Page:Encyclopædia Britannica, Ninth Edition, v. 17.djvu/278

Rh rad. xdep. AC_ cliff, lat. B(J = tan course. AC 951 log-rlO... 11-978180 BC 631 lop;... . 1-800029 B 56 26 7&quot; log tan.... 10-178151 BC63-1 log 1-800029 BC x sec B 56 26 7&quot;_ di=tancc B 56 26 7&quot; log sec- 10 0-257372 rad. AB dist. 114-1 log 2-057401 It is not necessary to work out a course to seconds ; scarcely so to minutes. Distances are invariably expressed in miles and decimal parts. . To find the Course and Distance to the Place of Destination. Suppose it be the Strait of Belle Isle. Choose a point of land, or assume a position, near the entrance, say 51 45 N. and 55 10 W. A line drawn on a chart of the Atlantic from the ship s position to the above will roughly indicate the course usually steered, which is a rhumb line cutting all the meridians at the same angle, and forming on the globe a spiral curve towards the pole. The excep tions are if it be east or west, north or south ; in the former case it would be a small circle round the pole, and in the latter it would be on the meridian (which is a great circle) directly to one of the poles. The direction of the line on the chart is found by means of parallel rulers conveying it to the compass, or by a protractor placed on a meridional line ; and the distance is measured on the scale of latitude at the side between the two places. Long distances are calculated by &quot;middle latitude,&quot; or &quot; Mercator s sailing.&quot; The former is most convenient in low latitudes and when the difference is small, but the latter is at all times more correct. Middle latitude may be made nearly as good a method by measuring from the chart half way between the two latitudes ; or the same result may be obtained by taking the latitude corresponding to the mean of the meridian parts, as the table of these embodies the whole principle upon which Mercator s charts are made. Continuing the day s work Lat. of ship 49 8 Merid. pts. 3394 29 Near Belle Isle 51 45 3640-90 2 37 = 157 Mid. lat. 50 27 30&quot;. 2)7035-19 Mean 3517 -59 which differs only one mile from the mean taken directly. To find the departure corresponding to the difference of longitude &amp;lt;55 10 -642 = 48 28 = 2908 ) in mean latitude 50 27 30&quot;: diff. long. 2908 x cos 50 27 30&quot; rad. = dep. 1851 -3 miles. &quot;With difference of latitude and departure to find the course : rad. xdep. 1851-3 diff. lat. 157 tan course 85 9 10&quot;; rad. x diff. lat. dep. = cot course. The course is evidently 85 degrees to the west of north. To find the distance : dep. 1851 3 x rad. sin course 85 9 10&quot;
 * 1858 miles.

To find the course and distance by Mercator s sailing. The differ ence between the meridional parts as above is 246 6, difference of longitude 2908 miles, and difference of latitude 157 ; whence rad. x diff. long, merid. diff. lat. = tan coxirse 85 9 10&quot;; cliff, lat. x rad. cos course = dist. 1858 miles. The result is the same by both methods when the difference of latitude is small. It is obvious that if we subtract the logarithm corresponding to the meridional difference latitude from that of the difference longitude increased by ten in the index, the remainder will be the logarithmic tangent of the course. In like manner, if the logarithmic secant of the course be added to the logarithm of the true difference latitude, and ten in the index rejected, the sum will be the logarithm of the distance. Under some circumstances, in high latitudes, the ordinary mode of reducing the ship s run during the twenty-four hours, by taking the difference between the northing and southing as the difference of latitude, and the excess of easting or westing as the departure, would not be correct. For example, suppose that from a position off the west coast of Ireland, in 52 N. and 11 W., a ship were to sail in such a manner as to make during the first eight hours 96 miles on a true N. W. course, during the next eight hours a similar distance due north, and during the last eight hours ISLE. 96 miles. Here we should have the two oblique courses producing 67 &quot;9 miles each, both of difference of latitude and departure, the departures .apparently cancelling each other and leaving the ship on the same meridian. But in reality the ship would not be on the same meridian, as the latter departure was performed in a higher latitude than the former, and they are of opposite names. Therefore this peculiar &quot;day s work &quot; requires to be divided into two distinct parts ; ach may be taken by inspection from the traverse table. The first course N&quot;. 