# NavList:

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

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Re: Selection of slide rules, compendium of messages
From: Gary LaPook
Date: 2011 Nov 26, 00:44 -0800
Here is a link to a previous posting going step by step solving the same problem using the sine-cosine method. The Hc found in 71° 06', a difference of 11.4'.

http://www.fer3.com/arc/m2.aspx?i=108339&y=200905

http://www.fer3.com/arc/m2.aspx?i=108365&y=200905

And here is a step by step procedure using the Bygrave formulas.

http://www.fer3.com/arc/m2.aspx?i=108342&y=200905

A simulation of solving for Hc using a 10 inch slide rule and the sine-cosine formulas is here:

http://www.fer3.com/arc/m2.aspx?i=108658&y=200906

The results are much worse than when using the Bygrave formulas as shown by the same type of simulation reported here:

http://www.fer3.com/arc/m2.aspx?i=108985&y=200907

`http://www.mccoys-kecatalogs.com/KEManuals/4080-3_1948/4080-3_1948.htmgl`

gl

--- On Fri, 11/25/11, Gary LaPook <garylapook@pacbell.net> wrote:

From: Gary LaPook <garylapook@pacbell.net>
Subject: [NavList] Re: Selection of slide rules, compendium of messages
To: NavList@fer3.com
Date: Friday, November 25, 2011, 11:49 PM

Paul Hirose worked out a sample problem using the Bygrave method on a 10 inch slide rule, using latitude 34° N, dec. 20° N and LHA 14°, see:
http://www.fer3.com/arc/m2.aspx?i=108411&y=200905

I decided to do the same sample problem on several slide rules. I also worked it out using a calculator so I know that the correct answer for Hc = 71° 17.4'. Paul got 71° 17' a difference of 0.4'.

Using a cheep Sterling plastic slide rule the Hc = 71° 18', difference of 0.6'.
Using a K&E 4081-3 I got 71° 18', difference 0.6'.
Using a K&E 4080-3 I got 71° 17', difference 0.4'.
Using a Pickett N803-ES I got 71° 20', difference 2.6'.
Using my Bygrave cylindrical replica I got 71° 18', difference 0.6'.
Using my flat Bygrave I got 71° 18', difference 0.6'.

So using a ten inch slide rule with the Bygrave method produces quite usable results but the two Bygraves produce better consistent accuracy.

A computer simulation confirms this, see:
http://www.fer3.com/arc/m2.aspx?i=108985&y=200907

gl

--- On Thu, 11/24/11, Gary LaPook <garylapook@pacbell.net> wrote:

From: Gary LaPook <garylapook@pacbell.net>
Subject: [NavList] Re: Selection of slide rules, compendium of messages
To: NavList@fer3.com
Date: Thursday, November 24, 2011, 1:21 PM

 The Stirling Decimal Trig Log-Log has the two tan scales which make it easier to use the Bygrave formulas. This is the first slide rule I ever had and I have had it since high school more than 40 years ago. there are several available on e-bay right now:http://www.ebay.com/itm/STERLING-Slide-Rule-DECIMAL-TRIG-LOG-LOG-1965-/300627077817?pt=LH_DefaultDomain_0&hash=item45fec52ab9gl--- On Thu, 11/24/11, Dave Walden wrote:From: Dave Walden Subject: [NavList] Re: Selection of slide rules, compendium of messagesTo: NavList@fer3.comDate: Thursday, November 24, 2011, 5:40 AMLook for a "T2" scale. A little hard to find, but \$10 on ebay occasionally.----------------------------------------------------------------NavList message boards and member settings: www.fer3.com/NavListMembers may optionally receive posts by email.To cancel email delivery, send a message to NoMail[at]fer3.com----------------------------------------------------------------
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