The best practical solution is just type the Lat-Lon of departure and destination into an electronic charting system (ECS) or a computer or calculator program and it will give the results immediately. The basic results are the distance between the two points and the initial heading of the route (in GC sailing the heading changes continuously along the route). To apply this route, however, we need waypoints along the route since the heading is changing. This is usually accomplished in these programs by telling them a longitude interval, and then the program tells you the Lat at each of these Lon intervals along the route–in other words, a set of waypoints.

Another parameter often given out is the vertex of the route, which is the Lat-Lon of the highest Lat along the route. Since the GC route only differs notably from the rhumb line route (RL) for high Lat at departure

*and*destination, we often find out that the vertex hits the ice, so we can’t use this route anyway!

If you do not have an ECS device to compute the basic data you can get all the answers by drawing a straight line from departure to destination on what is called a great circle chart or plotting sheet. Then just pull off the waypoints with dividers from the grid on the chart. The distance can also be summed up this way in steps along the route. Places like Captains Nautical Supply sell GC Plotting Sheets.

Without ECS or GC plotting sheets we are left to compute these things on our own. This can be done from the basic spherical trig equations used in cel nav, or we can use the sight reduction tables that celestial navigators use to solve routine position fixing from sextant sights. The best set of tables for this application is Pub. 229, and at this point we will have to assume you are familiar with these tables. This application is standard for the most part, but you will see we need a couple extra interpolations not usually called for, but in principle always a small improvement.

We just do a sight reduction but replace the Assumed Position (AP) with the Departure (dep), and replace the Geographical Position (GP) with the Destination (dest). In other words, the sight reduction process tells us the angular height (Hc) and azimuth (Zn) of a star viewed from the assumed position (a-Lat, a-Lon). And we know that the zenith distance (z = 90º - Hc) is the distance from the AP to the GP, so the GC distance is just z converted from angle to nmi at the rate of 1º = 60 nmi. Zn is then the initial heading (always poleward of the RL heading).

Let us work an example: What is the CG distance and initial heading from 13º 12’ N, 49º 35’ E to 15º 04.6’ N, 54º 49.2’ E? This brings up the point that we can use this procedure to figure the course and distance between two points, even if we do not care if there is any difference between CG and RL. This example is at low Lat and short distance, so there will not be much difference in the two solutions.

*Dear Reader: at this point you have to decide if you really care about this subject, because it gets more tedious from here on.*

a-Lat = dep. Lat = 13º 12’ N

a-Lon = dep. Lon = 49º 35’ E

dec = dest. Lat = N 15º 04.6’ (We write the N in front when calling it a declination.)

GHA = dest. Lon = 54º 49.2’ E = 305º 10.8’ (see below)

Now we spot a bit of a twist in the process. The GHA is measured 0 to 360 headed W from Greenwich, whereas Lon goes 0 −180 W and 0 − 180 E from Greenwich (Lon = 0). So we have to convert our dest. Lon into the equivalent meridian labeled as if it were a GHA.

(For example, 20º W Lon is just GHA = 20º, but 20º E Lon would be 360 − 20 = 340º GHA. To get to the meridian of 20º E, i have to go west 340º.)

So Lon 54º 49.2’ E = 359º 60’ - 54º 49.2’ is the same as GHA = 305º 10.8’

Then following regular sight reduction procedures, we figure the Local Hour Angle (LHA) = GHA + a-Lon(E), where we choose minutes of a-Lon so that they add to GHA for a whole degree, thus a-Lon = 49º 49.2’ so LHA = 305º 10.8’ + 49º 49.2’ = 354º 60’ = 355º

Next we choose an AP with whole degrees since the tables only have whole degrees of a-Lat and LHA. In this application we can just round them off to get a-Lat = 13º N and LHA = 355º, and we note that since our declination is also N, we have a Same Name solution.

Now we enter Pub 229 with a-Lat = 13 N, dec = N 15, and LHA = 355 same name to get: Hc = 84º 45.2', d = −26.9'* and Z = 067º, and since the Hc is so high, we need to get Z from a-Lat = 14 as well and it is Z = 057.6. In other words, when bodies are high in the sky, any small change in anything changes the bearing, so we will have to interpolate for 13º 12'. (d = altitude difference, and the * means a dsd correction is called for, but we are skipping this for now.)

ie Lat interpolation for consecutive declinations gives:

Zn = Z = 67.0 + [(12/60) x (77.7-67.0)] = 069.14 for dec = 15 at Lat 13º 12'

Zn = Z = 66.9 + [(12/60) x (66.9-57.6)] = 068.76 for dec = 16 at Lat 13º 12'

Now interpolate for the minutes of dec (which at 04.6/60 should be small)

Final Zn = 069.14 - [(4.6/60) x (69.14-68.76)] = 069.1º and that is the initial heading of the GC route we are after.

To find distance we have to finish getting an accurate Hc. With dec min = 4.6' and d = −26.9’ we get the Hc correction in three parts from Pub 229, but we can skip the dsd correction and just add the tens (1.5’) and units (0.5’) corrections to get 2.0’, which is negative, so we have Hc = 84 45.2 − 2.0 = 84 43.2.

Altitude difference (d) of 26.9 means tens = 20, units = 6, decimals = 0.9, and you get total correction as shown below.

Then z = 89 60 − 84 43.2 = 5 16.8, which converted to nmi = 300 + 16.8 = 316.8. But this is GC from the AP not from the departure, so we have to make a correction, which is best done by plotting.

After plotting we see we need to add another 9.5 nmi for a total GC distance of 316+9.5 = 324.5 nmi

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Alternatively, if you know these formulas and have a trig calculator, you can get the result this way:

lat = 13º 12' = 13.2º,

dec = 15º 04.6' = 15.077º

LHA = GHA + Lon = 305º 10.8' + 49º 35.0' = 354º 45.8' = 354.763º (we do not need to use AP when doing a direct computation.)

and sin Hc = sin (13.2) sin(15.077) + cos(13.2) cos(15.077) cos (354.763) = 0.995539

or solve the arcsin to get: Hc = 84.5862 and z = 90-Hc = 5.41378º then x 60 to get GC distance = 324.8 nmi. Our plot and table work was off a hair.)

And:

tan Z = cos(15.077) sin (354.763) / [ cos(13.2) cos(15.077) - sin(13.2) cos(15.077) cos(354.763) ] = -2.6172338039, and solving for the arc tangent:

or Z = Zn = 069.09º, which is what we got from the tables (069.1).

Notation note: cap Z = azimuth angle, lower case z = zenith distance, true bearing or azimuth = Zn.

The above formula is what the programs use. It is obviously much easier to use a calculator that has these formulas already programmed in. We offer a free calculator for this at www.starpath.com/navpubs.

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PS. The above note is for doing GC computations over large distances (without a computer!). If the run you need to compute is less than 500 miles or so at low latitude, you can get a good estimate more easily with mid-latitude sailing. That is:

Solve a right triangle with one side = dLat = 15.077 - 13.2 = 1.877 x 60 nmi = 112.62 nmi

and on the other side take the departure as dLon x cos (mid-lat)

convert to decimals

= (54.82 - 49.58) x cos [(15.077+13.2)/2] = 5.062º x 60 = 303.7 nmi.

Then the run is the hypotenuse = sq root (112.62^2 + 303.7^2) = 323.9 nmi compared to 324.8,

and the course is then E xx N, where xx = arc tan (112.6/303.7) so xx = 20.3º, and CMG = 90 -20.3 = 069.7 compared to 069.1

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Summary: after you pass all of your tests on these subjects, buy a calculator that will do all this for you... and much more.

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