In these days of shortened and abbreviated navigation, it is perhaps well to remember that we still have with us our old friend, Captain Weir’s Azimuth Diagram. There can be no question that for use aboard ship, where parallel rulers or a universal drafting machine is available, it is the most accurate graphic method yet devised for determining the azimuth of a celestial body, given the latitude, declination, and hour angle.
It is not generally realized, however, that this diagram will give a complete approximate solution of the astronomical triangle, save only in the case where the three sides or the three angles are given, and that from this property it may be made useful in many more ways than solely for finding the azimuth.
Among these various uses may be listed:
(1)Star identification; the diagram has been previously used to find the hour angle of the unknown star or planet, but it can also be used to determine the declination, without any ambiguity whatever as to sign.
(2)To determine the approximate altitude and azimuth of a given body, such as Venus, so that the navigator’s sextant may be pre-set, as for taking a planet sight in the day time. Also to find the approximate altitude and azimuth of stars, so that the sextant may be pre-set and thus avoid the necessity of bringing the star down to the horizon.
(3) To determine the time of the sun’s crossing the prime vertical, when the sun’s declination is of the same name and numerically less than the latitude.
(4)To determine the time that the sun will reach an altitude high enough to take a morning sun sight, or to determine the latest practicable time to take a sun sight in the afternoon.
(5) To determine the great circle course and distance between two points, of which the latitude and longitudes are given, where the distance is at least 1,800 miles; and to lay down on the Mercator chart the great- circle track.
(6) To plot radio bearings on a Mercator chart.
To shorten the labor involved, a slight modification of the Weir diagram seems advisable, which may be made on the present Weir’s diagram as issued to the service.
These changes are:
(1) From the equator, upward, cross out the figures on the horizon, and for them substitute the bearings continued to 180°. That is, in place of N. 85 E., the new figure would be 95° ; in place of N. 80 E., the new figure would be 100°; and so on until the 0° mark at the North point would be 180°.
(2) Then from the North point, write the bearings alongside the others, outside the horizon circle, from 180° at the North point, to 360° at the South point.
(3) Then along the inner side of the horizon mark the bearings in hours and minutes, each degree representing four minutes, and each 15° representing 1 hour from zero hours at the South point to 12 hours at the North point.
What we have done is to mark the horizon in degrees from 0° to 360° and in hours and minutes, from 0h to 12h. While the diagram can be used without these changes, they will greatly facilitate its use, as will be evident from what follows. Having made the changes we are ready to proceed.
To find the true' azimuth of a celestial body, given the observer’s latitude, the declination of the body, and the hour angle (time from crossing the meridian).
(1) From the latitude (on the meridian) follow the ellipse to its point of intersection with the hyperbola of the required hour angle and mark it lightly with a dot and a circle. This may be called the position of the object. (If the hour angle is less than six hours, this intersection will be on the same side of the equator as the latitude; if more than six hours it will be on the opposite side, as the amount over six hours must be measured beyond the equator to obtain the position of the required hour-angle hyperbola.)
(2) Mark the declination on the meridian. This may be called the position of the observer.
(3) With the parallel rulers or drafting machine, transfer the line joining these two positions to the center of the diagram; the point where the line cuts the graduated horizon is the true bearing, less than 180° if the body bears easterly, more than 180° if the body bears westerly.
It will be noted that it is not necessary to change the signs of the latitude and declination, as in the instructions on the chart. The new marking of the horizon eliminates this, and also gives the bearing in degrees from 0° to 360°, without further computation.
Example 1 (Fig. 1): Latitude 10° N., declination 20° N., hour angle 5h 0m E., required, the azimuth.
(1) Find the latitude ellipse, 10° N., on the meridian, and follow it to hour angle 5h (above the equator), and mark the intersection with a dot and circle. (Point A).
(2) Mark the declination 20° N., on the meridian (B).
(3) Lay the parallel rulers with the edge through the two dots made (A and B). Then move the rulers to the center of the diagram, keeping them parallel with this line, and the edge will intersect the horizon at azimuth 72° (because bearing is easterly).
Example 2 (Fig. 2): Lat. 20.5° S., Dec. 22.5° S., hour angle 2h 56'” (W-), required, the azimuth.
