# NavList:

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

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Re: Manufacture new Bygraves?
From: Hanno Ix
Date: 2009 Jul 10, 16:01 -0700
Gary: thank you for the hint.

I can see why Thomas' scales would have an advantage, though. I wonder if you could remake your scales with the "gap" between the lines?

Regards

--- On Fri, 7/10/09, Gary LaPook <glapook@pacbell.net> wrote:

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=enmake 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=200902So 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"

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

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