Dyson Sphere

[CaptainChewbacca]

Ah, ok, thanks.

Question though, wouldn't the sheer size of this sphere indicate it used more material than your average solar system contained? Granted I have no numbers to support that, but just on the surface it doesn't seem to equal out.

Not at all. A Dyson Sphere essentially reweaves the combined material of a planetary system into a shell surrounding a star. It also has to be remembered that this shell will, by definition, be hollow. So while the structure's radius might extend out to 1AU, the Sphere itself would be only a few kilometres thick if that.

That's still alot of material. That sphere would have a radius of 93 million miles, giving a surface area of 36,228,846,481,197,496 square miles, which is 5,7966,154,369,916,000 square kilommeters (If I carried my significant figures properly). Even if its only a kilometer thick, that's still alot of material, moreso than I think would be present in a system, unless that system had multiple brown dwarfs.

[CaptainChewbacca]

Edit: I got my formula wrong, its 173,898,463,109,748,000 square kilometers.

[Wyrm]

The surface area of a sphere is A = 4πr². Because a Dyson Sphere is at most a few kilometers thick, the thickness of the shell is insignificant compared to the radius, and therefore, the real volume of the shell is about V ≅ Ad, where d is the thickness of the shell. Let's assume that the shell is at least as thick as the Enterprise itself, 642.51 m. Then, r = 1 AU = 1.49598e11 m implies V ≅ 1.80693e+26 m³. At a density of 3000 kg/m³ (3 g/cc), mass is 5.42079e+29 kg. In comparison, the sun is 1.988435e30 kg, so this is just over 1/4 solar mass.

[Lost Soal]

This is the external view of the hatch as the Enterprise is being pulled in.


Internal with them still some distance away.


And closer.


And finally, with them almost through the door


If someone can figure out scaling from these shots you get get a good estimate as to its thickness.

[vivftp]

Ok, this will be a pretty rough scaling due to the small size of everything, but it should at least give us an approxamate figure for the spheres thickness.

As shown in the pictures Lost Soal posted, the tractor beams which brought the Enterprise into the sphere originated just inside the outer hatch, so I used this picture as my starting point:



I then figured that since we know where the tractor beams start, we could trace them back to the hatch, and this is what I wound up with:

http://img82.imageshack.us/img82/2087/d ... ingvc0.png

Let me describe what everything here is:

The green lines are the tractor beams so we can trace them back

The purple lines are a continuation of the end of the hatch so we can track it back to match up with the origin point of the tractor beams.

The tiny blue line on the saucer section is just for the thickness of it (8 pixels) to use as my reference point.


So tracking everything back, I wound up with the top tractor beam having an origin point further away than the bottom tractor. So I'll just scale both and we'll see what we wind up with.

My reference point is the thickness of the Enterprises saucer section (it's a rough figure since the ship is so tiny in the picture). According to my copy of The Art of Star Trek, the saucer section is 205 feet thick, but let's just use 200 since it's a rough scaling to begin with. At 8 pixels that means each pixel is 25 feet.

For the top tractor beam, it goes back 220 pixels, or 5,500 feet. For the bottom, it goes back 161 pixels, or 4025 feet.

That works out to 1676.4 meters thick using the top tractor and 1226.82 meters thick using the bottom tractor.

I dunno how well it jives with the other views Lost Soal posted though.

[vivftp]

Ok, just because I think I did something wrong with the scaling there, I decided to do one taking perspective into account (ie. looking down the "tunnel" towards the hatch), and came up with slightly different figures:

http://img201.imageshack.us/img201/169/ ... ng1bs0.png

In this pic, I used the yellow lines to guage the proper perspective, based on the things on the walls of this tunnel, and from that I was able to work out the angle leading from the end of the hatch back to the hatch (which is the purple lines).

From this, the top tractor beam works out to be 3216.25 feet, and the bottom one works out to be 2959.25 feet. Or 980.313 and 901.9794 meters respectively.

