i LI BRARY O F CONGRESS, ' ! UNITED STATES OF AMERICA.* .-.-.- ... . . ._ NAVIGATION AND NAUTICAL ASTRONOMY. PREPARED FOR THE USE OE THE U. S. NAVAL ACADEMY. SECOND EDITION. NEW-YORK : X>. VAJSC 3STOSXJK^JSTI>, 19£ BROADWAY. 1865. Entered, according to Act of Congress, in the year 1865, by D. VAN NOSTRAND, in the Clerk's Office of the District Court of the United States, for the Southern District of New- York. John A. Gray & Green, Printers, & 18 Jacob Street, New-York. NOTICE. This Treatise was originally prepared by Prof. Chauvenet to be used in manuscript by the students of the Naval Academy. With Bowditch's Navigator, oral instruction, the use of instruments, and computation of examples, it constituted the course of instruction in Navigation and Nautical Astronomy. In its printed form, some subjects are more fully discussed, others introduced, and various suggestions given on points of practice. In this edition examples are supplied, which will serve both as illustrations, and as forms for the arrangement of computations. Those in Nautical Astronomy are mainly adapted to the Ephemeris for 1865. It has been my purpose, as I should find time from incessant offi- cial duties, to prepare a more complete work, or to supplement it with a treatise on the practice of Navigation. J. H. C. COFFIN, Prof, of Astronomy, Navigation, and Surveying. Naval Academy, August 1, 1865. GREEK LETTERS. A a Alpha, N v Nu, B /3 Beta, S I Xi, r r Gamma, o Oniicron, A d Delta, n re- Pi, E s Epsilon, p P Eho, z c Zeta, S a Sigma, H v Eta, T T Tau, e e Theta, r u Upsilon, i t Iota,

that an^le. p « opposite) S F . g2 In the Tables, the columns of distance, difference of lati- tude, and departure, might be appropriately headed, respect- ively, hypothenuse, side adjacent, and side opposite. 4. The first two of equations (3) afford the solution of the most common elementary problem of navigation and survey- ing, viz. : Problem 1. Given the course and distance, to find the difference of latitude and departure, the distance being so small that the curvature of the earth may be neglected. These equations also afford solutions of all the cases of Plane Sailing. (Bowd., pp. 52-58.)* 5. Problem 2. Given the course and distance, to find the difference of latitude and departure, when the distance is so great that the curvature of the earth cannot be neglected. Solution. Let the distance C A (Fig. 1) be divided into parts, each so small that the curvature of the earth may be neglected in computing its corresponding difference of lati- tude and departure. * The first and sixth are the most important. 10 NAVIGATION. For each such small distance, as c a, l — d cos (7, p — d sin C. Representing the several partial distances by df 15 d^ c? 3 , &c, the corresponding values of I and p by ? 19 l 2 , Z 3 , &c, and jt? 1? jp 2 , j9 3 , &c, and the sums respectively by [cZ], [7], [j?], we have ?i+4+ ?3 + &C. == (^H-c^ + c^ &c.) cos (7, J Pi+i>a+#i + &c.= (d^ + dk + dk &c.) sin (7/ or, [l]= : [rf] cos (7, l>] = [*] nn a Since the distance between two parallels of latifliae is the same on all meridians, the sum of the several partial differ- ences of latitude will be the whole difference of latitude ; As in Fig. 1. CB = EA=: the sum of all the sides, c 5, of the small triangles ; and we shall have generally, as in Prob. 1, whatever the distance, <#, l — d cos (7. We also have p == d sin (7, if we regard p as the sum of the partial departures, each being taken in the latitude of its triangle ; so that the differ- ence of latitude and departure are calculated by the same formulas, when the curvature of the earth is taken into ac- count, as when the distance is so small that the curvature may be disregarded ; or, in other words, as if the earth were a plane. But the sum of these partial departures, b a of Fig. 1, is evidently less than C E, the distance between the meridians left and arrived at on the parallel C E, which is nearest the equator ; and greater than B A, the distance of these meri- dians on the parallel Bl A, which is farthest from the equa- TRAVERSE SAILING. 11 tor. But it is nearly equal to F G, the distance of these meridians on a middle parallel between C and A ; and exactly equal to the distance on a parallel a little nearer the pole, and whose precise position will be subsequently determined. (See Problem 10, Mercator's Sailing.) We take then i a = \ (i + i'), or, more exactly, Z = -%(Z + Z f ) + AZ, as the latitude for the departure, p. 6. Middle Zatitude Sailing regards the departure, p, as the distance between the meridian left and that arrived at on the middle parallel of latitude ; or takes Z = \ (Z + Z'). T R AT ERSE SAILING. 7. If the ship sail on several courses, instead of a single course, she describes an irregular track, which is called a Traverse. Problem 3. To reduce several courses and distances to a single course and distance, and find the corresponding dif- ferences of latitude and departure. Solution. If in Fig. 1 we regard C as different for each partial triangle, and represent the several courses by 6\, (7 2 , <7 3 , &c, we evidently have l x =± d x cos Ci, p x = d x sin dj l % = d 2 cos C 2 , p 2 = d 2 sin (7 8 , l z = d z cos <7 3 , p z — c7 3 sin <7 3 , &c. &c. [*] = ?!+ k+ k> &c, l>] = l>i + J P* + J ft> &c, and or, as in the more simple case of a single course, The ivhole difference of latitude is equal to the sum of the partial differences of latitude ; The whole departure is equal to the sum of the partial de- partures. 12 NAVIGATION. This applies to all cases, if we use the word sum in its general or algebraic sense. If we represent byZ n the sum of the northern diffs. of latitude, " " L s " " southern " M " P xo " " western departures, " P e " " eastern we have as the arithmetical formulas, [ I ] = L n ~ L s of the same name as the greater, [p] = P w ~I> e « which accord with the usual rules. (Bowd., p. 59 and p. 264.) The Traverse Form (p. 60 and pp. 266 to 286) facilitates the confutation. The course, (7, and distance, [c?], corresponding to [/] and [j:>], may be found nearly by Plane Sailing.* 8. The departure may be regarded as measured on the middle parallel, either between the extreme parallels of the traverse, or between that of the latitude left and that ar- rived at. In a very irregular traverse it is difficult to deter- mine the precise parallel ; but, except near the pole, and for * C and [d~\ are not accurately found, because [_p], the sum of the partial departures of the traverse, is not the same as p y the departure of the loxo- dromic curve connecting the extremities of the traverse. Thus suppose a ship to sail from C to A by the traverse C B, B A, her departure will be by tra- verse sailing d e + m n ; whereas, if the ship sail directly from C to A, the depart- ure will be o p, which is greater or less than d e + m n, according as it is nearer to, or farther from the equator. Thus we should obtain in the two cases a different course and distance between the same two points. In ordinary practice, how- ever, such difference is immaterial. PARALLEL SAILING. 13 a distance exceeding an ordinary day's run, the middle lati- tude suffices. (Bowel., p. 59, note.) It is easy, however, to separate a traverse into two or more portions, and compute for each separately. PARALLEL SAILING. 9. The relations of the quantities (7, d, Z, and p are ex- pressed in equations (3). When the difference of longitude also enters, then some further considerations are necessary. Problem 4. To find the relations between Z, the latitude of a parallel, p, the departure of two meridians on that parallel, and D, the corresponding difference of longitude. Solution. In Fig. 3, let PAA',PC C be two meridians. AC = jp, their departure on the parallel A C, whose latitude is A O A' = -O A B = Z, and whose radius is B A = r. A' C = Z>, the measure of A P C, the difference of longitude of the same meridians, on the equator A' C' a whose radius is O A' = A = JR. A C, A! C are similar arcs of two circles, and are there- fore to each other as the radii of those circles ; that is, A C : A' C = B A : O A', or p:D = r: JR. In the right triangle O B A, B A = O A X cos O A B, or r = R cos Z; (4) that is, the radius of a parallel of latitude is equal to the radius of the equator multiplied by the cosine of the latitude. Substituting (4) in the preceding proportion, we obtain 14 NAVIGATION. p : D = cos L : 1, or p — D cos Z, D — p sec X, (5) which express the relations required. (Bowd., p. 63.) These relations may be graphically represented by a right plane triangle (Fig. 4), of which I) is the hypothenuse, L, one of the angles, p, the side adjacent that angle. The Traverse Table, or Table of Right Triangles, may therefore be used for the computation (Bowd., p. 65, u by inspection"). MIDDLE LATITUDE SAILING. 10. Problem 5. Given the course and distance and the latitude left, to find the difference of longitude. Solution. By plane sailing, I z=z d cos 6 T , p = d sin C; (3) by Arts. 2 and 6, JJ = Z + l, Z Q '= J (Z'+Z) = Z+} l ; (6), and by equation (5), D=psecZ , (7) or JD — d sin (7 sec Z . (8) Equations (3), (6), and (7) or (8) afford the solution re- quired. The assumption of Z = -J (X' + i), or the middle latitude, suffices for the ordinary distance of a day's run; but for larger distances, and where precision is required, we must take (Art. 5) MIDDLE LATITUDE SAILING. 15 in which A L is a small correction to be added numerically to the middle latitude. A formula for computing it is given in Prob. 10, under "Mercator's Sailing." Its value in the most common cases is given in Bowd., p. 76, and in Stanley's Tables, p. 338. 11. Strictly, the middle latitude should be used only when both latitudes, L and Z', are of the same name, as is evi- dent from Fig. 1. If these latitudes are of different names, and the distance is small, \ (L + L'), numerically, may be used; or we may even take p = Z, since the meridians near the equator are sensibly parallel. If the distance is great, the two portions of the track on different sides of the equator may be treated separately. Thus, in Fig. 5, the track C A is separated by the equator into two parts, C E and E A. For C E, we have p x = — Z tan (7, C'E^Z^ =p l seG$Z i = — Z tan C sec i Z nearly. For E A, we have A'A = ^=.2/, p 2 == U tan (7, EA' =D 2 =p 2 sec£Z', =Z' tan C sec ^ U nearly. Whence we obtain C A' or D — A + -A- Instead of the middle latitudes \ Z and -J- Z', we may use more rigidly (| Ii+A Z) and (|Z' + A L'). When several courses and distances are sailed, as is ordi- narily the case in a day's run, p and I are found as in trav- 16 NAVIGATION. Fig. 6. erse sailing, and then D by regarding p as on some parallel midway between the extremes of the traverse. (Art. 8.) (Bowd., p. 264.) 12. The relations of the quantities involved in middle latitude sailing, namely, (7, d, p, I, X , and D, are represented graphically by combin- ing the two triangles of Plane Sailing and Parallel Sailing, as in Fig. 6, in which C = A C B, = AE. By these two right triangles, all the common cases classed under Middle Latitude Sailing (Bowd., p. 68) maybe solved, if we add the formulas, 13. Other problems may be stated, which never occur in practice; as, for example, — Problem 6. Given the course and distance, and the differ- ence of longitude, to find both latitudes. Solution. We have, c, d, and D being given, p = d sin C, I = d cos (7, cos Z = |, L={L Q —AL)— £/, Z'=(Z_JZ) + H A L being taken from the table (Bowd., p. 76) correspond- ing to X . MIDDLE LATITUDE SAILING. 17 Examples in Middle Latitude Sailing. L and \ represent the latitude and longitude of the place sailed from. L and ?,', the latitude and longitude of the place arrived at. X /L It 71 COURSE. DIST. 1. 39 30S. 74 20E. 41 28 S 70 30 E. S.W.byW. 210 2. 46 24 N. 47 15 W. 49 15 N. 42 21 W. X. E. £ E. 270 3. 51 10 S. 168 37 E. 48 31 S. 158 42 E 4. 22 18 S. 57 28E E. by S. 317 5. 23 15 S. 13 35 W E. 255 6. 20 5N. 154 17 W. 18 28 S E. S E. $ E. ... 7. 56 N. 29 34 W S. 47° E. 168 8. 45 16 S. 3 46E. 43 10 S. 5 22 W 9. 57 ION. 178 51 W N. 6fpts. E. 290 10. Required the bearing and distance of Cape Race from Cape Hatteras. [tan C = j cos (i + J L) <&—l sec C] Cape Hatteras, Cape Race, I I* 35 15 1ST. 46 39 " = 11 24 " = =40 57 D = 75 31 W. Tab.LIV. 53 5 " 684 , 22 26 E. = 1346 logD = 4- 17 =41 14 3.1290 lcos(Z + d Z) 9.8762 ar co log I 7.1649 <7=N.55°57'E.?.tan<7 0.1701 log I 2.8351 1. secC 0.2518 tf=1222' log d 3.0869 Note. — The logs of I and D may be obtained from the Table of " loga- rithms of small arcs in space or time " in the American Ephemeris and in Chauvenet's Lunar Method by regarding ' and * as ° and '. 