The Geometry of the Marcq Position-Lines

Their book also includes an explanation of that phase of the Sumner or double-altitude problem now known as the “new navigation.” This the reader can take or leave, as he pleases. Whilst admitting its neatness, the writer is averse to the diagram part of the method, which somehow seems to him out of place in a ship, except perhaps in fine weather, with the advantage of a commodious chart-room, and this latter is unfortunately not always to be had in the merchant service. Like “lunars,” the “new navigation ” is of the fancy type—all very well with gentle zephyrs, but out of place in oil-skin weather.

Other officers, without going so far as to condemn graphic processes altogether, are strongly averse to defacing their charts by an unnecessary net-work of lines. Thus Lord Ellenborough, who, as a retired commander of the Royal Navy, is entitled to speak with some authority, wrote in the Nautical Magazine of August, 1901:

The principle to which I wish to draw attention is a very simple one. It is to spore the chart at the expense of the work-book. In the problems I have referred to, and doubtless in many others, it will often be found preferable to expand or contract the perpendiculars by means of the traverse tables, as the case may require, and to draw diagrams in the work-book upon what is practically a plane chart scale, than to be per­petually covering the chart in use with superimposed sketches.

If, when juniors are employed in assisting a senior, they are allowed to draw diagrams upon the chart itself, the latter must soon become a hope­less maze of lines.

At present there seems to be no settled agreement, amongst those who have written upon the subject, as to the best method of making use of the intercepts to fix the actual position. Messrs. Brent, Walter and Williams,¹ to whom is due the credit of first introducing the method into England, favored the employment of protraction, with the assistance of squared sectional paper. Mr. J. R. Walker² protracts to scale on plane paper, adding an analytical solution by means of right-angled triangles, of a some­what artificial character, which, as he very candidly admits in his preface, is not likely to find much favor with the “ practical man.” Again, in a work published some years later, Lieutenant Simpson- Baikie³ throws over the graphic methods altogether, making use of certain special tables, based upon the formulae for errors in latitude and longitude due to a small change of altitude, with which we were not wholly unfamiliar even in the pre-Sumner days.

It is, then, a leading feature of Mr. Simpson-Baikie’s procedure that it dispenses with the necessity for a diagram. In place of it, however, we have to discriminate between a great variety of cases, some twelve in all, depending upon the respective quadrants which contain the bodies observed, the relative magnitude of the azimuths, etc., which is distinctly puzzling. The author indeed seems in the end to entertain some uncertainty as to the adequacy of his own precepts, for he writes: “If still in doubt about the reasons for these corrections, draw the figure’’ so that he is apparently driven back by the force of circumstances upon the very diagram which he was at first disposed to repudiate.

'Ex-Meridian Tables and the New Navigation. London, George Philip and Son. 1886.

² The New Navigation. Portsmouth, Griffin and Co. 1901.

^s Tables for Working Combined Altitudes by Saint-Hilaire’s Method. London, Imray Lawrie and Co. 1907.

One other work, by Lieutenant Cross, R. N. R.,^[1] ought not to be left out of account. It was published towards the end of 1911, and though in size not much more than a pamphlet, gives quite one of the clearest and best accounts of the subject that have as yet appeared. Like Mr. Walker, -the author concludes the work by protraction to scale, six miles to one-eighth of an inch being the ratio employed.

Possibly the combination of a rough diagram with some simple form of table might supply a not unsatisfactory solution, and it has occurred to the writer that it might be based upon what is undoubtedly one of the simplest problems of practical navigation, ⁴‘ Having given the course and distance from a point given in posi-

Fig. i.

tion, to find latitude and longitude intercepts,” a short process effected in a few moments by means of the traverse table. The given point of departure is of course the assumed position for which altitudes are calculated, the latitude and longitude deduced being those of the true place of ship.

Let us suppose that O is the D. R. position, and that by means of calculated zenith distances a, and b have been found as intercepts, which represent the differences between the calculated zenith dis­tances and the observed ones for the bodies employed.

Then if perpendiculars are drawn through A, B, the extremities of the intercepts, the intersection of these lines in P gives the true place of ship. Now, since the angles OAP, OBP are by con­struction right angles, it is obvious that a circle may be described about the quadrilateral figure OAPB. Also the angles OAB, OPB in the same segment will be equal to each other, as also the angles

OPA, OB A.

If, therefore, the triangle OAB is solved for the values a, b ot the intercepts, and an assumed value for AOB, the angle included between the intercepts, we at the same time obtain values of ABO, or 4>, and OPB, or 6.