[RedImperator]

It is also reasonable to surmise that under normal circumstances the Dyson Sphere is balanced with respect to the gravitational and tidal forces of the primary to keep it stable with respect to the star. A Death Star blast, on the other hand, would deliver force which was not accounted for in the design calculations and could seriously disrupt its structure and orbital stability.

The Sphere would have to have active stabilization systems, because a solid shell of matter would have zero net gravitational interaction with the star (or any other body inside the sphere, meaning those cities and clouds and oceans we saw had to be held down by some kind of artificial gravity system). There's no way to simply balance a solid Dyson shell so it won't eventually drift into its star.

All that said, it's unlikely in the extreme that the designers anticipated a 1E38j blast concentrated on one point (there's no natural phenomonon, either in real life or the vast pantheon of bizarre Trek space anomalies, capable of doing that), which might provide enough KE to knock the Dyson Sphere so hard out of position the stationkeeping thrusters can't compensate.

But I'd still lay my money on the Death Star punching a huge hole in it which leads to the whole structure coming apart. A Dyson Shell's own mass puts enormous stress on all parts of its structured. Put a big enough hole in it, and the load will become unbalance and cause more failures nearby, that would propagate through the shell until it had shattered into pieces small enough to hold together on their own. It would be a pretty spectactular sight, actually.

We do know that Scotty's retirement shuttle, the Jenolan, left a nasy gouge on the surface when it crashed.

<snip>

So I don't think the Death Star will have too much difficulty in destroying it.

While I don't disagree with your ultimate assessment, that has to be some softer surface layer of material that got damaged, not the Sphere's primary structure. If the Dyson Sphere's structure was that soft, its own mass would cause it to tear itself apart without any intervention on the part of the Death Star. Whatever the Dyson Sphere is made of, by physical necessity it has to be the strongest stuff ever seen in Star Trek, maybe in all media sci-fi. The problem for it in this scenario is that the Death Star is the most powerful weapon ever seen in media sci-fi (going by energy output).

[Mad]

All that said, it's unlikely in the extreme that the designers anticipated a 1E38j blast concentrated on one point (there's no natural phenomonon, either in real life or the vast pantheon of bizarre Trek space anomalies, capable of doing that), which might provide enough KE to knock the Dyson Sphere so hard out of position the stationkeeping thrusters can't compensate.

The momentum from a 1E38 J blast (assuming the Death Star uses massless particles as mentioned in the AotC:ICS) would be equivalent to Earth hitting the Dyson Sphere at about 56 km/s. (Earth orbits the Sun at about 30 km/s.)

Considering the size and age (the star was dying, after all; it wouldn't have been dying when the Sphere was constructed) of the Dyson Sphere, random debris flying through space would eventually get caught into its orbit or crash into it, depending on its angle. A rogue planet (perhaps quite a bit bigger than earth) could be a possibility over time.

If the Dyson Sphere was designed to withstand the impact of such debris, then it's very likely that it can handle the momentum.

If the surface can't survive such an impact, then the momentum issue is rather moot.

Coming back to the earth-sized object at 56 km/s, the object would have a KE of about 1E34 J. A bigger object's impact could rival that of the superlaser.

Still, the Death Star's superlaser should be able to blow out chunks of the structure. The Sphere likely would not have been designed to withstand thermal energy of that magnitude even if it could survive kinetic impacts on the same scale.

[Darth Wong]

You do realize that even if you disregard heating entirely, the localized mechanical stress produced by the superlaser impact would be orders of magnitude greater than that produced by a rogue planet crashing into the sphere, right?

[Mad]

You do realize that even if you disregard heating entirely, the localized mechanical stress produced by the superlaser impact would be orders of magnitude greater than that produced by a rogue planet crashing into the sphere, right?

Yeah. Not that it matters much, as I'm just pointing out that the arguments on the momentum of the blast knocking the Sphere around as a whole are rather moot here from any perspective.

[Raven]

Yeah. Not that it matters much, as I'm just pointing out that the arguments on the momentum of the blast knocking the Sphere around as a whole are rather moot here from any perspective.

