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FiahOwl
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« on: November 09, 2011, 04:45:59 PM » |
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Forever a revolution. Our Content (To help you map your own stars!)
Feel free to suggest some links that I should use. I would be very happy to include them in this thread. Star Information Planetary Mapping/ Modeling Generating Planets Planet Classification Inspiration Spaceflight Calculations
Calculate the life of a star 10^10*<star mass>/<star luminosity>
So a 1.17 Star mass star would have the following:
10^10*1.17/1.62
Which equals 6.5 Billion years on the main sequence.[Found somewhere on the internet] A worldbuilder's Guide to Banks Orbitals Size of a Banks Orbital Ring:
A ring designed to produce a 24 hour day and 1 gravity on its inner surface has a radius of 1.89 X 106 kilometres. Given that g = the acceleration on the inner surface t = the time the Orbital takes for a complete turn r = the radius of the orbital, then r ∝ g (r is proportional to g) r ∝ t2
Required ring-wall height:
Minimum for good containment of a 1 bar atmosphere at 1 gravity is 100 kilometres, where: h = height of the rimwalls g = gravity p = pressure then h ∝ 1/g h ∝ p Twice-yearly eclipses:
The amount of time for a total eclipse of the sun by the far side of the ring is given by:
t = ω/(2rΩsinθ) where: ω = the width of the Orbital r = the radius of the Orbital Ω = angular velocity of the Orbital about the star θ = the tilt of the Orbital
Or to put it another way, given a "standard" orbital (Earth-normed) with a width of 1000 kilometres moving about a sun just like ours and at 1 a.u., the time of the eclipse is 21.8 minutes. To vary that, t ∝ ω t ∝ 1/r t ∝ 1/Ω or t ∝ year length t ∝ 1/sinθ
Calculating the orbital's year given its distance from the star and the mass of the star is done in the same way as for a planet.
Apparent width of the Ring-arch in the sky at the zenith, for a 'one standard gee' Orbital 1000 km broad from rim to rim is 0.9 minutes of arc (by comparison Sol or Luna covers 30 minutes, or half a degree, as seen from Old Earth).
Required Materials for a typical 1000 km wide orbital:
1.6 X 1022 kilograms magnanotube fibres (a layer less than a few micrometers thick) 8.9 X 1020 kilograms nickel-iron (kamacite & taenite) 10 metres thick 3.2 X 1022 kilograms foamed diamondoid 2 kilometres thick 3.2 X 1022 kilograms corundumoid plus silicates & other minerals 0.5 kilometres thick 1.2 X 1021 kg water 100 m thick
Total costs: energy for creation of 16 exatonnes of magmatter, mass of 1 large rocky & carbonaceous moon, mass of 1 midsized icy moon
[Source]
Bode's Law Determining the orbital distance of planets from their star depends on a fairly intricate mechanism known as Bode's Law. According to Bode's Law, planetary orbits follow a recognizable mathematical pattern of development; this system replicates it. Roll 1d6 to create a “seed” number. (The Sol system’s seed number is 3.) Beginning with 0 and then the seed, run a series of doublings out for as many planets as your system has. (For the Sol system, that series is 0, 3, 6, 12, 24, 48, and so on.) Now roll the die again, and add that constant to the seed series. (The Sol Bode's constant is 4, which gives 4, 7, 10, 16, 28, 52, and so on.) Now divide the new series by 10, and that’s your planetary orbit pattern in AU. (Again for the Sol system, we get 0.4, Mercury; 0.7, Venus; 1, Earth; 1.6, Mars; 2.8, the asteroid belt; 5.2, Jupiter, and so on.) Even the Sol system pattern breaks down with Neptune, so you can vary the Bode's result if you like.[From a book]
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« Last Edit: March 13, 2012, 04:56:48 AM by FiahOwl »
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FiahOwl
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« Reply #1 on: November 09, 2011, 04:46:44 PM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 12:02:59 PM by FiahOwl »
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FiahOwl
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« Reply #2 on: November 09, 2011, 04:46:55 PM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 12:00:10 PM by FiahOwl »
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FiahOwl
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« Reply #3 on: November 09, 2011, 04:47:04 PM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 11:59:44 AM by FiahOwl »
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FiahOwl
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« Reply #4 on: November 09, 2011, 04:47:13 PM » |
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Fiah's historical systems These are some of the first systems I've ever made.The Ares SystemA Binary System With 2 Planets. Very old, and one of my first systems ever. The HD 821344342 SystemA Binary System With 2 Stars and Several Planets. AheraHome System of the Therosans. HaliusAn Ice planet with 4 moons.
