NavList:

   

Reply

On 10/2/2023 6:36 AM, NavList Community wrote:
> As you can see, in the diagrams, the Moon jumps around too much in
> azimuth for great accuracy in this situation, because your USNO Clone
> only gives Zn to the nearest 3'.

If you need extra precision in az/alt, try JPL Horizons:

https://ssd.jpl.nasa.gov/horizons/app.html#/

It gives these values for azimuth, unrefracted altitude, and angular
diameter (arcsec) during the partial eclipse 2014-10-14 18:00:00 UTC
near Albuquerque (N 35 W 107 +1.6 km):

160.618593 44.507534 1826.018 Moon
160.959776 44.912635 1924.145 Sun

(That time and place because I have notes from a previous computation.)

Calculate position angle (Sun to Moon, zero toward zenith) and
separation angle with any suitable method. Less precision is adequate
for PA since the separation is so small.

148.97 PA (Moon about 7 o'clock from Sun)
.4720 separation

semidiameters from Horizons:
.2536 Moon
.2672 Sun

The body centers and limb intersections form two "mirror image"
spherical triangles. We only need one. The sides of the triangle are the
semidiameters and the separation angle.

Angle B is the angle, at the Sun center, that subtends half the "cookie
bite" the Moon takes from the limb. Side b is Moon SD (.2536), side c is
separation angle (.4720), and side a is Sun SD (.2672).

Calculate B with any of the formulas from Chauvenet paragraph 22:

https://archive.org/details/manualofspherica01chauiala/page/38/mode/2up?view=theater

I get B = 24.3. Add and subtract that from the Moon PA (149.0) to get
the PA at each edge of the bite (124.7 and 173.3). The bite does not
include PA 180, so a lower limb altitude observation is feasible at 1800
near Albuquerque.

The triangle is so small, I think we could omit the formally correct
spherical solution and just solve the triangle on a plane.

--
Paul Hirose
sofajpl.com

   

Reply