11. A ship sails from Cape Frio south-easterly until her 18 NAVIGATION. departure is 3173 miles, and then, by observation, is in lati- tude 34° 30' S. ; required the course, distance, and longi- t tude. [tan C—t d=lsecC D — p sec (i + A Z)] O / O / Cape Frio, 23 1 S. 41 59 W. U — 34 30 « p — 3173 E. logp 3.5015 I— 11 29 " logZ 2.8382 Z = 28 45 " <7:=S. 77 45 E. L tan (7 0.6633 JI= 17 " 1. sec (7 0.6733 d — 3247 logce 3.5115 Z + AZ= 29 2" 1. sec 0.0583 D — 61 54 E. log D 3.5698 A' = 19 55 E. 12. A ship in latitude 39° 8' N., longitude 33° 45' W., sails N". 51° 5' E. 1014 miles ; required her position. Z' = 49° 45' N. r = 15° 16' W. 13. A ship in latitude 56° 46' S., longitude 170° 0' E., sails E. N. E. until she is in latitude 50° 10' S.; what is the dis- tance sailed, and what is her longitude ? d — 1035 miles, X ! == 163° 9' W. 14. A ship in 42° 42' N., 12° 49' W., sails 645 miles N. Wy., and is then in 49° 30' 1ST. ; required the course and longi- tude in. C = N. 50° 46' W. A' = 24° 51' W. 15. A ship sails from Port Jackson in New-Holland N. 40° W. until the departure made is 300 miles ; what is her position ? Z= 33° 50' S. Z'= 27° 52' S. X = 151° 18' E. A' =c 145° 28' E. .16. A ship in latitude 18° 50' N., and longitude 153° 45' W., sails S. E. | E., 3656 miles; what is her position? This example conies under Art. 11. MIDDLE LATITUDE SAILING. 19 \l = d cos G -A = L tan (7 sec (| L + A L ) D 2 =: U tan C sec \% E + A Z')] C = S. 41 E. 1. cos C 9.8024: 1. tan C 0.0858 d— 3656' logtf 3.5630 log L 3.0531 J = 38° 40' S. log I 3.3654 1. sec (U + J i) 0.0080 Z=18 o 50'N. D 1 = 23°22'E. log A 3,1469 L = 19° 50' S. 1. tan C 0.0858 iL+JL= 9° 25'+ 1° 31'= 10°56'N. log L 3.0756 ■iX'+J-^'= 9° 55'+ 1° 37'= 11° 32' S. 1. sec Q £'+ A L) 0.0089 i) i = 23°22'E. D a = 24°40'E. log D 2 3.1703 D == 48° 2' E. A = 153° 45' W. 7J = 105° 43' W. If J i and ^ i r are neglected, the resulting value of U will be 105° 57' TV". If the computations are made with the middle latitude, 0° 30 S., // will be 106° 39' W., or in error nearly 1°. 17. Find the latitudes of two places, whose longitudes are 12° 49' W. and 24° 51' W., their distance 645 miles, and the course from the first to the second N". 50° 46' W. (Problem 6.) c = N. 50 46 W. 1 cos 9.8010 1. sin 9.8891 d = 645 log 9.8096 log 9.8096 l=z 6 48 N. log 9.6106 log^ 9.6987 z> = 12 2 W. 46 12 N. or S. log 9.8585 4 = 1. cos 9.8402 a r=x — 6 46 6 tf. or 46° 6' S. **= 3 24 X. 3 24 N. L- 42 42 N. or 49 30 S. U = 49 30 X. or 42 42 s. Examples in Traverse Sailing. A ship from the position given at the head of each of the following traverse forms sails the courses and distances 20 NAVIGATION. stated in the first two columns ; required her latitude and longitude. 1. August 8, noon— Lat. by Obs., 35° 35' N. Long, by Chro. 18° 38' W. Courses. DlST. N. S. E. W. N. N. E. i E. / 50 44.1 / 23.6 ' S.|W, 46.2 45.7 6.7 S. by E. i E. 16.5 15.8 4.8 N. E. 38 26.9 26.9 S. S. W. i w. 41.8 37.8 17.9 192.5 71.0 99.3 55.3 24.6 S. 4J E. 41.5 28.3 30.7 38 = = D. August 9, noon — Lat by Acct., 35° 7' IS. Long. " 18° 0' W. 2. September 25, noon— Lat. by Obs., 49° 53' S. Long, by Acct., 158° 27' E. Courses. DlST. N. s. E. w. Pts. / / ' / / S. 4* E. 45.3 28.7 35.0 S. 5£ E. 19.5 10.0 16.7 S. 7 W. 38 7.4 37.3 S. 6i W. 25.7 8.7 24.2 S. 3 W. 51.2 42.6 28.4 N. 7i E. 13 1.9 12.9 N. 5f E. 10 4.3 9.0 73.6 202.7 6.2 97.4 89.9 S. 1 W. 93 91.2 D = 16.3 = 26 September 26, 8 a.m. — Lat. by Acct., 51° 24' S. Long. " 158° 1' W. In this example the courses are expressed m points, which is the preferable method. When the reductions are the same for all the compass courses, we may find the difference of latitude and depar- MERCATOR S SAILING. 21 ture for these compass courses, and the course and distance made good. The traverse is thus referred to the mag- netic meridian instead of the true. The course made good may then be corrected for variation, etc. ; and with this cor- rected course and the distance made good the proper diifer- ence of latitude and departure may be found. 3. September 16, 6 p.m.- -Lat by Obs., Long, by Chro. 50° 16\S. Comp. Course. DlST. N. s. E. w. S. W. i s. 25 ' 19.3 ' 15.9 s. s. w. 30 27.7 11.5 S. by W. 18 17.7 3.5 s. 43 43 S. by E. i E. 255 24.7 6.2 S. E. i S. 33 26.5 19.7 174.5 25.9 30.9 (map.) S. 2° W. 159 158.9 5.0 Var'n, &c, 18° W. (true) S. 16° E. 159 152.8 43.8 or S. by E. £ E. 70 = = D September 17, noon — Lat. by Acct., 52° 49' S. Lon^. " MERCATOR' S SAILING. 14. Middle Latitude Sailing suffices for the common pur- poses of navigation ; but a more rigorous solution of pro- blems relating to the loxodromic curve is needed. These solutions come under " Mercator's Sailing." Pkoblem 5. A ship sails from the equator on a given course, C, till she arrives in a given latitude, X, to find the difference of longitude, D. Solution. Let the sphere (Fig. 7) be projected upon the plane of the equator stereographically. The primitive circle A B C . . . ,M is the equator. 22 NAVIGATION. Fig. 7. P, its centre, is the pole (the eye or projecting point being at the other pole).* The radii, PA, PB, PC, &c, are meridians making the same angle with each other in the projection as on the sur- face of the sphere.* The distance P m, of any point m from the centre of the projec- tion, =tan|(90 o — X), the tangent of \ the polar distance of the point on the surface which m re- presents, the radius of the sphere being 1.* This curve in projection makes the same angle with each merid- ian, as the loxodromic curve with each meridian on the surface.* A M is the whole difference of longitude D. If we suppose this divided into an indefinite number of equal parts, A B, B C, C D, &c, each indefinitely small, and the meridians P A, PB, PC, &c, drawn, the intercepted small arcs of the curve A b c .... m may be regarded as straight lines, making the angles P A 5, P b c, P c d, &c, each equal to the course C ; and consequently the triangles PAJ,PJc,Pcrf, &c, similar. We have then PA:PJ = PS:Pc = Pc:P(?, &c, or the geometrical progression, PA:PJ :Pc:....Pra. If then _D = the whole difference of longitude, d = one of the equal parts of 2>, ~j will be the number of parts, and -j + 1 the number of meridians PA,P5....Pm, * Principles of stereographic projection. mercator's sailing. 23 or the number of terms of the geometrical progression : and, employing the usual notation, the first term a = PA = l, the last term l — Ym — tan £(90° —Z), the ratio r s= ^— . PA To find this ratio, we have in the indefinitely small right triangle AB5, tan BAJ = cotPAJ=:^-r, or „ PA — Vh cot 6 = 5 , a whence P A— P 6 == d cot (7/ Pft = PA-dfcot C, and, since PA=I, P5 r = p-: = l — df cot (7. Then by the formula for a geometrical progression, (Algebra, p. 240,) we have tan \ (90°— Z) = (l — c? cot C) 7 . (8) Taking the logarithm of each member, we have logtan|(90°— Z)=^log(l— tfcot C). (9) But we have in the theory of logarithms 71/ 71? 71/ (JVaperian) log (1 + ^)=^ — — + — — — + &c and |- ^,2 ^,8 ^ 4 "1 ( Common) log (1 +7i) = m rc — — + «- •— j" + & c — K 10 ) in which the modulus m =.434294482. Hence, putting n = — c? cot (7, log (1 -6? cot £7)= m [-d cot (7-i d 2 cot 2 C-\ d 2 cot 3 <7-&c. . . .], and substituting in (9) and reducing, log tan i (90°— Z)= -m x D [cot tf+£ cZ cot 2 G + id 3 cot 3 6'+&c. ...]. C 11 ) 24 NAVIGATION. This equation is the more accurate the smaller d is taken, so that if Ave pass to the limit and take e?=0, it becomes perfectly exact. The broken line A b e. . . .m then becomes a continuous curve, and our equation (11) becomes log tan \ (90°— L) = — m X D cot C ; whence 2> = logtani(90°-Z) tan (7, (12) But in this equation D is expressed in the same unit as tan (7, that is, in terms of radius. (Trig., Art. 11.) To reduce it to minutes we must multiply it by the radius in minutes, or r'=3437'.74677. Substituting the value of m, we shall have (in minutes), 3437'. 74677, . - /aa0 rw n .484294482 l ^ Un * < 90 ~ Z > tan (7 ' Z> = - To avoid the negative sign, w^e observe that 1 1 tan £ (90°— X): cot i(90°-Z) tan £(90° + Z)' or that —log tan i (90°— Z)=log tan £ (90° + i). Hence we have, by reducing, Z> = 7915 , .70447 log tan (45° + £Z) tan C. (13) Note. — Problem 5 may be more readily solved, and equation (13) obtained by aid of the Calculus. In Fig. 1, suppose c a to be an element, or infinitesimal part, of the loxodromic curve C A : cb will be the corresponding element of the meridian, and b a x sec X, the element of the equator ; L being the latitude of the indefinitely small triangle cab. Fig. 1. mercator's sailing. 25 By articles 5 and 10, using the notations of the Calculus, we have d X = cos Odd d p = tan C d X d D = sec Ldp = tan (7 sec XdX, in which C i3 constant. By integrating the last equation between the limits X = and X = Z, we shall have r L X = tan C\ sec X d X, the whole difference of longitude required in Problem 5. To effect the integration, put sin X = x, then by differentiating d x d X = — , and multiplving by sec L cos L t j t- d x d x sec X d L = sec Z d X = 1 — x cos 2 X 1— sin 3 X' d x Resolving into partial fractions, we obtain sec X d X = 4- — v and Ll + x 1 — xj J o L secXdX = i[log(l +x)-log(l-x] = log|/L^ ' 1 — X . . /l+sin = iogy - — - " 1 — sic X -sinX -sinX = log tan (45° + 1 X) Trig. (154). Whence we have D = log tan (45° + i X) tan C. But in this the logarithm is Naperian, and D is expressed in terms of the radius of the sphere. To reduce to common logarithms, we divide by ,m =.434294482, and to minutes by multiplying by r'= 3437'.T4677, and obtain D = 1915'. 70447 log tan (45° + } X) tan C y as in (13). 15. To facilitate the practical application of the formula just obtained, put ilf=7915 , .70447 log tan (45° + £ L) ; (14) 26 NAVIGATION. and let M be computed for each minute of L from up- ward, and its values given in a table. We shall thus form the Table of Meridional Parts or of Augmented Latitudes, such as Bowditch's Table III. This formula accords with that given in the Preface. (Bowd., Pref. p. iv.) In practice, then, we have only to take the value of M corresponding to X, and D is then found by the formula, D=M tan C. (15) M has the same name, or sign, as L. Example. To find tne meridional parts, or augmented latitudes, for each minute, from 30° to 33° ; log 791 5'.70447= 3.898490. L. 45° + JZ. log tan. I. log tan. log M. M. 30 30 20 30 40 31 31 20 31 40 32 32 20 32 40 60 60 10 60 20 60 30 60 40 60 50 61 61 10 61 20 0.2385606 .2414830 .2444154 .2473580 .2503108 .2532741 .2562480 .2592328 .2622286 9.377599 .382887 .388129 .393326 .398480 .403591 .408660 .413690 .418680 3.276089 .281377 .286619 .291816 .296969 .302081 .307150 .312180 .317170 1888.37 23.14 1911.51 7 23.21 1934.72 8 23.29 1958.01 9 23.38 1981.39 7 23.45 2004.84 9 23.54 2028.38 9 23.63 2052,01 9 23.72 2075.73 8 23.80 2099.53 33 61 30 .2652356 .423632 .322122 The second differences afford a check of the work. By interpolating into the middle, M can be found for each 10' ; and then, by simple interpolation, for each 1'. In the first step, one eighth of the second difference is to be sub- tracted. The following is an example : hercator's sailing. 27 M L M 30 1888.37 30 11 1901.09 1 1889.53 12 1902.24 2 1890.68 13 1903.40 3 1891.84 14 1904.56 4 1893.09 15 1905.72 6 1894.15 16 1906.87 6 1895.31 17 1908.03 1 1896.46 18 1909.19 8 1897.62 19 1910.35 9 1898.77 20 1911.51 1899.93 &c. 16. Problem 6. A ship sails from a latitude, L, to an- other latitude, U, upon a given course, C / find the differ- ence of longitude, D. Solution. Let M be the augmented latitude corresponding to L, M' " " " " JO. The difference of longitude from the point, A, where the track crosses the equator to the 1st position, whose latitude is L, will be D =31 tan C; and to the second position, whose latitude is U, D=M' tan C ; and we shall have D=D-I>={M'-M) tan C; (16) or, when M r in which C is one of the angles, m = CE, the side adjacent, D = E F, the side opposite. The triangle of " Plane Sailing" has the same angle (7, with ;— C B, the adjacent side, and ]) = BA, the opposite side. Fig. 8 represents these two triangles combined. By them, all the common cases under Mercator's Sailing can be solved, either by computation or by the Traverse Table. (Bowd., p. 79.) The relations between the several parts involved are I — d cos (7, L'=L + l, p^dsmC, m =M'—M, D — m^nC, X f ~X + D; and since p = 1 tan (7, I : m =p : D. 18. Problem 7. Given the latitudes and longitudes of two places, find the course, distance, and departure. (Bowd., p. 79, Case I.) Solution. L and U being given, we take from Table III. JfandJf'. We have I = Z'-Z, m = M'-M, D ^ X'- X- by Mercator's sailing, tan C = — ; and by Plane sailing, d = I sec (7, p = I tan (7/ (18) mercator's sailing. 