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As an example, I'll do a great circle course from Los Angeles to Tokyo
with HO 229 sight reduction tables. After that I'll reverse the
calculation, which is what the original poster wanted.

34N 118W  Los Angeles
36N 140E  Tokyo

Use Los Angeles as the "observer" and Tokyo as the "star". The "LHA"
is the longitude of Tokyo, measured west from Los Angeles: 360 - 118 -
140 = 102 degrees.

Looking in the tables for LHA 102 and "latitude same name as
declination", we find Hc = 10.9 deg and Zn = 306. The latter of course
is the great circle bearing to Tokyo. Distance to Tokyo = 90 - 10.9 =
79.1 degrees = 4750 miles.

Now let's work that backward: depart Los Angeles on course 306, follow
a great circle for 4750 miles. The computed endpoint should be Tokyo's
coordinates. Let's see.

To solve this problem, consider Los Angeles the "pole" and the north
pole the "observer". Therefore, we use the actual latitude of Los
Angeles as before. On the other hand, "LHA" is different: at the
"pole" (Los Angeles), measure clockwise from the "observer" (north
pole) to the "star" (Tokyo). In other words, use the initial great
circle course (306) as the "LHA".

For "declination", convert 4750 miles to degrees: 4750 / 60 = 79
degrees (to the nearest whole degree). Subtract that from 90 to
convert distance to "declination". So dec = 11 deg and is the same
name as lat. (If distance were greater that 90 deg, it would have
contrary name.)

Enter HO 229 with LHA 306, lat 34, dec 11 same name. The table gives
Hc = 36 (to the nearest degree). That's a direct readout of the
endpoint latitude. Longitude requires a little more work. The table
says Zn = 102. This is measured at the "observer" (north pole)
clockwise from the "pole" (observer) to the "star" (Tokyo). In other
words, Zn = 102 means the destination is 102 degrees west of Los
Angeles. This corresponds to longitude 140 east. So the calculation
produced the correct latitude and longitude.

Intermediate points on the great circle can be found on the same page
by varying the declination. Since dec 90 is zero distance from the
starting point, and dec 11 is all the way to the destination, dec 50
would be the approximate midpoint of the great circle, etc.

I used HO 229, but any of the sight reduction intercept methods can
solve these problems. HO 299 includes instructions for interpolating a
more accurate solution instead of simply working to the nearest degree
as I did. But I suspect the extra work wouldn't result in a
significantly shorter voyage. Certainly it's more convenient to
navigate between waypoints when they're at whole degree intersections.

   

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