Let us assume that the angle BPO, or 8, is known. Then, since the direction of OB, the intercept, is known (running directly towards or away from the body observed), the direction of BP, at right angles to OB, is known also.                                                                               _          _

We have therefore only to apply the angle 8 to this dilection, and we have a line of bearing for OP, which reversed is the course from O, the assumed, to P, the true position.

Also it is obvious that OP is the diameter of the circle.

For the length of OP, we have

OP — b cosec 9 —a cosec 4>,

either of which equations determines the distance OP.

These operations suggest the form of table required. Let us suppose that we assume a definite numerical value, say 24 miles, for the larger intercept a. Then the table might be ai ranged as follows, say for an assumed value of 50°, that angle representing the inclination of the two intercepts to one another:

Let us suppose that the two intercepts were 24 and 10. The value of 8 would be found to be 23° 36'. Applying this angle to the known bearing of the line of position for the intercept 10, we obtain the line of bearing of OP, that is, the course.

Thus, suppose in the diagram that OB runs N. 29⁰ W. Then BP, at right angles to this line, is S. 61° W., so that PB runs N. 61⁰ E.

Adding 23L5, the value of 8, to this value, we obtain N. 84°.5 E. for the course from P to O. And OP, from the table, is 25 miles.

Thus, the course and distance from O to P equal S. 84°.5 W., 25 • miles. Taking 24 as the limit for the larger intercept, and working for each 2° of angle, say from 26° to 154°, we would find the tabulation of these values for every possible pair of intercepts a somewhat formidable matter, involving a good many thousand entries. But with a very small additional amount of trouble to the navigator, the tables constructed for 24 miles, as the value of the larger intercept, might be made to apply to all cases, and then only a few pages of tabular matter would amply suffice.

From a moment’s consideration of the diagram it will be seen that, so lofig as the angle A OB is constant, the angles 8, <j> will retain the same values if the intercepts a, b are multiplied by a common factor m, so that they become ma, mb, m being greater or less than unity.

If then the larger intercept has any value other than 24, we may find the value for the smaller intercept with which to enter the standard table constructed for 24 miles, as follows:

Let o', b' be the new values for larger and smaller intercepts, x the tabular value for the smaller intercept, with which to enter table. Then

That is, the value for smaller intercept must have the same ratio to the actual smaller intercept that 24 has to the actual larger intercept.

Thus, suppose 11,5 are the actual values. Then

x = c ₌ ¹²⁰ = 11 nearly.

11 ³                          11

The appropriate value for the smaller intercept, with which to enter table, being thus selected, 8 is taken out unaltered.

With regard to the length of OP, since

OP =

AD _ sin AOB ’

and since, if a, b are changed to ma, mb, AB becomes mAB, it follows that the value of OP from table must be multiplied by m, that is, by the ratio of the actual larger intercept to 24, the standard value for which the table is constructed.

In the small auxiliary table, therefore, should be given at the head of each column a factor by which the value of OP taken from the table must be multiplied, F being simply the ratio of the actual larger intercept to 24.

The following extract will show the form of the proposed table:

TABLE.

Assuming then that the intercepts have been obtained by com­paring the true or observed with the calculated zenith distances, the several steps will be as follows:

1. With actual larger intercept as argument, look out in Table A the value for smaller intercept to be taken as argument in standard Table B.
2. From Table B take out the value of 6 to be applied to the direction of the position-line given by the smaller intercept, and also the value of OP.
3. Apply the angle 6 to find course, and multiply the tabular value of OP by F, to find actual value for distance.
4. With this course and distance calculate latitude and longi­tude of P by traverse table.

The data following, being details of observations taken, at sea more than sixty years ago, culled from an ancient work-book kindly placed at the disposal of the writer by Admiral Sir Walter Hunt-Grubbe, of the British Navy,^[2] may serve to illustrate the procedure.

September 23, 1850. D. R. position at 2^h 48“ G. M. T., Lat. 38° iST., Long. 1° E.

Fig. 2.

Between the observations the ship ran N. 73⁰ W., 28 miles. This was allowed for in the correction of the first altitude, which was thus reduced to the position of the second D. R. place of observa­tion (38° N. i° E.).

From 0 lay off the intercepts, from the forenoon observation OB N. 65° W., 7 miles, from the afternoon observation N. 56° E., 14.5 miles.

From Table A we take out the number 11.6 corresponding to smaller intercept 7, larger 14.5, ^ancl ^a^^so f^act°^r F = .6, by which OP as taken from the tables, is in the end to be multiplied.

