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    Re: Corrections for latitude when taking sights
    From: Brad Morris
    Date: 2019 Mar 13, 15:29 -0400
    Hello Brian

    I misunderstood your earlier post.  In an effort to more properly understand, I have resorted to HO71, Azimuths of the Sun.  The following is a numerical example of what I believe you are stating [if not, please provide a cogent example]

    For this example, I pretend to be at N50°.  The sun is at 23°S.  It is summer in the southern hemisphere.  Popping into HO71, I find the table for 50° with body declination (the sun) contrary name to latitude.  This occurs on 10 June, 3 July, 11 Dec and 2 January.  As the requirement is that it is summer in the southern hemisphere, we are restricted to 2 Jan.  The sun rises at 8hr02mn and sets at 3hr58mn apparent time, for a day length of 7hr56mn.  The azimuth of the sunrise is 127°26' east of north and the azimuth of the sunset is 127°26' west of north. 

    You now indicate that we should shift 90° in latitude.  50°N - 90° = 40°S.  I pop back into Azimuths of the Sun,  finding the table for 40°, declination same name.  Unsurprisingly, we find the entry for 2 Jan.  The sun rises at 4hr37m and sets at 7hr23mn, for a day length of 14hr46mn.  The azimuth to the sunrise is 59°20' east of south and the azimuth of the sunset is 59°20' west of south.

    Please explain what is wrong here.  At no time of the day will the sun ever be at an azimuth of 90° from your position. Or from the position 90° difference in latitude. It cannot.  The sun's declination ranges ~23° N/S.  But our position is 50°N (or 40°S, I am having a hard time with the explanation).  Therefore how will the azimuth ever be 90° ?


    Here is a second example.  I am at 50°N and the suns declination is 23° N.  Summer in the northern hemisphere so 3 July.  Azimuth to the sunrise 52°34' east of north.  Azimuth to the sunser 52°34' west of north.

    Shift 90° in latitude.  Latitude 40°S, contrary declination yields sunrise 120°40' E of S and sunset 120°W of S.

    The sun cannot be 90° or 270° from your position.  It will always be towards the S in the N hemisphere or towards the N in the S hemisphere.


    Please publish a mathematical example to clarify your remarks and help me to understand your assertion. 

    Thank you 

    On Wed, Mar 13, 2019, 10:17 AM Brian Walton <NoReply_Walton@fer3.com> wrote:

    Dave et al,

      You were asking what use could be made of sunrise/set tables and corrections to them for latitude.

       Fact:  the rise and set time of the Sun at a place 90° different in latitude to your latitude, is also the time the Sun is on an azimuth of  090 or 270 at your position, that is, on the Prime Vertical.  To see the Sun on the PV it must be summer.

       Look up your correction table, and find the correction to be made to your sunrise, to get the sunrise time at a place 90° different in latitude. Check out the azimuth of the Sun at your position, at that time, using the NavList Data app.  If it is not 090 or 270, it is because your correction table is rather crude.  It doesn’t matter much, but it tells you when to get your Sungun out.

       What use is a shot on the PV?  The trigonometry now becomes sin Alt=sin Dec/sin Lat.  Set your Sungun up to that figure, with normal corrections reversed, and go outside just before the time of PV. Need I go on?

       .I will.  If you were doing a time sight 150 years ago, you could enter your epitome and take logsinlat away from logsindec, and look up that figure in the same logsin table and take out an angle(Hc) and time(LHA) You have just precalculated your shot just as well as you did with sandwich fixes.  When you take the real shot time,  you compare your LHA to Greenwich ( maybe need equation of time in an old style almanac) and get Long.

       Those old guys were smart.


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