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I followed The Moon's topocentric position to lern how to calculate the moon's position.

  1. I have calculated topocentric RA and Decl, but I don't know how to convert it to the moon's horizon coordinates, so is there a formula or resource?
  2. I have calculated geocentric hour angle, so after calculating topocentric RA and Decl, should I calculate the topocenteric hour angle before converting topocentric RA and Decl to the moon's horizon coordinates?
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    $\begingroup$ See section 6 of your resource "Hour angle, altitude and azimuth" The horizon coordinates are the altitude and azimuth. $\endgroup$
    – James K
    Commented Oct 22 at 21:21
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    $\begingroup$ @JamesK Thank you, but what about number 2? $\endgroup$
    – Ahmed Dyaa
    Commented Oct 22 at 22:01
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    $\begingroup$ Sorry, I didn't get it, in my resource he calculated the HA by these formula HA = LST - RA = 272.3377_deg he computed it by the geocentric RA, so after I compute the topocentric RA should I recomoute the hour angle with the topocentric RA or not? $\endgroup$
    – Ahmed Dyaa
    Commented Oct 22 at 22:26
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    $\begingroup$ I just asked ChatGPT and this it's answer, When you correct the geocentric RA to topocentric RA, you're accounting for the observer's position on the Earth's surface, which means the topocentric RA is now specific to your location. This adjustment can slightly change the object's position in the sky from the observer's perspective. Since the Hour Angle depends directly on the RA, and you've now updated the RA to the topocentric value, you must recompute the HA with this topocentric RA to get an accurate value for the Moon's current position. $\endgroup$
    – Ahmed Dyaa
    Commented Oct 23 at 10:32
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Ahmed Dyaa
    Commented Oct 23 at 13:59

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Your question implies that you know how to calculate the Moon's geographical position (GP): the latitude and longitude of the point on the Earth it's directly above, so, first, find the distance between the Moon's GP and the observer using the law of cosines for spherical triangles applied to the triangle with vertices at the north pole, observer's position and Moon's GP.
Letting "a" be that angular distance, "b" be 90 deg minus observer's latitude, and "c" be 90 deg minus the Moon's GP, and the opposite angles for a,b,c be A,B,C; a = acos((cos(b)cos(c)-sin(b)sin(c)cos(A)). (A is the difference between the two longitudes.) (Remember that acos is double valued.)

The moon's geocentric altitude is 90 deg minus a. Its geocentric (and topocentric) azimuths can best be found by applying the law of sines for spherical triangles: sin(A)/sin(a) = sin(C)/sin(c), so asin(C) = sin(A)sin(c)/sin(A), (also double valued.)

That finishes all that's needed to know about geocentric coordinates. To correct for tropocentric altitude, apply the horizontal parallax factor (HP), which comes with your almanace's RA and declination, usually. From your reference's formula the parallax correction is HP*cos(geocentric altitude), which is subtracted from the geocentric altitude.

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  • $\begingroup$ Thanks for your answer, but I tried the formulas and the results is wrong, the used formula: double a = 90 - Math.Acos(Math.Cos((90 - latitude) * Deg2Rad) * Math.Cos((90 - eclipticLatitude) * Deg2Rad) - Math.Sin((90 - latitude) * Deg2Rad) * Math.Sin((90 - eclipticLatitude) * Deg2Rad) * Math.Cos(Math.Abs(eclipticLongitude - longitude) * Deg2Rad)) * Rad2Deg; then double _geoAltitude = 90 - a; then double _topAltitude = _topAltitude - lunarParallax * Math.Cos(_topAltitude * Deg2Rad);. Note: 1- Deg2Rad = Math.PI / 180.0 and Rad2Deg = 180.0 / Math.PI, 2- I used the formulas on c# script. $\endgroup$
    – Ahmed Dyaa
    Commented Oct 25 at 11:59
  • $\begingroup$ I just tried the the formulas in section 6 of the mentioned resource and I have already computed the Sidereal Time at the Greenwich meridian at 00:00, the LST for the time meridian and The HA then I converted the HA and Decl to rectangular (x,y,z) coordinate, then computed the azimuth and altitude but the results was wrong, and I tried these resource geoastro starting with computing sidereal time at Greenwich then LST at longitude then local HA then I converted them to altitude and azimuth and the results are correct. why? $\endgroup$
    – Ahmed Dyaa
    Commented Oct 25 at 12:01
  • $\begingroup$ Can you add detail about the wrong result? Your code excerpt , without context, up to the first "then" looks good. Your "a" is the geocentric altitude. The topocentric altitude (your _topAltitude" I believe) should be that geocentric altitude minus the parallax correction, HPcos(a). $\endgroup$
    – stretch
    Commented Oct 27 at 22:03

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