering nor materials' science, but I think the point is that unless the actual structure and physical make up are endemic to the adventure, then it's a non-issue. The thing works, and there are billions... nay... trillions of animals and sentient species to discover, and perhaps an equal number of places within said sphere to explore.

Grav plates are a fact of the OTU. Ergo someone more advanced who could build one of these behemoth condos can probably whip up a more advanced form of the science and give it an engineering application. Meaning again, the thing works without having to introduce extraneous Egyptian pyramid analogy technology for building something 4000 years later.

My only worry with such a structure has been heat buildup. The Earth rotates, and turning a fresh side to the sun is major factor in keeping the planet habitable. But, here again the Dyson Sphere technology is probably sufficiently advanced enough to deal with heat build up and sun exposure.

Having said all that, it is interesting to note what kind of explorative and adventuring opportunities there might be in a Dyson sphere that;s built strictly for collecting solar energy or whatever else. Maybe the thing was habitable at one time, but is now an empty burnt out shell, like that one in GT.

Just some more musings here.

 

[HG_B]

Which was kind of the purpose of my thread. When I think of one of these babies, ...

Since the sheer size is overwhelming, as a GM I'd probably have only a small portion semi-inhabitable and the whole thing ready to fail. That way the pc's need to get the adventure done and not have the structure unduly influence the game universe later. Another way is a worm hole to the place (created as a byproduct of the energy & mass that disappears after the structure fails.

 

[Blue Ghost]

I think that's reasonable and probably true of most people who've contemplated the Dytson sphere option. There may be more that can be done, but I'm too tired now write anymore