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« Last Edit: January 27, 2012, 12:01:28 PM by FiahOwl »
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FiahOwl
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« Reply #5 on: November 09, 2011, 04:47:21 PM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 12:02:39 PM by FiahOwl »
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FiahOwl
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« Reply #6 on: November 09, 2011, 04:47:31 PM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 12:02:22 PM by FiahOwl »
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FiahOwl
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« Reply #7 on: November 12, 2011, 11:28:51 AM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 12:01:49 PM by FiahOwl »
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FiahOwl
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« Reply #8 on: November 20, 2011, 08:23:25 PM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 12:02:04 PM by FiahOwl »
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FiahOwl
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« Reply #9 on: December 04, 2011, 05:55:43 PM » |
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[RESERVED] Go on, Good Sir. Nothing to see here.
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« Last Edit: January 27, 2012, 12:02:12 PM by FiahOwl »
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FiahOwl
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« Reply #10 on: December 06, 2011, 01:54:47 PM » |
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All the BBCode used to make this, it would boggle anyone's head.
Does anyone like the way the category "Star Information" is made?
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« Last Edit: December 20, 2011, 02:20:32 PM by FiahOwl »
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smjjames
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« Reply #11 on: December 06, 2011, 03:09:33 PM » |
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I think it's pretty good.
Also, maybe a calculator for finding a stars correct luminosity vs it's tempurature? I don't know how the tempurature vs luminosity calculation works. Although for now, the star tempurature only changes the color.
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FiahOwl
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« Reply #12 on: December 06, 2011, 03:52:50 PM » |
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The Star generator does that. Input the mass and it gives all the details.
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smjjames
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« Reply #13 on: December 06, 2011, 04:35:49 PM » |
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Cool, although it doesn't work very well, or accurately. Tried with Barnards Star and it didn't get the other settings correct, half the time it just bugs out. Tried to get it to do Eta Carinae and it wouldn't work very well.
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smjjames
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« Reply #15 on: December 06, 2011, 04:52:39 PM » |
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I thought Barnards Star was on the main sequence, maybe not. Eta Carinae certainly isn't.
Edit: Wikipedia doesn't specifically say if its a main sequence, but the calculator there did get the spectral type pretty close, M6 compared to the M4 that it actually is.
It did also get the siuns temp a bit off, but I guess it can be used to take a stab at the luminosity until Dan gets the calculations for it in US.
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« Last Edit: December 06, 2011, 05:05:05 PM by smjjames »
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FiahOwl
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« Reply #16 on: December 07, 2011, 02:52:04 PM » |
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I thought Barnards Star was on the main sequence, maybe not. Eta Carinae certainly isn't.
Edit: Wikipedia doesn't specifically say if its a main sequence, but the calculator there did get the spectral type pretty close, M6 compared to the M4 that it actually is.
It did also get the siuns temp a bit off, but I guess it can be used to take a stab at the luminosity until Dan gets the calculations for it in US.
Or maybe Dan's sun temperature is wrong??!!
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smjjames
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« Reply #17 on: December 07, 2011, 03:01:26 PM » |
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I think he's only off by around a hundred K anyways, the calculator thing went and rounded up to the nearest hundred for some reason.
