Graphic Interpolation in Azimuth Tables

The following exposition has been pre­pared for those who are familiar with navi­gation, not for the novice. It is a demon­stration through the solution of several problems, rather than an academic explanation. It is assumed that the reader is familiar with the abbreviations used.

The method utilizes cross-section paper, of which a convenient kind called “cross­section paper for time plotting” is issued by the Hydrographic Office, in sheets 16 by 21 in. Along the 21-in. side the inches are subdivided in tenths and along the 16-in. side, in sixths. The sheet is normally held so that the 21-in. side is horizontal. The illustrations of the following problems were prepared with this kind of cross-sec­tion paper.¹

¹ Editor’s Note.—The cross-section background did not reproduce when the cuts for these figures were made. In examining them it must be remembered that they should be on cross-section paper.

Problem I

To obtain a curve of azimuths of the sun over a period of 2.5 hours, for use in com­pensating magnetic compasses and swing­ing ship. Assumed conditions are:

Lat. 12°—35' S. (Magnetic equator off Peru in 1932).

L.A.T.’s from 2:30 p.m. to 5:00 p.m. of a day in March such that Dec. of the sun varied from 6°-19' N. to 6°-21'.4 N. during the 2.5 hours.

Limiting latitudes are 12° S. and 13° S., and declinations are of contrary name, therefore the azimuths will be found on pages 168-69 of the Red Azimuth Tables (H. 0. 71).

Preliminary steps.—The background is prepared on a piece of cross-section paper about 8 in. or 9 in. wide (Fig. 1) by strik­ing in four heavy vertical lines spaced 0.6 in., 5.4 in., and 0.6 in. apart, respec­tively, forming two pairs of vertical lines. The lines of each pair represent the in­clusive full degrees of declination (in this case, 6° N. and 7° N.) and are so marked at their tops. Each 0.1 in. separating them represents 10 minutes of declination. The left-hand pair represents one of the in­clusive degrees of latitude and the right- hand pair the other (in this case, 12° S­and 13° S.). They are so marked. It will be observed that the two 6°-declination lines and the two 7°-declination lines are separated laterally by 6 in., so that each 0.1 in. of lateral separation represents 1 minute of latitude.

Now determine from the azimuth tables, by inspection, the greatest and least full degrees of azimuth occurring between latitudes 12° S. and 13° S. for declinations of 6° N. and 7° N. and L.A.T.’s between 2:30 p.m. and 5:00 p.m. These are found to be 99° and 117°, but for the conditions of the problem afternoon azimuths are reckoned from south to west. That is, 180° must be added to azimuths taken from the tables, resulting in actual limiting azimuths of 279° and 297°. In Fig. 1 both the actual azimuths and those taken from the tables are shown, the latter being parentheses.

The background is now completed M marking a series of horizontal lines to represent the degrees of azimuth between the limiting azimuths. In Fig. 1 the scale
used is .5 in. equals 1° of azimuth. For

FIG 1

extreme accuracy the scale of 1 inch equals 1° of azimuth should be used.

Transcribing the data.—Azimuths are now read off from the tables and plotted 0n the diagram for each even tenth minute L-A.T. between 2:30 p.m. and 5:00 p.m. for example, along the left-hand vertical line (which represents 6° N. declination and 12° S. latitude) are plotted the azimuths in the 6°-declination column on page 168, commencing with 293-44 (113-44) for 2:30 p.m., and ending with 279-13 (99-13) for 5:00 p.m. In a similar manner, the equivalent azimuths from the 7°-declination column are plotted along the next to the left-hand vertical line; and azimuths from page 169 are similarly plotted along the two right-hand vertical lines.

Interpolating for declination.—Between each pair of vertical lines draw short inclined lines joining the points that mark azimuths for the same L.A.T. In each series of inclined lines cut the topmost line with a vertical so positioned as to represent the declination at 5:00 p.m., which was 6°-21'.4. This vertical is there­fore offset 0.214 in. to the right of the vertical 6°-declination line. The point where it cuts the inclined line is marked C in the case of the left-hand upper inclined line, and D in the case of the right-
hand upper inclined line.

