Yup - I was writing this too quickly and didn't double check the math involved. I was more concerned with the methodology, as to whether or not it would work properly. :)
Thanks Craig! -----Original Message----- From: [email protected] [mailto:[email protected]] On Behalf Of [email protected] Sent: Monday, October 10, 2011 1:12 AM To: The GURPSnet mailing list Subject: Re: [gurps] Planetary movement and checking the Math Couple problems with your numbers, here. On Sun, Oct 9, 2011 at 11:34 PM, Alaconius <[email protected]> wrote: > Hello Folks, > Hi, Alaconius, It has always been a sort of holy grail for me to try and find ways to > represent ship movement within a star system, along with planetary movements > concurrent with ship movement. One of those pesky things I've not been able > to pin down was planetary movement along an ellipse. I'm hoping that > perhaps even THAT issue will be overcome as of tonight. > > See, here's the problem. The velocity of a planet within its orbit along > an ellipse, is variable. The further away from the sun the planet is, the > slower its velocity becomes. The closer to its sun the planet it is, the > faster its velocity becomes. Kepler's second law of planetary motion states > that "A line joining a planet and the Sun sweeps out equal areas during > equal intervals of time". This in turn, gives me an idea for how to > determine where a planet is supposed to be at any given time... > > Suppose for example, you have a planet whose period is 365.24 days. You > know that the planet is 1 AU's distance (on average) from the Sun, and that > its orbital eccentricity is say, .02, then you know for instance, that at > its shorest axis, the planet's distance is 147,093,602 km, while its > furthest Distance is 152,097,053. Area of that particular Elipse is equal > to: > > Pi * A * B where A is the distance from the center of the Ellipse along its > shortest axis, and B is the Distance from the Center of the Ellipse along > its longest Axis. > > Or > > Pi * 147,093,602 * 152,097,053 km or 70,285,292,258,994,195.79 square > kilometers. > Okay, so far so good... > Now, if there are 365.24 days to that planet's orbit, then there are 86400 > seconds to the year in question. Stop. Here's your first stumbling block: A _day_ is 86,400 seconds long (24 hours times 60 minutes times 60 seconds). A year is 31,556,736 seconds (365.24 * 86,400) > The area swept out by the planet's revolution around its star would > therefore result in an area being swept out equal to its total area divided > by 86400 seconds or 813,487,178,923.54 square kilometers per second. > Divided by 31,556,736 seconds, or 2,227,267,492.40 sq km per second > All that remains then, is to determine what triangle will produce an area > equal to the value of 813,487,178,923.54 square kilometers where you know > two of the three sides of the triangle's length if you know the included > angle between those two sides. Once you know the included angle between two > sides, you can determine the last side of the triangle, and compute the area > of the triangle where it is equal to 1/2 Base times Height. > > Does this make sense? Have I missed anything? If this works, then I may > finally have a means for plotting the location of any given space craft for > use in a GURPS campaign (be it GURPS TRAVELLER, GURPS TRANSHUMAN SPACE or > even GURPS SPACESHIPS. > > Thoughts? Comments? > Otherwise, I think you're okay. But I'm only working on 4 hours sleep (getting the baby back to sleep) -- Craig Roth [email protected] http://www.google.com/profiles/craig.roth _______________________________________________ GurpsNet-L mailing list <[email protected]> http://mail.sjgames.com/mailman/listinfo/gurpsnet-l _______________________________________________ GurpsNet-L mailing list <[email protected]> http://mail.sjgames.com/mailman/listinfo/gurpsnet-l
