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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Zero-Altitude Shot
From: Paul Hirose
Date: 2019 Sep 6, 21:14 -0700

```On 2019-09-03 22:01, Bob Goethe wrote:
> Image1 is a copy of my worksheet, with extraneous fields erased.  The key
field in question is field 8.5, the Hc of -1° 16.0'.
> This seems to make sense to me when I work it out with Pub. 249.
> My question has to do with direct calculation of this sort of problem with
the Bygrave equations.  In the nature of things, one will seemingly never get
a negative value for Hc with these equations as they stand.
> My direct calculation values for Z and Zn agree with what I got from Pub.
249, seen in Image2.  But the Hc comes out positive instead of negative when
I reduce it with Bygrave.
>
> Attached File: http://fer3.com/arc/imgx/Image2.png
> Attached File: http://fer3.com/arc/imgx/Image1.png

I haven't looked in detail at your Bygrave implementation, but in the
calculation of X you subtract W where I would add. I get Hc = 12.7486.

My version of the Bygrave formulas:

W = arctan(tan dec / cos LHA)

If LHA is between 90 and 270, replace W with its supplement: W = 180 - W

X = 90 - lat + W (if lat and dec same name)
X = 90 - lat - W (contrary name)

If X is not 0 to 180, add or subtract 180 to make it so. If the
adjustment is necessary, altitude is negative.

Z = arctan(tan LHA cos W / cos X)

If (altitude < 0) or (X < 90), replace Z with its supplement:
Z = 180 - Z
Do the adjustment only when EXACTLY ONE of the conditions is true.

Hc = arctan(tan X * cos Z)
Apply negative sign if applicable.

NOTES

Ignore negative signs in trigonometric functions. All cosines and
tangents are positive. The only possible negative number is X, and then
it is immediately adjusted to a positive number.

Latitude and declination are always positive, "same name" or "contrary
name."

Z is the conventional azimuth angle in the range 0 to 180, measured east
or west from the elevated pole (north pole if assumed latitude is north).

The cosine of 90 or 270 is zero, and therefore a forbidden "divide by
zero" can occur in the W or Z formula. In that case, skip the formula
and make the result 90.

My Bygrave solution is valid for every combination of LHA, latitude, and
declination, including negative altitude. For example, here are
parameters for a sunset observation.

95.3057 = LHA
40 = lat
5 = dec (same name)

Bygrave solution:

W = arctan(tan 5 / cos 95.3057)
= 43.3154

LHA is between 90 and 270, so adjust W.

W = 180 - 43.3154 = 136.5855

X = 90 - 40 + 136.5855
= 186.5855

That's outside the range 0 - 180, so subtract 180. The adjustment means
altitude is negative.

X = 6.5855

Z = arctan(tan 95.3057 cos 136.5855 / cos 6.5855)
= 82.7620

No Z adjustment is necessary since (altitude < 0) and (x < 90).

Hc = arctan(tan 6.5855 cos 82.7620)
= .8333 = 50.00′

Apply a negative sign as indicated in the computation of X. This is the
nominal altitude of the Sun center at rise or set. The Bygrave
calculation of this negative altitude was accurate.
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