The angle between intercepts is 121⁰, the sum of 56° and 65°. Entering Table B with this value, we have for the smaller intercept

0= i8°.4, OP — 36-8-

For the course: At the point P apply i8°4 to N. 25⁰ E., the direction of the position-line for smaller intercept, obtaining S. 6°.6 W., as the direction of PO.

Again, OP, multiplied by .6, the value of F from Table A, gives 22 miles for the distance.

Thus true course and distance from O to P is N. 7° E. 22 miles nearly.

Without drawing .any lines we may at once mark upon the chart the corresponding point.

Or, utilizing the traverse table, wc have

Diff. Lat.                                         Departure.

22' N.                                2/7 = 3/5 E., Diff. Long.

Thus,

Lat. P = 38° 22' N„ Long. P=i° 3-'5 E.

0

Combination of Single Position-Line with Bearing of Terrestrial Object.

Quite one of the most useful applications of the Marcq posi­tion-lines is the combination of a single line deduced from the altitude of a heavenly body with the true bearing of a terrestrial object, such as a headland or light-house. In such cases the “ course and distance ” process offers so simple a method of treat­ment that no special table is required, and with the aid of a rough diagram the solution by means of the ordinary traverse table be­comes at once obvious.

Without any verbal rules the method of procedure in this case will be best exemplified by means of a practical example.

Example 7, from the work by Mr. Simpson-Baikie, to which allusion has already been made, will answer the purpose. The following are the data :

December 21, 1907, when the sun’s bearing was S. 49i° ^E-> ^the true altitude was 40° 10', the bearing of Mitiikoi light-house (Lat. 8° is' N., Long. 73° E.) being S. 20° E. Required, the position of ship.

It should be remarked that Mr. Baikie very wisely adopts the practice of working with the latitude and longitude of the point observed as the assumed position. By this device the work of de­ducing the true latitude and longitude is very much simplified.

-Z-

Fig. 3.

In this case the altitude found by calculation with the assumed latitude and longitude was 40° 26'.5 giving zenith distance 49 23'.5, which, compared with the actual zenith distance found by observation, of 49⁰ 50', gives the intercept as i6'-5 further off.

Let 0 be the position of light-house in Lat. 8° 15' N., Long. 73⁰ E. The position of ship is somewhere on the line OA, drawn through 0 N. 20° W. From O lay off OB N. 49 V ^W- directly away from sun. In OB take OM 16.5 miles, and at M draw MP at right angles to OM, intersecting OA in P, the position of ship. Then

OP = OM sec 29°.5,

the angle 29°-5 being the inclination of the two position-lines.

Entering the traverse table with 16.5 ^as Diff. Lat., and 29 .5 as course, we find distance to be 19 miles, which is the required dis­tance from light-house to ship.

And by hypothesis the course is N. 20° W., the reversed bearing of light-house.

Thus, the course and distance from light-house to ship is N. 20° W. 19 miles. Entering the traverse table, we have for these values

* Diff. Lat. I7'.9 N., Dep. 6'.5, giving Diff. Long. 7' W. nearly.

Whence

Lat. =8° 15' N. + 18' N. = 8° 33' N.                                     .

Long. =73° E.-7' W. = 72° 53' E.

The Case in which an Intercept Vanishes.

<0

Fig. 4.

A case, in many respects analogous to that just considered, arises when one of the intercepts vanishes, or when it is so small that it may be left out of account.

Such an instance we find in Example IV of Walker, the data being as follows:

October 16, 1899. D. R. 53° 55' N. 4⁰ W.

Bearing.                                           Intercept.

Altair,                       S. 55⁰ W. 13.4 nearer.

fi Andromedae, S. 67° E-                 -6 further off.

From O, the assumed position, lay off OA S. 67° E. Then if we neglect the small intercept, little more than half a mile, for /? Andromedae, the place of the ship lies upon the line MON running N. 23⁰ E. and S. 23⁰ W.

Again, from 0 lay off OB S. 55⁰ W. Upon this line take OQ 13.4 miles, and through Q draw QP at right angles to OB, inter­secting OM in P, the position of ship.

From the traverse table we have, for Diff. Lat. 13.4, course 32⁰ (difference between S. 55⁰ W. and S. 23⁰ W.), the distance 16 miles, which is the required distance from O to P. And the course from 0 to P is S. 23⁰ W.

• Under 16 miles S. 23° W. we have for Diff, Lat. 14.7 miles, for departure 6.2 miles, which gives 10'.5 Diff. Long, nearly.

Lat. 53⁰ 55' N.                                  Long. 4⁰ 00' W.

14'.7 S.                                                             io'.s W.

Lat. P 53⁰ 4F.3 N.                            Long. P 4⁰ io'.5 W.

