Graphic method.—Navigators have found some inconvenience in plotting Sumner lines in the way usually recommended, namely, on a chart of the ship's locality or one of the same latitude; first, because such a chart on a sufficiently large scale is not always at hand; second, because such a use of the chart rapidly wears it out; and third, because all officers aboard ship who have to work out the ship's position have not the facilities in their quarters for spreading out a good-sized chart and working upon it with parallel rulers.
To obviate these disadvantages, the method here given enables the plotting to be done graphically upon a small-sized sheet of paper and with an ordinary ruler.
Suppose that the lines are defined by the latitude and longitude of two points of each; that is, for example, that, in working time sights, two latitudes have been assumed and the longitude found corresponding to each.
Now consider that these lines are plotted on a chart which is constructed upon a sheet of elastic rubber. It is evident that if, while holding it fast in the direction of the meridians, we stretch this rubber along the lines of the parallels in a uniform manner until the length of each minute of longitude is made to equal a minute of latitude, the chart, while losing its accuracy as portraying actual conditions on the earth's surface, still correctly represents the positions of the various points in terms of the new coordinates which have been created, namely, those in which a minute of latitude is equal to a minute of longitude. Thus, if on the true chart a point is m minutes north and n minutes east of another, on the stretched one it will still be m minutes north and n minutes east, the only difference being that the minutes of longitude will now be of a different length; and if on the original chart the two Sumner lines intersect at a point m minutes north and n minutes east (on the original scale) of some definite point of one of the lines, the intersection on the stretched chart will lie m minutes north and n minutes (of the new scale) to the east of the same point.
A stricter mathematical conception of the stretched chart and its properties may perhaps be obtained by considering the chart of the locality to be projected (with the eye at the zenith) upon a plane passing through one of the meridians and making an angle with the plane of the horizon which is equal to the latitude; each minute of longitude will then be increased by multiplying it by the secant of the latitude, and thus become equal to a minute of latitude.
A consideration of the properties of this hypothetical chart enables us to lay down the following rule:
If two or more Sumner lines be plotted by their latitude and longitude upon any sheet of paper, using a scale whereon latitude and longitude are equal regardless of the latitude of the locality, the intersection of those lines, measured by coordinates on the scale employed, correctly represents the intersection of the lines as it would be measured upon a true chart.
It follows from this that we may plot Sumner lines upon any piece of paper, measuring the coordinates with an ordinary scale ruler, and assigning any convenient length for the mile; the larger the scale the more accurate will be the determination. Or, what is even more convenient, we may employ "profile paper," whereon lines are ruled at right angles to each other and at equal distances apart, in which case no scale ruler is needed.
One caution must be observed in using this method; all longitudes employed on the paper for any purpose must be those of the scale, namely, one minute of longitude equals one minute of latitude. For instance, if the two Sumner lines be taken at different times, in bringing the first up to the position of the second by the intermediate run, that run must be laid down to scale; that is, the easting and westing must appear as so many minutes of longitude, not so many miles. To do this enter the traverse table with course and distance run, and pick out latitude and departure; then, by means of the middle latitude, convert departure into minutes of longitude and bring the first line to the second by laying off so many minutes of latitude north or south, and so many of longitude east or west.
In the case where the Sumner is defined by one position and its line of direction, it is not correct to lay down the angle to the meridian on the hypothetical chart, for all angles are distorted thereon. The best way is to find another position on the line by assuming a second latitude ten or twenty miles removed from that of the point given; then enter the traverse table with the angle that the line makes with the meridian as a course, and abreast the latitude taking out the departure; convert departure into difference of longitude, and plot the second point by its coordinates from the first.
The figures show the Graphic methods of plotting Sumners— (1) on a Mercator chart, (2) by scale, and (3) on profile paper:
Computation, using Bowditch's Tables.—The finding of the intersection of two Sumner lines by computation may be divided into two cases:
Case I. When one line lies in a NE.-SW. direction, and the other in a NW.-SE. direction.
Case II. When both lie in NE.-SW., or both in a NW.-SE. direction.
Suppose that the lines are defined by the latitude and longitude of two points of each.
For the simplification of the problem it will be convenient to consider the lines projected on a plane passing through one of the meridians and making an angle with the plane of the horizon equal to the latitude, the properties of which were explained under the graphic method; this saves the necessity of converting minutes of longitude into miles of departure before the solution and converting them back again afterwards; as all points are thus projected in corresponding relative positions, the results are as exact as if the longer method had been followed.
CASE I. One line NE.-SW., and the other NW.-SE.
Suppose the two lines, projected as described, are as shown in the figure, A1 A2 and B1 B2; for the present assume that the two points, A1 and B1, have a common latitude. Drop the perpendicular PO from the intersection; then the latitude of the intersection will be a distance OP above the common latitude of A1 and B1, and its longitude will be a distance A1O to the right of A1 and B1O to the left of B1.