The momentum of the DS beam is 3.3e29 kg*m/s, which would only push a 2e30kg Dyson Sphere toward the star at .17 m/s.

However, the momentum of the DS beam (disregarding thermal effects and mechanical stress caused by thermal effects) would put more than 40 billion TPa stress on the impact site, which is far more than the minimum needed to hold the sphere together.
This is assuming a beam duration of 10 seconds, and beam width of 1km. It's probably less than this, increasing the stress on the sphere, but haven't seen ANH in a while, and I don't know how wide a superlaser beam is.

The Death Star wouldn't knock the Dyson Sphere into the sun, because it would carve the thing into pieces long before that happened.

[Mad]

This is assuming a beam duration of 10 seconds, and beam width of 1km. It's probably less than this, increasing the stress on the sphere, but haven't seen ANH in a while, and I don't know how wide a superlaser beam is.

The beam duration was a fraction of a second (impact time), though I'm not sure what the diameter of the beam was.

The Death Star wouldn't knock the Dyson Sphere into the sun, because it would carve the thing into pieces long before that happened.

Exactly. Those suggesting that the momentum from the superlaser would cause the Dyson Sphere to fall into the star (if the surface could somehow survive such a hit) certainly understand the scale of momentum that could be imparted by the superlaser but fail to understand the scale of the Dyson Sphere.

But we have no reason to believe that the Dyson Sphere's surface can withstand the impact of such a weapon. So we go back to the point where the Death Star starts poking holes in it and the Sphere collapses.

[Darth Wong]

Of course, it's also possible that the Dyson Sphere is far less massive than that. If it is indeed that massive, it begs the question of why its builders would abandon it. A dying sun should be no impediment to people who can construct a solar mass worth of metallic structure. They should be able to just dump vast amounts of hydrogen onto the star to turn back the clock on its life cycle.

[Wyrm]

I tried my hand at analyzing Lost Soal's piccies.

The entry hatch to the Dyson sphere is likely a parallel octogonal prism, which has each face parallel to each other. In Lost Soal's last picture, the face at the outer side (ie, including the hatch proper and its lip) measures 11 cm on my screen while the Enterprise itself measures 0.7 cm from top to bottom. Given that the Enterprise's height is 173 m, this makes the outer side 2,718.6 m. The inner side (if we believe that the entry lock is a parallel prism) is the same width.

The first image is rather undesirable given the low resolution, and we cannot discount a "virtual aperture" effect for viewscreens, but it has the desirable feature that we face the hatch dead-on. This view give an on-screen width of 6.1 cm for the inner side of the hatch and 3.7 cm for the outer side of the hatch. Given a "standard viewing distance" of 19 in (48.26 cm), the outer part of the hatch subtends 2 x 2.195° of the line of sight, and the inner part of the hatch subtends 2 x 3.616° of the line of sight. We choose these angles because it gives us a right angle to work with. The third angle for the outer part of the hatch is 87.805°, and the third angle for the inner part of the hatch is 86.384°. The law of sines gives us distances of 35,388.765 m and 21,465.32 m for the outer part and inner part of the hatch respectively. This makes the hatch 13,923.445 m deep. (!)

The second image shows the hatch at an angle, undesirable as it makes some parts of the faces closer than others, and we don't get a whole view of the inner part of the hatch. However, we do see two "prongs" exending into the inner hatch. On the previous image, the prongs extend a total of 1.6 cm into the 6.1 cm of the inner hatch (a measurement immune to virtual apertures), therefore the distance between these two prongs is 2,005.5 m. We can calculate the foreshortening this way:

Half the on-screen length of the inner edge of the near prong is 1.3 cm (subtended angle: 0.02693091 radians) and the far prong is 1.15 cm (subtended angle: 0.02382475 radians). If we suppose that the true length of these parts of the prongs are equal, then the far prong is 1.0000788 times as far as the near prong. This is close enough to parallel. Corresponding structures on the outer part of the hatch are close enough in length to judge them to be near-parallel.