In the same way, cut the lowest inclined line of each series with a vertical so posi­tioned as to represent the declination at 2:30 p.m., which was 6^o-19', obtaining points A and B.

Draw the nearly vertical line AC, cut­ting each inclined line of the left series, and the nearly vertical line BD, cutting each inclined line of the right series. The intersections are enclosed in Fig. 1 with small circles. The vertical distances be­tween consecutive inclined lines represent 10-minute increments of L.A.T., and each of the above mentioned intersections there­fore represents the correct azimuth at the represented minute of L.A.T. for the indicated full degree of latitude and the exact declination that existed at that mo­ment. In brief, interpolation for declina­tion has now been accomplished.

Interpolating for latitude.—Join the points C and D, and likewise the points A and B, with nearly horizontal straight lines. In Fig. 1 these lines are dashed. Similarly join other corresponding intersections, as shown. Mark these long in­clined lines with the minutes of L.A.T. that each represents. The latitude is 12°-35', there­fore offset laterally 3.5 in- from C and draw a vertical cutting the 5:00 p.m. L.A.T- line at F. Similarly, offset laterally 3.5 in. from A and draw a vertical cutting the 2:30 p.m. L.A.T. line at E- Join EF. The intersections of EF with the various L.A.T- lines (intersections are en­closed with circles and marked with pointers in Fig. 1) rep­resent the desired azimuths for the exact latitude and declination for each indicated minute of L.A.T. This completes the interpolation for latitude- Interpolation for L.A.T.—A time-azi­muth curve of conventional kind can now be plotted from the data thus obtained- Figure 2 represents a portion of such a curve. Where a number of azimuths must be obtained with great accuracy, such a curve probably would be advantageous, but for most practical purposes it would not be needed. The interpolation for L.A.T. ordinarily can be satisfactorily accomplished by eye. For example, in Fig- 2 there is shown a refined interpolation for 4:12:20 p.m. (L.A.T.), giving an azimuth

Another method of precise interpola­tion for L.A.T. is illustrated on Fig. 1, to the right of the 12°—35' latitude line, and is marked Fig. 1a. It illustrates the same interpolation as above, that is, for 4:12:20 p.m. (L.A.T.). Dotted horizontal lines are drawn to the right from the azimuth for 4:10 p.m. (L.A.T.), namely, 283-17 and from the azimuth for 4:20 (L.A.T.), namely, 282-29. A strip of cross-section paper is then laid across these dotted lines so that ten intercepts in the cross-section paper are included between the dotted lines, permitting ready interpolation for L.A.T. In Fig. 1a the strip of cross-section paper has been cut down in length to a single inch in order to avoid blocking out important features of Fig. 1. Only its uppermost edge is used in the interpolation. The arrow represents the L.A.T. of 4:12:20 p.m.

General comment.—It will be observed at the inclined lines of Fig. 1 are not evenly spaced, but there is an appearance. Uniformity about them, changes occurring gradually. This is a valuable feature, because any noticeable irregularity in the spacing or slopes of the inclined lines usually denotes an error in transcribing data the tables, helping to prevent mistakes. Another valuable feature is that the latitude line can be drawn in at the moment, using the ship’s actual latitude instead of the latitude where it was expected she would be when preparing the diagram.

Problem II

The method has a more complicated use of interest to navigators who wish to study the performance of a gyrocompass over a period of time while the ship steams stead­ily on her course. Yet this complication is as nothing to that necessary for handling the same problem arithmetically, as will be rendered evident by the following prob­lem:

To obtain azimuths of the sun over a period of 2.5 hours on a ship steaming north at 20 knots from a point in lat. 12°- 55' S. at 2:30 p.m., L.A.T. In 2.5 hours at 20 knots on course north the ship will cover 50' of lat., reaching lat. 12°-05' S. at 5:00 p.m. L.A.T. The mid-latitude is 12°- 30' S., about the same as in Problem I.

These data have been selected so that the solutions of Problems II and I are identical up to the point where the latitude line is about to be drawn in. In Problem I the latitude line is only slightly inclined, being parallel with the two final declina­tion lines AC and BD. This is because there is only one latitude involved. But in Problem II the latitude is changing continuously, with the result that a sharp­ly-sloping latitude line is formed (see Fig. 3).