29 /, m, and C are north or south according as U is north or south of X. D, p. and (7 are east or wes£, according as X' is eas£ or westf of A. If the two places are on opposite sides of the equator, we have numerically Mercator's sailing is rarely used except in this case, and when the differences of latitude and longitude are consider- able. There are two limits of its accuracy :— 1. Table III. contains the augmented latitude only to the nearest minute or mile.* • 2. It is computed on the supposition that the earth is a sphere. Some works on Navigation, as Mendoza Rios and Riddle, contain a table of augmented latitudes, in which the true form is taken into consideration.! Examples. 1. Required the course and distance from Cape Frio to 34° 30' S., 18° 30' E. Cape Frio, 23 1 S. 41 59 W. M - 1420 S. £'== 34 30 S. W = 18 30 E. M* = 2208 S. log D 3.5598 I = 11 29 S. D = 60 29 E. M = 788 S. logm 2.8965 c =s. 11 45 E. 1 sec C log/ 0.6733 2.8382 1. tan C 0.6633 d = 3247' logd 3.5115 * The most convenient unit for nautical distances is the geographical, nautical, or sea mile, which is 1' of the earth's equator, or 6086.43 feet. Regarding the earth as a sphere, this is also 1' of any great circle, f The formula for the terrestrial spheroid is M— 7915'.70447 log tan (45° +£ L) — 22'.98308 sin Z + 0'.01276 sin 3 L+ &c. Delambre has shown that a table of meridional parts constructed for the sphere may be used for the spheroid by using as the argument the geocen- tric latitude instead of the true latitude. 30 NAVIGATION. 2. Required the course and distance from Cape Frio to Lizard Point, England. Cape Frio, 23 1 S. 41 59 W. M = 1420 S. Lizard Pt., 49 58 N. 5 12 W. M' — 3471 N. log D 3.3438 1= 72 59 N. Dr = 36 47 E. M = 4891 N. log m 3.6894 <7=:N.24 17 E. 1. sec C log* 0.0402 3.6414 1. tan C. 9.6544 d = 4804' logd 3.6816 3. A ship in latitude 18° 50' 1ST., longitude 153° 45' W., sails S. 4^ points E., 3656 miles ; what is her position ? , , d = 3656 log d 3.5630 L = 18 50 N. A = 153 45 W. M = 1151 JST. 1 cos C 9.8024 I = 38 40 S. log I 3.3654 L' — 19 50 S. M' — 1215 S. 1. tan C 0.0858 M — 2366 S. log m 3.3740 log D 3.4598 D - 48 3 E. X = 105 42 W. 19. Other problems might be stated than those com- monly given ; as, for example, — Problem 8. Given the latitude left, the course and both longitudes, to find the latitude arrived in. Solution. We have D = A'— X, by Mercator's sailing m —D cot C (N. or S. as is C), by Table III. M corresponding to L, M' = M+m, and again by Table III, U corresponding to M'. Problem 9. Given the difference of longitude and differ- ence of latitude of tico places, and the course between them, find both latitudes. Solution. We have m = M— M — D cot C. But 3F= 7915'.70447 log tan £ (90° + Z') M = 7915'.70447 log tan £ (90° + Z), mercatok's sailing. 31 consequently, log tan \ (90° +27) - log tan * (90° +L) = 79 f 5 ^ 7 - (19) Put logcot^^^, * (20) then equation (19) gives tani(90° + Z') , . t^-WTT) = cot *• By PL Trig. (109) tan i (x + y) sin x -f sin y tan \ (x — y) sin a; — sin y' In this, if we take x + y = 90° + L' x — y — 90° + X, ' we have or putting cc = 90 o +i (Z'+i), i = | {11 + L) the middle latitude, »== 90° + i , and y = £ (Z'-Z)-= J Z, and tan j (90° + Z') _ cos L Q + sin j- Z __ tan" J (90° + L) ~~ cos Z — sin £ 2 ~~ COt " whence T cot ^ + 1 . , f cos Jj = — — sin 4- /, cot 9 — 1 z ' which, by PL Trig. (151), reduces to cos X = tan (45° + 0) sin £ I. (21) We have also The solution is effected by equations (20), (21,) (22). 32 NAVIGATION. Example. The difference of longitude of two places is 5 10 E. 3 28 N". N. 32 59 E; ar. co. log 6.10151 log 2.49136 log cot 0.18776 log 8.78063 log tan 1.15900 log sin 8.48069 W cos 9.63969 the difference of latitude, the course find the latitudes. (Constant) 7915'.70447 C= 32° 59' log cot = 0.06034 0=41° 2' 45°+ = 86° 2' %l = 1° 44' Z =s 64° 8' N". or 64° 8' S. Z = 62° 24' 1ST. or 65° 52' S. L' = 65° 52' N. or 62° 24' S. This problem cannot be solved with precision when Z is near 0. 20. Problem 10. To find the correction of the middle latitude in Mid, Lat. Sailing. (Tab. Bowd., p. 76 ; Stan- ley, p. 338.) Solution. In Mid. Lat. Sailing we have cosZ = -g. (23) in which precision requires that Ave take Z = i(Z' + Z) + AZ; A Z being a correction of the middle latitude, which it is now proposed to find. In plane sailing p = I tan C, in Mercator's sailing D = ?n tan (?, which substituted in (23) give l_ ■m COS Z : (24) MERCATOR'S SAILING. 33 whence 1—2 sin 2 i i ==— , and sin £ i = |/ m ~ Z . (25) Now for different values of Z and X' we may find (in minutes) I = U — X, the middle latitude, i m == ^ (X' + Z), m=7915'.70447 [log tan (45° + | Z') -log tan (45° + £ Z)],* and then i by (24), or if small by (25), from which sub- tracting Z m we have A Z, which is required. In computing m, logarithms to 7 places should be used when the difference of latitude is less than 12°. The correction of the Middle latitude computed for differ- ent middle latitudes and differences of latitude may be given in a table, as on page 76 (Bowd.) It becomes too large to be conveniently tabulated, when the latitudes are of differ- ent names, or the middle latitude is very small. Example. Find the correction of the Middle Latitude, when the lati- tudes are i = 12°, Z ! = 18°. o / (45° + £ Z') = 54 I. tan 0.1387390 (45° -f | Z) — 51 I. tan 0.0916308 const, log 3.89849 i {Z' + Z) = 15 0.0471082 log 8.67310 I = 6 == 360' ar. co. log 7.44370 Z — 15 7 I. sec 0.01529 AZ = 7 21. The loxodromic curve on the surface of the earth and its stereographic projection (Fig. 7) present a peculiarity * log 7915.70447 = 3.8981896. Another formula, requiring only 5 place logarithms, is m = 6875'.493 [q + 1 ? 3 + \ q* + \ q 1 + - .] in which q = sin ^ 1 sec % (L' +L). 34 NAVIGATION. worthy of notice. Excepting a meridian and parallel of latitude, a line which makes the same angle with all the me- ridians which it crosses would continually approach the pole, until, after an indefinite number of revolutions, the distance of the spiral from the pole would become less than any assignable quantity. It is usual to say that such a curve , meets the pole after an infinite number of revolutions. Still, however, it is limited in length. For we have for the length of any portion, by plane sailing, d == (Z'—Z) sec C. If Z =: o and Z' = 90° = J, the whole spiral from the equator to the pole will be, with radius == 1, d =— sec C. If Z = - 90° = -J, and Z' = 90° =J, we have, as the entire length from pole to pole, d = re sec C, If also C =: 0, or the loxodromic curve is a meridian, d = 7r, a semicircumference, as it should be. So also the length of the projected spiral A b c . . . (Fig. 1) from A to m can readily be shown to be (calling this length 6) ; 6 = Mm sec C = [1 — tan (45° — £ Z)] sec (7, or, (H. Trig. (.51),) * = ^f^; and its length from the equator to the pole — taking Z == 90° ; d = sec C. a meecatoe's chart. 22. On a Mercator's chart, the equator and parallels of latitude are represented by parallel straight lines ; and the A MERCATOR S CHART. 30 meridians also by parallel straight lines at right angles with the equator. Two parallels of latitude, usually those which bound the chart, are divided into equal parts, commencing at some meridian and using some convenient scale to repre- sent degrees, and subdivided to 10', 2', 1', or some other convenient part of a degree, according to the scale em- ployed. Two meridians, usually the extremes, are also divided into degrees and subdivided like the parallels of latitude, but by a scale increasing constantly with the latitude : so that any degree of latitude on such meridian, instead of be- ing equal to a degree of the equator, is the augmented de- gree, or augmented difference of 1° of latitude, derived from a table of " meridional parts." (Bowd., Table III.) The meridian is graduated most conveniently by laying off from the equator the augmented latitudes ; or from some parallel, the augmented difference of latitude for each degree and part of a degree, — using the same scale of equal parts as for the equator. Thus, on such a chart, the length of 1° in lat. is 60' of the equator, IC cc cc 10° cc 61' cc cc cc cc cc 20° cc 64' cc cc cc cc cc 30° cc 69' cc cc cc cc cc 40° cc 78' cc cc cc cc cc 50° cc 93' cc cc cc cc cc 60° cc 120' cc cc cc cc cc 70° cc 176' cc cc 90°. When A X > % A, X 2 is negative, and the vertex, C , is be- yond B, as in Fig. 10, instead of being between A and B, as in Fig. 9. In (30) we have Z v positive, or of the same name with the greater latitude, since numerically X 2 < 90°. The vertex, which is here used, is that which is nearest the point B, or the place whose latitude is numerically the greater. For this in (27) A X < 90°. There are, however, two vertices, which are diametrically opposite, as C and C' of the great circle C' E C in Fig. 10. For the vertex C, we have in (27) A X> 180°, or in the third quadrant, and in (30), Z v of a different name from Z 2 . 2d. To find any number of points, C, C", C"\ &c, C 19 C 2 , C 3 , &c, we may assume at pleasure the differences of longi- tude from the vertex C P C, C P C", C P G", &c. It is best to assume them at equal intervals of 5° or 10°. Let X ! = C P C, U = (90°- P C), the lat, of C, a" = C PC", Z^^-PC"), " C", X'"=: C P C", Z m = (90°- P C ,,; ), " C"\ &C. &G. then the right triangles C P C r , C P C'\ C P C", &c, give tan U = tan L v cos X\ tan L" = tan Z v cos X", \ (31) tan Z'"= tan Z v cos X'", &c. J 42 NAVIGATION". Or we may assume values of Z', Z", L"\ &c., and find the corresponding values of A/, A", A'", &c., by the formulas cos X' z=z tan J7 cot L v cos A" =i tan L" cot i y h (32) cos A"'= tan X'" cot Z v &c. J from which we shall have two values of A for each value of Z. Having thus found as many points as may be deemed sufficient, we may plot them upon the chart, and through them trace the required curve. 29. Another method con- sists in finding the longi- tude of the point E (Fig. 10), where the great circle intersects the equator, and then by the right triangles E C c\ E C" c", E &' c"\ &c, the latitudes and lon- gitudes of C, 0", C", &c. Let X L = E a, the longitude of A from E ; A 2 = E 5, the longitude of B from E ; A = X 2 — X x = a&, the difference of longitude of B from A ; Li and Z 2 , respectively the latitudes of A and B ; i v — Q C , the latitude of the vertex, and also the measure of QE C , the inclination of the great circle to the equator. From the right triangles E A a, E B S, we have tan A a tan B b Fig. 10. or tan E = tan L v = sin E a tan L x * sin E V tan Z.> sin I, sin 7 * GREAT-CIRCLE SAILING. 43 whence sin ?i t sin /t 2 tan L x tan L 2 ? and by composition and division, sin ^ + sin \ tan L 2 + tan Z x sin X 2 — sin a x tan L 2 — tan Z x * By PL Trig., (109) and (126), this becomes tan £ (% 2 + /Ij) sin (Z 2 4- L x ) tan \ {\ — \) sin (L. 2 — L x )* But A = A 2 — A x , and if we put A'X = -| (A 2 -}- A x ) we have sin (Z 2 4- A) tan J 'A sin (L 2 — L x ) A x = J'A ■ A 2 = J'A + iA tan £ A £A (33) a (34) From these we find A 19 or A 2 , which applied to the longitude of A, or of B, gives the longitude of E, the intersection with the„ equator. This point, it may be observed, is nearer A than B ; and is outside of A, when the two places are on the same side of the equator ; and between A and B when they are on different sides. For finding the inclination of the circle to the equator, we have above tan L v = tan i x cosee A x tan L v = tan X 2 cosee A 2 in which L v will have the same name, or sign, as Z 2 . To find any numbers of points, C, C", C", &c, we may as- sume at pleasure, as in Art. 28, the differences of longitude, A', A", A'", &c, from E ; and from the right triangles E CV, E C" c" E C" c"\ &c, find the corresponding latitudes L\ U\ U'\ &c, by the formulas tan U = tan L v sin A' tan Z"= tan L v sin A" j> (35) tan Z"= tan L v sin A'" &c. J The great circle intersects the equator at two opposite 44: NAVIGATION. points. The intersection, E, given by these formulas is that which is nearest A, the place whose latitude is the smallest. This method is preferable to that of Art. 28, only when the two places are on different sides of the equator, and the intersection with the equator is between them. In this case, X l and A 2 , as well as L^ and Z 2 , will have opposite signs. 30. Problem 12. To find the great-circle distance of two given points. Solution. Let A and B (Fig. 9) be the two given points. In the triangle P A B are given, as in Problem 11, PA = 90 o -Z n P B = 90°-Z 2 , A P B = A> to find AB = c?, the distance required. 