We now need to calculate the distance to each of the inner edges of the prongs. The on-screen separation between the inner edges of these two prongs is 13.8 cm (subtended angle: 0.2840262 radians). The law of cosines states that c² = a² + b² - 2ab cos γ, here γ = 0.2840262 radians (screen subtended angle of the distance between the prongs.) We know that the far prong (which we will call distance b) to be 1.0000788 times that of the near prong (distance a). Therefore, b = (1.0000788) a. c is the distance between the inner edges of the prongs. Therefore,

c² = a² + b² - 2ab cos γ
c² = a² + ((1.0000788) a)² - 2a((1.0000788) a) cos γ
c² = a² + (1.0000788)² a² - 2(1.0000788) a² cos γ
c² = [1 + (1.0000788)² - 2(1.0000788) cos γ] a²
c²/[1 + (1.0000788)² - 2(1.0000788) cos γ] = a²
c/0.28308363658 = a

Thus, a = 7,084.479 m and b = (1.0000788) a = 7,085.037 m. This confirms that the POV is very near the centerline of the hatch, and gives us the approximate distance to the inner part of the hatch. The outer part of the hatch is 8.5 cm on-screen (subtended half-angle: 0.08783804 radians), which means that it must be 15,435.252 m from us. We then subtract off the distance b (the further) and get 8,350.215 m. Not as much, but still impressive.

Okay, the two measurements give answers within an order of magnitude, so I am going to use it for now, and take the lesser of the two (8,350.215 m). As you shall see, however, the results are rather incredible, and I use that word rather precisely.

The given diameter of the Dyson Sphere is approximagely 200 million km (2e11 m; which is about 1/3 larger than the Earth's orbit about the Sun, not "nearly" -- you lose, Riker). Plugging in our V ≅ 4πr²d = 4π (2e11 m)² (8,350.215 m) = 4.197275856e+27 m³. Multiplying by our density of 3000 kg/m³, we get 1.2591827568e+31 kg... about 6.333 solar masses! Gah!

That must have been a pain to put together. Going by Anders Sandberg's estimate of the total amount of usable stuff in the solar system, 1.82e26 kg, this momma required the deconstruction of about 69,186 nearby system like our solar system. Assuming a galactic density of 1 star/pc³, this required emptying out a region at least 41.05 pc on a side.

Of course, it's also possible that the Dyson Sphere is far less massive than that. If it is indeed that massive, it begs the question of why its builders would abandon it. A dying sun should be no impediment to people who can construct a solar mass worth of metallic structure. They should be able to just dump vast amounts of hydrogen onto the star to turn back the clock on its life cycle.

Doesn't work that way. Dumping a bunch of hydrogen onto the star simply increases the pressure on the core, coaxing more energy production to support the added pressure, without adding to the fuel availible at the core. That shortens the star's life. In order to turn back the clock, they need to inject hydrogen into the core and stoke out the helium ash.

Still, if they can move 6-and-a-third solar masses about, one wonders why they just don't deconstruct the sphere and move it to a new star.

[vivftp]

Wyrm, I'm not so sure that works out. Look at the original pic I posted just below Lost Soals pictures. On the bottom we can see those small black sections, and we can see them in Lost Soals pictures in greater detail (pic 3 and 4).

In my pic the Enterprise is just barely through the inner hatch and we can see that it's larger than each of those dark sections. In Lost Soals pictures we can see that there aren't too many of those dark sections leading from the inner hatch to the outer hatch, so going by this view I don't think we can conclude it's anywhere near 8.3km.

Not to say your math is wrong, it could easily be they used different models which differed enough to account for our different results... I guess the question is if that's the case, which do we go with?

[vivftp]

Hmm, to add to the above, I probably should've used the Enterprises total height rather than just the saucer, since it's more easily measurable. So in the second analysis I posted, it works out that 1 pixel = 10.17 meters. That works out to 1,282.23 meters for the top tractor beam, and 1,221.17 meters for the lower tractor beam.

