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Forum » SpaceEngine » Science and Astronomy Discussions » Science and Astronomy Questions
Science and Astronomy Questions
WatsisnameDate: Wednesday, 14.08.2013, 00:43 | Message # 181
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Right, we generally either go by the galactic center or, in ambiguous cases, point of strongest emission. For very distant galaxies it matters little because their angular sizes are quite small.




 
themohawkninjaDate: Tuesday, 20.08.2013, 16:35 | Message # 182
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Does the corona of the Sun (and any other star for that matter) cause a redshift in the light output of the star, so that there would be a slightly redder "ring" of light around the edges of the star than towards the center of the star (edges and center referring to the surface of the star from the perspective of the observer) as the light would have had to have traveled through more of the corona at the observed edges than the observed center? Basically a stellar version of why the midday sky is blue, and the sunrise/sunset sky is red?
 
HarbingerDawnDate: Tuesday, 20.08.2013, 17:48 | Message # 183
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No. The corona is extremely thin, and does not scatter light in the same way that the atmosphere does. Also, that is not "redshift". It's Rayleigh scattering, which is an entirely different and unrelated phenomenon.




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SpaceEngineerDate: Wednesday, 21.08.2013, 09:22 | Message # 184
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Darkening and "reddening" of a stellar limb at the edges have another nature:

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

In short, at the center you can see deeper and hotter layers of a photosphere, while at edges you can't due to increased path that light ray must travel inside the photosphere. So center appears hotter and brighter than edges.





 
SalvoDate: Friday, 11.10.2013, 14:17 | Message # 185
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Hard question, what would be approximately the mass of a Stanford torus with the radius of 3 km?




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HarbingerDawnDate: Friday, 11.10.2013, 14:34 | Message # 186
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Quote Salvo ()
Hard question, what would be approximately the mass of a Stanford torus with the radius of 3 km?

That would depend on exactly how it's constructed. So there's not enough information to give a single answer.





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SalvoDate: Friday, 11.10.2013, 16:23 | Message # 187
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Quote HarbingerDawn ()
That would depend on exactly how it's constructed. So there's not enough information to give a single answer.


I used 50000 tons for my calculations, could it be realistic? smile






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Edited by Salvo - Friday, 11.10.2013, 16:24
 
HarbingerDawnDate: Friday, 11.10.2013, 17:35 | Message # 188
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In your drawing you have a radius of only 1.5 km. Which did you want, 1.5 or 3?

Quote Salvo ()
I used 50000 tons for my calculations, could it be realistic?

I suspect it would be much more than that.





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WatsisnameDate: Friday, 11.10.2013, 18:23 | Message # 189
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Mmm, I think determining a 'realistic' value for the mass is not going to be a straightforward thing; as Harbinger says it will depend greatly on the design and the material used.

We could probably ballpark it through the following procedure:
-know what size of torus you want
-know what surface gravity you want on the ring
-determine the angular velocity to meet that criteria (looks like you've already done it, though I did not check the numbers yet)

Then, suppose some density of material used in the construction, and assume that most of the mass of the torus is contained in the torus itself (as opposed to some central hub or the spokes connecting/supporting it).
-Make a reasonable guess as to the thickness of the torus' shell (on the order of a meter maybe?)
-Calculate the volume of this shell.
-Multiply this volume by the density.





 
SalvoDate: Friday, 11.10.2013, 19:56 | Message # 190
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Quote HarbingerDawn ()
Which did you want, 1.5 or 3?


Sorry, I meant diameter of 3 km, so the radius it's 1.5

Quote Watsisname ()
-know what size of torus you want
-know what surface gravity you want on the ring
-determine the angular velocity to meet that criteria (looks like you've already done it, though I did not check the numbers yet)

Then, suppose some density of material used in the construction, and assume that most of the mass of the torus is contained in the torus itself (as opposed to some central hub or the spokes connecting/supporting it).
-Make a reasonable guess as to the thickness of the torus' shell (on the order of a meter maybe?)
-Calculate the volume of this shell.
-Multiply this volume by the density.


Got it, I will think about it smile
However, I didn't meant to do such an accurate a calculation, just a approximation to avoid hard shi-fi's fans to say something like "that rotation is unrealistic!" or "that would never generate a 1g force!" biggrin





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Edited by Salvo - Friday, 11.10.2013, 19:59
 
midtskogenDate: Friday, 11.10.2013, 20:47 | Message # 191
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Quote Salvo ()
I used 50000 tons for my calculations, could it be realistic?

If you build it from rice paper, perhaps.

Since the 1g force will work totally different in your construction, comparing it with an earthly building might not be very accurate. But consider: The weight of the Taipei 101 building is 700,000 tonnes. But Taipei 101 is only 500 metres tall. You want something that is nearly 10 km long. And possibly more spacious. You might be off by a factor of 100.





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WatsisnameDate: Friday, 11.10.2013, 21:15 | Message # 192
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Okay, your angular velocity looks good: 0.081rad/s ~ 4.6 degrees/s ~ 0.025pi/s. That will give you 1g acceleration for the torus out at 1.5km.

To find the volume of the shell of the torus we'll just do the triple integral over the structure:


Where R is the radius of the torus (1.5km), and (rb - ra) is the thickness of the shell. If we take the thickness as one meter then the volume is ~30,000 cubic meters. If we take the average density as 8000kg/m3 (comparable to that of steel) then the mass is on the order of a few hundred million kilograms, or a few hundred thousand tons. So your 50,000 ton estimate seems very believable. :)

Edit: Scratch that last bit; I was being dumb. But plug in some numbers and you'll see a believable estimate should be one or two orders of magnitude larger. If the living space is on the order of 10 meters wide with 1m thick walls then the thing is going to have a mass on the order of millions of tons.





 
HarbingerDawnDate: Friday, 11.10.2013, 21:25 | Message # 193
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But that would only get you the mass of the outer shell framework. It doesn't take into account everything inside it: atmosphere, soil, water, plants, buildings, etc, which would contribute a significant amount of mass. Aside from that, I'm skeptical that something of that size, when fully outfitted and occupied, could have only half the mass of a Nimitz-class aircraft carrier.




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WatsisnameDate: Friday, 11.10.2013, 21:30 | Message # 194
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I goofed; it's going to weigh a lot more than that because it's the square of the difference of radii. So if the tube is say, 10 meters in radius (a comfortable living space), then the mass is on the order of megatons.




 
SalvoDate: Friday, 11.10.2013, 21:55 | Message # 195
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Quote Watsisname ()
So if the tube is say, 10 meters in radius (a comfortable living space), then the mass is on the order of megatons.


It's a "tube" of 4 meters in radius (0.5 of shield), but there are some "living" areas, big as a normal house

Quote HarbingerDawn ()
It doesn't take into account everything inside it: atmosphere, soil, water, plants, buildings, etc, which would contribute a significant amount of mass.


Right, but it doesn't contains very much, except 2 thousand of people (about 150 000 kg), an "atmosphere" and 12 small spaceships smile

Anyway, thank you guys for your help smile





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Edited by Salvo - Friday, 11.10.2013, 21:57
 
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