It may not be great, but it'll help in figuring out luminosity.
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atomic7732
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« Reply #18 on: December 07, 2011, 03:06:05 PM » |
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Fiah, the topic isn't very descriptive as to what you have in it... I was wondering what this was.
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FiahOwl
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« Reply #19 on: December 07, 2011, 03:25:28 PM » |
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Fiah, the topic isn't very descriptive as to what you have in it... I was wondering what this was.
Fixing it...
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FiahOwl
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« Reply #20 on: December 07, 2011, 03:29:01 PM » |
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Fiah, the topic isn't very descriptive as to what you have in it... I was wondering what this was.
Fixing it... Done!
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atomic7732
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« Reply #21 on: December 07, 2011, 04:06:48 PM » |
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I meant the actual topic. Not the first/original post
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FiahOwl
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« Reply #22 on: December 07, 2011, 04:19:36 PM » |
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I meant the actual topic. Not the first/original post
OH.
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FiahOwl
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« Reply #23 on: December 08, 2011, 09:51:31 AM » |
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Determining the orbital distance of planets from their star depends on a fairly intricate mechanism known as Bode’s Law. According to Bode’s Law, planetary orbits follow a recognizable mathematical pattern of development; this system replicates it. Roll 1d6 to create a “seed” number. (The Sol system’s seed number is 3.) Beginning with 0 and then the seed, run a series of doublings out for as many planets as your system has. (For the Sol system, that series is 0, 3, 6, 12, 24, 48, and so on.) Now roll the die again, and add that constant to the seed series. (The Sol Bode’s constant is 4, which gives 4, 7, 10, 16, 28, 52, and so on.) Now divide the new series by 10, and that’s your planetary orbit pattern in AU. (Again for the Sol system, we get 0.4, Mercury; 0.7, Venus; 1, Earth; 1.6, Mars; 2.8, the asteroid belt; 5.2, Jupiter, and so on.) Even the Sol system pattern breaks down with Neptune, so you can vary the Bode’s result if you like.
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FiahOwl
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« Reply #24 on: December 13, 2011, 03:07:28 PM » |
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Here are the most used phrases in the OP. 
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« Last Edit: December 20, 2011, 02:12:35 PM by FiahOwl »
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mudkipz
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« Reply #25 on: December 13, 2011, 05:07:03 PM » |
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cool what program did you use to make that image?
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Darvince
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« Reply #26 on: December 13, 2011, 05:09:11 PM » |
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wordle.net
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FiahOwl
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« Reply #27 on: December 20, 2011, 02:26:59 PM » |
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I just finished the update!
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smjjames
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« Reply #28 on: January 04, 2012, 10:30:30 AM » |
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I've got a question as far as Bodes law. I want to create a fictional Kepler-22 (star system, not the forumer), however I'm not sure how to do bodes law with one planetary body already set?
Edit: And um, putting the radius (had to convert it to diameter) to 2.38 that of Earths makes the mass to 0.32 Jupiters, assuming Earth desnity, does that seem correct? *goes off to do more research on it....*
Edit2: Looks like I may have been doing the steps and conversion the wrong way, causing the number to balloon twice as much as it should have.
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« Last Edit: January 04, 2012, 10:59:09 AM by smjjames »
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FiahOwl
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« Reply #29 on: January 04, 2012, 11:19:26 AM » |
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I've got a question as far as Bodes law. I want to create a fictional Kepler-22 (star system, not the forumer), however I'm not sure how to do bodes law with one planetary body already set?
Edit: And um, putting the radius (had to convert it to diameter) to 2.38 that of Earths makes the mass to 0.32 Jupiters, assuming Earth desnity, does that seem correct? *goes off to do more research on it....*
Edit2: Looks like I may have been doing the steps and conversion the wrong way, causing the number to balloon twice as much as it should have.
What's the SMA from the parent star? I'll work out a formula.
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