45 &quot;W&quot;. 96 miles will produce latitude 53 7 54&quot;, middle latitude 52 34, which with departure 67 &quot;9 will give difference longi tude 111 6 W. The second course simply adds 96 miles to the latitude ; the third, like the first, increases the latitude by 67 9 miles making it 55 51 48&quot;, and the middle latitude of the last section 55 18, which with departure 6 7 &quot;9 will give difference longitude 119 &quot;2 ; this being east, and preponderating over the former by 7 &quot;6 miles, the ship will be 7 &quot;6 miles east of the meridian she first sailed from, or in 10 52 24&quot; W. Great Circle Sailing. The course between the ship on May 10th and the Strait of Belle Isle, as ascertained by middle latitude and Mercator s sailing, is a straight line on Mercator s chart, but is in reality a curve upon the globe, and the nearer the positions are to either pole the greater the curvature and consequent increase of distance. The shortest distance between any two points on the globe is an arc of a great circle, the centre of which is also the centre of the sphere. The equator, ecliptic, and all meridians are great circles, between which an infinite number of others may be drawn in every conceivable direction, each dividing the surface of the globe into two equal parts. It is only of late years that this subject has been revived, since ships have made long voyages entirely under steam, thereby having the ability to steer any course desired. The practice of this mode of sailing requires but a small general knowledge of spherical trigonometry, and to that extent only will spherical problems be here introduced ; though a clear understanding of the doctrine of the sphere is very desirable for all navigators. The formulfe by which spherical problems are solved must here be postulated. They will be proved in TRIGONO METRY. In all spherical triangles there are six parts, three sides and three angles, the sides as well as the angles being measured by degress; any three being given the remainder may be found. The angles measured between heavenly bodies, or their altitudes, are treated as spherical sides. When two places are on the same parallel of latitude, the shortest course is not east or west (except on the equator), as the parallels are small circles having th pole as their common : centre. In finding the true course and shortest distance, by calculating a segment of a great circle, the earth is treated as a sphere, this being sufficiently accurate for the purpose. The advantage of sailing on a great circle instead of a rhumb is best de monstrated by stretch ing a thread across a globe, cutting the two places under considera tion ; while by calcula tion it can be found numerically. Returning to the day s work on May 10th, let us suppose it desired to find the great circle course and distance to Belle Isle. The co-latitude of each place gives the two sides, and the difference of longitude expresses the angle between them at the pole. Hence the two sides and included angle are given ; required the third side and the angle at ship, which will be the distance and course (see fig. 14). P represents the pole, S the ship, and B Belle Isle. It is necessary first to find the angles PSB and PBS. The formula is sinjdiff. sides x cotj contained angle ^, (liff&amp;gt;of other two angles; sin i sum of the two sides cos^diff. sides x cotS contained angle _ tan i nm ?f tn ?rrg^ , cos 4 sum of sides Instead of subtracting a sine or cosine, it is easier to add the co secant or the secant and reject the index. The result is sin J(PS - PB)[ = 1 18 30&quot;] x cot JP[ - 24 14 ] _ . = 39 33 30&quot;] 6 J& Hence 70 51 30&quot; + 4 33 15&quot; = 75 24 45&quot; is the greater angle B, opposite the greater side ; and 70 51 30&quot;- 4 33 15&quot; = 66 18 15&quot; is angle S, the ship s course at commencing, say, N. 66 &quot;W. At the termination it will be S. 75 W. Having all the angles, we find the third side by the rule that the sines of the angles are proportional to the sines of the opposite sides. sin B 75 24 45&quot; sinPB38-15xsinP4828 . sin BS 30 24 15&quot;. sin S 66 18 15&quot; The distance by Mercator s sailing was 1858, or 33 8 miles longer.