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Topic: A question on Open Source... (Read 20724 times)
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Neutrino 123
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I don’t think the ideal gas law is appropriate for this situation. The gas would need to lose its energy via radiation very quickly, or else we would see glowing material flying off, instead of just disappearing.
An anisotropic explosion could indeed account for some more of the Death Star’s mass (though it alone could not be decisive or we would see it in the explosion). In fact, we already know that “something” about the destruction was anisotropic due to the usual ring o’ blowin’ big things up that we see.
I don’t remember much about the Death Star interior except for the scene where obiwan was standing near the long pylon to use a terminal. I assumed the pylon was filled with stuff…important stuff…
Are any pictures of these available? A quick google search yields nothing…
There is of course uncertainty in the precise radius of Endor, but it could go either way compared to the Earth. However, I would say more likely that Endor is larger and less dense then the Earth, which was struck by the Moon (then a really big asteroid), and absorbed some of the Moon’s iron, a very dense substance.
For Death Star sizes, see below images. The planet radii are clearly much different on the first, and in the second, nowhere on the big circle is there an area of apparently constant slope like the area near the super star destroyer (which is scaled to 900km deathstar for the circle and 17.6km super star destroyer in the small image).
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-Neutrino 123 (pronounced Neutrino One-Two-Three)
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Death 999
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We did. You did. Yes we can. No.
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It's not in orbit, it's sitting on some sort of force field (presumably a different one than the one on the ground). Anyway, after being thrown out of an orbit at velocities far in excess of escape velocity... the debris is not in orbit anymore.
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Neutrino 123
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Sorry, I was a bit sleepy typing the E-mail, so wasn’t very clear or even fully articulated. The ideal gas law is not directly applicable if there is any outgoing radiation, since that takes away energy (thus making the process non-adiabatic as well). This may be significant if it was a metal that was vaporized, since radiation is proportional to T^4.
Still one can assume none is and use the ideal gas law combined with the adiabatic condition to obtain (if I remember my thermodynamics correctly) constant = V*T^(f/2) where f is the number of degrees of freedom. f depends on the configuration of the gas molecules: 3, 5 ,or 6. If dealing with a monoatomic gas, it is 3, diatomic 5, and other 6 (I think).
Since the radiation involves temperature, we should see it decline steadily as the volume decreases, but we really don’t. It goes from white-hot to hardly illuminated while not expanding much at all. He is a picture with some circles:
P.S. Looking at the matter more closely, I am again unsure if the heat could transfer quickly enough to vaporize the entire asteroid. The temperature gradient would be large, but the hot part would already be vaporized, so would just have one collision to transfer energy before it bounces away, and not all molecules would originally bounce in the right direction to transfer heat. Thus, the kinetic energy shockwave would break it apart while the molecules supposed to be transferring heat would just fly off. We don’t see debris, so maybe we are just not close enough, or maybe it is just turbolaser magical properties...
Also, for the Death Star interior: didn’t you refer to episode IV in your earlier post? That should be the one with the original Deathstar...
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-Neutrino 123 (pronounced Neutrino One-Two-Three)
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Death 999
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We did. You did. Yes we can. No.
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The ideal gas law is not directly applicable if there is any outgoing radiation, since that takes away energy (thus making the process non-adiabatic as well). This may be significant if it was a metal that was vaporized, since radiation is proportional to T^4.
It is directly applicable. It is an equation of state. If the temperature dropped due to other factors, that will change the temperature independently. You mean the adiabatic approximation is invalid, then? You already quite correctly pointed out that radiative cooling was so slow as to be irrelevant. That means it's adiabatic.
If you don't believe it's fast enough, I'll get quantitative: you have the gas at the boiling point of the metal, initially confined to around a volume 30 meters on a side. It is thus glowing-hot. How much volume must it expand into to drop to a temperature too low to glow red? Answer: a volume roughly T(boiling metal) / T (red hot) larger, which I guesstimate to be a factor of 30 (though this would be hotter than the surface of the sun). That corresponds to a linear expansion increase of only THREE TIMES.
This expansion occurs very quickly, as you may have noticed.
P.S. Looking at the matter more closely, I am again unsure if the heat could transfer quickly enough to vaporize the entire asteroid. The temperature gradient would be large, but the hot part would already be vaporized, so would just have one collision to transfer energy before it bounces away
I don't see why this is the case. If the beam transferred energy along a line, then almost all of the initially vaporized particles would have nowhere to escape to, and be forced to dissipate their energy to the surrounding metal, as heat and as a compression wave as it expanded.
