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PostPosted: Fri Apr 07, 2017 8:43 am 
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Agemegos wrote:
Traveller wrote:
Why in the heck did I think you misspelled it?

Damn.

Sorry.

None of us is as young as we used to be.


Ain´t that the truth? All that braining isn´t as easy as it used to be.

_________________
Space isn't remote at all. It's only an hour's drive away if your car could go straight upwards. Sir Frederick Hoyle
Earth is the cradle of humanity, but one cannot live in a cradle forever. Konstantin Tsiolkovsky
Man has earned the right to hold this planet against all comers, by virtue of occasionally producing someone completely bat**** insane. xkcd #556
Just like people, stars can be very important without being terribly bright. Phil Plait, "Bad Astronomy"


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PostPosted: Fri Apr 07, 2017 7:57 pm 
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Now... binaries. Also, more complex hierarchical systems.

The distinction between Close, Near, Far, Distant binaries is useful, but kind of artificial. Since we already have a nice and infinitely scaleable system of orbital shells, I´d much rather randomly determine the orbital radius that way. And I think I have found a way to do so.

You roll 2d6+x, then d66 (i.e. use one d6 for the "tens" digit and another for the "ones" digit) to determine the value of the X. For a result of 11 on the d66 roll, X equals -92, meaning the end result ranges from -90 to -80. For increasing rolls of the d66, X increases by +5 each, up to +83 (end result from +85 to +95).

This is, admittedly, a bit cumbersome, but it is also as close as I could manage to an equal distribution of probabilities for all results from -85 to +90: Every result divisible by 5 in that range has a probability of 8/1296 (0.62%), all the others within that range of 7/1296 (0.54%).
Depending on how much granularity is desired, a simple d66 roll assigning orbital shells from -85 to +90 could be substituted.

If, on the other hand, it would be better for distribution to be more bell-shaped, we could replace the d66 roll with rolling (6d6-21)*5, with or without rolling 2d6-7 for "fine-tuning", giving us a distribution from -75 to +75 (-80 to +80 with fine-tuning) instead.

I´m torn here. Does anyone else have an opinion on the matter?
Remember, orbitals are on a logarithmic scale, a difference of 10 doubles the orbital radius, 0 is the "sweet spot" (1 AU for a G2V star) - so a range from -80 to +80 means radii can range from 0.004-ish AU (600,000 km) to 256 AU (38.4 billion km).


For hierarchical systems... EDG´s New Revised Stellar Generation Tables seem to suggest that distances between components of a hierarchical system are more often than not *really* big - anywhere from 1,000 AU to, looking at real-world systems, sizable fractions of light-years. That´s the kind scale where actual distances do not really matter in practice, because travel between components will take place by jump. So I am tempted to simply declare this the norm, and for practical purposes consider the components separate systems that happen to be located in the same map hex.
Again: Opinions? Or does anyone happen to know more about the "habits" of hierarchical systems?


Lastly for the moment, there is one thing that has me stumped.
I know that, in a binary system, there is a certain minimum difference between the orbital radii of the binary and any planets of either star (or circumbinary planets) that needs to be maintained for the planetary orbits to be - i.e. one orbital radius has to be X times the other. I´ve variously seen X stated to be 3.5 or 4. Let´s assume 4 right now for simplicity´s sake. So if a binary pair orbit each other at 5 AU, each can have planets in orbits with radii of 1.25 AU or less, and circumbinary orbits will be stable at 20 AU or further.
At least, my common sense (such as it is) tells me, that should be the case if both stars of a binary pair have equal mass. But what if both stars have different masses?
Could a planet around the more massive star A have an orbit further out from A than (in the above example) 1.25 AU, because B exerts less gravitational influence and A exerts more? And would a planet around B have to be closer than 1.25 AU for those same reasons? I would appreciate some input here.

_________________
Space isn't remote at all. It's only an hour's drive away if your car could go straight upwards. Sir Frederick Hoyle
Earth is the cradle of humanity, but one cannot live in a cradle forever. Konstantin Tsiolkovsky
Man has earned the right to hold this planet against all comers, by virtue of occasionally producing someone completely bat**** insane. xkcd #556
Just like people, stars can be very important without being terribly bright. Phil Plait, "Bad Astronomy"


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PostPosted: Sat Apr 08, 2017 1:05 am 
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Sir Chaos wrote:
Or does anyone happen to know more about the "habits" of hierarchical systems?


