Traverse

The several courses and their corresponding distances Sailing, or being properly arranged in the table, find the true difference Compound of latitude and departure answering to each in the traverse Courses. table, remembering that the true difference of latitude is to be put into a N. or S. column according as the course is in a northern or southern direction, and that the departure is to be put in E. or W. column according as the course is easterly or westerly. Add together these several quantities in each of the columns, and set the sum down at the bottom. The difference between the sums in the N. and S. columns will be the true difference of latitude made good, of the same name with the greater; and the difference between the sums of the E. and W. columns is the departure made good, of the same name with the greater sum. Look in the traverse table for a true difference of latitude and departure agreeing as nearly as possible with those above; then the distance will be found on the same line, and the course at the top or bottom of the page, according as the true difference of latitude is greater or less than the departure, since in the former case the course is less than 45° or 4 points, and in the latter case greater.

Having found the latitude, find also the meridional difference of latitude; and to the course and meridional difference of latitude in a latitude column, the corresponding departure will be the difference of longitude, which, applied to the longitude from, will give the longitude in.

It is also easy to resolve a traverse by construction; and we now show how this may be done, although it is scarcely ever practised at sea.

Describe a circle with the chord of 60° as radius, and in it draw two diameters at right angles to each other, at whose extremities are to be marked the initials of the cardinal points, N. being uppermost.

Lay off each course on the circumference, reckoned from its proper meridian; and from the centre to each point draw lines, which are to be marked with the proper number of the course.

On the first radius lay off the first distance from the centre, and through its extremity, and parallel to the second radius, draw the second distance of its proper length; through the extremity of the second distance, and parallel to the third radius, draw the third distance of the proper length; and so on until all the distances are drawn.

A line drawn from the extremity of the last distance to the centre of the circle will represent the distance made good; and a line drawn from the same point perpendicular to the meridian, produced if necessary, will represent the departure; and the portion of the meridian intercepted between the centre and departure will be the difference of latitude made good.

To construct for the difference of longitude we must find by the table the meridional difference of latitude, and lay it off on the meridian, and then complete the triangle similar to that whose sides represent the true difference of latitude; distance and departure as usual.

Ex. 1.-A ship from Fayal, in Lat. 38° 32' N., and Long. 28° 36′ W., sailed as follows:-E.S.E., 163 miles; S.W.W., 110 miles; S.E.S., 180 miles; and N. by E. 68 miles: required the latitude and longitude in, the course, and distance made good.

N

72

B

By Construction.-With chord of 60° describe the circle NESW
(fig. 23), the centre of which represents the place the ship sailed
from. Draw two diameters NS,
EW, at right angles to each other,
the one representing the meri-
dian, and the other the parallel
of latitude of the place sailed
from. Take each course from the
line of rhumbs, lay it off on the W
circumference from its proper
meridian, and number it in order,
1, 2, 3, 4. Upon the first rhumb
C1, lay off the first distance 163
miles from C to A; through it
draw the second distance AB
parallel to C2, and equal to 110
miles; through B draw BD equal
to 180 miles, and parallel to C3;
and draw DE parallel to C4, and equal to 68 miles. Now CE being
scale, will measure 281 miles. The arc Sn, which represents the
joined, will represent the distance made good, which, applied to the
course, being measured on the line of chords, will be found equal
to 41. From E draw EF perpendicular to CS produced; then
CF will be the difference of latitude, and FE the departure made
good, which, applied to the scale, will be found to measure 210 and
equal to 262, the meridional difference of latitude; and through G
186 miles respectively. On CF produced lay off to the scale CG
draw GH parallel to FE, meeting CE produced in H. Then GH is
the difference of longitude; and, when applied to the scale, will be
found to measure 230 miles.

E

D

G

H

Fig. 23.

Although the above method is that usually employed at sea to find the difference of longitude, yet, as it has been already observed, it is not to be depended on, especially in high latitudes, long distances, and a considerable variation in the courses; in which case the following method becomes necessary :

RULE 2.-Complete the traverse table as before, to which annex five columns. Now, with the latitude from, and the several differences of latitude. find the successive latitudes, which are to be placed in the first of the annexed columns; in the second, the meridional parts corresponding to each latitude are to be put; and in the third, the meridional differences of latitude.