That would in turn be transferred to heat.
, and not all molecules would originally bounce in the right direction to transfer heat. Thus, the kinetic energy shockwave would break it apart while the molecules supposed to be transferring heat would just fly off.
And if you accelerate an object hard enough just by pushing on it, the compression wave is converted into heat...
I would say the best explanation for the visual effect is that the turbolaser deposited energy through a significant volume of the asteroid (medium coupling permits it to deposit not only on the surface), which thus evaporated fairly uniformly (though certainly violently). This would minimize large debris fragments which could hit the ship or its fighters.
Also, for the Death Star interior: didn’t you refer to episode IV in your earlier post? That should be the one with the original Deathstar...
Uh, yeah. I meant 6. Oops?
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« Last Edit: September 18, 2006, 05:15:59 pm by Death 999 »
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Holocat
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My following facts:
1a: The Ur-Quan won their war against the alliance of the free stars, though they were stalemated for a long time.
1b: The sa-matra was destroyed by a improved planeteering device, its raw form believed to be a planeteering tool of some kind, perhaps a planetary destroyer, It was maneauvered in at great cost only after being stripped of most of its guard.
1c: Their ships can slag cities from orbit, and have sensor capablity to search ocean floors, deep under ice crusts, etc.
2a: The empire lost their war against a set of much less supplied, manned or strategically brilliant rebels who used converted passenger liners rather than real war-ships to do their capship fighting.
2b: The first Death Star was lost to a single (or maybe two?) fighter-carried missle(s) due to an engineering mistake that no one with an ounce of competence would make.
2b-2: The second Death Star was lost because the ground forces protecting its shield were defeated by stone-age carebears assisted by a very small rebel ground force.
2c:... and despite having most of the imperial fleet guarding the star in a startling ambush who's very design was to crush the rebel fleet, they managed to lose their flagship (the executor) the death star, and suffered devastating tactical defeat, at the hands of said same motely rebels.
My conclusion? The Ur-Quan will win. 'Cause they're competent. 
Edit: I tried to report this post to a moderator via link because it's just another Star Wars/Trek/Fleet/Cat/Dog/Whatever VS. Star Yak/De/Yak/De/Yak and has absoultely nothing to do with the open source question somewhere before this started, but the link said I can't report my own posts, as it doesn't make any sense. Oh well. :3
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« Last Edit: September 16, 2006, 03:27:09 am by Holocat »
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Deus Siddis
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He was joking, that was the punch-line.
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Neutrino 123
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Okay, a precise volume calculation can't be done with my present materials, and even a good estimate can't be done since I don't know if the block coefficient applies to light, standard, normal, or maximum displacement (I estimated using draft as 10 meters)). However, I was able to make a rough estimate of 116,200 cubic meters. The Iowa battleships had 2,700 crew during WWII and 1,500 in more recent service.
Here is the picture I used to make the rough estimate:
Ironically, I had put one big box in the picture for the initial calculation and got 120,000 meters, but then decided that that wasn't accurate enough, so I spent the next 20 minutes doing the more detailed work above that got essentially the same value... 
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-Neutrino 123 (pronounced Neutrino One-Two-Three)
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Neutrino 123
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Death 999, I posted the relevant equation. Temperature and volume are not related inversely, but by constant=V*T^(f/2) where f=3 for monatomic particles (such as pure metals, it is larger for more complex molecules). Thus, using your own estimate of a temperature drop of 30x and making the minimum assumption of pure metal, the volume change would need to be ~164 times greater, or assuming that it’s a sphere, the radius would need to be ~5.5 times larger. We only see a small amount of expansion, linearly much less then a factor of two.
Why would the beam deliver energy in a line within the asteroid? It should deliver energy to a single sport on the asteroid at the surface. Many of these particles would be able to escape outward, along the path of the beam, and absorb even more energy from the beam as they do this. Also, if the energy was deposited uniformly, we should see a spot of white-hot material in the first picture on the other side rather then just around the edge as the blast moves through.
One can heat something by compressing it, but the compression could also simply break it into smaller pieces, like smacking a cracker into a table. It all depends on the compression...
Finally, where are the support beams in episode VI? I am not sure where to look.