1st Method. By (27), (28), (29), and (30), we may find Z„, the latitude of the vertex, and X x and A 2 , the longitudes of the two given places from the vertex. Then from the right triangles, A C P, B C P, (Fig. 9), putting A C = d i and BC = c? 2 , we have tan c?! = tan X x cos L v tan d 2 = tan A 2 cos L v \ (36) and d = d Y + d 2 . d reduced to minutes Avill be the distance in geographical miles. When A 2 is negative, which happens when the vertex is beyond B (Fig. 10), d 2 is also negative, and d is numerically the difference of d l and d 2 . 2d Method. By (33) and (34) we may find X l and A 2 the longitudes of the two given places from E (Fig. 10), the in- tersection, and L v the angle which the circle makes with the equator. Then from the right triangles, E a A, E b B, we have, putting d f = E A, d"= E B, tan dl = tan X v sec L v 1 tan d"~ tan A a sec L v \ (37) and d=d"-d'. J GREAT-CIRCLE SAILING. 45 When £ x and L 2 are of different names, so also are a x and A 2 , and d is numerically the sum of d' and d" . Zd Method. By Sph. Trig. (4) we have cos d c= sin j^ sin Z 2 + cos X x cos L 2 cos A, and, putting Jc cos = sin X 2 ) _ * sin 4> = C os A cos X or tan ^ = cot A cos A, 1 COS , sin Z 2 sin (Z 2 + 0) COS a = 2 » from which is in the same quadrant with X. 30. Problem 13. To find the course on a great-circle route. Solution. If A (Fig. 9), the point whose latitude is nu- merically the smaller, is the point of departure, it is required to find the angle P A B : if B is the point of departure, then the angle P B A. 1st Method. The position^ of the vertex having been found by (27), (28), (29), (30), we have from the right triangles, A C P, B C P, cos A = sin L v sin X x ) cos B == sin L v sin A 2 j ^ ' in which A < 90° ; and B < 90°, when the vertex is between A and A (Fig. 9), but > 90°, when the vertex is beyond B (Fig. 10). 2c? 3fethod. Having found the intersection and angle with the equator by (33) and (34), we have from the right tri- angles E a A, E b B (Fig. 10), cos A = sin L v cos X x cos B == — sin L v cos JL Sd Method. By Napier's Analogies we have 46 NAVIGATION. tan £ (B + A) tan i (B-A) cos \ {L^—L x ) sin i (Lz+LJ sin \ (A— L x ) cos \ (Z 2 + L x ) A=i(B + A)-l(B B = * (B + A) + | (B cot J A cot % X A) A) (41) When A and B are on opposite sides of the equator, ^ (i 2 — i x ) is numerically half the sum, and % (J0 2 + X x ) is half the difference of the two latitudes. 31. When the courses are found by this last method, the distance may be found by or, tan £ d •■ tan £ d : sin j(B + A) : sin i (B— A) cos^(B + A) : cos i (B-A) tan $ (i 3 — ij), cot i (i 2 +ii), (42) The 1st is preferable, when % (Z 2 -f .Z^) is near 0, and conse- quently J (B-f A) is near 90° ; the second when % (X 2 — L x ) and consequently \ (B — A) are near 0. (Sph. Trig. 74.) 32. Example. To find the great circle from San Francisco to Jedo. (Formulas 27, 28, 29, 30.) San Francisco, Jedo, Lat. Z 2 = 37 48 N. L x = 35 40N. Z 2 +Z 1= 73 28 Z a — L x = 2 8 A = 97 38 i A = 48 49 JA= 1 57 \=\X—A\= 50 46 ^=£ A— .J A = 46 52 Long 122 22 W. 140 E. 1. cosee 0.0183 1. sin 8.5708 1. cot 9.9420 1. tan 8.5311 1. sec \ 0.1990 L sec ^0.1651 1. tan L x 9.8559 1. tan Z 2 9.8897 Vertex, Z„=48° 37' N. Long.169 14 W. I. tan Z„ 0.0549 0.0548 GREAT-CIRCLE SAILING. 47 Long, from " Vertex. L cos A. 1. tan L. ± 5 ±10 ±15 ±20 ±25 ±30 ±35 ±40 ±45 ±50 0.0000 9.9983 .9933 .9849 .9729 .9573 .9375 .9134 .8842 .8494 .8081 0.0549 .0532 .0482 .0398 .0278 .0122 9.9924 .9683 .9391 .9043 .8630 Lat. 48 37 N. 48 30 48 10 47 37 46 50 45 48 "44 30 42 55 41 38 44 36 7 Longitudes. 169 14 W. 164 14 159 14 154 14 149 14 144 14 139 14 134 14 129 14 124 14 119 14 W. 169 14 W. 174 14 179 14 W. 175 46 E. 170 46 165 46 160 46 155 46 150 46 145 46 140 46 E. (Vertex). (36). (39) 1. sin A x 1. sin A 2 1. sin L v 1. cos d 9.8891 9.8632 9.8753 9.7444 1. cos C» 9.7385 1. tan A x 0.0880 1. tan A 2 0.0283 L cosi v 9.8202 1. tan d x 9.9082 1. tan d 2 9.8485 d, = 39° 0' Course, C l N. 54° 27' E. from JTedo. d 2 = 35 13 C 9 N. 5Q 48 W. from San Francisco. Distance, d = 74° 13' = 4,453 miles. Distance by Mercator's sailing, 4,689 miles. 33. To follow a great circle rigorously requires a con- tinual change of the course. As this is difficult, and indeed in many cases is practically impossible, on account of currents, adverse winds, &c, it is usual to sail from point to point by compass, thus making rhumb-lines between these points. When the ship has deviated from the great circle which it was intended to pursue, it is necessary to make out a new one from the point reached to the place of destination. It is a waste of time to attempt to get back to an old line. 34. As the course, in order to follow a great circle, is practically the most important element to be determined, mechanical means of doing it have been devised. Towson's Chart and Table is used much by English navigators. 48 NAVIGATION. Chauvenet's Great-Circle Protractor renders it as easy as taking the rhumb-line course from a Mercator chart. Charts have been constructed by a gnomonic projection, on which great circles are represented by straight lines ; but by these computation is necessary to find the course. 35. A great circle between two points near the equator or near the same meridian differs little from a loxodrornic curve. But when the differences both of latitude and of longitude are large, the divergence is very sensible. It is then that the great circle, as the line of shortest distance, is preferred. But it is to be noted, that in either hemisphere the great- circle route lies nearer the pole, and passes into a higher latitude, than the loxodrornic curve. Should it reach too high a latitude, it is usually recommended to follow it to the highest latitude to which it is prudent to go, then follow that parallel until it intersects the great circle again. 36. A knowledge of great-circle sailing will often enable the navigator to shape his course to better advantage. Let A B (Fig. 11) be the loxodrornic curve on a Mercator's chart, AC B, the projected arc of a great circle. The length on the globe of the great circle A C B is less than that of the rhumb-line A B, or of any other line, as A D B, between the two. But A C B is also less than lines that may be drawn from A to B on the other side of it, that is, nearer the pole ; and there will be some line, as A D' B, nearer the pole than the great circle, and equal in length to the rhumb-line. Between this and the rhumb-line may be drawn curves from A to B, all less than the rhumb-line. If the wind should prevent the ship from sail- Fig, n. SHAPING THE COURSE. 49 ing on the great circle, a course as near it as practicable should be selected. If she cannot sail between A B and A C, there is the choice of sailing nearer the equator than A B, or nearer the pole than A C. The ship may be near- fing the place B better by the second than by the first, al- Ithough on the chart it would appear to be very far off from the direct course. This may be strikingly illustrated by the extreme case of a ship from a point in a high latitude to another on the same parallel 180° distant in longitude. The great-circle route is across the pole, while the rhumb-line is along the small circle, the parallel of latitude, east or west ; the two courses differing 90°. Any arc of a small circle drawn between the two points, and lying between the pole and the parallel of latitude, will be less than the arc of the parallel. Hence the ship may sail on one of these small circles nearly west, and make a less distance than on the Mercator rhumb, or parallel due east. This is, indeed, an impossible case in practice, but it gives an idea of the advantage to be gained in any case by a knowledge of the great-circle route. It is possible in high latitudes that a ship may have such a wind as to sail close-hauled on one tack on the rhumb- line, and yet be approaching her port better by sailing on the other tack, or twelve points from the rhumb-line course. 37. The routes between a number of prominent ports recommended by Captain Maury are mainly great-circle routes, modified in some cases by his conclusions respecting the prevailing winds. SHAPING THE COURSE. . 38. The intelligent navigator, in selecting his course to a destined port, will not only have regard to the directness of the route, but will take into consideration obstructions and dangers which may be in his way ; prevailing winds 50 NAVIGATION. and currents ; and, in case of a threatening storm, the course to be taken to avoid its greatest violence, or being driven on a lee shore. Good charts and books of sailing directions afford all requisite information respecting obstructions and dangers in the most frequented seas. Exploring expeditions from England, France, and the United States have of late years added greatly to this branch of nautical knowledge. The labors of Maury, and his recent colaborers in Eng- land and France, have greatly increased our knowledge of prevailing winds in large portions of the ocean. The care- ful observations of intelligent navigators are much needed still further to develop it. A few of the stronger currents, such as the Gulf Stream off the coast of the United States, are well known. But more extended observations are wanted. Currents are often indicated by the difference in the temperature of their wa- ters from that of those surrrounding them ; so that the thermometer, as well as barometer, has become an import- ant instrument to the navigator. The works of Redfield, and especially of Reid and Pid- dington, afford much information respecting storms and tor- nadoes. That class of storms called cyclones is particularly deserving of attention. . These branches of physical geography are well worthy of study by those engaged in navigating the ocean. CHAPTER II. REFRACTION. —DIP OF THE HORIZON.— PARALLAX.— SEMIDIAMETERS. REFRACTION. 39. It is a fundamental law of optics, that a ray of light passing from one medium into another of different density is refracted, or bent, from a rectilinear course. If it passes from a lighter to a denser medium, it is bent toward the perpendicular to the surface, which separates the two me- dia ; if it passes from a denser to a lighter medium, it is bent from that perpendicular. Let M and N (Fig. 12) represent two media each of uniform density, but the density, or refracting power, of N being the greater; a b c, the path of the ray of light through them ; P 5, the normal line, or perpendicu- lar, to the separating surface at b. If a J is the incident ray, b e is the refracted ray ; P b a is the angle of incidence / P b a 1 is the angle of refraction. If c b is the incident ray, b a is the refracted ray, and P b a! and P b a are respectively the angles of incidence and refraction. Moreover, these angles are in the same plane, which, as it passes through P 5, is perpendicular to the surface at which the refraction takes place ; and we have for the refraction Fig. 12. 52 NAVIGATION. a r b a ~ P b a — P b a\ or the difference of direction of the incident and refracted rays. A more complete statement of the law for the same two media is, that - — TJrnr-7 — m > a constant for these media ; sin V o a or, the sines of the angles of incidence and refraction are in a constant ratio. This law is also true when the surface is curved as well as when it is a plane. 40. If the medium N instead of being of uniform density is composed of parallel strata, each uniform but varying from each other, the refracted ray b c will be a broken line ; and if, as in Fig. 13, the thickness of these strata is indefinitely small, and the density gradually increases in proceeding from the surface 5, b c will become a | n curved line. But we shall still have for any point c of this curve, c a! be- Fig. 13. j ng a tan g ent to it, sin P h a a constant for the particular strata in which c is situated. This law, which is true for strata in parallel planes, ex- tends also to parallel spherical strata, except that the nor- mals P i, P' c are no longer parallel, but will meet at the centre of the sphere. But the refraction takes place in the common plane of these two normals. 41. The earth's atmosphere presents such a series of pa- rallel spherical strata, denser at the surface of the earth, and decreasing in density, until at the height of fifty miles the refracting power is inappreciable. REFRACTION. 