[Wyrm]

Perspective can play very funny tricks, vivftp. Have you ever seen a TV/movie moment where the background rushes back while the character in the foreground remains only jitters a bit? That trick is made by starting the camera a far ways back with a narrow focus, then rolling the camera forward while zooming out. The depth of the shot increases, but the character remains the same size on-screen. In order to analyze distances in a photo, you need to know where the camera is.

It's clear in your image, vivftp, that the entryway is at least as deep as it is wide. The on-screen width of the hatch is 2.75 cm and it's length is about 3.90 cm, but the face-on view of the hatch reveals that the hatch is just about as long as it is wide. This defect between the two measurements can only be explained by a POV that is looking at the hatch 45° away from normal. That means the line of sight intersects the plane of inside far wall of the hatchway at a distance equal to that of width of the hatch. (The face of the hatch forms a 90° angle with the inside far wall and the plane of the face forms a 45° angle with our line of sight to the far wall, so the remaining angle is 45°, forming a 1,1,√2 right triangle.) My figure of 2,718.6 m for the hatch can only be an overestimate if the Enterprise has already made it through the hatch, which it hasn't by this time. Since we know that the entryway is at least as deep as the hatch is wide, and we know that the hatch is at least as wide as 2,718.6 m, then we know that the entryway is at least as deep as 2,718.6 m.

This sanity check says that both your figures of 901.9794 m to 1676.4 m are dead wrong.

[RedImperator]

Of course, it's also possible that the Dyson Sphere is far less massive than that. If it is indeed that massive, it begs the question of why its builders would abandon it. A dying sun should be no impediment to people who can construct a solar mass worth of metallic structure. They should be able to just dump vast amounts of hydrogen onto the star to turn back the clock on its life cycle.

Would that actually work? I thought the major problem in old main sequence stars was a buildup of helium "ash" in the core to the point that hydrogen fusion became impossible due to all the helium contamination. Dumping more hydrogen into the core would bring down the helium concentration, but at the cost of increasing the star's mass tremendously.

It's also possible the builders were capable of maintaining the star on the main sequence indefinitely, but abandoned the Dyson Sphere for some other reason and left the star to its fate. Data was just speculating, and of course Trek's writers completely forgot about the Dyson Sphere afterwards so we have no idea what the Federation researchers who undoubtedly crawled all over the thing did or didn't discover about it or them.

At any rate, I don't see how the shell could have been anything but metallic, unless the sphere builders created some sort of lightweight material with enormous compressive strength. You've identified one of the main paradoxes involved in solid shell Dyson spheres: if you can build one, you don't need it.

EDIT: And vivftp, I turned your format-busting pictures into links. It's extremely annoying to everyone in the thread to have to scroll back and forth just to read posts. Next time, reduce their size.

[OmegaGuy]

We do know that Scotty's retirement shuttle, the Jenolan, left a nasy gouge on the surface when it crashed.

*snip image*

So I don't think the Death Star will have too much difficulty in destroying it.

Well wouldn't it have to be stronger than that just to keep itself structurally intact? I'm not exactly sure how to do the math to calculate that though. Can anyone help?

[NecronLord]

Well wouldn't it have to be stronger than that just to keep itself structurally intact?

Structural Integrity Fields. It's got one hell of an energy source to power them, after all.

[Lost Soal]

Does anyone else other than the feds actually use them? I don't recall them ever being mentioned on DS9 for either the station or any of the Klingon ships.

[NecronLord]

Does anyone else other than the feds actually use them? I don't recall them ever being mentioned on DS9 for either the station or any of the Klingon ships.

Yes. They do. DS9 has one 'Looking for Par'mach in all the wrong places.' I don't know of any mention of klingons having them, but it's not at all unlikely.

[vivftp]

Does anyone else other than the feds actually use them? I don't recall them ever being mentioned on DS9 for either the station or any of the Klingon ships.

Well it's not exactly the same application, but it's said when a Borg cube enters a transwarp conduit it projects a SIF ahead of the ship to counteract gravimetric shear. The specific name of the episode in which this was mentioned escapes me at the moment.

[Stark]

Gravimetric shear?

[vivftp]

Gravimetric shear?

*shrug*
That's the term 7 used, whatever it means.