Find the angles a and ? from the traverse table (Table 2, Bowditch), they being taken out with the difference of latitude between the two points of the same line in the column Lat. and the difference of longitude in the column Dep. (Do not overlook the fact that we are dealing now with the plane of projection and that a and ? are not the angles made by the Sumner line with meridians on the earth's surface.)
(I) For solution by table of distance run between two bearings (Table 5 B, Bowditch); observe that the case is the same as if a ship were steaming along the line A1 B1 and took the first bearing of the point P when at A1, at an angle from the course equal to 90°—a, and the second bearing when at B1 at an angle from the course equal to 90°+?, with an intervening run equal to the difference of longitude A1 B1; or, she may be considered as steaming from B1 to A1, in which case the first angle is 90°—? and the second 90°+a. Picking out of Table 5 B, corresponding to the angles given, the quantity in the second column, we shall have the ratio of the distance of passing abeam, OP, to the distance A1 B1; multiply the difference of longitude by the ratio, and we shall have the actual length of OP. Then entering the traverse table with this as a latitude and a as a course, we find in the departure column the distance A1O by which the longitude of OP is defined; it is recommended also to pick out B1O, using the angle ?, which affords a proof of the correctness of all work done after the finding of a and ?.
(2) For solution by traverse tables only, referring again to the figure, the method is to find by trial and error some latitude such that its departure corresponding to a, plus its departure corresponding to ?, equals the difference of longitude A1 B1; then the point will be defined by the latitude and by its longitude from either A1 or B1.
Longitude A1 B1 = 22.8'.
First draw a rough sketch to illustrate the direction of coordinates.
Notice that A1, is west of B1. The line through A1 runs NW.-SE. That through B1, NE.-SW. The intersection is therefore south of both, east of A1, and west of B1.
To solve by Table 5 B: First bearing, 90°—a,=39°; second bearing, 90°+?.= 137°. Corresponding ratio, 0.43, multiplied by 22.8'=9.8' lat. (The angles 90° —? and 90°+a would have given the same ratio, 0.43.) Then (Table 2) with a=51°, lat.=9.8', dep.=12.1'; and with ?=47°, lat.= 9.8', dep.= 10.5'.
Hence, intersection:
9.8’ of lat. 49° 40’=40° 30.2’ N.
12.1’ E of long. 6° 55.3’=6° 43.2 W } check.
10.5’ W. of long. 6° 32.5’=6° 43.0’ W.} check.
To solve Table 2:
Assuming Lat. 5’ 8’ 10’ 9.9’
__ __ __ __
Dep. for 51° 6.2 9.7 12.3 12.2
Dep. for 47° 5.3 8.5 10.7 10.6
___ ___ ___ ___
Sum 11.5 18.2 23.0 22.8
Hence, intersection:
9.9’ S. of 40° 40’=49° 30.1’.} check.
12.2’ E. of 6° 55.3’=6° 43.1’.} check.
10.6’ W. of 6° 32.5’=6° 43.1’. } check.
The results of the two methods substantially agree.
CASE II. Both lines NE.-SW. or both NW.-SE.
Consider the lines as drawn in the figure, and continue the assumption that A1 and B1 have a common latitude. The differences from the first case by both methods simply involve a change of signs.
(I) By Table 5 B. If the ship is steaming from A1 toward B1, the first angle from the keel line is 90°—a, and the second, 90°—?; if steaming from B1 toward A1, the first angle is 90°+?, and the second 90°+a; in other words, either add both angles to 90° or substract both from 90° and enter with the smaller angle as the first bearing.
(2) By Table 2. It now appears that OA1—OB1 = A1 B1; in other words, the values must be so found that the difference of the corresponding departures equals the difference of longitude, instead of their sums, as before.
Let us now find the intersection of the lines:
In this B1 is west of A1, the lines both run NE.-SW. and ? is the greater angle; therefore intersection lies to the N. and E. of both points.
(I) By Table 5 B. First course, (90°+9°)=99°; second course, (90°+59°)=149°; ratio, 0.67X1.0’=0.7’; or, first course, (90°-59°)=31°; second course (90°-9°)=81°; ratio 0.67, as before.
a=9°, lat.=0.7’, dep.=0.1’; and ?=59°, lat.=0.7’, dep.=1.2’.
Hence, intersection: 0.7’ N. of 49° 30’ = 49° 30.7’
0.1’ E. of 5° 24.8’ = 5° 24.7’} check.
1.2’ E. of 5° 25.8’ = 5° 24.6’ } check.
(2) By Table 2.
