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    Re: azimuth of polar star
    From: Gary LaPook
    Date: 2011 Jan 19, 10:23 -0800
    If you calculate some test azimuths of Polaris by using the Bygrave procedure you will immediately see why it is so insensitive. The first step of the computation is finding an intermediate value ( I labeled it "W", Bygrave called it "y") which is  arc tan W = (tan declination / cos hour angle).

    Since the current declination of Polaris is 89° 18.9', its tan is 83.639. The cos of 1' is 0.999 and the cosine of 89° 59' is 0.0003 so 'W" is limited to the very narrow range of 89.315°  to 89.999° ( 89° 18' 54.0" to 89° 59' 59.2" ) no matter what the hour angle is. Since "W" is the largest factor in the calculation of azimuth this narrow range greatly limits the range of the azimuth and makes it very insensitive to changes in hour angle and latitude.

    Below is a list of the keystrokes for doing this on a calculator from my website at

    https://sites.google.com/site/fredienoonan/other-flight-navigation-information/modern-bygrave-slide-rule

    gl


    --------------------------------------------------------------------------------------------------------------

    CHECKING YOUR COMPUTATIONS

    An easy way to check the computation on a Bygrave is to do the same computation on a calculator since this allows you to check the intermediate steps.

    Just use the standard Bygrave formulas in the three step process following along on the form I have posted.

    First calculate co-latitude and save it in a memory in the calculator. If you are using a value for hour angle that is not a whole number of degrees you might want to make the conversion to decimal degrees and save it in a memory since it will we used twice. If you are using whole degrees then this step is not necessary.

    Then you calculate "W" using the formula:

    tan W = tan D / cos H

    and sum it to the memory where you have saved co-latitude which is then X and then make any adjustment necessary to convert X to Y. (If you are just making trials you can avoid this step by your choice of the trial values.) There is no reason to store W itself since it is not used again. You can then convert W to degree and minute format to compare with the Bygrave derived value.


    Then you compute azimuth angle using the formula:

    tan Az = (cos W / cos Y ) x tan H.

    If you want you can also convert Az to degree and minute format to compare with the Bygrave.

    The last step is to calculate altitude with the formula:

    tan Hc = cos Az x tan Y.

    Then convert to degree and minute format to compare with the Bygrave result.



    (When entering values in the format of degrees minutes seconds, change decimal minutes to seconds, 6 seconds per tenth of a minute, in your head  before punching in the assumed latitude, declination  and hour angle if necessary.)

    Using whole degrees for declination, assumed latitude and hour angle, using a TI-30 with only 3 memory locations the key strokes are:

    ---------------------------------------------------------------------

                 (co-latitude = 90 - Assimed latitude)

    90
    -
    Assumed Lat
    =
    STO 1  (co-latitude stored in memory 1)

    ---------------------------------------
                   
                  
    (tan W = tan D / cos H)


    Declination  

    tan
    /
    H
    cos
    =   
    inv
    tan    (computed W)
    SUM 1  (X now stored in memory 1)(change X to Y if necessary)

    --------------------------------------


          
    (tan Az = (cos W / cos Y ) x tan H)

    cos    (of W from prior step)  

    /
    RCL 1  (recalls Y from memory 1)
    cos
    x
    H
    tan
    =
    inv
    tan (computed Azimuth angle)



    ------------------------------------

          
    (tan Hc = cos Az x tan Y)


    cos    (of Az from prior step)  

    x
    RCL 1  (recalls Y from memory 1)
    tan
    =
    inv
    tan  (computed altitude, Hc)


    2nd
    D.D - DMS (changes Hc in decimal degrees to degrees, minutes and seconds)



    DONE



    --- On Tue, 1/18/11, P H <pmh099---.com> wrote:

    From: P H <pmh099---.com>
    Subject: [NavList] Re: azimuth of polar star
    To: NavList@fer3.com
    Date: Tuesday, January 18, 2011, 11:32 AM

    From what I would guess (and also from reading the site with my rather limited knowledge of German) it is intended to increase the accuracy of determining true north from looking at the direction of Polaris.  This could perhaps be used for checking the gyroscopes.

    As for the sensitivity of the computed azimuth to inaccuracies in input data, I did the following calculation:

    2011 Jan 18, UT = 00:00:00
    N 40.0,  W 120.0:   Az = 37.4'
    N 40.5,  W 120.0:   Az = 37.6'
    N 40.0,  W 121.0:   Az = 38.0'

    Five minutes later from the same reference position:
    N 40.0,  W 120.0:   Az = 36.5'

    And now with all three "errors" present:
    2011 Jan 18, UT = 00:05:00
    N 40.5,  W 121.0:   Az = 37.5'

    It appears that the (in)sensitivity of azimuth to input data is greatly aided by the star's closeness to the Pole.  After all, if it were exactly at the Pole, we'd have Az=0 always, regardless of time and location.  It very well may be (I am not sure) that a mechanical computer (such as a slide rule) could match the requisite 1' sensitivity - that is a separate question, which other NavList members are in a better position to answer.


    Peter Hakel



    From: Ronald van Riet <ronald{at}van-riet.nl>
    To: NavList@fer3.com
    Sent: Tue, January 18, 2011 8:41:10 AM
    Subject: [NavList] azimuth of polar star

    A slide rule (well not quite, but close enough to be called a slide rule) exists that was used to determine the azimuth of the pole star: the P.A.44 (http://www.rechenschieber.org/pa44.html web page in German).

    Does anyone know of a reason why the azimuth of the polar star is important enough to develop a dedicated slide rule for it?

    Also, they claim that with a half degree of latitude, a full degree of longitude and about five minutes time accuracy, the azimuth could be determined with a 1 minute of arc precision. Could that be a valid claim? It seems a bit too precise with the imprecise inputs...

    thanks

    Ronald
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