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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

    You can download a manual for the K&E 4080

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

    gl





    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=item45fec52ab9

    gl

    --- On Thu, 11/24/11, Dave Walden <waldendand@yahoo.com> wrote:

    From: Dave Walden <waldendand@yahoo.com>
    Subject: [NavList] Re: Selection of slide rules, compendium of messages
    To: NavList@fer3.com
    Date: Thursday, November 24, 2011, 5:40 AM

    Look for a "T2" scale. A little hard to find, but $10 on ebay occasionally.
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