By the aid of plotting sheets now issued by the Hydrographic Office, the plotting of position lines has become a comparatively easy task and at the same time, because of the large scale on which these plotting sheets are constructed, the accuracy of the fix has been increased. Yet an opinion prevails that the point of intersection of two position lines should be calculated as well as plotted. In the case of finding by calculation the intersection point of position lines computed by the St.-Hilaire method, three or four methods have been devised. Among these may be mentioned those by Dr. Fulst and Professor Goodwin as being the simplest and perhaps the least complicated.
Dr. Fulst’s method was very thoroughly explained in the issue of the U. S. Naval Institute Proceedings of September, 1913, by Lieut. Commander DeAquino of the Brazilian Navy, while Professor Goodwin's method and table were published in the January and February, 1913, issues of the Nautical Magazine. In the present article attention is called to a method of calculating the fix of two lines, that, in the opinion of the writer, offers more convenience and greater facility than the methods just referred to. In addition to ordinary traverse tables only a short two-page special table is required.
The method consists in determining the D. Lat. and the D. Long, between the assumed, or dead reckoning, position and the final fix. Thus, in Fig. 1, if the angles α1 and α2 represent the azimuths h1 and h2 the corresponding intercepts, the point of intersection Z of the two position lines EZ and CZ is found by determining BD, the D. Lat., and DZ, the Dep. To find BD and DZ from the known quantities α1α2 and h1h2, it is convenient to make use of an auxiliary triangle RZF formed by extending the position line CZ to meridian BF. It will be noticed that if, in this auxiliary triangle, a perpendicular is drawn from Z towards the meridian, the angle RZD=α2, and the angle DZF = α1 from which follows that the sum of these angles or RZF is equal to the angle Α, or the sum of azimuths.
The procedure of finding the D. Lat. and Dep. may now be accomplished as follows:
In the right triangle BCF, Fig. 1, we have
BF—h1 sec α1.
In the right triangle BER, we have
BR = h2 sec α2.
Hence,
FR = BF — BR = h1 sec α1 — h2 sec α2
In the triangle RZF, we have
RZ: FR = sin RFZ: sin FZR.
Now, the angle
RFZ = (90°- α1)
and the angle
FZR = (α1 + α2) = Α.
Therefore,
RZ: FR = sin (90° — α1): sin Α
whence,
RZ = FR sin (90° — α1)/ sin Α = (FR cos α1)/ sin Α.
Substituting in this equation the value of FR previously obtained, we have
RZ = (h1 sec α1 — h2 sec α2) cos α1/ sin Α.
But, sec α1, X cos α1, = 1; hence,
RZ = (h1 — h2 sec α2 cos α1)/ sin Α.
To find the departure DZ a perpendicular is dropped from Z toward RF. In the right triangle RDZ
DZ = RZ cos α2.
Substituting in this equation the value of RZ found above, we get
DZ = (RZ cos α2. (h1 — h2 sec α2 cos α1)/ sin Α.
But, cos α2 X sec α2= 1; hence,
DZ = (h1 cos α2. — h2 cos α1)/ sin Α
whence,
DZ = cosec Α (h1 cos α2 — h2 cos α1).
To find the D. Lat., BD, Fig. 1, we have in the right triangles BER and RDZ
BR = h2 sec α2;
RD = RZ sin α2
BR + RD = h2 sec α2, + RZ sin α2
or,
BD = h2 sec α2 + RZ sin α2.
Substituting the value of RZ previously found, we get
BD = h2 sec α2 + (h1 — h2 sec α2 cos α1) sin α2 / sin Α.
Multiplying, the equation becomes
BD = h2 sec α2 + (h1 sin α2 — h2 tan α2 cos α1) / sin Α.
Transposing,
BD = h2 sec α2 — (h2 tan α2 cos α1) / sin Α + (h1 sin α2) / sin Α.