The latitude at 2:30 p.m. L.A.T. (12°- 55' S.) is only 5' less than 13° S., amounting to .5 in. on the diagram of Fig. 3. Set off .5 in. horizontally left from point B and draw a vertical cutting the 2:30 p.m. L.A.T. line at E.

At 5:00 p.m. L.A.T. the latitude (12°- 05' S.) is 5' more than 12° S. Set off .5 in. horizontally right from point C and draw a vertical cutting the 5:00 p.m. L.A.T. line at F. Draw the straight line FE which becomes the latitude line for the conditions of Problem II. Its intercepts with the L.A.T. lines indicate the corresponding azimuths, as in Problem I.

It will be noted that in Fig. 3 the dec­lination used is 6°-20' S., this being about the mean of the two declinations used in Fig. 1. For most practical purposes it will be sufficiently accurate to use the mid­
declination, as comparisons of Figs. 1 and 3 will indicate.

Problem III

When the ship is steaming on a course having an easterly or westerly component, the change of longitude causes a change in the relation between zone time and L.A.T. Therefore, when the course involves changes in both latitude and longitude it is necessary to evaluate the effect of change in longitude as well as that of change of latitude. For example, consider the following problem:

To obtain azimuths of the sun over ^a period of 2.5 hours when the ship is steaming at 25 knots on course 323° from a point in lat. 12°-55' S. at 2:30 p.m. L.A.T.

The speed of 25 knots and course of 323° were selected because they result in ^{a r}ate of change of latitude of 20 knots. This renders the latitude part of the solution identical with that of Problem II.

By a layout on the chart (or by com­putation) it will be found that the change of longitude in 2.5 hours at 25 knots on course 323°, starting in latitude 12°-55' S. will be 38.5 minutes. This amounts to 2 minutes and 34 seconds of time. The board so that 14—50-27 of the zone (plus ship’s course has a westerly component, resulting in a gain in mean time of 2’-34”

over apparent time. Thus, if the zone (Phis 5) time of 2:30 p.m., L.A.T., is 14- ®~27, then the zone (plus 5) time of ^;90 p.m., L.A.T., will become 17-23-01. To express this relation graphically, pre­fer⁶ two time scales on separate strips of cross-section paper (see Fig. 4). A convenient scale is 1 in. equals 20 minutes of time. Mark one strip for the 10-minute intervals of L.A.T. between 2:30 p.m. and 5:00 p.m., thus: 2:30, 2:40, 2:50, 3:00, 3:10, etc. Mark the other for 10-minute intervals of zone (plus 5) time between 14-50-27 and 17-23-01, thus: 1450, 1500, 1510, 5) scale aligns with the vertical line passing through 2:30 of the L.A.T. scale, and 17-23-01 of the zone (plus 5) scale aligns with the vertical line passing through 5:00 of the L.A.T. scale. The conversion of time is then accomplished simply by projecting from the desired moment on the zone (plus 5) scale along a perpendicular to the L.A.T. scale, to find the L.A.T. with which to enter the azimuth diagram to find the corresponding azimuth.

This procedure neglects the change in L.A.T. due to change in equation of time because during 2.5 hours this change is too small to affect azimuths of the sun.

If the change in longitude were easterly the zone-time scale would become the shorter and must be used as the base scale, the L.A.T. scale then being inclined above it. For example, if in Problem III the ship’s heading had been 37° instead of 323°, the problem would have been identi­cal except that the 2 minutes and 34 seconds change of zone time would be sub­tractive. The zone (plus 5) time of 5:00 p.m., L.A.T., would then become 17-17-53 instead of 17-23-01. Figure 5 illustrates this situation. One would then project perpendicularly from the zone-time scale to find the equivalent L.A.T.

General comment.—Even when there is no change in longitude (hence no change in time) the above-described scales for converting zone time to L.A.T. are help­ful. Since time intervals are then of equal length, the two scales would be parallel (see Fig. 6).