233. THE RHUMB – LINE SAILINGS

Parallel Sailing. When a vessel sails on any course other than due N. or due S., she moves some distance easterly or westerly. The value of this movement in an E.-W. direction is called Departure (usually abbreviated to dep.) and it may be represented by the arc of a parallel of Lat. cut off between the meridians of the points of the initial and final positions on the Rhumb-Line, measured in nautical miles.

If a vessel sails along the equator between two places then the departure in Long., (d.long.), between the two places in minutes of arc, will be numerically equal to the departure between the two places in nautical miles, because at the equator, minutes of Longitude are equal to minutes of Lat. or nautical miles. But in any other Lat. when a vessel sails along a parallel of Lat. (i.e. when her course is either due E. or due W.). The departure measured in nautical miles will always be numerically less than the d.long. in minutes of arc between the initial and final positions, because of the convergence of the meridians towards the poles.

In fig. 61-1/2 XY is the arc of a parallel of Lat. and is thus the distance along the parallel between the meridians through X and Y. AB is the distance along the equator between the same meridians. XY is therefore the departure between X and Y and AB the d.long. between X and Y. It will be apparent that the nearer the parallel is to the pole (in other words, the higher the Lat.) the shorter XY, becomes, but the d.long. does not alter, XY must therefore bear to AB some relation depending on the Lat.. To find this relation, consider the sections of the sphere DXY and CAB. They are parallel and equiangular, therefore: –

XY = DX

AB = CA

But in triangle DCX, ∠DXC = ∠XCA = Lat. of X, and as triangle DCX is a right-angled triangle,

from our knowledge of trigonometrical ratios: – Cosine  ∠DXC  =   DX

                                                                                                              CX 

Therefore DX  = CX x Cosine ∠DXC

But ∠DXC  = ∠XCA (Lat.) and CX  = CA (as each is a radius of the Earth)

Therefore DX = CA x Cosine Lat.

But XY  = Departure, and AB  = d.long., so we get the basic Parallel Sailing formula which is: 

Departure    = cos.lat.                          ——————–①

d.long.

From which, by transposing and inversion, we can also get: –

departure. = d.long. x cos.lat.           ———————②

d.long.   = sec.lat.                                 ———————③

Departure.

d.long. = Departure x sec.lat.            ———————④.

These four formulae are known collectively as the Parallel Sailing Formulae. Parallel Sailing originated in the very early days of Navigation when there was no accurate means of determining Longitude. A vessel sailed due N. or due S. in a known Longitude until she reached the Lat. of her destination, then she sailed due E. or due W. along the parallel of this Lat. to her destination. Parallel Sailing is important to us today in expressing the relationship between departure, d.long. and Lat., all of which are used in more modern forms of Rhumb-line Sailings.

234. EXAMPLES OF PARALLEL SAILING

 

235. PLANE SAILING.

If a vessel sails along any Rhumb line other than a meridian or a parallel of Lat. the distance sailed, the d.lat. between the first and final positions, and the departure, may be regarded as the sides of a plane right-angled triangle with the course angle opposite to the line, which represents the departure (fig. 61-3).

This plane right-angled, triangle is called, the Plane Sailing triangle, but it should be realised that it is not a triangle on the Earth’s surface; it is simply an artifice which shows the relation between a Rhumb–line course and distance, d.lat., and departure. The course angle in Plane Sailing triangles, and indeed in all Rhumb-line sailing triangles, being right-angled triangles, must be an acute angle (i.e., less than 90º). For this reason, the Course steered on the compass, usually expressed in the 360º notation, must be converted to a quadrantal one to get the correct course angle to use in the Sailings formulae.

In other words, the course angle must be expressed from either N. or S. towards E. or W. A quick sketch will usually solve this course angle problem. For instance, a course steered of 158º is obviously in a SSE’ly direction, and the triangle completed by drawing a vertical line for the d.lat. and a horizontal dep. one for the departure. The course angle will then be seen to be 180º-158º  = 22º (or S22ºE in the old quadrantal notation).

A course of 210º is SW’ly, and completing the triangle shows the course angle to be 210º-180º  = 30º. Similarly, a course of 320º is NW’ly and the angle in this case is 360º-320º  = 40º.

Examples of finding the correct course angle are illustrated in fig. 61-4.

