in the direction NE, so that, PM being perpendicular to MD, the angle of reflection PMN is equal to that of incidence PMS, or the inclination of the incident ray SMD is equal to that of the reflected ray NMC, and also angle GNM is equal to FNE. From these relations of the angles, it is easily proved that the angle E, of the triangle MNE, is equal to twice the angle F of the triangle NMF. But angle F is equal FMB, as GN is parallel to MB; hence the double of angle CMB, which is measured by twice the arc BC, is the measure of angle E. In Hadley's quadrant the arc AB is an octant, that is, the eighth part of a circle, and therefore it contains only 45°; the sextant differs from the quadrant, merely in having its limb AB a sextant, or the sixth part of a circle. The sextant is sometimes furnished with a small telescope, to show with more precision when the image of one of the objects coincides with the other. PRELIMINARY PROBLEMS. 494. PROBLEM I.-Given the distance sailed as determined by the reckoning, and the error of the log-line and sand-glass, to find the true distance. I. When only the log-line is incorrect-as the correct length of the knot or 50 feet to the incorrect length, so is the incorrect distance to the true distance. II. When only the sand-glass is incorrect-the number of seconds run by the glass is to 30 seconds, as the incorrect distance to the true distance. III. When both the log-line and sand-glass are incorrect -the number of seconds run by the glass is to 30 seconds, as the distance found by I. to the true distance." Let k, k' the true and incorrect length of a knot, S, s' = the true and incorrect number of seconds, d, d' the true and incorrect distances; ·d= k then, for I. k: k' =d': d, and d = k II. s':s=d': d, and d = s 1 50 k'd. d' =30 s' d k! For if k: k=d': d", then d" = k d = Distance 1. 245 miles 2. 156 3. 126 3 k 5 s EXAMPLE. The distance by reckoning is 92 miles, the length of the knot 51 feet, the seconds by the sand-glass 28; what is the true distance? ... Length of a Knot. d'. EXERCISES. The true distance is required from the data in the first three columns: d', and s': s=d": d, × 92100.5 miles. 495. PROBLEM II.-Given the magnetic course, that is, the course per compass, and the variation, to find the true course. "Apply the variation to the magnetic course towards the left when the variation is W.; and towards the right when E.' Let the number of points to be added towards the left be reckoned negative, and those to be added towards the right, positive; and if Ve, Vw, denote the easterly and westerly variation, and C and C' the true and compass course, then CCV, or CC'+ Ve. It must be observed that the signs + and used here have no reference to addition or subtraction, for they refer only to the direction of the variation. EXAMPLE. What is the true course when the course is NW., and the variation 2 points W. C=C'-Vw= NW.-2=WNW. EXERCISES. Find the true courses from the magnetic courses, and variations giyen in these exercises: Magnetic Course. 1. NE. 2. SW. 6 W. 3. N. 6 E. 4. SSW. W. 5. W. 6 N. N. Variation. compass 1 W 2 W 3 E. 21 E. 11 W. 11 E. Answer. True Course. NE. 6 N. SW. b S. NE. SW. W. W. SSE.E. 496. PROBLEM III.-Given the true course, and the variation, to find the magnetic course. This problem is solved exactly as the last, only the variation is applied in the opposite direction to the true course. The true courses and variations in the exercises to the preceding problem may be taken as data for exercises to this problem, and the corresponding magnetic courses will be the answers. 497. PROBLEM IV.-Given the compass course, the variation, and lee-way, to find the true course. Apply the variation, then apply the lee-way in a direction from the wind, that is, to the left, when the vessel is on the starboard tack, and to the right, when on the larboard tack." As the lee-way for the starboard tack is applied to the left, like the westerly variation, it may be reckoned negative; and that for the larboard tack, positive; then, if L denote the lee-way, and V the variation, C=C±V ± L. When V and L are both + or both, their sum may be applied according to its sign; and when they have different signs, their difference may be applied according to the sign of the greater. EXAMPLE. The magnetic course is NE. 6 E. on the larboard tack; required the true course, the variation being 2 points W., and the lee-way 5 points. C=C'—V+L=NE.6E.—2+5=NE.6E.+3= E. EXERCISES. Find the true course in the following exercises, the compass course, lee-way, and variation being given : Canton, lat. Quito, Variation 2 W. 2 W. 3 W. 5 E. lat.= 498. PROBLEM V.-Given the latitudes and longitudes of two places, to find their difference of latitude and longitude. Lee-way 2 11 21/ "When the latitudes are of the same denomination, find their difference; but when they are of different names, take their sum; and the remainder in the former case, or the sum in the latter, will be the difference of latitude. Find the difference of longitude in the same manner as that of latitude, observing, that when the longitudes are of different names, and their sum exceeds 180°, it must be subtracted from 360°, and the remainder will be the difference of longitude." EXAMPLE. What is the difference of latitude and longitude of Quito and Canton? Answer. True Course. NE. 6 N. SW. W. 23° 8' 9" N. long. Difference of lat. 23 21 26 =113° 2′45′′ E. 78 45 15 W. 191 48 0 Dif. of long. = 168 12 0 The difference of longitude in miles is estimated on the equator. EXERCISES. Find the difference of latitude and longitude of the places stated in each of the following exercises: 1. Liverpool, lat. 53° 24′ Ñ., long. 2° 59′ W., and New York, lat. 40° 42′ N., long. 73° 59′ W. Ans. Dif. of lat. 762 miles, dif. of long. 4260 miles. 2. Valparaiso, lat. 33° 0' S., long. 71° 38′ W., and Manilla, lat. 14° 36' N., long. 120° 51' E. Ans. Dif. of lat. 2856 miles, dif. of long. 10,051 miles. 499. PROBLEM VI.-Given the latitude and longitude of the place left, and the difference of latitude and longitude made by the ship, to find the latitude and longitude of the place reached. Apply the difference of latitude and longitude respectively to the latitude and longitude left by addition or subtraction, according as they are of the same or different denominations. When the longitude and difference of longitude are of the same name, and their sum exceeds 180°, subtract it from 360°, and the remainder is the longitude of a contrary denomination from that left." EXAMPLE. The latitude and longitude of the place left are 24° 36′ N., and 174° 40′ W. respectively, and after sailing SW. for some time, the difference of latitude and longitude were found to be 245 miles and 384 miles; what is the latitude and longitude in? Lat. left = 24° 36′ N. Lat. in =20 31 N. Long. left Long. in 174° 40′ W. 6 24 W. 4 W. 181 360 =178 56 E. EXERCISES. Find the latitude and longitude arrived at in the following exercises : |