Page:Encyclopædia Britannica, Ninth Edition, v. 17.djvu/279

NAVIGATION 267 Should the sum of any two sides exceed 180, the supplement oi each must be used, and the supplement of the angles found. If a side be over 90 the logarithm found will be sine of its supplement or the cosine of its excess over 90. It will be readily seen by the frequent alterations necessary in the course, to keep near the great circle, that the ship passes the pole, continually changing its bearing, as she would any other distant object on the globe. It is always desirable to ascertain the highest latitude a ship will pass when sail ing on a great circle ; this is done by imagining a perpendicular let fall upon the arc; the position is then found by right-angled spherical trigonometry, the rules for which are very brief. Exclud ing the right angle, the triangle consists of five parts, three sides and two angles. Two of those parts at least must be given to find the rest. If the two parts given and the one required are joined, the central one is called the "middle part." If they are not joined (that is, an angle omitted between two sides, or a side between two angles, forming part of the equation) the part which is quite separate from the other two is called the "middle part." The solution is then by Napier s rules, " which have been embodied in the following rhymes : " The product of radius and middle part s sine Equals that of the tangents of parts tht combine, And also the cosines of those that difjoin." " Instead of the angles and hypothenuse Their difference from ninety or complements use." "We know that the perpendicular will fall upon the arc between B and S, because both the angles already found are acute; thus the triangle is divided into two parts, each having one side and one angle known. Take the hypothenuse PS and angle S, to find the length of the perpendicular PK. Here are opposite sides and angles, therefore sin PS 40 52 x sin S 66 18 15" . ,, Q ,,o AQ , on ,, , - , - ,, - =sm PR 36 48 30 ", which is the rad complement of 53 11 30", the highest latitude the ship will attain. To ascertain the longitude of that point it is necessary to find the angle RPS, which may be done according to the above rules in three ways, by varying the side or angle which is made the middle part. Thus, rad. + log cos PS - log cot PSR + log cot SPR ; rad. +log cos PSR = log cos PR +log sin S PR ; rad. + log cos SPR = log tan PR + log cot PS . As the part required is in each formula coupled with one which is known, the difference between that and the sum of the other two will give the angle SPR = 30 7 45" ; which, added to the longitude at ship, will give 36 49 45" as the longitude of the highest point. As the original oblique triangle BPS has been divided into two distinct right-angled triangles, having the perpendicular PR common to both, the parts which form BPR may be treated in a similar manner for the purpose of finding the angle BPR and perpendicu lar PR, either in preference to the first, or as a proof. Other points on the arc may be found by taking a part of the angle at the pole, with the side and angle previously known, and thus forming another oblique spherical triangle, in which two angles and an in cluded side are given to find the rest ; or by assuming a distance on the arc (in degrees and minutes) with the angle and side before known. In that case two sides and an included angle will be given to find the other side (co-latitude), and the part of the angle P to indicate the longitude of the assumed point. Given two angles and an included side, to find the other two sides : sin i diff. of angles x tan 4 side ~ = I = tan J diff. of other sides ; sin sum of angles cos i diff. of angles x tan i side ,, ., . , --f , = tan * sum or other sides. cos Js sum ot angles The second case where two sides and an included angle are given has been done. When the polar angle (difference of longitude) does not much exceed 50, especially in low latitudes, an approximate curve may be drawn on Mercator s chart by means of a pencil attached to a wire or batten making the curve to pass through three or four known points, after which it can be used from day to day as a guide to the ship s course ; she may not keep on it, but she can keep near it. This curve becomes very incorrect when the angle is large. One other example of finding the course and distance by each method will be given. The position has been chosen to show the great distance an arc of a great circle would carry a ship to the southward of her ordinary course, and also as touching upon the difficulty always experienced in finding a side or angle which is near 90, in consequence of the very slight change in the value of large sines and small cosines. A ship oif the Cape of Good Hope in 34 29 S. and 18 24 E. is bound to Portland in Victoria. Assume a position outside the latter port to be in 38 20 S. and 141 40 E. The difference of latitude is 3 51 = 231 miles, the meridional difference of latitude 28714, and difference of longitude 123 16 = 7396. rad. x diff. long. 7396 merid. ditf. lat. 287"14 =tan COU1 Se 87 46 37 " > diff. lat. 231 x rad. A, - = distance 5955 ! ; cos course 87 46 37 diff. lat. 231 x sec course 87 46 37 rad. = distance 5955 1. These two statements produce similar figures, because they are worked by the same course ; an error of 10 seconds would produce over 7 miles in the distance, as the logarithm cosine and loga rithm secant (arithmetical complements of each other) change very rapidly in large angles. When the difference of latitude between two places is compara tively small, it is better to find the departure and use that in find ing the distance. The table of meridional parts can be used for that purpose in two ways. true diff. lat. 231 x diff. long. 7396 merid. din-Hat. 287 "14 . 5950 4. For this _ the logarithms of numbers alone are used. The middle latitude is best found by seeking the latitude corresponding to the mean of the meridional parts, which in this instance will be 2350 13 and 36 26 . The latter is the true central position on the chart, between the two parallels of latitude, the same as if measured on the side of the chart. Radius bears the same proportion to the difference of longitud e as the cosine of middle latitude does to the departure = 5950 4. Then rad. xdep. 5950 4 diff. lat. 231 =tau COUrse 87 46 3G rad. x diff. lat. 231 or - i - -. . - = cot course, dep. 5950-4 dep. 5950 "4 x rad. ... anc l - OTO Aa , c, ,,= distance 5954 9. sm course 87 46 36 Having found by Mercator and middle latitude that a ship from the position off the Cape would, by steering S. 87 46 E. for the whole distance, traverse 5954 - 9 miles before reaching the assumed point near Portland, it is now required to find an arc of a great 15. circle passing through those points. The co-latitude of each place and the difference of longitude are given, that is, two sides and an included angle; required the third side and angle C (fig. 15). 1st position 34 29 ; co-lat. = PC 55 31 18 24 E 2d 38 20 =PV5140 141 40 E 107 11 Half sum 53 35 30" 3 51 123 16 = P 61 38 = half P Half difference 1 55 30" siajdiff. sides l55 3Q"x cot JP61"38 _ tan , diffiof ^g^ r ir 28 " sin ^ sum of sides 53 35 30 cos i diff. of sides x cot aP cos i sum of sides ten ^ gum of Qther ^ ang Consequently angle V is 43 34 5" and angle C is 40 59 9", the larger angle being opposite the larger side. Having all the angles, the remaining side is found by the proportion which the sines of angles bear to their opposite sides ; thus sinPCxsinP |~ sinVPxsinP"! sin V sin C J = sinVC9012 . This example has been chosen because the side VC is so near to 90 that it can only be treated with certainty by letting fall a perpendicular from "P to the arc, as represented by the dotted line PH ; which also gives the latitude of the highest point the ship would reach if she sailed on that arc. Thus the oblique-angled triangle, in which all the angles were known, is divided into two right-angled spherical triangles, each possessing one side and