Then to each course, and corresponding meridional difference of latitude, find the difference of longitude by Ex. 4, chap. iii., which place in the fourth or fifth columns, according as the course is easterly or westerly; and the difference between the sums of these columns will be the difference of longitude made good upon the whole, of the same name with the greater.

Remarks.

1. When the course is north or south, there is no difference of longitude.

2. When the course is east or west, the difference of longitude cannot be found by Mercator's Sailing; in this case the following rule is to be used:

To the nearest degree to the given latitude taken as a course, find the distance answering to the departure in a latitude column; this distance will be the difference of longitude.

Ex. 2.-A ship from Lat. 78° 15' N., Long. 28° 14' E., sailed the following courses and distances, viz. :-W.N.W. 154 miles, S.W. 96, N.W.W. 89, N. by E. 110, N.W. N. 56, S. by E.E. 78. The latitude in is required, and the longitude, by both methods; the bearing and distance of Hacluit's headland, in Lat. 79° 55 N., Long. 11° 55' E., is also required.

Traverse Sailing, or Compound Courses.

CHAP. V.-OF PARALLEL SAILING.
When the course is 8 points or 90° from the meridian,
—i.e., due E. or W.,—the true difference of latitude be-
comes =0, and the rules we have investigated in chaps.
iii. and iv. fail to give any result. In this case the ship
sails on a parallel of latitude. We have already proved in
chap. i., that, neglecting the earth's oblateness, the arc of
a parallel, in any given latitude, intercepted between two
meridians, is equal to the corresponding arc of the equator,
in other words, the difference of longitude, multiplied by the
cosine of the latitude.

Whence we derive these three formulæ for parallel sail-
ing:-

By Gunter's Scale.-The extent from 90° to 34°, the complement of the given latitude on the line of sines, will reach from 60 to 33.6 on the line of numbers.

There are two lines on the other side of the scale, with respect
to Gunter's line, adapted to this particular purpose, one of which
is entitled chords, and contains the several degrees of latitude; the
other, marked M. L., signifying miles of longitude, is the line of lon-
gitudes, and shows the number of miles in a degree of longitude in
each parallel. The use of these lines is therefore obvious.

Ex. 2.-Required the compass course and distance from A to B.
Given lat. A=17° 30' S.
lat. B 17 30 S.

Long. A= 9° 12′ E.
Long. B 10 42 E.

Problems in parallel sailing may be solved by construc-
tion; for it is evident that we have only to construct a right-
angled triangle whose hypothenuse is the difference of lon-
gitude, one of the sides the distance, and the angle between Variation 1 E., and deviation as in the table on p.

is side and the hypothenuse the latitude. Also it is evi-
dent, that in a traverse table, if we consider the latitude
a course, and the difference of longitude a distance, the
distance will be a true difference of latitude.

Ex. 1.-Required the number of miles contained in a degree of longitude in latitude 55° 58′.

The true course is due E.

Also, long. A = 19° 12 E.
long. B 10 42 E.
Diff. long.

L cos lat....

= 1 30 90 E.
90
17° 30'

13.

= 1.954243
9.979419

-10

85.8 m. 1.933662

Log. diff. long.

Log. dist........

draw CB perpendicular to AB; cz

line of equal parts, will be found equal to 33·5, the miles required.

3 left of N.

right of N., or E.N.E.E.

3 left of N.

By Calculation, as under:

21

Latitude

56° 0'
L cos lat. from.......
Log. distance on required parallel 232 miles =

=

9.74756

2.36549

Sailing.

