NavList:

From: Gary LaPook <glapook@pacbell.net>
Subject: [NavList 9069] Re: Manufacture new Bygraves?
To: NavList@fer3.com, glapook@pacbell.net
Date: Friday, July 10, 2009, 2:06 PM

It isn't a major problem but does require different procedures.

I started this thread back in February:

http://groups.google.com/group/NavList/browse_thread/thread/ac94da3086897276?hl=en

make sure you open up the first post to show quoted text.

Most of you questions will be answered..

Some portions of that thread.



When declination is less than 55' on my version (less than 20' on the 
original) you can't compute "W" because you start the process with 
declination on the cotangent scale. In this case, Bygrave says to use the 
same process as when the azimuth exceeds 85º, you simply interchange 
declination and latitude and compute altitude. But Bygrave didn't tell us how 
to calculate azimuth in this case. In my testing I have found a method that  
produces quite accurate azimuths. You simply skip
 the computation of "W" and 
simply set "W" equal to declination. The  worst case I have found is that the 
azimuth is within 0.9º of the true azimuth but most are much closer. If the 
declination is less than one degree and the latitude is also less than one 
degree, follow this procedure and also assume a latitude equal to one degree. 
After you have computed the Az you then follow the same procedure discussed 
above for azimuths exceeding 85º by interchanging the latitude and 
declination and then computing  Hc.


http://www.fer3.com/arc/m2.aspx?i=107414&y=200902

So your MHR-1 doesn't provide for cases where the declination is less than 
20', the lowest mark on the cotangent scale. Bygrave says to use the same 
procedure in this case as in the case where the azimuth is near
 90?, simply 
interchange the declination and the latitude and then compute the altitude 
and this works fine and you get accurate altitudes. But the azimuth that you 
derive in this process is not the correct azimuth and is thrown away and not 
used for plotting the LOP just as in the case of azimuths near 90?. Bygrave 
gives no instruction for computing the azimuth in this case. It cannot be 
computed in the normal way since the first thing you need to to is find 
declination on the cotangent scale and values less than 20' are not on the 
Bygrave or MHR-1. This is an important special case since the sun's 
declination is in this range for several days around each equinox. I 
developed an approximation that works well giving azimuths within one degree 
of he correct value and usually much closer. I simply skip the first step, the 
derivation of "y" (Bygrave's terminology), "W" (my terminology). I simply set 
"y"
 equal to declination and then proceed normally.

http://www.fer3.com/arc/m2.aspx?i=107947&y=200904



http://groups.google.com/group/NavList/browse_thread/thread/20ac6296103b4c7e?hl=en

gl

Hanno Ix wrote:

Gary and Thomas: Could please one of you educate me/us somewhat about the "equinox problem" you discussed? Regards