POV-Ray : Newsgroups : povray.binaries.images : Ping Pong : Re: Ping Pong Server Time
2 Aug 2024 02:20:29 EDT (-0400)
  Re: Ping Pong  
From: alphaQuad
Date: 24 Mar 2008 11:55:00
Message: <web.47e7dc6241aa233d87aa2f830@news.povray.org>
"Kirk Andrews" <kir### [at] tektonartcom> wrote:
> No one's going to add some moons?
> Or a giant space craft of some kind?

I was kind of waiting for Kepler and everything in their place, but hey, go for
it (or help in the translation).


double SolveKepler(double M, double ec) {
  double E, d, delta;
  E = M;
  d = E - ec * sin(E) - M;
  while (fabs(d) > 1.0E-8) {
    delta = d / (1.0L - ec * cos(E));
    E = E - delta;
    d = E - ec * sin(E) - M;
  }
  return E;
}


Orbital************ string/calc/backtoString
 /* days, .DailyMotion,  .MeanLongitude, .Perihelion,
 *.Eccentricity, .Semimajor, .AscendingNode,.Inclination
 */
  double M, e, E, r, v, o, p, i;
  M = atof(data) * atof(gettok(data,2,32));
  p = atof(gettok(data,4,32));
  M += atof(gettok(data,3,32)) - p;
  e = atof(gettok(data,5,32));
  E = SolveKepler(M, e);
  r = atof(gettok(data,6,32)) * (1.0 - e * cos(E));
  v = 2.0L * atan(sqrt((1.0L + e) / (1.0L - e)) * tan(E / 2.0L));
  o = atof(gettok(data,7,32));
  i = atof(gettok(data,8,32));
  v = v + p - o;
 e = sin(v);
 p = sin(o);
 o = cos(o);
 v = cos(v);
 M = cos(i);
 sprintf(data, "%f %f %f",
      r * (o * v - p * e * M),
      -r * (e * sin(i)),
      r * (p * v + o * e * M)
   );



I have about 30 planet moon/maps and there is orbital elements to do for about
100 solar system objects with a default moon map.

Everything is accurate except I may have fudged some of the moon meanlongitudes
from Mean anomaly at epoch (Mo) which was the only thing that confused me and
hopefully we will get some assistance on that. I am not holding my breath.


I almost forgot the final position is:
Eclipmatrix = 1.0 0.0 0.0 0.0 0.917491720124239 0.39775487866708 0.0
-0.39775487866708 0.91749172012424

which is calced evertime you run the program:
alias obliquity {
  var t = julian(time) - 2451545.0
  t = t / 36525
  return (-23.496932 + 0.86 * sin((0.01532 * (t + 4.4))))

  e = A+Bsin(C(T+D))



Last 5 million yrs obliquity of the equator on the moving

For the next one million years the range will be only

Yet it remains an approximation in which the amplitude
of the wave remains the same, while in reality
as seen from the results of the Milankovitch cycles,
irregular variations occur.

but the low value may

}

popmatrix(Eclipmatrix,0.0 1.0 0.0) ecliptic pole

Eclipmatrix =
rotate ID matrix <radians(obliquity),0,0)


All of this could be wrong but is what I have assumed.
Today obliquity = -23.437924

ra dec of stars are final in their current position, which means you'll be able
to point a camera at a star


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