Very preliminary.

A Draft Report on Constructing and Calculating Almucantar Circles, in
two parts. Okay, three.

Almucantar circles are "nested but not concentric circles" used to
denote the latitude of celestial events in an Astrolabe. A quick
history review of Astrolabes: Astrolabes were a Hellenistic invention
first made between 220 and 150 BC. They were improved by Muslim
astronomers, and in the late 14th Century, Geoffrey Chaucer compiled
_A Treatise on The Astrolabe_. Astrolabes worked "well enough for the
purpose" that they stayed in use until the 17th century, when
"reflecting instruments" made the work of celestial observation and
navigation "simpler."

Astrolabes work by means of stereographic projection of the circles of
latitude from the sphere of the heavens to the plane of the chart: the
above mentioned "Almucantar" circles, from the Arabic for "Sundial". Constructing these circles is a rather straight forward exercise in
the use of compass and rule.

Part 1: Stereographic Projection through Geometric Construction:
The "easiest" way is to construct the circles with a compass and rule.
Tedious, but "easy".
[insert diagrams]
Thus it was done in Chaucer's time. Simple, eh no?

Part 2: Calculating the Stereographic Projection.
Calculating it is a bit more complicated, particularly when your
numbers do not work out the first several times.
From my notes:
The unit sphere in three-dimensional space R3 is the set of points (x,
y, z) such that x2 + y2 + z2 = 1. Let N = (0, 0, 1) be the "north
pole", and let M be the rest of the sphere. The plane z = 0 runs
through the center of the sphere; the "equator" is the intersection of
the sphere with this plane.
For any point P on M, there is a unique line through N and P, and this
line intersects the plane z = 0 in exactly one point P'. Define the stereographic projection of P to be this point P' in the plane.
In Cartesian coordinates (x, y, z) on the sphere and (X, Y) on the
plane, the projection and its inverse are given by the formulas
The transformation equations for a sphere of radius R are given by
x = k cos sin( - 0) (1)
y = k[cos 1sin -sin 1 cos cos( - 0)], (2)
where 0 is the central longitude, 1 is the central latitude, and k=(2R)/(1+sin 1sin +cos 1 cos cos( - 0)). (3)
The inverse formulas for latitude and longitude are then given by
= sin^(-1)(cos c sin 1+(y sin c cos 1)/ ) (4)
= 0+tan^(-1)((x sin c)/( cos 1 cos c-y sin 1sin c)), (5) where
= sqrt(x^2+y^2) (6)
c = 2tan^(-1)( /(2R)) (7)
and the two-argument form of the inverse tangent function is best used
for this computation.
For an oblate spheroid, R can be interpreted as the "local radius,"
defined by
R=(R_e cos )/((1-e^2sin^2 )cos ), (8)
where R_e is the equatorial radius and is the conformal latitude.

In regular perspective a point in three dimensions, say P = (x, y, z)
is mapped to two-dimensions on the (x,y) plane with the following
coordinates: P* = ( ax/a-z , ay/ a-z) where a is the distance position
along the z axis where the eye sits (I.e. where one is looking from).
If we let the eye sit at (0, 0, 1) and take a sphere with radius 1
centered at ( 0, 0 , 0), and the x-y plane, then the x-y plane will
divide the sphere into two hemispheres. Notice that the point (0,0,1),
where the eye sits, would be mapped to the point (0/0, 0/0) and thus
would be represented at ¥ and all points near this would be mapped
very far away.

So, let us now take any sphere and mark off the poles on the sphere,
which are diametrically opposite to one another. Next we identify the
plane "E" which divides the sphere into two equal hemispheres and
whose normal runs through both the poles. Now we take any point P on
this sphere (except the poles which we know map to infinity) and we
want to project this point P onto the plane "E" we defined above. We
draw a line that contains a pole, the point P itself, and the plane E
and where this line intersects with E we call P*.