53 In Fig. 14, the concentric, circles M N" represent sections of these parallel strata, formed by the vertical plane passing through the star S and the zenith of an observer Z E S at A. The normals C A Z at A, and C B E at B, are in this vertical plane. S B, a ray of light from the star S, passes through the atmosphere in the curve B A, and is re- ceived by the observer at A. Let A S' be a tangent to this curve at A ; then Fig. 14. the apparent direction of the star is that of the line A S' ; and the astronomical re- fraction is the difference of directions of the two lines B S and A S'. This difference of directions is the difference of the angles EBS,EDS', which the lines SB,S'A make with any right line C B E, which intersects them. If, then, r represent the refraction, we have r = E B S-E D S'. Also, E B S is the angle of incidence, and Z A S', the appa- rent zenith distance, is the angle of refraction ; and we have sin E B S sin ZAS' == m, a constant ratio for a given condition of the atmosphere and a given position of A ; but varying w T ith the density of the atmosphere, and for different elevations of A above the sur- face. For a mean state of the atmosphere and at the sur- face of the earth, experiments give m = 1.000294. The principles of Arts. 39 and 40, applied to this case, 54 NAVIGATION. show that astronomical refraction takes place in vertical planes, so as to increase the altitude of each star without affecting its azimuth. The refraction must therefore be subtracted from an observed altitude to reduce it to a true altitude; or h = h'-r, in which h is the true altitude, h\ the apparent altitude, r, the refraction. These laws are here assumed. The facts and reasoning on which they depend belong to works on optics. (Bowd., p. 153 ; HerschePs Astronomy, p. 37 ; Lardner's Optics.) 42. Problem 14. To find the refraction of a star. Solution. In Fig. 14, let 2 = Z A S', the apparent zenith distance of the star, r = EBS-EDS',the refraction, w=ZCE; then EDS' = ADC = ZAS'-ZCE^2-w, E B S % E © S' f r w a — u + n , sin E B S sin (z — u 4- r) and . „ . Q , = = m, sm ZAS sin z or sin \z— (u— r)~\ == m sin z\ (43) which is of the same form as (309) of Plane Trig., sin (z + a) = m sin z, the solution of which gives tan (z + ia) he tan * a. Putting a =z — (ii—r), its value in (43), we have tan [z-i {u-r)] = j— J tan J (u—r) ; whence tan | (u-r) = y-j-^ tan [z— £ (w-r)]. (44) REFRACTION. 55 In this u and r are both unknown, but we may note that each is a very small angle, being when the zenith distance is 0, and increasing with the zenith distance. As it is ne- cessary to make some supposition respecting them, let us assume that they vary proportionally, and that u 7 = ?' a constant, reserving it for observations to test the rigor of this assumption. Equation (44) then becomes tan \ (q-1) r = ~=^ tan [>-£ (y-l>r], or, since i {q— 1) r is quite small, * (3.-1) r sin 1" =\=^ tan |>-| (q-1) r] ; whence 1 +m 1 — m tan (z— \ (q— 1) r). (q — 1) sin V 1+m Since observation is to determine q, we may as well con- sider that it determines the whole of the constant factor into which q enters. 2 1 — ra Put then n = ■ (q — 1) sin 1" 1+m' and the formula reduces to r — n tan (z— p r), which is known as Bradley's formula. Suppose at two given zenith distances z' and z n the refrac- tions r' and r ,r are found by observations in a mean state of the atmosphere, then we have the two equations, r' = n tan (z! —p r'), r /r = ntan ty'—pr"); 56 • NAVIGATION. and the two unknown quantities n and p may be found by proper transformations, or by successive approximations. By comparing pairs of observations in this way at various zenith distances, the values of n and p come out very nearly the same, except at very low altitudes ; so that the hypothe- sis that <7, and therefore n and p are constant, is found to be nearly, though not rigorously exact. The values that have been found are, with the barometer at 29.6 inches, and the thermometer at 50°, n ~ 57 /; .036, p = 3 ; and the formula by which Tab. XII. (Bowd.) has been com- puted is r =z 57".036 tan (z — 3 r). (46) In computing by this formula, we must find an approximate value of r, by assuming first r = 57".036 tan z, and substitute the value thus obtained in the second member of the proper equation. Example. Find the refraction for the altitude 30°. log 57".036 — 1.75615 . . . . . . . . 1.75615 2=60° l.tanO.23856 z— 3 r = 59°55'4" 1. tan 0.23712 r= 98".8 log 1.99471 r = 98 /r .46 log 1.99327 3r = 4' 56".4 r = l'38 /r .46 43. Laplace, from a more profound investigation of this problem, obtained a more complicated formula, which agrees better with observations. Bessel has modified and improved Laplace's formula. His tables of refractions are now considered the most reliable. They are found in a convenient form for nautical problems in Chauvenet's Method of Equal Altitudes, Table III. The mean refractions in this Table are for the height of the REFRACTION. 57 barometer, 30 inches, and the temperature 50° of Fah- renheit.* 44. The mean values of n and p in Art. 42 correspond to the height of the barometer, b = 29.6 inches, the thermometer, t = 50° Fahrenheit. Xow, the refraction in different conditions of the atmos- phere is nearly proportional to the density of the air ; and this density, the temperature being the constant, is propor- tional to its elasticity ; that is, to the height of the baro- meter. Hence, if b is the noted height of the barometer (in inches), r, the mean refraction of Tab. XII., A r, the correction for the barometer, r -\- A r ~b then 29.6' By this formula the correction for the barometer in Tab. XXXVI. is computed. Again, the elastic force being constant, the density in- creases by T £o part for each depression of 1° Fahrenheit. Hence, if A 1 r = the correction for the thermometer, t = the temperature in degrees of Fahrenheit, A ' r = ^m^( r + A ' r ) ( 48 ) which reduces to A 1 r = ^ a0 . ^ r oOU -f- t * Chauvenet's Astronomy, Vol. I. pp. 127-172, contains a thorough in- vestigation of the problem of refraction, especially of Bessel'? formulas. 58 NAVIGATION. by which the correction for the thermometer in Table XXXVI. is computed. Bessel's formulas are more rigid, but more complex. 45. Problem 15. To find the radius of curvature of the path of a ray in the earth? s atmosphere. Solution. By the radius of curvature for any point of a curve, is meant the radius of the circular arc, which most nearly coincides with the curve at that point. If we consider the curvature of the path of a ray to be uniform from B to A (Fig. 14), it is the same as considering the curve B A itself to be a circular arc, and the problem is reduced to finding the radius of this arc. Let C be the centre of the arc A B, B! ~ C A, the radius of curvature, H = C A, the radius of the earth. Z E S Fig. 14. Since S B and S' A are tangents to the curve at B and A respectively, they are perpendicular respectively to the radii C B, C A; hence, AC'B = r, the difference of direc- tions of S B and S' A. As A B is a very small arc, we may put AB=:i2' sinr, and, since they are very nearly equal, we may also put AD=AB=:jft'sinr. REFRACTION'. 59 In the triangle ADC, AD AC" R' sin r sin A C D sin ADC or whence R sin (z — u) ' R sin u sin (z — u)' sin r or nearly enough, since u and r are small, B! = — - 5. sin s* r" But in the preceding problem u = qr, jP = i^— 1)== 3 > whence 2—^f u = 7 r ; 7R so that J?'= sin 3' (49) (50) (51) which is the required formula, nearly. 46. When z = 0, or the star is in the zenith, JR = cc ; that is, the path is a straight line. When z = 90°, or the object is in the horizon, H r = 1 E ; that is, near the earth's surface a ray of light nearly horizon- tal moves in a curve, which is nearly the arc of a circle whose radius is seven times the radius of the earth. This, however, is in a mean condition of the atmosphere. The curve is greatly varied in extraordinary states of the atmosphere, or by passing near the earth's surface of differ- ent temperatures ; in very rare cases even to the extent of becoming convex to the surface a short distance. 60 NAVIGATION DIP OF THE HORIZON. Fig. 15. 47. Problem 16. To find the dip of the horizon. Solution. Let A (Fig. 15) be the position of the observer at the height BA = A, above the level of the sea ; A H, perpendic- ular to the vertical line, C A, re- presents the true horizon. The most distant point of the horizon visible from A is that at which the visual ray, H" A, is tangent to the earth's surface. The apparent direction of H" is A H', the tangent to the curve A H" at A. /H == H A H' is the dip of the horizon to be found. Let C be the centre of the earth, C, the centre of the arc H" A. H", C, C, are in the same straight line, since the arcs H" B, LT A are tangent to each other at H", C A, C A, are perpendicular respectively to AH, A H' ; hence C A C'= II A H'= A H, the dip. Let JR = C B, the radius of the earth ; then R + h = C A, 7 B = C A = C B.% the radius of curvature of H"A, We have, then, in the triangle C A C, by PL Trig. (268), siniJH^j/^ • **)(**) . 7 B.(R + h) » and, since h is comparatively very small and may therefore be omitted alongside of i?, DIP OF THE HORIZON. 61 or, putting sin \ A H—\ A H sin 1", AH= * i/Jl=*^/J-Vh. (52) sin 1" V 7 E sin 1" r 7E Taking B = 20923596 feet (Herschel, p. 126), we find the constant factor 2 J 8 = 59".040, sinl" ' 7i2 (53) and J,ff=59 ,/ .040 4 / A, log J j? = 1.77115 + | log h, h being expressed in feet, which is nearly the formula for Tab. XIII. (Bowd.) Since . „ y — ^ is constant, depending only upon the radius of the earth, J H is proportional to Vh y or the dip is proportional to the square root of the height of the observer above the level of the sea. 48. Were the path of the ray, H A, a straight line, we should have A f H=KAW=WCA, and in the triangle H /; C A cos A' Hz E JK+ff whence, and 2 sinVj#2?=^=|, nearly, AJECz h 2J2' sin I' or with h in feet, A' H= 63".77l Vh. Comparing this with A H= 5 9". 040 Vhl we find A H= AH- 4".731 Vhzzz A'lT-.OU A' H y (54) 62 NAVIGATION. or that the dip is decreased by refraction by .074, or nearly tV of it. But from the irregularity of the refraction of horizontal rays (Art. 46), the dip varies considerably, so that the tabu- lated dip for the height of 16 feet can be relied on ordinarily only within 2'. When the temperatures of the air and water differ greatly, variations of the dip from its mean value as great as 4' may be experienced. In some rare cases, varia- tions of 8' have been found. The dip may be directly measured by a dip-sector. A series of such measurements carefully made, and under different circumstances, both as to the height of .the eye, temperature and pressure of the atmosphere, and temperature of the water, are greatly needed. Prof. Chauvenet (Astron. L, p. 176) has deduced the fol- lowing formula, which it is desirable to test by observations : t — t n or in seconds, AH- A'H-24:02l" • in minutes, AH— A'H—Q'.6l in which A'H t is the temperature of the air, t that of the water, by a Fahrenheit thermometer. When the sea is warmer than the air, the visible horizon is found to be below its mean position, or the dip is greater than the tabulated value ; when the sea is colder than the air, the dip is less than its tabulated value. (Raper's N~av., p. 61.) This uncertainty of the dip affects to the same extent all altitudes observed with the sea horizon. 49. Near the shore, or in a harbor, the horizon may be ob- structed by the land. (Bowd., p. 155.) The shore-line may then be used for altitudes instead of the proper horizon. Tab. DIP OF THE HORIZON. 63 XVI. (Bowd.) contains the dip of such water-line, or of any object on the water, for different heights in feet and dis- tances in sea miles. It is computed by the formula q X D=:-d+ 0.56514 -j 7 a (55) in which h is the height in feet ; c?, the distance of the object in sea miles ; Z>, the dip in minutes. 50. Problem 17. To find the distance of an object of known height, which is just visible in the horizon. Solution. If the observer is at the surface of the earth at the point H /; (Fig. 15), a point A appears in the horizon, or is just visible, when the visual ray A W just touches the earth at H". Let h = B A, the height of A, d = H" A, the distance of A. As this arc is very small, we have d= H" C A sin 1' X C A = 1 R X H" C A sin l\ since by (51) C A = 1 B. From the three sides of the triangle C C A by PL Trig. (268), Fig. 15. sin iH'C'A = f /\h(R+\K) 42 R 2 or nearly and i H" C A sin 1 W; 84 £? H' C' A sin 1' = i/III. y 21 r 64 NAVIGATION. This, substituted in the expression for c?, gives *-**VS«y (■£•**); (») In this, <#, A, and J? are expressed in the same denomina- tion. But if h and B are in feet, in statute miles, d = -^— -^ j/ f — B A J, in geographical miles, d = -^ |/ f- B h). Taking B ■== 20923596 feet as before, we find in stat. miles <#= 1.323 V\ orlogc?