Now, sec α2 = 1/cos α2 ; hence,
BD = h2/ cos α2 — (h2 tan α2 cos α1) / sin Α + (h1 sin α2) / sin Α.
Since tan α2 = sin α2/cos α2, by substituting this value and factoring, we get
BD = h2 [1/cos α2 — (sin α2 cos α1)/(cos α2 sin Α)] + h1 sin α2/ sin Α .
Placing the terms within the parenthesis under a common denominator, the equation becomes
BD = h2[sin (α1 + α2) — sin α2 cos α1]/(cos α2 sin Α) + h1 sin α2/ sin Α.
But the angle Α = (α1 + α2) ; hence,
BD = h2[sin α1 + cos α2 + sin α2 + cos α1 — sin α2 cos α1]/(cos α2 sin Α) + h1 sin α2/ sin Α .
Canceling, the equation becomes
BD = h2 sin α1 / sin Α + h1 sin α2/ sin Α.
Whence,
BD = cosec Α (h2 sin α1 + h1 sin α2).
The formulas found may now he used in calculating the D. Lat. and Dep. of the point of intersection of any two lines of position computed by the St.-Hilaire method. Assuming that α1 and h1 represent, respectively, the first azimuth and intercept, α2 and h2 the second azimuth and intercept, as in the foregoing demonstration, the formulas are:
D. Lat. = cosec Α (h1 sin α2+ h2 sin α1)
Dep. = cosec Α (h1 cos α2 — h2 cos α1).
While the formulas appear easy of manipulation, certain rules must be followed to meet and comply with the terms involved. Thus, by examining the two formulas it will be noticed that each intercept is multiplied by the sine or cosine of the opposite azimuth.
To meet this condition it is necessary to exchange the intercepts before multiplying, considering h1 with α2 and h2 with α1.
By the use of Table 1, which gives the product of cosecants of angles from 20° to 90° and for intercepts from 1 to 18 miles, and the Traverse Tables the actual numerical values of D. Lat. and Dep. are determined. But in order that these values may be correctly named, it is necessary to “ swing ” the azimuths to the adjacent quadrants. The transfer of the azimuths and the interchanging of the intercepts may be performed as one operation, care being taken when transferring each of the azimuths to refer its numerical value to the north or south point, and to transfer the azimuths in opposite directions and the nearest way.
Example 1.—Assumed position Lat. 50° 14' N., Long. 27° 19' W. The azimuth and intercept at first sight are, respectively, S. 30° E. 9 miles ( + ); at second sight S. 42° W. 7 miles ( + ). Find Lat. and Long, of fix.
Proceed as follows: First transfer the azimuths to the adjacent quadrants as indicated by arrows in Fig. 2. In this case the first azimuth is S. 30° E. and when transposed becomes S. 30° W. The second azimuth is S. 42° W. and when transposed becomes S. 420 F. Thence,
For intercept 9 miles the azimuth is S. 42° E.
For intercept 7 miles the azimuth is S. 30° W.
In this case the sum of azimuths, designated Α, is 42° +30° = 72°. If Α exceeds 90° its supplement is used. Each intercept is now multiplied by the cosecant for Α. To facilitate this multiplication, Table 1 is used, entering with intercept at top and with A in the side column. Thus,
For intercept 9 and 72° the product is 9.5.
For intercept 7 and 720° the product is 7.4.
These factors are now entered in a traverse as distances with azimuths as courses:
Azimuths | Factors | D. Lat. | Dep. | ||
N | S . | E | W | ||
S 42° E | 9.5 | — | 6.4 | 7.1 | — |
S 40° W | 7.4 | — | 3.7 | — | 6.4 |
D. Lat. = 10.1' S. Dep. = 0.7' E.
In picking out the D. Lat. and Dep. from ordinary traverse tables, reverse the data by entering the D. Lat. in the Dep. column; or, read the tables from the bottom when azimuths are less than 45°, and from the top when greater than 45°. This is necessary on account of the change made in names of original azimuths.