So in our example above with the unit sphere centered at the origin,
if we took the point P = (x, y, z) on the sphere and then take the
south pole which has coordinate (0, 0 ,-1) we can parameterize the
line that runs through both these points as:
Pole + t(P - Pole) = (0,0,-1) + t( x,y,z - 0,0,-1)
= (0,0,-1) + (tx, ty, tz + t)
= (tx, ty, -1+ tz + t).
So, when this line intersects the x-y plane (ie z =0) and t = 1/z +
1and so using this formula for t, our point is mapped to:
P* = (x', y', 0) where x' = tx = x/ z + 1 and y' = ty = y / z + 1 {https://www.math.ubc.ca/~cass/courses/m309-01a/montero/math309project.html}
I did not understand it either.

What was worked out from other sources was a "simpler" calculations
for H(n) and H(s) where H(n) is the leftmost point of the Horizon
Circle (Latitude North = 0) when projected onto the line of the
Equator, and H(s) being the right most point of the Horizon Circle
when projected on the Line of the Equator.

H(n) = Radius * tan((Latitude North - Observers Latitude)/2)
H(s) = Radius / tan((Latitude North - Observers Latitude)/2)

Where Radius is the radius of your 'circle of construction'. {A radius
of 2"[50mm] will work for an astrolabe ~7.5 inch[190mm]}

From those is easy to calculate the diameter of each latitudinal
circle, their midpoint, and the location of the midpoint on the line
of the equator.

Graphing the almucantar circles would use the values in column F
"midpoint from H(n) (X)" as the center, with the radius taken from
column G "Radius".
Column Element
A Observer Latitude is 48
B H(n)
C H(s)
D dia
E midpoint
F midpoint from H(n) (X)
G Radius
A B C D E F G
48 0.000 1.801 1.801 0.900 0.900 0.450
0 -0.890 4.492 5.383 2.691 1.801 0.900
5 -0.788 4.011 4.799 2.400 1.612 0.806
10 -0.689 3.608 4.297 2.148 1.460 0.730
15 -0.592 3.264 3.856 1.928 1.336 0.668
20 -0.499 2.965 3.464 1.732 1.233 0.617
25 -0.407 2.703 3.110 1.555 1.148 0.574
30 -0.317 2.470 2.787 1.393 1.077 0.538
35 -0.228 2.261 2.488 1.244 1.016 0.508
40 -0.140 2.071 2.211 1.105 0.966 0.483
45 -0.052 1.898 1.950 0.975 0.923 0.461
50 0.035 1.739 1.704 0.852 0.887 0.443
55 0.122 1.591 1.469 0.734 0.857 0.428
60 0.210 1.453 1.243 0.621 0.832 0.416
65 0.299 1.324 1.025 0.512 0.811 0.406
70 0.389 1.202 0.813 0.406 0.795 0.398
75 0.480 1.086 0.606 0.303 0.783 0.392
80 0.573 0.975 0.402 0.201 0.774 0.387
85 0.669 0.870 0.200 0.100 0.769 0.385
90 0.768 0.768 0.000 0.000 0.768 0.384

Conclusion:
This is the most "tedious" of the calculations. But once set up in a spreadsheet or program, it simplifies the generation of tables for an
observers specific latitude(E.G the Barony of Aquatera, An Tir, as
here. Or for Cambridge, Oxford, Ravenna, Jerusalem, Pennsic, Clinton
or any other "local" large event where knowing the local time might be considered useful.)

There remain the determining of the azimuth lines, as well as the
Climate Circles (the Tropic of Cancer, the Equator and the Tropic of
Capricorn)
=== herendeth the lesson

tschus
Nikolai
--
Nikolai Petrovich Flandropoff
Whimiscal Order of the Ailing Wit
Scribe & Zampollet to Clan MacFlandry
Loose Canon, An Tir Heavy Opera Company
Semi-offical TASS correspondent (That makes me - the Demi-Tass)

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