(1) Follow lat. ellipse of 20.5° S., to hour-angle hyperbola 2h 56m (south of the equator) and mark the point (A).
(2) Mark the declination 22.5° S., on the meridian (B).
(3) Lay the parallel rulers with their edge through the two dots made (A and B), and transfer the line to the center of the diagram. The edge of the parallel rulers will then cut the horizon at azimuth 259° (since the body bears westerly).
To find the time of crossing the prime vertical; when declination of sun is same name and less than the latitude:
(1) Place the parallel rulers through the declination on the meridian keeping them parallel to the equator.
(2) The time hyperbola through the point of intersection of the latitude ellipse with the edge of the parallel rulers will be the hour angle at the time of crossing the prime vertical.
Example 3 (Fig. 3): Latitude 30° N., declination 20° N., required, the hour angle of the sun at the time of crossing the prime vertical.
(1) Place the parallel rulers through declination 20° N., on the meridian, parallel to the equator (Point B).
(2) Follow the ellipse of latitude 30° N., to the edge of the parallel rulers. The time hyperbola at this point is 3U 24m (Point A). Therefore the sun will cross the prime vertical at about 8:36 a.m. or at about 3:24 p.m. (disregarding equation of time and assuming that zone time is the same as local time).
Given the altitude, azimuth, time of observation, and dead reckoning position: to identify an unknown celestial body:
(1) Along the horizon read the azimuth, converted from degrees into hours and minutes.
(2) Follow the latitude ellipse to the time hyperbola representing the azimuth converted into time. Mark this point lightly with a pencil (A).
(3) Mark the altitude of the body on the meridian (B).
(4) Place the parallel rulers through the two points marked (A and B), and transfer them to the center of the diagram. The edge of the rulers will intersect the horizon at the approximate hour angle of the body observed. Record this hour angle.
(5) Follow the latitude ellipse to the time hyperbola representing the hour angle as determined in (4) above. Mark this point (A).
(6) Place the parallel rulers through the center of the diagram and the azimuth on the horizon. Transfer them to the point (A) determined in (5) above. The point at which the parallel rulers intersect the meridian will be the declination of the unknown body north or south according to whether the point on the meridian is north or south of the equator (Point B).
(7) With the hour angle determined in and the declination determined in (6) the name of the star may be determined by any of the standard or abbreviated methods.
Example 4 (Figs. 4a and 4b): Given: observed altitude of unknown body to be 43.5°; azimuth by standard compass 305° (T); dead reckoning position latitude 34°-00 N., longitude 60° W.; local sidereal time 9h 12“ ; required, the name of the body.
(1) Convert the azimuth 305° (T) to hours and minutes, reading it on the horizon without computation or (—) 3h 40ra.
(2) Follow the latitude ellipse of latitude 34° N., to the time hyperbola of 3h40m and mark it (A).
(3) Mark the altitude 43.5° on the meridian (the altitude always being the same name as the latitude (B).
(4) Place the parallel rulers through the two points thus marked (A and B) and transfer the line to the center of the diagram. The edge of the parallel rulers will intersect the horizon in the hour angle of the body (—) 4h 02". This is the hour angle to be combined with the local sidereal time to find the right ascension of the star thus—
(5) Follow the ellipse of latitude 34° N. to the time hyperbola 4h 02" and mark the point (A) (Fig. 4b).
(6) Place the parallel rulers through the center of the diagram and the bearing 305° (T) on the horizon. Transfer this line to the point (A) marked in (5). The edge of the parallel rulers will intersect the meridian at the point 47° N., which is the declination of the unknown body (B).
(7) Having the right ascension as determined in (4) above of 5" 10" and the declination of 47° N., scan the star list and see that the star is Capella.
To determine the altitude and true azimuth of a known heavenly body at any given time (given the time of observation, the dead reckoning position, the name of the body and consequently its right ascension and declination) :
(1) Find the hour angle of the body at the time it is proposed to observe it, by means of the local sidereal time and the body’s right ascension.
(2) Then with the D.R. latitude, the hour angle determined in (1) and the declination, determine the azimuth as previously explained.
(3) Follow the latitude ellipse to the time hyperbola representing the azimuth converted into time, and mark the point (A).