Also, to Holocat: the Ur-Quan might not win since they are very bureaucratic. Yoda could just reword their Path of Now and Forever doctrine using the force (or something), and then they would have to just talk about flowers or help clear Yoda’s swamps to make more productive gardening land. The question is, would they consider computerized copies sufficient, or would rewording the original initiate the change?
Anyway, I suppose the original question in this topic has been answered, but this discussion should really be moved into that thread with the Ur-Quan vs. Star Wars story where it is more relevant.
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-Neutrino 123 (pronounced Neutrino One-Two-Three)
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Death 999
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We did. You did. Yes we can. No.
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Death 999, I posted the relevant equation. Temperature and volume are not related inversely, but by constant=V*T^(f/2) where f=3 for monatomic particles (such as pure metals, it is larger for more complex molecules).
Oops, I mis-considered the situation -- pressure can change! PV = NkT does still apply, it just isn't helpful.
Thus, using your own estimate of a temperature drop of 30x and making the minimum assumption of pure metal, the volume change would need to be ~164 times greater, or assuming that it’s a sphere, the radius would need to be ~5.5 times larger. We only see a small amount of expansion, linearly much less then a factor of two.
Well, Actually, I only chose that temperature drop because I was trying to drive in a point. Nickel evaporates at 3187 celsius. Glowing red cuts off at around 750 celsius. Ratio? 4. Taking that to the 3/2 power yields an expansion factor of 8 -> length scale factor... 2.
Anyway, it's not like it's an expanding balloon. It's got a denser, hotter core, and a diffuser cooler outer region. This makes the expansion factor smaller (the hottest gas escapes most quickly, cools most quickly, so it's alreay gone; other superheated solid elements are trapped but evaporate when they can, which means rapid cooling occurring to a small portion of the asteroid at once).
And sorry, but that image does not show an expansion factor 'much less than 2'. Note that your neat little red circle does in fact have a lot of stuff sticking out of it. Also, that asteroid was nothing like spherical, and if it has the usual 'potato' shape, it'd probably be kind of thin in the z direction. Only one dimension close to the red circle radius...
Why would the beam deliver energy in a line within the asteroid? It should deliver energy to a single sport on the asteroid at the surface.
1) my super-1337 forensic skills (very similar to your endor-moon-gravity-measuring skillz) revealed that all SW weaponry effects were very similar to that of planted charges, not consistent with surface insertion of energy. This raises the question of 'how'.
2) If the particles in the turbolaser are relativistic, they can proceed through matter a substantial distance before interacting (we use this effect today, in radiation therapy). While the turbolaser's visual effect proceeds less than C, it has been observed that things would blow up before the visual streak arrived, which suggests that it is either a tracer or a peak in a density wavelet which proceeds at a slower pace than each component of the beam.
Many of these particles would be able to escape outward, along the path of the beam, and absorb even more energy from the beam as they do this. Also, if the energy was deposited uniformly, we should see a spot of white-hot material in the first picture on the other side rather then just around the edge as the blast moves through.
I see a white-hot spot on the far side; don't you?
In any case, what we know for sure is:
1) The blasted asteroid does not release chunks back towards the Star Destroyer that are heavy enough and high-enough velocity to be a threat to the Star Destroyers
2) The whole asteroid would have been a threat.
3) The blasted asteroid vanishes quite thoroughly quickly enough to be inconsistent with radiative dissipation.
There can be small fragments which were ejected too fast to see, but they should all be sent away from the Star Destroyer. A large portion of the asteroid was vaporized. Star Destroyers' kinetic shielding is, at least under certain circumstances, not good. As for the cooling, we have an adequate mechanism, there is no need to reach for technobabble.
Finally, where are the support beams in episode VI? I am not sure where to look.
When they're flying through the Death Star, they are always surrounded by... stuff. Heavy stuff. Lots of it.
Neutrino: In the end, and in the beginning, I've just been taking exception to your one-sided assessment of "1 GCS destroys SSD LOL" and "Sources? Saxton is worthless, st-v-sw is good".
Holocat: The spaceborne forces were only defeated by foes of comparable technological level. Ground forces are a different question altogether. Note that the Ewoks took massive casualties, far in numeric excess of the absolute number of Imperial casualties.
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« Last Edit: September 18, 2006, 05:31:07 pm by Death 999 »
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