=:0.12l72-h£log A, ) , ,v ingeog. " c7= 1.148 Vh, or log c?= 0.05 994 + i log A. ) The first of these is nearly the formula given by Bow- ditch for computing Table X. (Bowd., Preface.) 51. Were the visual ray, H" A, a straight line, we should have from the right triangle C H" A y H"A = |/(C A 2 -ir'C 2 ), or d f = V {2 B + h) h; or nearly d' = V 2 B X VL Introducing the same numerical values as before, we have in statute miles d' = 1.225 Vh. Comparing this with the expression above, we see that the distance is increased about y 1 ^ part by refraction. This, however, is subject to great uncertainty. 52. If the observer is also elevated at the height of B' A' (Fig. 16), and sees the object A in his horizon, then its dis- tance is A' ir -fir a, PARALLAX. 65 or the sum of the distances of each from the common horizon, H". By entering Table X. with the heights of the observer and the ob- ject respectively, the sum of the cor- responding distances is the distance of the object from the observer. The distances in this table are in statute 5280 miles. Multiplying them by to geographical miles. =s .86751, reduces them P AE ALL AX . 53. The change of the direction of an object, arising from a change of the point from which it is viewed, is called parallax ; and it is always expressed by the angle at the object, which is subtended by the line joining the two points of view. (Hersch. Ast., Art. 70.) Thus in Fig. 17, the object S would be seen from A in the direction A S ; and from C in the direction C S. The angle at S, subtended by A C, is the difference of these directions, or the parallax for the two points of view, C and A. 54. In astronomical observations, the ob- server is on the surface of the earth ; the con- ventional point to which it is most convenient to reduce them, wherever they may be made, is the earth's centre. (Hersch. Ast., Art. 80.) In those problems of practical astronomy which are used by the navigator, we have only to con- 66 NAVIGATION. sider this geocentric parallax, which is the difference of the direction of a body seen from the surface and from the cen- tre of the earth. It may also be defined to be the angle at the body subtended by that radius of the earth, which passes through the place of the observer. Thus, in Fig. 17, if C is the centre of the earth, and A the place of the observer, the geocentric parallax of a body, S, will be the angle S = ZAS-ZCS, at the body subtended by the radius C A. If the earth is regarded as a sphere, C A Z will be the vertical line through A, and will pass through the zenith Z. Then will the plane of C A S be a vertical plane ; Z A S, the apparent zenith distance of S as observed at A ; Z C S, its geocentric or true zenith distance ; and Z A S > Z C S. Thus we see that this parallax takes place in a vertical plane, and increases the zenith distance, or decreases the altitude, of a heavenly body without affecting its azimuth. 55. This suffices for all nautical problems except the com- plete reduction of lunar distances. For these and the more refined ob- servations at observatories, the spher- oidal form of the earth must be con- sidered. Then, as in Fig. 18, the radius C A does not coincide with the normal or vertical line C A Z, but meets the celestial sphere at a point Z', in the celestial meridian, nearer the equator than the zenith, Z. We may remark here that A C" E, the angle which the vertical line makes with the equator, is the latitude of A ; and Fig 18. PARALLAX. 67 ACE, the angle which the radius makes with the equator, is its geocentric latitude. 56. Problem 18. To find the parallax of a heavenly body for a given altitude. Solution, In Fig. 17, let p = S, the parallax in altitude ; z — Z A S, the apparent zenith distance of S, corrected for refraction ; H == C A, the radius of the earth ; d h= C S, the distance of the body, S, from the centre of the earth. Then from the triangle C A S, we have sin C S A = ^-h sin CAS, ii? sin z .. or sin p = — -= — , (58) If the object is in the horizon as at H, the angle A H C is called its horizontal parallax ; and denoting it by 7r 3 we have from (58), or from the right triangle C A H, sin 7T = ~ (59) which substituted in (58) gives sin^> = sin n sin z. (CO) If h = 90°— z, the apparent altitude of the object, we have sin^? = sin n cos h; (61) or, nearly, since p and it are small angles, p = n cos h. (62) 57. The horizontal parallax, tt, is given in the Nautical Almanac for the sun, moon, and planets. From Fig. 17 it is obviously the semidiameter of the earth, as viewed from the body. As the equatorial semidiameter is larger than any other, so also will be the equatorial horizontal parallax. 68 NAVIGATION. This i3 what is given in the Almanac for the moon. Strictly it requires reduction for the latitude of the observer, and such reduction is made at observatories, and in the higher order of astronomical observations. 58. Tables X. A and XIV. (Bowd.) are computed by for- mula (62). Table XXIX. contains the correction of the moon's alti- tude for parallax and refraction corresponding to a mean value of the horizontal parallax, 57' 30". It should be used, however, only for very rough observations or a coarse ap- proximation. Tab. XIX. contains the difference of 59' 42" and the com- bined correction of the moon's altitude for parallax and re- fraction. The numbers taken from this table subtracted from 59' 42", give the correction of an apparent altitude for parallax and refraction. To this may be applied the reduc- tion of the refraction to the actual condition of the atmos- phere (Art. 42). If, instead of the equatorial hor. parallax, we enter the table w T ith the augmented parallax of Chau- venet's Lunar Method (Tab. III.), we shall obtain the re- duction, not to the centre of the earth, C (Fig. 18), but to the point, C, where the normal line through A intersects the axis of the earth. Table XIX. of Bowditch was arranged especially for one of the Lunar methods in that work, so that the reductions of the distance should all be additive. APPARENT SEMIDIA METERS. 5 \ The apparent diameter of a body is the angle which its disk subtends at the place of the observer. Problem 19. To find the apparent semidiameter of a heavenly body. Solution. In Fig. 19, let M be the body; d = C M, its distance from the centre of the earth ; APPARENT SEMIDIAMETERS. 69 d r = AM, its distance from A; r = MB, its linear radius or semidiameter ; 5 = MCB, its apparent semi- diameter, as viewed from C ; J = M A B',its apparent semi- diameter, as viewed from A (B and B' are too near each other to be distinguished in the diagram) ; iJ-CA, the earth's radius. 1st. For finding s, the right triangle C B M, gives sin s = y r (63) Were the body M in the horizon of A, or Z A M = 90°, its distance from A and C would be sensibly the same, so that the angle s is called the horizontal semidiameter. In (59) we have for the horizontal parallax, R „ R sin n = ? , which, substituted in (63), gives Fig. 19. or d : sin s = r- sin 7r, Jtc or nearly, since s and n are small, s== n n - (64) (65) ^ is constant for any particular body, as it is simply the ratio of its linear diameter to that of the earth. (Hersch. Ast., p. 544.) For the moon (Hersch. Ast., p. 214), ■^ = 0.2729, n.\ ( 66 ) 70 NAVIGATION. s = 0.2729 7T, and log s = 9.43600 + log By this formula the moon's horizontal semidiameter may be found from her horizontal parallax. The Nautical Almanac contains the semidiameters as well as the horizontal parallaxes of the sun, moon, and planets. 2d. For finding s', the apparent semidiameter as viewed by an observer at A on the surface of the earth, the right triangle A B' M gives sins=^. (67 In the triangle CM A, sinMAC CM sin M C A ~~ A M> or, putting and h = 90°— Z A M, the apparent, h! = 90°— Z C M, the true altitude of M, cos h d , cosA'-rf" < 68 whence, ,, ,cos^' cosh which, substituted in (67), and by (63), gives . r cos h . cos h Bin s = -= 77 = sin s — T7, a cos A cosA ' or approximately, s' = s — =7, (69) by which s' may be found when s and A are known. Since h < A', cos h > cos A', and consequently s r > s ; that is, the semidiameter increases with the altitude of the body. The excess A s = «'— s, is called the augmentation. The moon is the only body for which this augmentation is sensible. APPARENT SEMIDIAMETERS. 71 60. Problem 20. To find the augmentation of the moorf s horizontal semidiameter. /Solution. From (69) we find A , cos h — cos h! A S = S 5 — 5 r . , cos A' which, by PI. Trig. (108), becomes, As 2 sin j (N + h) sin \ Qi' — h) cos h' h! — h =p, the parallax ; since it is small, we may put 2 sin \ (h' — h) = 2 sin \p =p sin 1" == 77 cos A sin 1* ; and, in computing so small a quantity as A s, we may take h for \ (A' + A), and cos A for cos h'; and then A s = s 7T sin 1" sin A, or, since (65) For the moon S =E^ A s — -^ 7T 2 sin 1" sin A. r E 0.2729; then A s = .000001323 tt 2 sin A. (70) If we take 7r = 57' 20", which is nearly its mean value, we have A s = 15".65 sin A, (71) which agrees nearly with the formula for Tab. XV. (Bowd.) The augmentation may differ 2" from this mean value. Tab. II. of Chauvenet's Lunar Method contains this aug- mentation for different values of s, as well as of A, computed by a more precise formula. CHAPTER III. TIME. 61. Transit. The instant when any point of the celestial sphere is on a given meridian is designated as the transit of the point over that meridian. 62. Hour-angle. The hour-angle of any point of the sphere is the angle at the pole, which the circle of declination pass- ing through the point makes with the meridian. It is pro- perly reckoned from the upper branch of the meridian, and positively toward the west. It is usually expressed in hours, minutes, and seconds of time. The intercepted arc of the equator is the measure of this angle. 63. Sidereal Time. The intervals between the successive transits of any fixed point of the sphere (as, for instance, of a star which has no proper motion) over the same meridian would be perfectly equal, were it not for the variable effect of nutation, (Hersch. Ast., Art. 327.) This correction, arising from a change in the position of the earth's axis, is most perceptible in its effect upon the transit of stars near the vanishing point of that axis, i. e. near the poles of the heavens. Hence, for the exact measurement of time, we use the transits of some point of the equator, as the vernal equinox. This point is often called the first point of Aries. Its usual symbol is c p. 64. The interval between two successive transits of the vernal equinox is a sidereal day ; and such a day is regarded as commencing at the instant of the transit of that point. TIME. 73 The sidereal time is then O h O m s . This instant is sometimes called sidereal noon. The effect of nutation and precession in changing the time of the transit of the vernal equinox is so nearly the same at 'two successive transits, that the sidereal days thus defined [are sensibly equal. It is unnecessary, then, except in refined discussions, to discriminate between mean and apparent sidereal time. 65. The sidereal time at any instant is the hour-angle of the vernal equinox at that instant, and is reckoned on the equator from the meridian westward around the entire cir- cle, that is, from to 24 h . It is equal to the right ascension of the meridian at the same instant. 66. Solar Time. The interval between two successive transits of the sun over a given meridian is a solar day, and the hour-angle of the sun at any instant is the solar time of that instant. In consequence of the motion of the earth about the sun from west to east, the sun appears to have a like motion among the stars at such a rate that it increases its right as- cension daily nearly 1°, or 4 m of time. With reference to the fixed stars, it therefore arrives at the meridian each day about 4 m later than on the previous day ; consequently, solar days are about 4 m longer than sidereal days. 67. Apparent and Mean Solar Time. If the sun changed its right ascension uniformly each day, solar days would be exactly equal. But the sun's motion in right ascension is not uniform, varying from 3 m 35 s to 4 m 26 s in a solar day. There are two reasons for this, — 1st. The sun does not move in the equator, but in the ecliptic. 2d. Its motion in the ecliptic is not uniform, being most rapid at the time of the earth's perihelion, about January 1, and slowest at the time of the aphelion, about July 2. 