A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Position-Finding
From: Hanno Ix
Date: 2014 Nov 5, 20:28 -0800
there might be an opportunity for another RIC!
Doniol can be re-written such that only one table is needed - the haversine table. There is one condition, though, that requires adopting "s2s" in stead of Hc: s2s is my abbreviation for "Star to Ship" or "Ship to Star" meaning the spherical distance between the ship's location and the substellar point. We know that sin(Hc) = cos(s2s). Now, I can show to you that Doniol assumes the following form which uses only a single trig function if one adopts this s2s convention:
hav(s2s) = hav(L-d) - [ hav(L-d) + hav(L+d) - 1 ]* hav(LHA).
I don't need to stress the great advantages of the haversine - they are all well known. These advantages plus the reduction of the number of tables to one, the haversine table, will justify the proposed change of convention from Hc to s2s I think. s2s has the additional nicety that "smaller" means "closer" - a welcome fact for teaching and memorizing. You know, I always found it peculiar that one explains CelNav by drawing circles on the globe but ends up not using their radius in actual work. The additional complexity of teaching “higher” is “closer” is a cumbersome but unnecessary detour and just a matter of tradition.
Now, here is an example / comparison: L =10; d =20; LHA = 40.
n = cos(L-d) = cos(10-20) = cos(20-10) = cos(10) = 0.9848;
m = cos(L+d) = cos(30) = 0.8660;
a = hav(LHA) = hav(40) = 0.1170;
So, sin(Hc) = n - (n+m)* a = 0.9848 - (0.9848 +0.8660) * 0.1170;
sin(Hc) = 0.7683;
Hc = asin(0.7683) = 50deg 12';
and s2s = 39deg 48';
b. Above formula:
hav(L-d) = hav(10-20) = hav(20-10) = hav(10) = 0.0076;
hav(L+d) = hav (30) = 0.0670;
hav(LHA) = hav(40) = 0.1170;
So hav(S2S) = 0.0076 - [ 0.0076 + 0.0670 - 1]* 0.1170;
= 0.0076 - [ - 0.9254] *0.1170 = 0.1159;
and s2s = ahav(0.1159) = 39deg 48';
As you can see both agree. One sign rule here is the universal arithmetic rule of
subtracting negative numbers which amounts to inverting both minus-signs -
a case arising occasionally. And yes, there are the facts that hav is always positive
and that hav(a-b) is always equal to hav(b-a). Not hard to remember, is it?
So, Greg, what do you think?