About all I know about multiple systems is that they are always structured as binary trees with objects at the leaves and never at the nodes. And the separation at any node is always at least four times the separation at either sub-node and sometimes much more. The obvious way to generate these at random is recursively.

Quote:
Lastly for the moment, there is one thing that has me stumped.
I know that, in a binary system, there is a certain minimum difference between the orbital radii of the binary and any planets of either star (or circumbinary planets) that needs to be maintained for the planetary orbits to be - i.e. one orbital radius has to be X times the other. I´ve variously seen X stated to be 3.5 or 4. Let´s assume 4 right now for simplicity´s sake. So if a binary pair orbit each other at 5 AU, each can have planets in orbits with radii of 1.25 AU or less, and circumbinary orbits will be stable at 20 AU or further.
At least, my common sense (such as it is) tells me, that should be the case if both stars of a binary pair have equal mass. But what if both stars have different masses?
Could a planet around the more massive star A have an orbit further out from A than (in the above example) 1.25 AU, because B exerts less gravitational influence and A exerts more? And would a planet around B have to be closer than 1.25 AU for those same reasons? I would appreciate some input here.

Yeah. I always see the radius of the zone of stable orbits stated as a multiple of the separation of the stars (3, 3.5, 4…), but it seems obvious to me that the figure ought to be related to the Hill Sphere radii of the stars. In the case of a large disparity of mass the usual rule-or-thumb would potentially suggest that things orbit the smaller star outside its Hill Sphere, which can't be right. I would use half the Hill Sphere radius at periapsis as the outer limit of any orbit around a star, and three (or four) times the separation as the minimum for a circumbinary.

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My SFRPG setting, Flat Black

© My posts on SFRPG must not be reproduced beyond the board except with explicit permission from me.


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PostPosted: Sat Apr 08, 2017 10:55 am 
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Agemegos wrote:
Sir Chaos wrote:
Or does anyone happen to know more about the "habits" of hierarchical systems?


About all I know about multiple systems is that they are always structured as binary trees with objects at the leaves and never at the nodes. And the separation at any node is always at least four times the separation at either sub-node and sometimes much more. The obvious way to generate these at random is recursively.


True. My general idea was to first treat each hierarchical system as a collection of binary pairs (plus a single solitar star, if it´s an odd number of stars), determine the separation of each pair; then "pair off" two binary pairs to determine the separation between them, using the wider of the two binary separations as the starting point to determine minimum separation between these two pairs; pair off that collection with another pair, using the widest separation among that whole gaggle as a starting point for minimum separation, and so until until all is taken care off.

Not that something this complicated is going to happen often, of course. With EDG´s tables, 1 in 36 systems will be trinaries, 1 in 86.4 (5/432) will be quarternaries; if you take rolling two 12´s in a row to mean there are 5+ stars in the system, that´s a 1 in 1296 chance. All of that in sectors with maybe 200 systems each.

Quote:
Yeah. I always see the radius of the zone of stable orbits stated as a multiple of the separation of the stars (3, 3.5, 4…), but it seems obvious to me that the figure ought to be related to the Hill Sphere radii of the stars. In the case of a large disparity of mass the usual rule-or-thumb would potentially suggest that things orbit the smaller star outside its Hill Sphere, which can't be right. I would use half the Hill Sphere radius at periapsis as the outer limit of any orbit around a star, and three (or four) times the separation as the minimum for a circumbinary.


Ah... complications... lovely! :mrgreen:

FWIW, I prefer 4 times the separation of the star, because in this system that means a stable orbit is possible 20 orbital shells away from the shell the binary pair occupies - a nice round number. I like nice round numbers.

Okay, off to go run some numbers through the Hill sphere formula.