74 NAVIGATION. To obtain a uniform measure of time depending on the sun's motion, the following method is adopted. A fictitious sun, called a mean sun, is supposed to move uniformly in the ecliptic at such a rate as to return to the perigee and apogee at the same time with the true sun. A second mean sun is also supposed to move uniformly in the equator at the same rate that the first moves in the ecliptic, and to return to each equinox at the same time with the first mean sun. The time which is measured by the motion of this second mean sun is uniform in its increase, and is called mean time. That which is denoted by the true sun is called true or apparent time. The difference between mean and apparent time is called the equation of time. It is also the difference of the right ascensions of the true and mean suns. The instant of transit of the true sun over a given meridian is called apparent noon. The instant of transit of the second mean sun is called mean noon. The mean time is then h m s . Mean noon occurs, then, sometimes before and sometimes after apparent noon, the greatest difference being about 16 m , early in November. 68. Astronomical Time. The solar day (apparent or mean) is regarded by astronomers as commencing at noon (apparent or mean), and is divided into 24 hours, numbered successively from to 24. Astronomical time (apparent or mean) is, then, the hour- angle of the sun (true or mean) reckoned on the equator westward throughout the entire circle from h to 24 h . 69. Civil Time. For the common purposes of life, it is more convenient to begin the day at midnight, that is, when the sun is on the meridian below the horizon, or at the sun's lower transit. The civil day begins 12 h before the astro- nomical day of the same date; and is divided into two TIME. 75 periods of 12 h each, namely, from midnight to noon, marked A.M. (ante-meridian), and from noon to midnight, marked P.M. (post-meridian). Both apparent and mean time are used. The affixes A.M. and P.M. distinguish civil time from as- tronomical time. During the P.M. period, this is the only distinction, — the day, hours, &c. being the same in both. 70. Sea-Time. Formerly, in sea-usage, the day was sup- posed to commence at noon, 12 h before the civil day, and 24 h before the astronomical day of the same date ; and was di- vided into two periods, the same as the civil day. Sea-time is now rarely used. 71. To convert civil into astronomical time, it is only necessary to drop the A.M. or P.M., and when the civil time is A.M., deduct l d from the day and increase the hours by 12b. To convert astronomical into civil time, if the hours are less than 12 h , simply affix P.M. ; if the hours are 12 h or more than 12 h , deduct 12 h , add l d , and affix A.M. Examples. Ast. Time. Civil Time. dhme dhms 1860 May 10 14 15 10 = 1860 May 11 2 15 10 A.M. 1862 Sept. 8 9 19 20 = 1862 Sept. 8 9 19 20 P.M. 1863 Jan. 3 23 22 16 = 1863 Jan. 4 11 22 16 A.M. 1863 Jan. 4 3 30 = 1863 Jan. 4 3 30 P.M. 72. The hour-angle of the sun (true or mean), at any me- ridian, is called the local (apparent or mean) solar time. The hour-angle of the sun (true or mean) at Greenwich at the same instant is the corresponding Greemcich time. So also the hour-angle of °p at any meridian, and its hour- angle at Greenwich at the same instant, are corresponding local and Greenwich sidereal times. 76 NAVIGATION. 73. Tlxe difference of the local times of any two meridians is equal to the difference of longitude of those meridians. Demonstration. In Fig. 20, let P M, P M' be the celestial merid- ians of two places ; P S, the declination circle through 9^ the sun (true or mean) ; M P S, the hour-angle of the sun at all places whose meridian is P M, will be the local time (apparent or mean) at those places ; so also M' P S will be the corresponding local time at all places whose meridian is P M 7 ; and M P M'= MPS- W P S will be the difference of longitude of the two meridians. If P °p is the equinoctial colure, M P °p and M' P °p will be the corresponding sidereal times at the two meridians ; still, however, M P M'= MPf-MTT. The proposition is true, then, whether the times compared are apparent, mean, or sidereal. The difference of longitude is here expressed in time. It is readily reduced to arc by observing that 24*= 360° 1 rio_ 4m ; . \ or <^ 1' = 4 s IB =: 15' J l "" 15 Iu comparing corresponding times of different meridians, the most easterly meridian is that at which the time is greatest. 74. If (Fig. 20) PM is the meridian of Greenwich, M P S is the Greenwich solar time, and M P M' the longitude of the meridian P M'. TIME. 77 MPM'=MPS-M'PS; so also MPM'=MPf-M'Pfj or, the longitude of any meridian is equal to the difference between the local time of that meridian and the correspond- ihg Greenicich time. 75. If we put r o =MPS, the Greenwich time, T = M' P S, the corresponding local time, A = MP M', the longitude of the meridian, P M', we have X = T — T, ) ,» . and T =T+^ J V ; in which A is + for west longitudes, and T and T are sup- posed to be reckoned always westward from their respective meridians from h to 24 h ; that is, T and T are the astro- nomical times, which should always be used in all astro- nomical computations. 76. Usually the first operation in most computations of nautical astronomy is to convert the local civil time into the corresponding astronomical time (Art. 71). The Greenwich time should never be otherwise expressed than astronomically. On this account it would be conven- ient to have chronometers intended for nautical or astronomi- cal purposes marked from h to 24 h , instead of h to 12 h as is now customary with sea-chronometers. * 77. The second operation often required is to convert the local astronomical time into Greenwich time. For this we have (72), which numerically is ( + when the longitude is west) • ~~ \ — when it is east, and, in words, gives the following Rule. Having expressed the local time astronomically, add. the longitude, if west ; subtract it, if east: the result is the corresponding Greenwich time. 78 NAVIGATION. TIME. Examples. 1. In Long. 76° 32' W., the local time being 1861, April Id 9 h 3 m 10 s A.M., what is the Greenwich time ? Local Ast. T. == March 31 d 21 h 3 m 10 3 Longitude — +568 G. T. = April "1 2 9 18 2. In Long. 30° E., the loc. time being March 20 d 6 h 3 m A.M., what is the G. T. ? Loc. Ast. T. = March 19 d 18 h 3 m Long. = — 2 G. T. s= March 19 16 3 3. In Long. 105° 15' E., the loc. time being Aug. 21 d 4 h 3 m P.M., what is the G. T. ? Loc. Ast. T. = Aug. 21 d 4 h 3 m Long. = — 7 1 G. T. == Aug. 20 21 2 4. Long. 175° 30' W., Loc. T. Sept. 30 d 8 h 10 m A.M. G. T. Sept. 30 d 7 h 52 m . 5. Long. 165° 0' E., Loc. T. Feb. l d 7 h ll m P.M. G. T. Jan. 31 d 20 h ll m . 6. Long. 72° 30' W., Loc. T. April 10 d 7 h 10 ra A.M. G. T. 10 d h m . 7. Long. 100° 30' E., Loc. T. June l d b m A.M. (or mid- night.) G. T. May 31 d 5 h 18 m . 8.- Long. 75° W., Loc. T. June 3 d h m M. (noon.) G. T. June 3 d 5 h m . By reversing this process, that is by subtracting the longi- tude if west, or adding it if east, we may reduce the Green- wich time to the corresponding local time. 78. When observations are noted by a chronometer regu- lated to Greenwich time, an approximate knowledge of the TIME. 79 longitude and local time is necessary in order to determine whether the chronometer time is A.M. or P.M., and thus fix the true Greenwich date. If the time is A.M., the hours must be increased by 12\ Examples. 1. In Long. 5 h W., about 3 h P.M., on Aug. 3 d , the Green- wich chronometer shows 8 h ll ra 7% and is fast of G. T. 6 ra 10 9 . What is the G. T. ? Approx. Loc. T. Aug. 3 d 3 h G. Chro. SMl m T Long. + 5 Correction — 6 10 Approx. G. T. Aug. 3 d 8 h G. T. Aug. 3 d 8 b 4 m 57 s 2. In Long. 10 h E., about l h A.M., on Dec. 7 d , the G. Chro. shows 3 b 14 m 13 8 .5, and is fast 25 m 18 9 .7, find the G. T. Approx. Loc. T. Dec. 6 d 13 h G. Chro. 3 h 14 m 13 3 .5 Long. — 10 Correction — 25 18 9 .7 Approx. G. T. Dec. 6 d 3 h G. T. Dec. 6 d 2 h 48 m 54 s .8 3. In Long. 9 h 12 m W., about 2 h A.M., on Feb. 13 d , the G. Chro. shows ll h 27 m 13 9 .3, and is fast 30 ra 30 9 .3, find the G. T. Approx. Loc. T.Feb. 12 d 14 h m G. Chro. ll h 27 ra 13 9 .3 Long. + 9 12 Correction — 30 30 9 .3 Approx. G. T. Feb. 12 d 23 h 12 m G. T. Feb. 12 d 22 h 56 m 43\0 The operations on the approximate times may be per- formed mentally. CHAPTER IV. THE NAUTICAL ALMANAC. 79. The American Ephemeris and Nautical Almanac " is divided into two distinct parts. One part is designed for the special use of navigators, and is adapted to the meridian of Greenwich. The other is suited to the convenience of astronomers, on this continent particularly, and is adapted to the meridian of Washington." 80. The Nautical part of this Ephemeris and the British Nautical Almanac give at regular intervals of Greenwich time the apparent right ascensions and declinations of the sun, moon, planets, and principal fixed stars, the equation of time, the horizontal parallaxes and semidiameters of the sun, moon, and planets, and other quantities, some of which little concern the navigator, but are needed by astronomers. 81. Before we can find the value of any of these quanti- ties for a given local time, we must first find the correspond- ing Greenwich time (Art. 77). When this time is exactly one of the instants for which the required quantity is put down in the Almanac, it is only necessary to transcribe the quantity as it is there given. When, as is mostly the case, the time falls between two Almanac dates, the required quantity is to be obtained by interpolation. And generally, except when great precision is desired, it is sufficient to use first differences only ; that is, regard the changes of the quantity as proportional to the small intervals of time, which are employed. THE NAUTICAL ALMANAC. 81 Thus, for a day the change of the sun's right ascension may be regarded as uniform, so that for l h it is ^\ of the daily change ; for 2 h , ^ T ; and in general for any part of a day it will be the same part of the daily change. Generally, then, if A represent the quantity in the Almanac, for a date pre- ceding the given Greenwich time ; J x , its change in the time T ; £, the time after the Almanac date for which the value of the quantity is required, expressed in the same unit as T ; and A, the required value ; we have, A = A,+ ^A lm (73) When A is increasing, A l has the same sign as A ; but when A is decreasing, A x has the opposite sign. 82. If the given time is nearer the subsequent than the preceding Almanac date, it may be convenient to interpolate backward. If, then, A x represent the quantity in the Al- manac for a subsequent Greenwich date, and t' the time be- fore the Almanac date, we have A^A-^A,. (74) 83. The Almanac contains the rate of change, or difference of each of the principal quantities for some unit of time. Thus, in the Ephemeris of the sun and planets the "Diff. for l h ," in part of that of the moon, the " Diff. for l m ," are given. If t or t r is expressed in the same unit of time as that for which the " Diff.," J 19 is given, formulas (73) and (74) become Thus, for using hourly differences, we w r ish the hours, 82 NAVIGATION. minutes, U M. T. of mer. pass. June 6 9 59.6 +1^ June 5 12 41.8 —4.45 in l d T1L82 (T.Tlin6 h — 13.2 J 1.18 + 1.1 ( 1.11 in ] .16 (1 .08 June 6 9 46.4 [ 1 June 5 12 43.0 or June 6 43.0 A.M. 94 NAVIGATION. In the case of the moon the hourly differences have been interpolated for half the longitude. 96. Problem 27. To find the right ascension or declines tion of the moon, or a planet, at the time of its transit over a given meridian on a given day. Solution. Find the local mean time of transit, as in Pro- blem 25 ; deduce the corresponding Greenwich time by ap- plying the longitude ; and for this Greenwich time take out the right ascension or declination, as in Problem 21. If the time of transit has been noted by a clock or chrono- meter, regulated to either local or Greenwich time, it should be used in preference to the time of transit computed from the Almanac. 97. Problem 28. To find the Greenwich mean time of a given lunar distance. Solution. The angular distances of the moon from the sun, the principal planets, and several selected stars, are given in the Almanac for each 3 h of Greenwich mean time. If d represent the given distance ; c? , the nearest distance of the same body in the Almanac preceding in time the given distance ; J 15 the change of distance in 3 h ; t, the required time (in hours) from the date of d ; by (75) we have approximately, using 1st differences only, whence, for the inverse interpolation, * = !