(4) Place the parallel rulers through the center of the diagram and the point on the horizon representing the hour angle. Transfer the line to the point (A) marked in (3) above. The intersection of the parallel rulers with the meridian will indicate the altitude of the body at the time it is desired to take a sight of Jupiter (RA 12“ 57" 09“, contrary name to the latitude the body is below the horizon.
(5) Having thus determined the approximate altitude and true azimuth, the body may be found in the heavens. This is particularly valuable if, for instance, it is desired to take a planet sight in the daytime to cross with a sun line. The sextant may be set to the approximate altitude, and directed in the proper direction at the proper time, and the navigator should be able to pick up the planet near the horizon with his sextant on a clear day.
Example 5 (Figs. 5a and 5b): It is desired to take a sight of Jupiter, (RA 12“ 57" 09s, dec. 4° 51' 1") at watch time 7“ 17" 40s. The LST is determined to be 16“ 40" 59.1s and consequently the hour angle is 3“ 33" 50.1s. lat. 25.5° N., long. 47° W., required, the approximate altitude and azimuth.
(1) Follow the ellipse of latitude 25.5° N., to the hour-angle hyperbola 3“ 34". Mark this point (A) (Fig. 5a).
(2) Mark the declination of Jupiter 4.9° S., on the meridian (B).
(3) Place the parallel rulers through the points marked (A and B). Transfer the line to the center of the diagram and the azimuth will be read on the horizon 247° or 7“ 30".
(4) Follow the latitude ellipse of latitude 25.5° N., to the time hyperbola 7“ 30" (below the equator) and mark the point (A) (Fig. 5b).
(5) Then place the parallel rulers through the center of the diagram and the point on the horizon representing the hour angle (3h 34'"). Transfer this line to the point previously marked (A) in (4) above. The point (B) at which the parallel rulers intersect the meridian indicates the required altitude of Jupiter 29.8°. (This was later calculated and found to be 29° 58' so the result was remarkably close.)
To find the approximate time at which the sun will reach an altitude sufficient to take a sight in the morning, or the latest time practicable to take a sight in the afternoon (given the dead-reckoning position and approximate declination of the sun):
(1) Assume an altitude at which the sight is to be taken, say 15°.
(2) From the Weir diagram or by other means determine the time of sunrise.
(3) With an hour angle one-half hour later than sunrise and the given latitude and declination, determine the azimuth, as explained above, and the altitude.
(4) Then if the sun changes from 0° to the altitude obtained in (3) in half an hour, by a rough interpolation it is possible to find the approximate hour angle at which it will reach an altitude say, 15°. The process for determining the altitude near sunset is exactly the same except that the time of sunset is used, instead of the time of sunrise. This is a rough method but it is sufficiently accurate for ordinary purposes.
To find the great circle course and distance between two points not less than 1,800 miles apart, given the positions of the two points:
(1) Find the difference in the longitude between the two points. Convert this from degrees to hours and minutes on the horizon circle without computation.
(2) On the ellipse corresponding to the latitude of the first position, mark its intersection with the time hyperbola corresponding to the difference in longitude.
(3) Mark the latitude of the second position on the meridian north or south as the case may be.
(4) Lay the parallel rulers through the points determined in (2) and (3) above and transfer them to the center of the diagram. The intersection of the edge of the rulers with the horizon circle will give the initial great-circle course. Read this course in degrees and also in hours and minutes.
(5) Follow the latitude ellipse of the first point to its intersection with the time hyperbola corresponding to the great-circle course and mark the point.
(6) Lay the parallel rulers through the center of the diagram and the point representing the difference of the longitude on the horizon. Transfer this line to the point marked in (4) above. The edge of the parallel rulers will then intersect the meridian at the point representing the complement of the great-circle distance in degrees. Subtract the reading on the meridian obtained in (5) from 90° and multiply the result by 60, to obtain the great-circle distance in miles.
Advantage may be taken of this method to lay down a great-circle track between any two positions on a Mercator chart, without the use of a great-circle sailing chart, and with practically as much facility as with one of those charts as follows:
(1) Determine the initial great-circle course as explained above (the distance being unnecessary).
(2) From the latitude and longitude of the initial point on a position-plotting sheet lay off on the initial course a distance equal to any convenient unit, say 200 miles. Pick off latitude and longitude of this point.
