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Guy wrote

> I was working out a problem here are the particluars
> DR 40 deg 9 min N
> 45 deg W
> object Regulus
> Dec 11deg 59. 9 min N
> LHA 16:25:48
> HC 58 Deg 20.4 Min
> Using HO 249 the Z is 150 and the ZN is 210 I also checked with the star
> section and the ZN was the same.
> When I did the calculations using
> Cos rasied to -1 (Zn) = sin (dec) - Sin (lat) x sin (Hc) / cos (Hc) x cos
> (lat)
>
> The answer came out to 148 deg 11.4 min which is close to the 150 Z however I
> though that formula gave me the Zn?

Yes and no, it is situational as the HC can be the same at some point before
and some point after noon (using the sun as an example). With the sun LAN
(local area noon, sun at its highest point) is also the upper meridian
passage. For any body, LHA = 0 for upper meridian passage.  To extract Zn
from Z,  determine whether the body is rising (pre meridian passage) or
setting (post meridian passage.

If rising, Zn = Z.  If setting, Zn = 360 - Z.

When in doubt, if the sine of LHA is positive, Zn = 360 - Z.  When the sine
of LHA is negative, Zn = Z.

My memory device is "plus minus."  When the sign of sine LHA is PLUS, Zn =
360 MINUS Z.

Bill

   

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