_________________
Space isn't remote at all. It's only an hour's drive away if your car could go straight upwards. Sir Frederick Hoyle
Earth is the cradle of humanity, but one cannot live in a cradle forever. Konstantin Tsiolkovsky
Man has earned the right to hold this planet against all comers, by virtue of occasionally producing someone completely bat**** insane. xkcd #556
Just like people, stars can be very important without being terribly bright. Phil Plait, "Bad Astronomy"


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PostPosted: Sat Apr 08, 2017 1:22 pm 
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I assume the Hill Sphere formula here was meant for two bodies with a vast disparity in mass?
https://en.wikipedia.org/wiki/Hill_sphe ... d_examples

Because if the less massive body has more than 3/8 the mass of the more massive body, the Hill Sphere radius is more than half the semi-major axis - meaning the LESS massive body would dominate the attraction of bodies even in some cases when they were closer to the MORE massive body than to itself. For example if the less massive body has 8/9 the mass of the more massive body, its Hill Sphere radius is 2/3 (cube root of 8/27) the semi-major axis, meaning it includes object twice as far from it than from the more massive body. Which kind of doesn´t make sense, when you think about it. Or at least *I* cannot make sense of it - and unfortunately these are the sorts of mass disparities that we´re going to find among many a binary pair.


It occurred to me, however, that the rules so far already allow for smaller zones of stable orbits around the less massive (or at least less luminous) star of the pair:
The rules use orbital shells whose radius depends on the parent star´s luminosity. So if star A of a binary pair was identical to Sol, and star B had a luminosity 0.25 times that of Sol, the orbital radius of the "sweet spot" orbital shell (the one that gets the same amount of energy from its star that Earth gets from Sol) around star B would be half that of Sol´s sweet spot, i.e. 0.5 AU. Since a difference of 10 orbital shells equal half/twice the orbital radius, the orbital radius of star A´s shell 0 and star B´s shell 10 would be identical - 1.0 AU.
So let´s say these two stars orbit each other at a distance of 4.0 AU, which would be orbital shell 20 for the more massive star, A. That means that, under the base model for stars of indentical mass, planets could orbit each at shell 0 or further in - 1.0 AU for star A.
Of course, given the disparity in mass, an orbit 1.0 AU out from the less massive star, B, wouldn´t be stable. But orbital shell 0 for star B isn´t at 1.0 AU, it´s at 0.5 AU.

So, a vastly simplified and probably mathematically blasphemous idea would be to simply assume that the reduction in zone of stable orbits due to lower mass and the reduction orbital radii of shells due to lower luminousity are roughly equivalent - so that "take the orbital shell of the more massive star at which the binary orbits each other, subtract 20 and use the result as the orbital shell of the less massive star that is its outer limit for stable planetary orbits" produces somewhat reasonable results.
Unfortunately, for all I can tell from looking at real star data, using luminosity as a proxy for mass in that rule appears to vastly overstate differences in mass - our hypothetical star B would probably be midway through spectral type K with 2/3 to 3/4 the mass of Sun. So the outer limit for stable orbit would be a lot smaller under this rule than in reality.
The rule also doesn´t account for the more massive star having a larger outer limit for stable orbits than it would if its companion had the same mass.
Both problems could be adressed, though, by reducing the minimum separation between the binary orbit and planets around each star by one shell for every X points of luminosity difference, up to a certain minimum.

_________________
Space isn't remote at all. It's only an hour's drive away if your car could go straight upwards. Sir Frederick Hoyle
Earth is the cradle of humanity, but one cannot live in a cradle forever. Konstantin Tsiolkovsky
Man has earned the right to hold this planet against all comers, by virtue of occasionally producing someone completely bat**** insane. xkcd #556
Just like people, stars can be very important without being terribly bright. Phil Plait, "Bad Astronomy"


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PostPosted: Sun Apr 09, 2017 2:48 am 
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Sir Chaos wrote:
I assume the Hill Sphere formula here was meant for two bodies with a vast disparity in mass?


Yep. Here is a derivation:

www.jgiesen.de/astro/stars/roche.htm wrote:
The satellite or moon (mass μ) is orbiting the star (mass M) with the same angular velocity ω at the distance R+r as the planet (mass m) at the distance R (permanent full moon position).

The equilibrum condition for the planet is:
m ω2 R = G m M/R2

ω2 = GM/R3

The satellite is dragged by the combined gravitational forces exerted by the star and the planet:

μ ω2 (R+r) = G μ M/(R+r)2 + G μ m/r2

Inserting ω2:

G μ M (R+r)/R3 = G μ M/(R+r)2 + G μ m/r2

M (R+r)/R3 = M/(R+r)2 + G m/r2

M (R+r)3 r2 = M R3 r2 + m R3 (R+r)2

m R3 (R+r)2 = M r2 (R3+3R2r+3Rr2+r3) - M R3 r2

m R3 (R+r)2 = M r3 (3R2+3Rr+r2)

For r<<R: (R+r)2 ≈ R2, and 3Rr+r2 ≈ 0. The equation simplifies:

m R5 = 3 M r3 R2
m R3 = 3 M r3

r = R [m/(3M)]1/3

[right]link to source[/right]

I indicated in bold the part where the derivation assumes that r is small compared to R to simplify the expression.

