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    Re: An All-Haversine Azimuth (from Lat, Dec and dLon alone)?
    From: Willi Strohl
    Date: 2021 Mar 16, 02:27 -0700

    Hello Tony,

    If you are interested in a method of finding a position line only by using haversine or archaversine functions, here is an alternative way to do so:

    The idea is to use two arbitrarily chosen latitudes (in the vicinity of your assumed position) and calculate the corresponding longitudes. So you get two geographically defined points. – The line, which goes through both points is the desired position line. - No calculation of the azimuth is necessary.

    Precedure:

    1. Select your celestial body, measure the height h with the sextant and record the time UT1 of the measurement (as usual).

    2. Estimate the appropriate Grt and the declination δ of the selected celestial body using the Nautical Almanac (as usual).

    3. Pick two latitudes ϕ1 and ϕ2  which are roughly in the vicinity of your currently assumed position. They should be about 5 to 20 nm apart from each other, depending on the scale of the nautical chart you are using.

    4. Calculate the longitude ∆λ1 which corresponds to the latitude ϕ1  by using the haversine formula:

    hav(∆λ1) = [hav (90-h) - hav (ϕ1-δ)] / [1 – hav (ϕ1+δ) - hav (ϕ1-δ)]

    ∆λ1 = archav[hav(∆λ1)] , note that there are always 2 possible solutions: ∆λ can be positive or negative!

    5. Calculate the corresponding preliminary  longitudes λ11 and λ12:

    λ11* = + ∆λ1Grt   à       If |λ11*| < 180° then λ11= λ11* ;  If λ11* < -180° then λ11= λ11* + 360°

    λ12* = - ∆λ1Grt    à       If |λ12*| < 180° then λ12= λ12* ;  If λ12* < -180° then λ12= λ12* + 360°

     

    6. Of λ11 and  λ12 pick the one which is closest to your assumed position. This is the corresponding longitude λ1 to the chosen latitude ϕ1.

    7. Repeat steps 4. to 6. in order to get λ2 using ϕ2 in the haversine formula (step4.).

    As final result you will get the two desired coordinate pairs ϕ1/ λ1 and ϕ2/ λ2 to draw your line of position.

     

    Important for the calculation:

    Latitudes ϕ and declinations δ can take values from -90° to +90°,

    (-) corresponds to southern latitudes or declinations

    (+) corresponds to northern latitudes or declinations

    Longitudes can take values from -180° to +180°

    (-) corresponds to longitudes west of Greenwich

    (+) corresponds to longitudes east of Greenwich

     

    Advantages of this method:

    + No exact dead reckoning position is required, you just need to know roughly where you are.

    + Only one haversine table and the nautical almanac are needed to solve the problem

    Disadvantages:

    - You need to do a manual division in step 4. for each coordinate pair (or use a calculator).

    - The procedure doesn’t work, if the celestial body has an azimuth of 180° or 0° (or close to it). If you chose Polaris for example, there will only be one possible latitude. The line of position in this case would be the latitude itself.

    Lycka till and best wishes

    Willi

       
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