The D. Lat. and Dep. is now applied to the assumed, or dead reckoning, position, in the usual way; thus,
D. R. Lat. = 50° 14' N. D. R. Long. = 27° 19' W.
D. Lat.= 10.1' S. D. Long.= 0.9' E.
Lat. of fix = 50° 3.9'N. Long, of fix = 27° 18.1' W.
Example 2. Assumed position Lat. 34° 37' S., Long. 8° 51' W. First azimuth and intercept = S. 28° W., 9.5 miles (—). Second azimuth and intercept = N. 48° W., 5 miles (—). Find point of intersection.
In this case both intercepts are negative or away from the observed bodies. Transfer azimuths to adjacent quadrants which gives S. 28° E. for intercept 5, and N. 48° E. for intercept 9.5 (see Fig. 3).
Α=180° — (62° + 42°) = 76°.
For intercept 5 and 76°, Table 1 gives 5.2.
For intercept 9.5 and 76°, Table 1 gives 9.8.
Azimuths | Factors | D. Lat. | Dep. | ||
N | S . | E | W | ||
S 28° E | 5.2 | — | 2.4 | 4.6 | — |
N 48° W | 9.8 | 7.3 | — | 6.6 | 6— |
D. Lat.=4.9' N. Dep. = 11.2'E.
D. Long. = 13.6' E.
Assumed Lat. = 34° 37' S. Long. —8° 51' W.
D. Lat.= 4.9'N. D. Long.= 13-6' E.
Lat. of fix = 34° 32.1' S. Long, of fix = 8° 37.4' W.
It will be noticed that Table 1 may be used also to pick out the D. Long, corresponding to the departure obtained. This is done by entering the Dep. in top column with latitude in vertical column to the right.
Example 3. Position by dead reckoning Lat. 16° 25' S., Long. 38° 1' W. Azimuth and intercept at first sight: S. 12° W., 10.3 miles (—); at second sight: S. 68° W., 5.5 miles ( + ). Find latitude and longitude of fix.
In this case one intercept is positive, the other negative. The latter is therefore laid off in the opposite direction or N. 12° E., as shown in Fig. 4. As before, each azimuth is transferred to the nearest adjacent quadrant; thus,
For intercept 5.5 miles the azimuth is N. 12° W.
For intercept 10.3 miles the azimuth is N. 68° W.
A=68° —12° = 56°.
Then,
For intercept 5.5 miles and 56°, Table 1 gives 6.6
For intercept 10.3 and 56°, Table 1 gives 12.5
Azimuths | Factors | D. Lat. | Dep. | ||
N | S . | E | W | ||
N 12° W | 6.6 | 1.4 | 2— | — | 6.5 |
N 68° W | 12.5 | 11.6 | — | — | 4.7 |
D. Lat. = 13.0' N. Dep. = 11.2'W.
D. Long. = 11.7' W
D.R. Lat. =16° 25’ S. D.R. Long.=38° 1’ W.
D. Lat. =13. N. D. Long.= 11.7 W.
Lat. of fix=16° 12’ S. Long. of fix =38° 12.7’ W.
After some practice these simple calculations should be performed without the aid of a diagram, but until proficiency is attained it is well to use a rough sketch representing azimuths and intercepts to avoid mistakes in the transfer of azimuths.
It will be noticed that Table 1 is limited to a distance of 18 miles. If intercepts exceed that length the excess is entered and the corresponding number is added to those for 18 miles. Thus, if the angle A is 46° and the intercept 26 miles, Table 1 is entered with 46° and 18. The resulting number is 25.0; then the number corresponding to 46° and 8 miles, which is 11.1, is picked out. The factor corresponding to 46° and 26 miles is therefore 25.0+11.1 = 36.1. The D. Lat. and Dep. obtained in each of the three examples given may be verified by actual measurement in the accompanying diagrams.