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— Brett Evill

My SFRPG setting, Flat Black

© My posts on SFRPG must not be reproduced beyond the board except with explicit permission from me.


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PostPosted: Sun Apr 09, 2017 9:19 am 
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So "r" is the Hill Sphere radius, and the formula I found was the one for the case where the mass of the smaller body is negligible compared to the mass of the larger body - or more precisely, where the smaller body´s Hill Sphere radius is negligible by comparison. Presumably those will be cases like the illustration in the Wikipedia article, where the smaller Hill sphere is entirely inside the larger one.

That still leaves me, at least, without a way to calculate the respective Hill Spheres of elements of a binary pair, since I cannot see how I could resolve the unsimplified formula for "r", or for "R" for that matter.

Oh, by the way, in the German version of the Hill Sphere Wikipedia article, I found the statement that, for prograde rotation, orbits are stable at up to 50% of the Hill Sphere radius, and for retrograde rotation, at up to 70%. So your suggestion of "half the Hill Sphere radius at periapsis" looks spot on. And if we assume that for two bodies of equal mass, their Hill Sphere cannot possibly extend more than halfway to their companion in the direction of that companion, 50% of the Hill Sphere radius is exactly the 1/4 of the orbital separation between the stars of the binary pair we´re already assuming as the maximum anyway.

In any case, at this point the math is getting so complicated and, to be honest, beyond my capabilities, that I think I´m just going to stick with a gap of twenty shells (i.e. a 1-to-4 ratio in orbital radii for an equal mass binary pair) in each direction, and let the smaller orbital radii of less luminous stars work as proxies for their smaller Hill Spheres.

_________________
Space isn't remote at all. It's only an hour's drive away if your car could go straight upwards. Sir Frederick Hoyle
Earth is the cradle of humanity, but one cannot live in a cradle forever. Konstantin Tsiolkovsky
Man has earned the right to hold this planet against all comers, by virtue of occasionally producing someone completely bat**** insane. xkcd #556
Just like people, stars can be very important without being terribly bright. Phil Plait, "Bad Astronomy"


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PostPosted: Mon Apr 10, 2017 6:02 pm 
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All right... at this point, there´s only one more thing to "nail down" so that we can build at least the skeleton of a star system. We have an inner limit for planetary orbits, we have rules for how stellar companions impair planetary orbits, now we need an outer limit for the planetary system.

The planetary system is (other than the occasional captured object) the end result of a process that begins with the protoplanetary disk. I´m going to assume that more massive stars have more massive disks on average, and I´m going to make the (probably overly simplistic) assumption that the mean size of the disk scales with luminosity. That last assumption is important because it allows us to determine the size of the resulting planetary system in orbital shells rather than in AU or km, which we would have to laboriously convert to orbital shells, and also to use the same roll regardless of what star type we are talking about.

That said, a roll of 4d6*5 seems reasonable - anything from 20 to 120 shells is possible, with most results in the 50-90 shell range. That number is added to the inner limit in order to determine the shell number of the outer limit.

The inner limit of the planetary system (these numbers are a slight modification of what I posted previously on this) is at -40 for all stars of type M4V or brighter; from M5V on down, +5 per subtype is added to reflect that the Roche limit, not black body temperature, becomes the constraining factor - so the inner limit of an M9V star is -15 instead of -40. There´s also a random roll that may add +5, +10 or +15 to the limit for the purpose of determining the innermost occupied orbit, to makes system like Sol´s possible where the closest planet is on shell -18.
White dwarves, for what it´s worth, have an inner limit of 0, right on the sweet spot, but what with how dead a white dwarf´s planetary system is, that sweet spot isn´t so sweet any more anyway.

So, within these rules, the Sol system would have an inner limit of -40, list most stars, but with the random roll its planetary system would start at -25; the roll for the size of the planetary system would be 15, result in a system spanning 75 shells - out to shell 50, in other words, which is at 32 AU from the star and thus just outside Neptune´s orbit on shell 49.


