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FiahOwl

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Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« on: November 09, 2011, 04:45:59 PM »

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]


« Last Edit: March 13, 2012, 04:56:48 AM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #1 on: November 09, 2011, 04:46:44 PM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 12:02:59 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #2 on: November 09, 2011, 04:46:55 PM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 12:00:10 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #3 on: November 09, 2011, 04:47:04 PM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 11:59:44 AM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #4 on: November 09, 2011, 04:47:13 PM »

Fiah's historical systems


These are some of the first systems I've ever made.


The Ares System
A Binary System With 2 Planets. Very old, and one of my first systems ever.

The HD 821344342 System
A Binary System With 2 Stars and Several Planets.

Ahera
Home System of the Therosans.

Halius
An Ice planet with 4 moons.

* The Ares System.ubox (125.66 KB - downloaded 20 times.)
* Ahera - Directors Edition.ubox (560.62 KB - downloaded 18 times.)
* HD 821344342 System.ubox (191.41 KB - downloaded 18 times.)
* Halius - 4 Moons.ubox (230.7 KB - downloaded 15 times.)
« Last Edit: January 27, 2012, 12:01:28 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #5 on: November 09, 2011, 04:47:21 PM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 12:02:39 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #6 on: November 09, 2011, 04:47:31 PM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 12:02:22 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #7 on: November 12, 2011, 11:28:51 AM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 12:01:49 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #8 on: November 20, 2011, 08:23:25 PM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 12:02:04 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #9 on: December 04, 2011, 05:55:43 PM »

[RESERVED]

Go on, Good Sir. Nothing to see here.
« Last Edit: January 27, 2012, 12:02:12 PM by FiahOwl » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #10 on: December 06, 2011, 01:54:47 PM »

All the BBCode used to make this, it would boggle anyone's head.

Does anyone like the way the category "Star Information" is made?


* 680px-Morgan-Keenan_spectral_classification.png (147.32 KB, 680x290 - viewed 406 times.)

* Plangenera.png (216.57 KB, 518x227 - viewed 329 times.)
« Last Edit: December 20, 2011, 02:20:32 PM by FiahOwl » Logged
smjjames

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Posts: 559


« Reply #11 on: December 06, 2011, 03:09:33 PM »

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

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #12 on: December 06, 2011, 03:52:50 PM »

The Star generator does that. Input the mass and it gives all the details.
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smjjames

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Posts: 559


« Reply #13 on: December 06, 2011, 04:35:49 PM »

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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FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #14 on: December 06, 2011, 04:49:41 PM »

http://web.archive.org/web/20030218125318/www.geocities.com/Area51/Corridor/8611/mseqstar.htm

Input the direct mass and you've got your realistic specs.

(It also says Main Sequence stars.)
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smjjames

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Posts: 559


« Reply #15 on: December 06, 2011, 04:52:39 PM »

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.
« Last Edit: December 06, 2011, 05:05:05 PM by smjjames » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #16 on: December 07, 2011, 02:52:04 PM »

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??!!


* 220Ma680x290.jpg (33.99 KB, 680x290 - viewed 407 times.)
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smjjames

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Posts: 559


« Reply #17 on: December 07, 2011, 03:01:26 PM »

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
Beta Team

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Posts: 2652


Like a scientific expert. With magic.


WWW
« Reply #18 on: December 07, 2011, 03:06:05 PM »

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

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #19 on: December 07, 2011, 03:25:28 PM »

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

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #20 on: December 07, 2011, 03:29:01 PM »

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
Beta Team

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Posts: 2652


Like a scientific expert. With magic.


WWW
« Reply #21 on: December 07, 2011, 04:06:48 PM »

I meant the actual topic. Not the first/original post
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FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #22 on: December 07, 2011, 04:19:36 PM »

I meant the actual topic. Not the first/original post

OH.
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FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #23 on: December 08, 2011, 09:51:31 AM »

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

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #24 on: December 13, 2011, 03:07:28 PM »

Here are the most used phrases in the OP.



* Planfeilda.png (12.72 KB, 670x176 - viewed 323 times.)
« Last Edit: December 20, 2011, 02:12:35 PM by FiahOwl » Logged
mudkipz

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Posts: 723


One does not simply walk into Space.


« Reply #25 on: December 13, 2011, 05:07:03 PM »

cool what program did you use to make that image?
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Darvince

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Posts: 865


Nuedarfarevar


« Reply #26 on: December 13, 2011, 05:09:11 PM »

wordle.net
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FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #27 on: December 20, 2011, 02:26:59 PM »

I just finished the update!
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smjjames

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Posts: 559


« Reply #28 on: January 04, 2012, 10:30:30 AM »

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.
« Last Edit: January 04, 2012, 10:59:09 AM by smjjames » Logged
FiahOwl

*****
Posts: 902


The last person standing after a battle is either Jackie Chan, or an Aeridané.


« Reply #29 on: January 04, 2012, 11:19:26 AM »

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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