(3) With the position determined in (2) above as an initial position and the destination as the second position, determine the second great-circle course. Then from the position determined in (2) above on the Mercator position-plotting sheet, lay off the new great-circle track to a third position. Pick the latitude and longitude of this position off the chart.
(4) Repeat the above procedure until the destination is reached. The complete great- circle track will now be laid down on the position-plotting sheets, and the distance will be equal to the sum of the several legs of the course.
(5) If the limiting parallel has been determined upon, the procedure is exactly the same except that when the limiting parallel is reached start at the destination and work backwards towards the starting point until the limiting parallel is reached. The course on the limiting parallel is of course east or west on the Mercator chart as the case may be.
Example 6: Find the great-circle course and distance between San Francisco and Honolulu.
San Francisco Lat. 37° 47’ 30” N. Long. 122° 27’ 49” W.
Honolulu Lat. 21° 17’ 57” N. Long. 157° 51’ 34” W.
(2) Find the hours and minutes on the horizon corresponding to 35.4° or 2h 22m.
(3) Follow the latitude ellipse 37.8° N. to the time hyperbola 2h 22m N., and mark the point (A).
(4) Mark the latitude of Honolulu 21.3° on the meridian (B).
(5) Transfer the rulers to the center of the diagram and read the initial course 252° or 7h 11m.
(6) Follow the latitude ellipse of San Francisco 37.8° N., to the time hyperbola of the great-circle course 7h 11m and mark it (A).
(7) Place the parallel rulers through the center of the diagram and the point on the horizon representing the difference of longitude 2h 22m N.
(8) Transfer the rulers to point marked (A) in (6). The edge of the parallel rulers will pass through the point 55.5° N. The complement of this is 34.5°. 34.5 X 60 equals 2,070, the great-circle distance between San Francisco and Honolulu.
To lay off the great-circle track between San Francisco and Honolulu.
(1) From Lat. 34° 47' 30" N., Long. 122° 27' 49" W., a rhumb-line distance of 300 miles on course 252° will put the ship in Lat. 36° 14.8' N., Long. 128° 25' W.
(2) From this position the great-circle course to Honolulu as determined from the diagram is 248°. Laying off 300 miles on the position-plotting sheet from Lat. 36° 14.8' N., Long. 128° 25' W„ on course 248° we get the position at the end of the second leg Lat. 34° 22.2' N„ Long. 134° 06.3' W.
(3) From the position determined in (2) above determine the next great-circle course as before to be 244°. Laying off 300 miles along this course from the last position gives the position at the end of this leg, Lat. 32° 10.7' N„ Long. 139° 28.8' W.
(4) Determine the next course 240°, by means of the diagram and plot the next position on the plotting sheet, Lat. 29° 40.7' N., Long. 144° 31.9' W.
(5) Determine the next course 238° by means of the diagram and the next position on the plotting sheet, Lat. 27° 01.7' N., Long. 149° 21' W.
(6) Determine the next course 235° and the next position, Lat. 24° 09.6' N., Long. 153° 54.3' W.
(7) From the last point, as it is close to Honolulu, lay off a rhumb-line course 232°, distance 278 miles, into Honolulu.
(8) Adding the various legs of the course, we find them to total 2,078 or very nearly equal to the great-circle distance between the two points, as calculated, 2,076 miles.
(9) What we have done in the foregoing is to head directly for the port of destination at the time of beginning each of several rhumb-line courses. At the end of each leg of the course, we are slightly to northward of the true great-circle track, but the error is negligible even with legs as long as those taken, 300 miles, and courses taken only to nearest whole degree.
This method has the advantage that the great-circle course can be laid down say at noon each day by the navigator. He is required by the Navy Regulations to report the course and distance to his destination each noon. Therefore if he, from each noon fix, determines the course and distance to his destination on the Weir diagram he has automatically determined the course for the next leg of his great-circle track.
The steps for laying down radio bearings on a Mercator chart by means of the Weir diagram are discussed in Dutton’s Navigation, and will not be taken up in detail here.
The foregoing explanations may seem long and involved but the actual work when done on the chart is remarkably simple and rapid, and anyone detailed to navigation duty will be well repaid for the hour or two necessary to master these additional uses of the Weir Azimuth Diagram.