Starting from the inner edge of the system, roll 1d6 to determine if the distance to the next orbit is going to be 3, 4, 5, 7, 8 or 9 shells. These distances more or less (some more, some less) correspond to 4:3, 3:2, 5:3, 2:1, 7:3 and 5:2 orbital resonances. Keep going until you make a roll that would put the next orbit beyond the outer limit. Don´t "skip" the parts of the system made unstable by companion stars when rolling; just roll and keep those orbits empty.

I´ve put the "gas giant limit" at shell 20 (4.0 AU for Sol); any planet on or beyond shell 20 is a gas giant. For the sake of simplicity I´m going to ignore the possibility of gas giants migrating so far in that they sweep the inner system empty and/or stay inside the gas giant limit.

If the two orbits directly inside shell 20 and the one directly outside shell 20 are occupied (i.e. not destabilized by a companion star), the two shells of the inner orbits become the boundaries of the asteroid belt. If the innermost of the those orbits is destabilized, the innermost shell that would be stable becomes the inner boundary. (e.g. if you have orbits at 12, 19 and 24, but a companion star at shell -5 destabilized the orbit at shell 12, shell 15 becomes the inner boundary of the belt)

Any other occupied orbit closer to the star than shell 20 is a rocky planet.

_________________
Space isn't remote at all. It's only an hour's drive away if your car could go straight upwards. Sir Frederick Hoyle
Earth is the cradle of humanity, but one cannot live in a cradle forever. Konstantin Tsiolkovsky
Man has earned the right to hold this planet against all comers, by virtue of occasionally producing someone completely bat**** insane. xkcd #556
Just like people, stars can be very important without being terribly bright. Phil Plait, "Bad Astronomy"


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PostPosted: Tue Apr 11, 2017 9:37 am 
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Note: At this point - or better yet, after the next step, determining the presence of moons - one could simply pick any one of the orbits of a system generated with these rules, declare its planet (or one of the moons of the planet) the mainworld, and continue with any version of Traveller worldgen from then on.

What I´m probably going to do, since habitable planets are the main focus of most games, is to concentrate on those systems most likely to have them. That is, rather than fully develop all 1,729 systems across nine sectors, focus first on those meeting the following conditions:
- F, G or K type stars
- no giant stars or white dwarves within the system that would have destroyed whatever life (and thus potential breathable atmospheres) the system may have held in the past
- no O, B or A type stars within the system, as such a system would no doubt be too young for complex life to have developed
- at least one occupied orbit around the "sweet spot" that is not disrupted by a companion star (I define "around the sweet spot" as the orbital shells from -5 to +5, rough the area from Venus´ orbit to Mars´ orbit for Sol)

That should cut down the list of systems to somewhere from 15 to 25% of the total.

Depending on how the planet generation rules end up, M type stars should be considerably less likely to have habitable planets, so those M type star that meet all the other above conditions (except for, obviously, being F/G/K type stars) would be mostly left for later - with the exception that those systems in a position to connect habitable systems, if they do not have a gas giant, would be checked for the presence of liquid water as a fuel source.


Edited to add:
A word about the time needed for star system generation... with light-to-moderate Excel support - I consider my Excel skill level to be barely average - generating the abovementioned 1,729 systems in nine sectors took perhaps 2-3 hours for placing the stars (i.e. rolling which hexes held stars, and marking them on a hex map in a cartography program), another 3-4 to determine number and type of stars in each system and their configuration, and perhaps another 5 hours (I´m not done yet) to determine all occupied orbits - i.e. everything there are rules for at the moment.
Not all of that effort scales with the area covered, like setting up the map and the Excel formulae, but by my estimate, a typical Traveller sector of around 300 system should take 2 to 2.5 hours to get this far - probably a bit less for someone with better Excel skills than mine.

_________________
Space isn't remote at all. It's only an hour's drive away if your car could go straight upwards. Sir Frederick Hoyle
Earth is the cradle of humanity, but one cannot live in a cradle forever. Konstantin Tsiolkovsky
Man has earned the right to hold this planet against all comers, by virtue of occasionally producing someone completely bat**** insane. xkcd #556
Just like people, stars can be very important without being terribly bright. Phil Plait, "Bad Astronomy"


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