Assuming Lat. 2.0’ 0.5’ 0.6
___ ___ ___
Dep. 9° 0.3 0.1 0.1
Dep. for 59° 3.3 0.9 1.1
___ ___ ___
Difference 3.0 0.8 1.0
Hence, intersection of 0.6’ N. of 49° 30’ = 49° 30.6’
0.1’ E of 5° 24.8’=5° 24.7’} check.
1.1’ E. of 5° 24.8’=5° 24.’} check.
In discussing these cases, we have assumed that there was one point on each line which had a common latitude; this would be the case in the two Sumner lines worked from time sights taken at the same time. It may occur, however, (a) that they have not a common latitude, but do have a common longitude, as in the case of two lines worked from ?’ ?” (latitude) sights taken at the same time; or, (b) that they have neither a common latitude nor a common longitude, as with one time sight and one latitude sight, or with two sights taken at different times.
(a) In the case where there is a common longitude, which will be rather a rare one, the problem is worked with OP as a longitude coordinate; the modification of the other method will suggest itself, the principal change rendered necessary being due to the fact that the angles from the course in Table 5 B will be complementary to what they were before, as we are now dealing with angles to the meridian instead of angles to the parallel.
(b) When there is no common coordinate of either latitude or longitude, the simplest way of solving is first to find some point on one line which corresponds in latitude with one of the points on the other line, then solve as before.
Thus, in the figure, given A1 A2 and B1 B2 find a and ?, then find the longitude of a point A3 corresponding to the difference of latitude between A1 and B1, on the course a; after which the intersection of A3 A2 and B1 B2 is found in the usual way.
The directions of the lines require us to follow Case I.
A1 is west of B3. The line through A1 runs SE.-
NW., and that through B3, SW.-NE. Therefore the intersection is south of A1 and B3, east of A1, and west of B3.
(I) By Table 5 B. (90° — a ) 48°; (90° +?)= 109°. Ratio, 0.81 X 9.9' = 8.0' lat.; a = 42°; lat. = 8.0'; dep. = 7.2'. ? = 19°; lat. = 8.0'; dep. = 2.7’.
Intersection: 8’ S. of 25° 30’ S.=25° 38’ S.
7.2’ E. of 115° 22’ E.=115° 29.2’ E. }check.
2.7’ W. of 115° 31.9’ E.=115° 29.2’ E.} check.
(2) By traverse table.
Assuming Lat. 6’ 8’
___ ___
Dep. for 42° 5.5 7.2
Dep. for 19° 2.1 2.7
___ ___
Sum 7.6 9.9
Intersection: 8’ S. of 25° 30’ S.= 25° 38’ S.
7.2’ E. of 115° 22’ E. = 115° 29.2’ E. }check.
2.7’ W. of 115° 31.9’ E. = 115° 29.2’ E.} check.
SUMMARY OF PROCEDURE.
By Table 5 B.
- Write down lines. If given by coordinates of two points, find a and ?
- If there are no points which have a common latitude, reduce one point of one line to latitude of some given point of the other.
- Write down difference of longitude.
- Draw rough sketch to illustrate direction of point of intersection.
- Enter Table 5 B:
Case I, angles (90°—a) and (90°+?) or (90°—?) and (90°+a).
Case II, angles (90°+a) and (90°+ ?) or (90°—?) and (90°—a).
Take out ratio from second column, and multiply by difference of longitude; this gives difference of latitude of intersection from the common latitude.
- Find departure corresponding respectively to a and ? with latitude; this gives differences of longitude to the point of intersection from the points of each line.
By Table 2.
- Write down lines. If given by coordinates of two points, find a and ?.
- If there are no points which have a common latitude, reduce one point of one line to latitude of some given point of the other.
- Write down difference of longitude.
- Draw rough sketch to illustrate direction of point of intersection.
- Enter Table 2, at pages a and ?; find by trial some latitude at which—
Case I, the sum of the corresponding departures equals the total difference of longitude.
Case II, the difference of corresponding departures equals the total difference of longitude.
These give differences of latitude and longitude to the point of intersection from the point of each line.
If the lines, instead of each being defined by coordinates of two points, are defined by coordinates of one point with the direction of the Sumner line as deduced from the azimuth of the body, it will be better not to consider the projection on the fictitious plane through the meridian, as there will then be no advantage in so doing. In this case, consider the angles of the lines with the meridian, as given, a and ?; reduce the difference of longitude A1 B1 to departure, and use this in miles instead of the A1 B1 in minutes; and when A1O and B1O are found, being in miles of departure, they must be reduced to minutes of longitude before being applied to the longitude of A1 and B1. The “Summary of Procedure" must be modified accordingly.
NOTE.—It is evident that an application of the trial and error method of the traverse table will furnish a solution for finding the distance of an object by two bearings.