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Just an aside here. Herman Dekker, you wrote:
"HA=cos^-1{(sin Ho - sin Dec·sin Lat) / (cos Dec·cos Lat)} "

I realize you were asking about a paper method in your last post, but getting back to a modern approach for a moment, and just to remind you...
We start, as you did, with:
   cos HA  = {(sin Ho - sin Dec·sin Lat) / (cos Dec·cos Lat)},
but we can get rid of all those parentheses...
   cos HA  = sin Ho / cos Dec / cos Lat - (sin Dec·sin Lat) / (cos Dec·cos Lat),
or 
   cos HA  = sin Ho / cos Dec / cos Lat - (sin Dec·sin Lat) / (cos Dec·cos Lat),
or since sin/cos is tan
   cos HA  = sin Ho / cos Dec / cos Lat - tan Dec · tan Lat.
And we can always replace any sin with a cos of 90° minus the original argument (replace Ho by ZD), so
   cos HA  = cos ZD / cos Dec / cos Lat - tan Dec · tan Lat.

That means we can calculate A, given by 
   A = cos ZD / cos Dec / cos Lat.
Then subtract from that B, given by 
   B = tan Dec · tan Lat.
Get the inverse cosine of the result, and that's the HA. In earlier decades, switching to tangents as I did here was frowned upon, since that meant you might need longer tables, but calculators have tan keys, and they're exactly as convenient to use as sin and cos. It's quicker. The underlying math, of course, is exactly the same. 

Frank Reed