5
$\begingroup$

With Astronomical Algorithms by J. Meeus (1998) I have obtained the relevant illuminated fraction of the disk and position angle.

But they're numbers (all Greek to me!) I know, Meeus explains that they're ratios of the lengths of the illuminated limb and areas.

Okay, so I have a ratio. How do I implement that together with the position angle to work out the direction of the illuminated limb and the amount illuminated?

I'm assuming that the ratio 1.0 is totally illuminated (i.e. Full Moon) and 0.0 is nothing (i.e. New Moon).

and then there is the position angle, I'm assuming that the direction in degrees is starting at 0 pointing upwards. Is that the crescent shape direction i.e. the horns are pointing away from the directional angle?

I would like to be able plot this so that I can show visually the moon's illuminated fraction and direction for any day in any given year.

As I said, I have the numbers, I'm trying to translate that into a more visually appealling and meaningful result.

Meeus says this is the selenocentric elongation of the Earth from the Sun (Phase Angle) - i.

the formula for illuminated fraction k is:

k = ( 1 + cos i ) / 2

tan i = (R sin l) / (D - (R cos l))

R = distance Earth-Sun D = distance Earth-Moon l = geocentric elongation

Meeus calls k the ratio of the lengths BC:AC the same as the ratio of the areas: NBSC:NASC

Where

Illuminated Fraction of Moon

$\endgroup$
  • $\begingroup$ If celestial north is up, this diagram shows an illuminated fraction of 0.35 and a position angle of 290$^\circ$. $\endgroup$ – Mike G May 8 '18 at 13:58
5
$\begingroup$

Your understanding of the illumination is correct. k is the ratio of the illuminated length BC to the diameter AC. New Moon has an illumination of 0, Full Moon an illumination of 1, and the quarter phase an illumination of 0.5.

The position angle is measured counterclockwise from north (celestial north, along a line of right ascension) to the bright limb C. You need to calculate the parallactic angle, q, which is the angle from "up" to celestial north. From chapter 14 (in the 2009 version) $$ \tan q = \frac{\sin H}{\tan \phi \; \cos \delta - \sin \delta \; \cos H} $$ where $\phi$ is the geographical latitude, $\delta$ is the declination of the Moon, and H is the hour angle. The diagram below shows a positive value for both q and PA.

Parallactic and Position Angles

$\endgroup$
  • $\begingroup$ You answered a question factually correctly. Thank you. However, I still want to implement this programmatically. I'm going to have to ask in Stackoverflow to figure out how to do this. $\endgroup$ – Danny F May 10 '18 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.