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    Re: Hav-Doniol re-visited
    From: Michael Bradley
    Date: 2020 May 4, 15:18 -0700

    Hello David

     …You asked about the meaning of the parameter C on Hanno Ix’s ‘Azimuth by Graphical Method’ diagram. No-one else has risen to the bait. I think I may have sorted i out,...

    Hanno has left us a clue. You can think of the label C and the c axis values as the clue, almost a craftsman’s mark. Some algebra is needed to spot what is going on, I hope it works for you.

     

    For a spherical triangle with angle LHA at the pole opposite the side (90-Alt), and azimuth angle z opposite the side (90 – Dec) we can use the Sine rule for the spherical triangle:

     

    Sin(LHA)/(90-Alt) = Sin(z)/(90-Dec) to find z, already having LHA, Alt, and Dec.

     '

    For ease of identifying the Hanno’s logic, we can convert this division calculation into a product form by cross multiplying 'Top of one side multipled by the bottom of the other side', to get : 

     

    (90-Dec)*Sin(LHA) = (90-Alt)*Sin(z)

     

    Hanno’s family of curves represents both the left hand side and also the right hand side of this cross multiplied Sine law. We can use Dec and LHA to “pick a curve” and read z from the point where the curve is intersected by Alt.

     

    Hanno has further transformed both sides of the cross multiplied Sine Law, by taking their Sines, reducing the label values to the range 0 to 1, his parameter C. He has actually graphed Sin(90-Dec)*Sin(LHA), the transformed left hand side, and will solve for z by using the same graph for getting z out of Sin(90-Alt)*Sin(z). This hurts nothing except our heads until we figure out how it works, but has the advantage of distorting the family of 100 curves, each for a separate value of C, so they together use the whole page, and are kept separate enough to be individually identifiable.

     

    So you still you identify ‘the graph of the sight’ from known values of LHA and Dec, and use that graph to look up the value z from the value of Alt

     

    I’ve checked my guess using approximate values from 3 sights:

     

    LHA 50, Dec 20, Alt 39 uses the curve labelled 0.72. Also Sin(90-Dec)*Sin(LHA) = 0.72

     

    LHA 31, Dec 09, Alt 38 uses the curve labelled (approx.) 0.51. Sin(90-Dec)*Sin(LHA) = 0.508

     

    LHA 122, Dec 50, Alt 25 using (180-LHA)   (Hanno’s instruction)

    ... ses the curve labelled 0.54. Also Sin(90-Dec)*Sin(180-LHA) =  0.54

     

    I may be wrong. Right or wrong, don’t worry about C, or the labels for the curves, Snopake them out if you wish. Just use them.

     

    I don’t recall anyone recently making the point that Hanno’s graphical solution for z works independently of the reduction method for ZD, and can give a very quick z with one sheet of paper, preferably laminated, some form of marker, and a ruler. 

     

    Good navigating        Michael Bradley  

       
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