12-11305

61 13 3673 W.
=9-8188250
3-5650209

48° 47'

3673

Ex. 4.-A ship from Cape Finisterre, Lat. 42° 52′ N., Long. 9° 17′ W., sailed due W. 342 miles; required the longitude in. By Construction.-Draw the straight line AB (fig. 25), equal to the given distance 342 miles, and make the angle BAC equal to 42° 52', the given latitude; from B draw BC perpendicular to AB, meeting AC in C; then AC applied to the scale will measure 466, the difference of longitude required.

CHAP. VI. OF MIDDLE-LATITUDE SAILING.

It has been already explained in chap. ii. that the departure is greater than the intercepted arc of the parallel of the higher latitude, and less than that of the parallel of the lower of two places between which a ship sails; but that there is an intermediate parallel, the arc of which is exactly equal to the departure. This parallel is supposed to pass through the middle point between the extreme latitudes; and hence the latitude of this point is called the middle latitude. The relations between course, distance, departure, and true difference of latitude are to be found as in chap. iii.; and the relation between the departure and difference of longitude is given by the above considerations, viz.,—

By Construction.-Make the line AB (fig. 26) equal to the given Whence also

distance; to which let BC be drawn per-
pendicular, with an extent equal to 522',
the difference of longitude; describe an
arc from the centre A, cutting BC in C;
then the angle BAC, being measured by
means of the line of chords, will be found
equal to 463°, the required latitude.

A

B

Ex. 6. From two ports in Lat. 33° 58′ N., distance 348 miles, two ships sail directly N. till they are in Lat. 48° 23′ N.; required their distance.

By Construction.-Draw the lines CB, CE (fig. 27), making angles with CP equal to the complements of the given latitudes, namely, 56° 2′ and 41° 37′ respectively. Make BD equal to the given distance 348 miles, and perpendicular to CP. Now from the centre C, with the radius CB, describe an arc intersecting CE in E; then EF drawn from the point E, perpendicular to CP, will represent the distance required; which being applied to the scale, will measure 278 miles.

By Calculation, as under:

Log. given distance

L cos lat. in.............

=

A

Distance x sin course= diff. long. x cos mid. lat. (B.) If ABC (fig. 29) be the triangle for plane sailing, where AB is the true difference of latitude, AC the distance, BAC the course, and BC the departure; at C make BCD equal to the middle latitude, and produce CD to meet AB produced in D; then CD is evidently the true difference of longitude, and all the problems c may be resolved and constructed by the two triangles which have a common side, -viz., the departure BC.

Also problems in middle-latitude sailing may be solved by the traverse table; for

the relations between middle latitude, difference of longi-
tude, and departure, are the same as those between course,
distance, and true difference of latitude, and therefore
may
be found at once by inspection from the table.

Ex. 1.-Required the compass course and distance from the Island of May, in Lat. 56° 12′ N. and Long. 2° 37′ W., to the Naze of Norway, in Lat. 57° 50′ N., and Long. 7° 27′ E.; variation 24 W.

Log. distance required..... 278.6 miles =

Ex. 7.-Two ships, in Lat. 56° 0′ N., distant 180 miles, sail due S.; and having come to the same parallel, are now 232 miles distant. The latitude of that parallel is required.

By Construction.-Make DB (fig 28) A equal to the first distance 180 miles, DM equal to the second 232, and the angle DBC equal to the given latitude 56°. From the centre C, with the radius CB, describe the arc BE; and through M draw ME parallel to CD, intersecting the arc BE in E. Join EC, and draw EF perpendicular to CD; then the angle FEC will be the latitude required; which, being measured, will be found equal to 43′ 53'.

Difference of longitude...............................10

D

n

By Construction.-Draw the right line AD (fig. 30) to represent the meridian of the May; with the chord of 60° describe the arc mn, upon which lay off the chord of 32° 59', the complement of the middle latitude from m to From D through n draw the line DC equal to 604', the difference of longitude; and from m C draw CB perpendicular to AD; make BA, equal to 98', the difference of latitude, and join AC; which, applied to the scale, will measure 343 miles, the distance sought; and the angle A being measured by means of the line of chords, will be found equal to 73° 24', the required course.

B

A

Fig. 30.

Middle Latitude

Sailing.

Oblique Sailing.