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Hc by azimuth diagram
From: Hanno Ix
Date: 2013 May 27, 21:44 -0700
From: Hanno Ix
Date: 2013 May 27, 21:44 -0700
It occurred to me that the azimuth diagram I proposed some time ago
can be used to estimate Hc as well. Unfortunately, my main computer is dead,
so I will give you just a short description today. Samples will follow when I can
make drawings again.
You will remember that the diagram performs graphical multiplications of sin() and cos().
Obviously, a sine scale () can be read also as cos() table, and vice versa, by inverting the scale,
i.e from 90 to 0 vs 0 to 90. So we add a sine scale parallel to the cos() scale - nothing more.
Now, since sin(Hc) = sin(D)*sin(L) + cos(D)*cos(L)*cos(t) we
can use the diagram to
solve for Hc as well by executing the multiplications with the azimuth diagram enhanced as
just described. I am sure you can fill in the further details yourself. It is actually a simple process.
I have executed the procedure a number of times and received accurate values + 30 min. This is
entirely sufficient, I think, to check the execution of a more elaborate method for gross errors. And
no
more resources than you already have are needed.
The linearity of the angular scales has made this diagram rather easy to use for both,
azimuth and Hc. Let me know what you think.
h