Page:Encyclopædia Britannica, Ninth Edition, v. 17.djvu/282

Rh 270 NAVIGATION intelligible. PS represents the sun s polar distance when observed, and M his position when he reaches the meridian. Consequently PS and PM are sides of an isosceles triangle, having the distance between the two positions as a base, and equal angles since the sides are equal. The thing sought is the length of ZM, or where PM is to be divided at noon ; the lower part being the true zenith distance and the remainder co-latitude. The data are the same as those in Towson s first example, which he there solves by his table only. Time from noon 28 m 14 s ; sun s true altitude 24 53 53&quot;, and declination 18 35 16&quot; S: to find the angles at the base we use the supplements of the legs, and get sec 71 24 44&quot; (180 - 108 35 16&quot;) x cot |P[ = 14 ra 7 s ] = tan88 52 25&quot;, the supplement of each angle at the base ; and sin ZS 65 6 7&quot; x sin PS 108 35 16&quot;. _ sin P 28* 14* As PZS is very obtuse, and the supplement is really required, it is taken out at once as the sine of 7 22 38&quot;, which is the angle MZS. The angle PSZ is next found; it is between two known sides, and the other angles are known, but not the opposite side : cos sum of sides x tan i sum of angles , ,.. r-j cos dm. sides which is 5 22 12&quot; ; taking this from PSM 91 7 35&quot;, we have the angle ZSM 85 45 23&quot;. Finally, BinZSxrinZSM sin ZMS the true meridional zenith distance, which taken from the polar distance gives co-latitude 43 47 54&quot;, showing that the sun Avas 18 45&quot; higher when on the meridian. When the sun is on the equator, both the angles at the base are 90, and the base MS will be the measure of the angle MPS. Norie s method is short, but the latitude by dead-reckoning is used, which may cause the whole operation to be repeated. Also the use of a &quot;log-rising&quot; and natural versed-sines is objectionable. No one possessing a copy of Towson s table need take the trouble to work the problem when in actual want of the latitude. Double Altitudes. Another mode of obtaining the latitude is by two altitudes of the sun or a star, with the interval between the observations carefully noted, or by simultaneous altitudes of two stars having a suitable difference in bear ing. These are called double alti tudes.&quot; The principle is the same in each case. The elapsed time in the one and the difference between the right ascensions in the other indicate the angular distance at the pole. Fig. 17 represents two observations of the sun on the same side of the meridian, the dark lines showing the parts which are given and the dotted lines those which are to be found. On October 2, 1882, the estimated latitude was 45 10 JST. and longitude 35 30 W. The first altitude was taken about 8.13 A.M. apparent time, and when corrected for errors of instrument, dip, refraction, and parallax was 20. The second alti tude, corrected, was 39 15, taken about 10.58 A.M. The elapsed time by a good watch was 2 h 44 m 33 s, which being mean time is reduced to apparent time by adding the proportionate increase in the equation taken from the Nautical Almanac, which for 2f hours will be two seconds additive. Therefore the change in apparent time was 2 h 44 m 35 s = SPs. Each altitude taken from 90 will give a zenith distance, 70 and 50 45 respectively = ZS and Zs. Having the apparent time at ship and the longitude, which converted into time will be 2 h 22 m , we take the declination of the sun from the Nautical Almanac for apparent noon and correct it for the Greenwich time of each obser vation, obtaining 3 36 56&quot; S. and 3 39 36&quot; S. Those amounts added to 90 will give the sun s polar distance at each period = PS and Ps. Thus four sides and one angle are known, which is all that is required to find the complement of the latitude PZ. &quot;We shall work out this by the most simple method, in order to show the principle upon which it is done. First, in the triangle sPS two sides and the included angle are given, required the other two angles. Add together the sides Ps (93 39 36&quot;) and PS (93 36 56&quot;), and find their half sum and half difference : sin diff. x cot P. , . . = tan J diff. of other angles = 3 36 ; sin sum cos diff. x cot = tan sum of other angles = 88 38 9&quot;. The half difference taken from the half sum gives the smaller of the two angles = 88 34 33, and the two added give the greater. Had there been no change in declination, the two sides including the elapsed time would with the side Ss form an isosceles triangle, and the first equation would be unnecessary. As the two sides were each more than 90 the angles found are the supplements of those required. One is sufficient for