Regarding the Plane-Sailing triangle as an ordinary right-angled triangle XYZ , then: –

sine X  =  ZY

                 XY Therefore                      ZY  =  sine          X  x  XY

But in the Plane-Sailing triangle:  ZY Departure, XY  = Distance sailed, X  = Course

Therefore departure  = sine Course x Distance             ______①

In ∠XYZ, cosine X = ZX, therefore ZX  = cosine X  x  XY

Therefore in the Plane-Sailing Triangle:

d.lat. = Cosine Course x Distance                                     ______②

In ∠XYZ, Tangent X  = ZX , so in the Plane Sailing triangle:

tan. Course = Departure                                                      ______③

                             d.lat.

In ∠XYZ, Secant X  = XY, therefore XY  = secant X  x  ZX

Therefore in the Plane-Sailing Triangle:

distance. = secant Course x d.lat.                                     ______④

 

These four formulae constitute the Plane-Sailing relationships, and with them the navigator can find the course and distance along a Rhumb-line track between two positions, or, having steered a certain course for a certain distance, he can find the departure he has made good and his final Lat.. It should be noted, however, that it is not possible to find the final Longitude by Plane-Sailing. When sailing obliquely across the meridians, the Lat. is changing constantly and difficulty arises in determining the correct Lat. to use in order to convert departure into d.long. To find the position of a vessel after she has sailed a given distance such that her Lat. and Longitude change, the navigator must use Mid. Lat. Sailing

236. MID. LAT. SAILING A.

In fig. 61-5, XY represents a Rhumb Line connecting two places in the Northern Hemisphere, X being on the parallel of Lat. AB and Y being on the parallel of Lat. AY. CD is a parallel of Lat. the exact arithmetical mean between XB and AY; in other words, if XB is Lat. 20º N. and AY is Lat. 60º N. CD would be in Lat. 40º N.

If a vessel sailed from A to Y her departure would be the distance along the arc AY, and if she sailed from X to B,  her departure would be the distance along the arc XB, but if she sailed along the Rhumb-line from X to Y, her departure would obviously be greater than AY but less than XB. The actual departure between X and Y must be the distance along some parallel between AY and XB, and this parallel is not, as might be expected, the mean parallel CD between AY and LB, but another parallel, MN1 which is called the Mid. Lat..

If the Earth were perfectly spherical, the Mid. Lat. would be greater than the mean, or average Lat. of two places, but as the Earth is not a perfect sphere, in some cases the Mid. Lat. can be less than the mean Lat. as the correction tables will show. The difference between the mean Lat. and the Mid. Lat. depends on the d.lat. and the mean Lat. of the two places, and is given in a small table Mean Lat. to Mid. Lat. in all nautical tables.  The correction taken from the table is applied to the mean Lat. to give the Mid. Lat.. For example, suppose a vessel sailed along a Rhumb-line track between Lat. 17º N. and Lat. 21º N. The mean Lat. would be 19º N, but from the correction tables for a mean Lat. of 19º, and a d.lat. of 4º we find the correction to apply to the mean Lat. is -60′, or –1º, so the Mid. Lat. in this case is 19º-1º-18º N.

Having found the Mid. Lat. in this way, the problem of Mid. Lat. Sailing is merely a matter of combining the formulae used in Parallel and Plane Sailing, substituting in the case of the basic Parallel Sailing formula the more accurate Mid. Lat. Sailing formula:

The normal navigational problems, namely: (a) finding the Rhumb-line track course and distance from one position to another, and (b) finding the final position given the course and distance sailed along a Rhumb-line track can be solved by means of Mid. Lat. Sailing as follows.

Course and Distance from one position to another: since the d.long. is known and the Mid. Lat. can be found: departure = d.long. x cosine mid.lat.

But           Tangent Course = departure

                                                        d.lat.

So            Tangent Course  = d.long. x cosine mid.lat.   ____________①

                                                                  d.lat.

And         distance = d.lat. x secant Course.                    ____________②

The Position from Course and Distance Sailed: since the course and distance are known and the Mid. Lat. can be found:

d.lat. = distance. x cosine Course.                                    ____________③

departure  = distance. x sine Course. – But d.long. = dep. x secant mid.lat.

