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Like the Metonic cycle, except for position in the sky, not phase.

Original question: If I look up at the moon tonight, how long before it is in the same apparent position, and same phase, again?

I realized I can calculate this myself, if only I know two piece of information. Unfortunately I do not know the name of the second.

My question should be answered by the product of:

  1. the Metonic cycle, and
  2. the cycle where the moon's azimuth and altitude recur on a given day

What is the name of this second cycle?

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You are looking for the Metonic cycle. This is almost exactly 19 years, and represents 235 synodic months.

Now you have edited the detailed question so that it is out of sync with the original Title. The other period you need is the Draconic month, which is the period between the moon crossing the ecliptic (in the same direction).

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  • $\begingroup$ Ok, so the Metonic cycle is when the same lunar phase will recur on a given date, correct? What about the lunar azimuth and altitude? $\endgroup$ – apraetor Jun 22 '16 at 18:47
  • $\begingroup$ Yes, they are pretty close too. To cite Wikipedia: 235 synodic months (lunar phases) = 6,939.688 days (Metonic period by definition). 254 sidereal months (lunar orbits) = 6,939.702 days (19 + 235 = 254). 255 draconic months (lunar nodes) = 6,939.1161 days. $\endgroup$ – Dr Chuck Jun 22 '16 at 19:37
  • $\begingroup$ Wouldn't ~0.57 of a day still be quite a big discrepancy if you're looking for the apparent position and phase to coincide? $\endgroup$ – apraetor Jun 22 '16 at 21:52
  • $\begingroup$ Thanks for your help in working through this, btw. I appreciate it. $\endgroup$ – apraetor Jun 22 '16 at 22:00
  • $\begingroup$ When you say coincide, you need to think about how accurately you want that to be. If you have planetarium software eg Cartes du Ciel, which is free, you can do a simulation. For example comparing the moon's position now and 38 years (ie 2 *19), there is about a 1 degree difference. $\endgroup$ – Dr Chuck Jun 23 '16 at 8:37

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