the present purpose; 180 -88 34 33&quot; -91 25 27&quot;, which is PSs opposite the larger side. That angle includes the two angles PSZ and ZSs, which must be separated in order to obtain PSZ. The side Ss, which is the spherical distance between the two positions of the sun, is given by the equation sin Ps x sin sPS. , = smSs 41 3 S3. sin Few In the second part of the problem three sides are given to find an angle, namely ZSs. Write under each other the values of the two sides (SZ and Ss) enclosing the required angle and also the opposite side Zs. Find the half sum of the three sides and the difference between it and the opposite side (Zs). Add together the logarithm cosecant of the first two and logarithm sine of the last two ; half the sum of those four logarithms (rejecting twenty from the index) will be the log. cosine of half the required angle. Thus ZSs =52 36 36&quot;, which taken from PSs will leave 38 48 51&quot; as the angle PSZ. The third part is similar to the first, two sides and an included angle given, to find the rest, principally the side PZ: sin j(PS - ZS) x cotjPSZ 7pc;, , &quot; A similar statement using two cosines instead of sines will give the half sum of the angles = 87 3 51&quot;. By subtracting the half difference the smaller angle is obtained, ZPS = 56 39 30&quot;, which by the figure should evidently be placed opposite ZS. Finally, sin ZS x sin PSZ. ^.^ ., ., ,_ = smPZ 44 49 58 , sin ZPS which is the complement of the latitude 45 10 2&quot;. If the first observation of the sun had been used to find the longitude by chronometer (the ship being stationary between the observations) the hour angle would coincide with the angle ZPS, and afford a test of its correctness. Simultaneous observations of two stars would be worked in a similar manner, with the advantage of having the difference of E.A., SPs, the spherical distance Ss, and the angles at S and s nearly permanent (requiring only a slight annual correc tion), consequently that part of the calculation might be used many times. Kerigan s tables have been mentioned as giving the dis tances between the principal stars for this purpose. When two altitudes are taken of the same star, the interval by a watch keeping mean solar time must be reduced to sidereal time, as the stars pass the meridian quicker than the sun by 9 s 8 per hour. The above illustration has been treated as if the ship were stationary, which at sea would seldom be the case. The bearing of the sun or star should be taken at the time of the first observation, and the run of the ship should be worked with reference to that bearing and also with regard to the change in longitude. Every mile which has been made good in a direct line towards the sun must be added to the first corrected altitude; if from the sun, it is subtracted. For every mile of longitude four seconds must be sub tracted from the elapsed time if the ship ran west or added if she ran east. The result will then be the same as though the ship had been stationary at the place of the second observation. Had the observations been taken on opposite sides of the meridian the calculations would be similar to the above, and the figure would differ only in having the triangle PsZ folded over, as it were, to the opposite side of the central line. Other Double Observations. The double altitude, though a very interesting problem in connexion with navigation, is at the present day of very little practical utility. When ships are provided with chronometers both latitude and longitude can be found more readily by two distinct observations of any of the heavenly bodies on different bearings. To find the longitude by means of a chronp- meter and also the variation of the compass should be one com bined operation under ordinary circumstances, therefore they will be described together. It is desirable that every navigator should accustom himself to act independently of an assistant. A pocket- watch with a distinct second hand being taken to the chronometer, the hours and minutes about to be shown by each may be noted in the &quot;deck-sight book,&quot; and if the observer then counts the half- second beats of the chronometer for five seconds previous to the even minute, the tenth beat will answ r er the purpose of the word top &quot; of the assistant, and more correctly. Let a similar com parison be made after the sights have been taken, and if there be any difference adopt the mean, or proportion. Take one or three bearings of the sun s centre with the azimuth or standard com pass in its usual place, noting the time, and then take three or five altitudes of the sun s lower limb. The watch being held in the- palm of the left hand, the fingers are free to move the tangent