So d.long. = dist. x sine Co. x sec.mid.lat.                      _____________④

238. THE TRAVERSE TABLE

Students will have noticed that the formulae for Rhumb Line sailing so far discussed do no more than solve an ordinary right-angled triangle, the hypotenuse of which is the distance, the vertical side the d.lat. and the horizontal one the departure. To enable the navigator to solve this triangle quickly, all volumes of nautical tables provide a Traverse Table.

The Traverse Table is an orderly collection of solutions to right-angled triangles, which may be used for solving any plane right-angled triangle, but its main uses in practical navigation are: –

to find the course and distance between two places, and

to find the position after sailing a given course and distance.

As these are the two principal functions of the Rhumb-line Sailings, it will be appreciated that the Traverse Table is, without any doubt, the most useful in any volume of nautical tables.

The Traverse Table tabulates d.lat. and departure for any distance up to 600 miles and for any course angle up to 90º. The table appears to extend only to 45º, but the angles between 45º and 90º are printed at the bottom of the pages and the columns headed d.lat. and dep. at the tops of the pages are labelled, respectively, dep. and d.lat. at the bottom of the pages. The reasons for this will become clear by studying the two triangles in fig. 61-10.

In any right-angled triangle, the two non-90º angles are complementary angles (i.e., their sum equals 90º), because the sum of the angles in any plane triangle is 180º. To solve a Sailing Triangle where the course angle is 20º and the distance sailed is, say,  100  miles, the Traverse Table (fig. 61-12) is entered at angle 20º, and abreast of  100  in the distance column will be found the values  94.0  for d.lat. and  34.2  for departure.

To solve a Sailing Triangle where the course angle is 70º and the distance 100 miles, the Traverse Table is entered at the same page, i.e., at the page for angle 20º (which is the complement of 70º- 90º minus 20º equals 70º). It will be seen that 70º is printed at the bottom of this pages and reading the columns from the bottom of the pages opposite the distance of  100  will be found the values  34.2  for d.lat. and  94.0  for departure. Comparison of these two triangles in (fig. 61-10) shows that in fact they are one and the same triangles that the triangle for course 70º is merely the triangle for course 20º turned round. The angles at the top and bottom of each page of the Traverse Table are therefore complementary and the table need only extend to 45º because the triangles for angles between 45º and 90º are the same as those for angles between Oº and 45º, only turned round so that, in this case, where the d.lat. for a course of 20º is  94.0,  for a course of 70º,  94.0  is the departure.

A navigator following a course between 000º and 090º T. is moving both N. and E.; his d.lat. is thus N. and his departure E.. Following a course between 090º and 180º T., his d.lat. is S. and his departure again E.. These facts indicate that, before using the Traverse Table the course angle must be expressed in relation to the cardinal points of the compass as described in section for finding the course angle to use in Plane Sailing triangles. Since the Traverse Table does no more than multiply the distance by the sine or cosine of an angle, as in the above example:

100  X  SIN. 20º  =  34.2 – 100  X  COS. 20º  =  94.0

it can be used for converting departure into d.long. or d.long. into departure. for in the Mid. Lat. Sailing formula, dep. = d.long. x cos. mid. lat. Using the Traverse Table for this operation, it is therefore sufficient to treat the d.long. as distance and the Mid. Lat. as the course, reading off the departure in the d.lat. column. The appropriate columns in the Traverse Table are bracketed d.long. and Dep. accordingly.

For example, to convert d.long. into departure, enter the Traverse Table at an angle equal to the Mid. Lat. (as the distances are small – less than 600 miles – when the Traverse Table is used, no appreciable error will result from using the arithmetical mean Lat. without correction to Mid. Lat.), run down the d.long. column and read off the departure in the dep. column opposite the appropriate d.long. To change departure into d.long. (the more usual problem), the table is again entered at an angle equal to the Mid. (or mean) Lat., and the d.long. read off in the d.long. (or dist.) column opposite to the appropriate departure in the dep. column. It should be borne in mind that, whereas the d.long. is fundamentally an angle measured in minutes of arc, departure is a distance measured in nautical miles.

The Traverse Table is fundamentally simple, but the facility in its use can only be acquired with practice, and students should take every opportunity to become thoroughly familiar with this, the most useful table in all navigation.

239. USE OF THE TRAVERSE TABLE

To convert d.long. into dep. at mid.(mean) lat. 43º 15.0, find the dep. for a d.long. of 160 at 43º and 44º and interpolate.

