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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: The Bygrave Slide rule
Date: 2009 May 29, 13:36 -0400

```I brought out the extended markings to help clarify the discussion!  I
realized that tan(89) does not equal the tan (91), yet the value denoted on
the scale is the same for either tan(89) and tan(91).  The same applies to
the outer scale, there is a sign change at 90 degrees, although the abs
values are the same.

While I remain interested in resolving what the scales actually represent, I
don't know that we have enough information yet. Peter has essentially
utilized trigonometric identities to "flip the problem over".  Neither
argument is compelling enough to state, hey, that's the answer.  No argument
accounts for the absolute value, yet that must occur.  The scales are
logarithmic, yet there are no logarithms of negative numbers.

Best Regards

-----Original Message-----
From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of NavList@fer3.com
Sent: Friday, May 29, 2009 5:46 AM
To: NavList@fer3.com
Subject: [NavList 8441] The Bygrave Slide rule

Brad's point is interesting - he points out that the reversed 90-180 degree
markings on both scales show that the scales are logs of abs(tan) and
abs(cos) respectively. Maybe this is just a 'kludge' for the annoying fact
that logs of negative numbers don't exist. It's left to the user to sort out
the signs when using logs to work with mixed-sign numbers.

However, I don't think Brad's point does impact on my postings, since it
applies whether or not the scales are 'thought of' as cotan/cos or
tan/secant.

In effect all I am saying is that I found the references to cotangent scales
in the papers by Ron and Gary were counter-intuitive, and the alternative
tangent/secant interpretation was easier to grasp, so I thought it worth a
posting. If no-one else agrees with me then I will go back into my cave.

I can't contribute an answer to Gary's puzzle about the apparent impossibility
of using the tangent/cosine method at the equinoxes (this puzzle is not
limited to the Bygrave solution). A quick look at the celestial triangle
tells me that if the sun is near the equator then the perpendicular from the
sun to the observers meridian is also near the equator, so the equation for
azimuth simplifies to tan(A)=tan(H)/cos(c), but this may already be what Gary
has rejected as not being very pretty. I wondered whether the discarded
azimuth in the alternative calculation might be a clue, but this is just
equivalent to swapping the observer with the sun. The altitude comes out the
same of course, but the discarded 'false' azimuth is then seen to be the
azimuth of the observer as seen from the sub-solar point, which isn't very
useful, even on March 21st.

regards
Peter M

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