More Comments

A thread of 2 messages, 5 December 1990 to 6 December 1990, in the order they were sent.

#1944 More Comments

From: Orcinus orca [email removed]
John H. Kim says:

>To generate the position of each star/planet/moon requires
>six orbital elements for each object of interest, including
>the object you're viewing from.  Furthermore, the formulae
>involved, while not horribly complicated, are a pain to solve.
>I know, I've done it before.  You'd also need other information
>like mass, luminosity/albedo, and on and on to cover the other
>points you mentioned.

>All this is waaaaaay out of the scope of the Traveller rules.
>If you're still seriously considering this and there is no
>commercial software currently available, I'd recommend you
>get a good book on astrophysics (a background in higher
>calculus as about all you need, although more is helpful)
>and devote at least a month of your evenings to the problem.

All this is true however a simplified first cut at the problem might be
to use p^2 = a^3 where p is the period and a is the semimajor axis.  For
example suppose there is a gas giant around which 3 moons orbit.  The middle
moon is where the players are and they want to know where the other moons
are and when the gas giant eclipse the sun(s).  Calculate or set a period for
the middle moon with respect to the sun (this takes into account the motion
of the gas giant ie our moon has a period wrt the stars of 27 days but the
time between full moons is 29 day its period wrt the sun).  If you know the
semimajor axis of the other moon in terms of the semimajor axis of the middle
moon you can calculate the periods (or vice versa).  An example:

    Moon      semimajor axis  period
     1            1/4           1/8
     2             1             1
     3             4             8

Therefore if the middle moon has a period of 8 days then the inner moon is
1 day and the outer is 64.  Alternatively, given the periods you can find the
semimajor axis.  With this knowledge, pick where you want the relative
positions to be initially then as time passes you can estimate where the new
positions will be and can say where in the sky things are.  Using this data
and John Kim's angular size equation you can calculate the percentage of the
orbit of the moon that is in the shadow of the gas giant and therefore how long
the eclipse lasts.  This can be done for each moon.  In fact if you assume
circular orbits (which can be show to be generally the case as high ecc. objects
tend to get there orbits corrected) you can sketch out the system in question
for some aritrary time (like 001-1100) and then calculate how many orbits or
partial orbits have elapsed for each object and figure its current position
for any future time.  This ignores lots of interesting physics but for the
purpose of the game its sufficient.  For more details on this calculation
look up Kepler's Laws in any basic physics or astronomy book.

If you have a Mac (or access to one) there is a program called Voyager that
would be of some use.  There are PC programs as well, I just can't remember
their names.

TLH

- - --
T.L.Hayes                  |
MIT/Lincoln Laboratory     |
Lexington, MA              |

#1949 Re: (1944) More Comments

[ Good idea from T. L. Hayes about using only orbit periods to
  determine eclipses deleted]

If you're only interested in an occasional eclipse or two,
it's a great idea.  The impression I got from the original
letter (I don't save these for much more than a day so I
can't check) was that he wanted to know what the night sky
would look like given a certain time and place.  If you use
T. L. Hayes's simplification for this, you get the boring
(in my opinion) result of all the planets/moons/suns lined
up in one line in the sky.  I like to imagine the view at
Regina as something like the Saturn mosaic JPL released,
with a moon here and there.  I suppose you could fake it.

I've got the Saturn mosaic as a very well done digitized
bitmap (640x480x256) if anyone's interested.  You can even
pick out some of the background stars (I could almost swear
one group is the Pleiades...)
- - --
John H. Kim
[email removed]
[email removed]