Enter the Traverse Table at angle 43º, and using the bracketed d.long. and dep. columns, opposite a d.long. of  160  will be found dep.  117.0.

Enter the T.T. at angle 44º, and opposite a d.long. of’  160  will be found a dep. of  115.1.

The departure for a d.long. of  160  at mean lat. 43º 15.0 is therefore  116.5.  To find the course and distance, search through the T.T. until a d.lat. of  50  corresponding to a dep. of  116.5  is found.

At angle 66º  a d.lat. of  50.0  corresponds to a dep. of  112.4  at Distance  123.

At angle 67º  a d.lat. of  50.0  corresponds to a dep. of  112.4  at Distance  128.

The difference between the two departures for  50.0  d.lat.  =  117.8 – 112.4  =  5.4.

To convert d.long. into dep. at mid.lat. 56º 17.0 N., find the dep. for a d.long. of  624  at angles 56º and 57º and interpolate for the  17.

Enter the T.T. at angle 56º, and using the bracketed dep./d.long. columns, search for a d.long. of  624 . It will be found that the d.long. column only extends to  600,  and when this occurs the procedure is to divide by 2 (or any other convenient figure) and multiply the extracted figure by the same amount. In this case, if we divide by 2, d.long.  312  in the table gives us dep.  174.5 which, multiplied by 2, gives us dep. 349.0 for the required d.long. of  624.

At angle 56º, d.long- 312  gives dep.                 174.5  x  2          dep.  349.0) diff.  9.2

At angle 57º                                                            169.9  x  2                     339.8)

For d.long. 624 at mid.lat-56º 17.0, dep.    =    17  x  9.2  from  349.0  towards  339.8

                                                                                   60

                                                                             =    346.3

To find the course and distance, the T.T. must be searched until a d.lat. of  328  corresponds to a dep. of 346.3.  It will be found that:

At angle 46º,       d.lat.  327.9  corresponds to a dep. of  339.5 at Distance  472.0

                               d.lat.  328.0                                                 339.6                         472.2

At angle 47º d.lat.  328.0    351.8   481.0

We require d.lat.  328  to correspond to a dep. of  346.3,  so – 

The dif. between the two departures. corresponding to d.lat.     328, =351.8 – 339.6                         = 12.2

The dif. between the req. dep. and the dep. for                                328  at 46º =  346.3 – 339.6         =   6.7

The required Course angle is  6.7  of the way between 46º and 47º           = N 46º 33.0 E.             12.2

The required Distance is   6.7  of the way between  472.2  and  481.0         = 477.0       

                                             12.2

From River Humber to Oslo Fjord, Course  =  046½º T., Distance = 477.0

(which compares with calculated course and distance on page 5).

Enter the T.T. at angle 38º, and reading the columns bracketed Dep./D.long, look for a dep. of  492.4  in the dep. column. It will be found that this column only extends to a dep. of  472.81. so divide the required dep. by 2 and multiply the extracted d.long. by 2.

Final position of yacht: Lat. 38º 00.2 N., Long. 10º 16.2 E.

(which compares with calculated position in Ex. 4)

240. PRACTICAL APPROACH TO THE SAILINGS

In general, for distance up to about 600 miles (the limit of the Traverse Table), the Traverse Table may be used to solve all Sailing triangles and the use of the arithmetical mean Lat. without correction to Mid. Lat., will usually suffice in these circumstances. It is, however, advisable to check with the Mean to Mid. Lat. Correction table to ascertain whether the correction is appreciable, if it is, the Mid. Lat. should be used for greater accuracy. When the distance involved is greater than 600 miles, the Sailing should be calculated by Mid. Lat. Sailing or Mercator Sailing (described in the next Navigation Lesson).

In cases where the conversion between departure and d.long. or vice versa is awkward, involving elaborate interpolation, a combination of the use of the Traverse Table and the calculation methods may be used if preferred, extracting the d.lat. and dep. from the Traverse Table and then converting the dep. into d.long. by formula.

It should be appreciated that courses found by the Sailings, whether calculated or extracted from the Traverse Table, are true courses; similarly, final positions found by the Sailings are D.R. positions, as in neither case has any allowance been made for current, tidal stream, or leeway.