The Tables A and B.

The special tables to be employed in connection with the pro­posed “ course and distance ” method are then two in number. Of these, Table A is only a small auxiliary table of some 260 entries, which has for its object the conversion of the actual in­tercepts resulting from observation to the appropriate arguments for the standard Table B.

•

This table has for its arguments the values of the greater and smaller intercepts, in the first case extending from 2 to 23, in the latter from 1 to 22.

At the head of each vertical column is given the value of the factor “ F,” by which the value of OP taken from the tables has to be multiplied, as has been already explained.

The proper argument having been determined from Table A for the smaller intercept, we enter Table B, and thence obtain (1) the angle 6, which is the angle of inclination for OP to the position­line of the smaller intercept, and (2) the distance OP, subject to multiplication by the factor F.

The value of the larger intercept is taken as constant at 24. The other arguments are the angle included between intercepts, pro­ceeding by intervals of 2° from 26° to 154° and the smaller inter­cept, as taken from Table A.

If the actual value of the larger intercept happens to be exactly or approximately 24, there is of course no occasion to refer to Table A, but we enter Table B with the actual value of smaller in­tercept. If one of the intercepts exceeds 24, we have only to divide each by 2 and proceed as usual, except that in compensation

TABLE A.

Auxiliary Table, for Finding Value for Smaller Intercept, to be Used as Argument in Table B.

OP, as taken from table, has to be multiplied by 2F, instead of F. The extracts which follow will perhaps illustrate sufficiently the structure of Tables A, B, and will also be found to contain all the details necessary for the solution of the practical examples given later, selected from the works of Messrs. Cross and Walker, to which reference has been already made.

TABLE B.

For Finding 0, the Inclination of OP to the Position-line for Smaller Intercept, and also the Value of OP.

LARGER INTERCEPT 24.

Angle included between intercepts

Practical Examples in Use of Tables.

Example III (Walker, p. 35).—D. R. position, 53⁰ 40' N. 115⁰ 15' E.

Bearings.                                        Intercepts.

N. 89° E.                                         18' nearer.

S. 43⁰ E.                                         22' further.

Angle included, 132°. Position-line for smaller intercept. S. i° E.

From Table A, with larger intercept 22, smaller intercept 18, we have argument for Table B 19.6, F=.92.

From Table B, with included angle 132⁰, smaller intercept 19.6, wefind(9=2i°.5, OP = 53-5.

Applying 21 “.5 to S. i° E., we have S. 20°.5 W. for direction of position-line at P. And OP = 53'-5 X .92 = 49'.2.

Thus, the course and distance from O to P equal N. 20°.5 E., 49', and the latitude and longitude of P will be found to be 54° 26' N., 115⁰ 44'.5 E.

Example V (Walker, p. 50).—D. R. position, 54⁰ N. 5° W.

Bearings.                                                   Intercepts.

Arcturus, S. 67° E.                               15/1 further.

Venus,                        S. 25⁰ E.             13/5 further.

Angle included, 42°. Position-line for Venus, N. 65° E.

From Table A, argument is 21.5, F = .62.

From B, 6 = 61° ; direction of PO, N. (65⁰+6i⁰) E. = S. 54° E.

And OP = 24'.7X.62=i5'-3.

Thus, course and distance from 54⁰ N. 5⁰ W. is N. 54° W. i5'-3> and

Lat. P = 54° 9' N., Long. P=s° 21'W.

Example (Cross, p. 22).—D. R. position, 34⁰ 47' S. 12⁰ 21' W.

Bearings.                                                   Intercepts.

Altair, N. 48° W.                                   5' further.

a Centauri, S. 28° W.                           9'.5 further.

Angle included, 104°. Position-line for Altair, N. 42⁰ E.

From Table A, argument is 12.7, P = .4.

. From Table B, 6 is 24°-S and OP runs N. 66°.5 E.

OP = 30'.7 x .4= 12'.3.

Course and distance, N. 66°.5 E. i2'.3.

Lat. P, 34⁰ 42' S., Long. P, 12° 7' W.

Example (Cross, p. 28).—D. R. position, 39⁰ 10' S. 6o° 45' E.

Bearings.                                              Intercepts.

Sun, N. 61⁰ E.                                       29' nearer.

Sun, N. 5⁰ E.                                         45' nearer.

In this case, both the intercepts lie outside the limits of our tables. Each, therefore, must be divided by 2, the tabular value of OP being multiplied by 2F instead of F, as already explained.

With arguments 22.5, 14.5 we have, from Table A, argument for Table B 15.5, 2p=i.88. Position-line for smaller intercept, S. 29⁰ E.