The Sailings, can, however, be used to find the Course to Steer between two places, or an E.P., if due allowance is made for current, tidal stream and leeway when appropriate, as shown in the following examples.

EX. No. 10: Find the Course to steer from Newhaven in Lat. 50º 47.0 N., Long.  000º 04 E. to Cherbourg in Lat. 49º 40.0 N., Long. 001º 36.0 W., allowing for an estimated set and drift of the tidal stream over the run of 070º x 12 miles.

In (fig. 61-13), N represents Newhaven and C Cherbourg. The course required to make good is the Rhumb-Line NC. But from our knowledge of chart-work we know that in order to counteract the tidal stream we must lay-off the estimated set and drift D.LAT. from the departure position, (and this is represented by the line HP) and from position P lay-off the course to steer to the destination C. The ∆ NOP can be solved quickly with the Traverse Table to find position P, since the set (∆-ONP) and the drift (side NP) are both known, and these may be considered the course and distance.

From the T.T., angle 70º Dist. 121 gives d.lat.  4.1, dep. 11.3.

Convert dep.  11.3 into d.long. at mean lat. 50¼º = d.long. 17.9

EX. No. 11 A motor cruiser sails from Newhaven (Lat. 50º 47.0 N., Long. 000º 04.0 E.) on a course of 155º T. Her speed is 7 knots and the tidal stream is estimated to be setting 245º at 1.5 knots.  Find her E.P. after four hours, and the course and distance she has made good.

In 4 hours,  motor cruiser sails 4 x 7  = 28 miles on course 155º T. (S. 25º E.)

Tidal stream drifts 4 x 1.5  = 6 miles, setting 245º T. (S. 65º W.)

This problem could be worked by solving first triangle NCD (fig. 61-13), to find the D.R. position D, and then solving triangle DEF to find the E.P. position E, but it is more convenient to combine the working to find the total d.lat. and the total dep., and thus solve the triangle NGE.

In (fig. 61-14), N is the initial position off Newhaven, and triangle NCD represents the Sailing triangle for her course and distance in 4 hrs., D being the D.R. position after this time. The line DE represents the set and drift of the current in the same time (i.e., 245º T. x 6 miles), and the triangle DEF (in which DF is the d.lat. and FE the dep.) when solved will give the E.P. position E. By using the combined d.lat’s. and deps. of these two triangles (NCD and DEF) however, we can not only find position E but also the direction and length of line NE which represents the course and distance made good in the four hours, because in the triangle NGE, GNE is the course made good, and NE is the distance made good. The figure shows that the d.lat’s. from the first two triangles NCD and DEF, NC+DF = NC+CG = NG in triangle NGE, and the deps. from the first two triangles, CD – FE = GF – FE = GE in triangle NGE. We now have the d.lat.(NG) and the dep.(GE) of the triangle NGE, from which we can find course angle GNE and distance NE.

From the T.T.      course 25º  x   distance  281  =  d.lat.   25.4 S.    dep.    11.8 E.

                               course 65º  x  distance   6      =              02.5 S.                05.4 W.

                                                                            Total   d.lat.    27.9 S.  Total    06.4 E.

Initial lat.    50º  47.1 N.

d.lat.                     27.9 S.

Final lat.     50º  19.1 N.

Initial lat.   50º 47.0 N.

                2) 101º 06.1

Mean lat.    50º 33.1 N.

Convert dep.  6.4  into d.long. at mean lat. 50º 33.1. By T.T.:

At angle 50º dep.    6.4      d.long.   10.0     Initial long.    00º      04.0 E.

At angle 51º                                         10.2     d.long.                          10.1  E.

At angle 51½                                        10.1     Final long.      00º      14.1  E.

To find the course and distance made good (∠GNE and dist. NE), find a course angle and distance where a dep. of  6.4  E. corresponds with a d.lat. of  27.9 S.

At angle 12º , dep.  6.4  corresponds with d.lat.     30.13  at distance     30.80

At angle 13º ,                                                                   27.72                            28.45

Course     =   12º  +  (2.23  x  60) =  S. 12º 56.0 E.  (167º T.)

                                            2.41

Distance  =  30.8 – (2.23  x  2.35)  =  28.6 miles.

                                            2.41

E.P. after 4 hours = Lat.50º 19.1 N., Long. 00º 14.1 E.

Course and Distance made good.  =  Course 167º T. Distance  28.6