(24 l1 , M may be taken out for l d later than stated in the previous precept, and interpolated for the excess of (A + /S) over 24 h ; and when (A + £) is negative, to retain its negative character, or else take out M Q for one day earlier. 3d Solution. From (89) we have t = £-[£ + .0027379 (£ + A)], (94) so that, when the Greenwich mean time (£+A) is sufficiently known, we may find for it the right ascension of the mean sun, (Art. 107) $, + .0027379 (£ + A), and subtract it from the given sidereal time : or, the mean time is equal to the sidereal time — the right ascension of the mean sun. So also we have from Art. 107 the precept:— the apparent time is equal to the sidereal time— the right ascension of the true sun. 106 NAVIGATION. Examples. 1. 1865, Jan. 30 (ast. day), in long 10 h m 52 9 .7 TV\, the sidereal time is 6 h 57 m 42. 8 4 ; find the mean time. h m s h m s L. sid. t. 6 57 42.4 L. sid. t. 6 57 42.4 — # (Jan. 30) —20 38 56.00 M (Jan. 30) 3 20 31.06 — Red. for long. — 1 38.71 Red. for long. — 1 38.44 Sid. int. 10 17 7.69 Red. of sid. t. — 1 8.43 Red. of sid. int. — 1 41.10 L. m. t. Jan. 30 10 15 26 .6 L. m. t. Jan. 30 10 15 26.6 2. 1865, Jan. 30, (ast. day,) in long. 10 h m 52 s .7 E. ? the sidereal time is 6 h 54 m 25\0; what is the mean time? h m • s h m s L. sid. t. 6 54 25^) L. sid. t. 6 54 25.0 — # (Jan. 30) —20 38 56.00 M (Jan 30) 3 20 31.06 — Red. for long. + 1 38.71 Red. for long. + 1 38.44 Sid. int. 10 17 7.71 Red. of sid. t. — 1 7.89 Red. of sid. int. — 1 41.10 L. m. t. Jan. 30 10 15 26.6 L. m. t. Jan. 30 10 15 26.6 3. 1865, Sept. 26, 9 b A.M., in long. 4 h m 52 9 W., the sidereal time is 9 h 37 m 40 s . 1 ; find the mean time. h m s h m s L. sid. t. 9 37 40.1 L. sid. t. 9 37 40.1 — S (Sept. 25) —12 17 15.89 M (Sept. 25) 11 40 48.98 — Red. for long. — 39.57 Red. for long. — 39.46 Sid. int. 21 19 44.64 Red. of sid. t. —1 34.64 Red. of sid. int. — 3 29.65 L. m. t. Sept. 25 21 16 15.0 L. m. t. Sept. 25 21 16 15.0 4. 1865, Sept. 25, 3 h P.M., in long. 8 h 16 m 25 s .3 E.,the side- real time is 15 h 32 m 41\6 ; find the mean time. RELATION OF HOUR-ANGLES AND TIME. 107 L. sid. t. — S (Sept. 25) - — Reel, for long. Sid. int. Red. of sid. int. 15 32 41.6 12 17 15.89 + 1 21.55 3 16 47.26 — 32.24 L. m. t. Sept. 25 3 16 15.0 L. sid. t. M (Sept. 24) Red. for long. Red. of sid. t. 15 32 41.6 11 44 44.89 + 1 21.33 — 2 32.80 L. m. t. Sept. 25 3 16 15.0 RELATION OF HOUR-ANGLES AND TIME. 109. Problem 33. To find the mean time of meridian transit of a celestial body, the longitude of the place or the Greenvrich time being known. Solution. In the case of the sun the instant of meridian transit is apparent noon of the place ; for which we have (84) T m — E, the equation of time, which can he taken from page I. of the Almanac, and inter- polated for the longitude, which in this case is also the Greenwich apparent time ; or from page II., and interpo- lated for the Greenwich mean time. When E is subtractive, the subtraction from the number of days can be performed. The apparent right ascension of any body at the instant of its meridian transit is also the right ascension of the me- ridian, or sidereal time. (Art. 65.) It suffices therefore to find the right ascension of the body, and, regarding it as the sidereal time, reduce it to mean time by Problem 31. The American Ephemeris contains the apparent right as- censions of two hundred principal stars for the upper cul- minations at Washington ; the British Almanac contains the positions of one hundred for the upper culminations at Greenwich. They are reduced to any other meridian, when necessary, by interpolating for the longitude. The right ascensions of the moon are given for each hour, and of the planets for each noon, of Greenwich mean time, 108 NAVIGATION. and may be found for a given Greenwich mean time by Problem 21. If, however, the longitude of the place is given, the local mean time of transit of the moon, or a planet, may first be found from the Almanac to the nearest minute or tenth (Probs. 25, 26) ; then for this mean time the right as- censions of the moon, or of the planet (Prob. 21), and of the mean sun (Prob. 24), may be computed. Subtracting the right ascension of the mean sun from the right ascen- sion of the moon, or planet, will give the mean time of transit (Prob. 32, 3d solution). If it differ sensibly from that previously obtained, the process may be repeated with this new approximation. If the time of transit has been noted by a clock, or chro- nometer, regulated either to local or Greenwich time, it should be used in preference to the approximate time of transit found from the Almanac in computing the right ascensions. The American Ephemeris contains also the right ascen- sions of the moon and principal planets at their transits of the upper meridian at Washington. They can be reduced to any other meridian by interpolating for the longitude from Washington. This solution will give the time of the upper culmination of a heavenly body. To find the time of a lower culmi- nation, 12 b may be added to the right ascension of the body, if sufficiently Avell known ; or, as is generally preferable, 12 h may be added to the longitude of the place. The in- stant of a lower culmination on any meridian will be that of an upper culmination on the opposite meridian. Examples. 1. Find the times of meridian passage of the moon and Jupiter for 1865, June 6 (civil day), in long 100° 15' W. (Example 1, Art. 95, p. 93.) RELATION OF HOUR-ANGLES AND TIME. 109 Approx. m. t. Long. G. m. t. 2) h m June 6 10 12.9 + 6 41.0 June 6 16 53.9 June June U h m 5 12 40.6 + 6 41.0 5 19 21.6 R. Asn. Red. for G. m. t. R. As'n at transit h m s 15 13 28.54-j- + 1 53.66 | 15 15 22.20 s 2.1088 105.440 6.326 1.898 h m s 17 39 52.S0- — 25.64 h 17 39 27.16 s - 1.340 "13.40 11.760 .402 80 So Red. for G. m. t. 4 59 3S.32 + 2 46.56 5 2 24.88 4 55 41.76 + 3 10.82 4 58 52.58 M. t. of transit, June 6 10 12 57.32 t. + 8.32 June 5 12 40 34.58 Diflf. from approx. — 1.42 - „, O0 ( Ch. of R. A. + .117 In o,o2- _ CLof ^ _ <0()9 M. t. of transit, June 6 10 12 57.43 110. Problem 34. To find the hour-angle of the sun for a given place and time. Solution. The hour-angle of the sun, reckoned from the upper meridian toward the west, is the apparent time reckoned astronomically (Art. 72). Its hour-angle east of the meridian is negative, and numerically equal to 24 h — the apparent time. A given mean or sidereal time must then be converted into apparent time ; for this, the longitude, or the Green- wich time, must be known approximately. 111. Problem 35. To find the hour-angle of the moon, a planet, or a fixed star, for a given p>lace and time. Solution. In Fig. 21, as described in Art. 104, T M is the right ascension of the meridian, and measures M P Y, the sidereal time. Let 110 NAVIGATION. PS be the declination-circle of the mean sun, then °f S is the right ascension of the mean sun, and cfr M P S is the mean time, and is measured by the arc of the equator, S M. Let P M' be the declination-circle of some other celestial body ; then Y M' is its right ascension, and M P M' is its hour-angle, and is measured by the arc M' M. From the figure, M' M =TM -Y M'= Y S + SM-Y M'. (95) If Y S is the right ascension of the true sun, S M will measure the apparent time. From (95), then, we have the following rule : — To a given apparent time add the right ascension of the true sun ; or to a given mean time add the right ascension of the mean sun, to find the corresponding sidereal time. Then from the sidereal time subtract the body's right ascen- sion ; the difference is the hour-angle west from the meri- dian. If it is more than 12 h , it maybe subtracted from 24 h : the hour-angle, then, is — , or east of the meridian. It is necessary to know the longitude, or the Greenwich time, sufficiently near to find the right ascensions of the sun and body. 112. Problem 3G. To find the local time, given the hour- angle of the sun and the Greenwich time. Solution. The hour-angle reckoned westward is itself the local apparent time, which may be reduced to mean or side- real time (Probs. 29, 30), as may be required. The Green- RELATION OF HOUR- ANGLES AND TIME. Ill wich time, or the longitude of the place, is needed only for this reduction. 113. Pkoblem 37. To find the local time, given the hour- angle of some celestial body and the Greenwich time. Solution. Find from the Almanac for the Greenwich time (Prob. 21) the right ascension of the body. Then, from (95), we have ¥M=YM'+M'M 3 from which, and Arts. 105, 107, we have the following rule, regarding hour-angles to the east as negative : — To the right ascension of the body add its hour-angle, the result is the sidereal time. From this subtracting the right ascension of the true sun gives the apparent time ; or the right ascension of the mean sun gives the mean time. The Greenwich time is needed for finding the required right ascensions. If the longitude of the place is given, but not the Green- wich time, we may first use an estimated Greenwich time, and then revise the computations with a corrected value, until the assumed and computed values sufficiently agree. Examples. 1. 1865, Jan. 16, 12 h 15 m 17 S .6, mean time in long. 150° 13' 10" W., find the hour-angle of the moon. h m s h m s L. m. t. Jan. 16 12 15 17.6 L. m. t. Jan. 16 12 15 17.6 Long. + 10 52.7 S 19 43 44.22 G. m. t. Jan. 16 22 16 10.3 Red, for long. + 1 38.71 ^sR.A.tJan. 16 22 lj ) 1148 31.61 + 1 9 .8584 Red. of L. m. t. + 2 0.79 fl8 .584 Red. for G. m. t. + 30.05 J 11 .150 L. sid. t. 8 2 41.32 1 .186 t .133 D's R, A. at date 11 49 1.66 D's hour-angle — 3 46 20.34 112 NAVIGATION. 2. 1865, Jan. 16 22 h 16 m 10\3, G. mean time, the moon's hour- angle is — 3 b 46 m 20 9 .3 ; find the local mean time. h m s D's hour-angle — 3 46 20.3 j)'s R. A. (Jan. 16 22 h ) +11 48 31.61 + K8584 f 18 .584~ Red. for G. m. t. + 30.05 J 11 .150 ] .186 L. sid. t. 8 2 41.36 I .133 —S (Jan. 16) -19 43 44.22 —Red. for G. m. t. — - 3 39.50 L. m. t. Jan. 16 12 15 17.6 Subtracting this from the G. m.. t. gives for the longitude 10 h m 52 b #7 ^y 3. 1865, Jan. 16, 12 h (nearly) in long. 150° 13' 10" W., the moon's hour-angle is — -3 h 46 m 20 s .3 ; find the local mean, time. h m s h m m Long. 10 52.7 D's mer. pass. Jan. 16 15 50.7 +1.74 2>'s h. ang.— 3 h 46 m .3 Red. for long. +17.4 . gh - g ( ch. of R. A. - 7 .0 Jan. 16 16 8.1 ' ( -ch. of JS Q + .6 - 3 52.7 1st approx. L. m. t. Jan. 16 12 15.4 Long. +10 0.8 1st approx. G. m. t. Jan. 16 22 16.2 D's h. ang. h m a — 3 46 20.3 D's R. A. (Jan. 16 22 h ) + ll 48 31.61 + l s (18 .8584 .584 Red. for G. m. t. + 30.11- i 11 .150 ch. in — 1 8 .6 .372 —.046 L. sid. t. 8 2 41.42 — S (Jan. 16) — 19 43 44.22 —Red. for G. m. t. — 3 39.50 — ch. in — 1 8 .6 + .004 2d L. m. t. Jan. 16 12 15 17.70 cor. for —I s . 6 —.04 Long. 10 52.7 2d G. m. t. Jan. 22 22 16 10.4 Diff. from 1st G. m. t. —1.6 3dL. m. t. Jan. 16 12 15 17.7 CHAPTER VI. NAUTICAL ASTRONOMY. ALTITUDES. AZIMUTHS. HOUR-ANGLES AND TIME. 115. Nautical Astronomy comprises those problems of Spherical Astronomy which are used in determining geo- graphical positions, or in finding the corrections of the in- struments employed. In general, they admit of a much more refined application on shore, where more delicate and stable instruments can be used, than is possible at sea, where the instability of the waves and the uncertainty of the sea-horizon present practical obstacles, both to precision in observations and to the accuracy of the results, which can- not be obviated. 116. In the problems which are here discussed the follow- ing notation will be employed : — L z= the latitude of the place of observation ; h = the true altitude of a celestial body ; z = 90°— A, its zenith distance ; d = its declination ; ^ = its polar distance ; t =: its hour-angle ; Z = its azimuth. Let the diagram (Fig. 22) represent the projection of the celestial sphere on the plane of the horizon of a place : — Z, the zenith of the place ; N Z S, its meridian ; 114 NAVIGATION. P, the elevated pole, or that whose name is the same as that of the latitude ; M, the position of a celestial body ; Z M H, a vertical circle ; and P M, a declination-circle, through M. Then, in the spherical triangle Fig. 22. P M Z, PZ = 90°— X, the co-latitude of the place ; P M = p — 90°— d, the polar distance of M ; ZM= 90°— A, the complement of its altitude, or its zenith distance ; Z P M = t, its hour-angle ; PZM = Z, its azimuth. The angle P M Z is rarely used, but is sometimes called the position angle of the body. This triangle, from its involving so many of the quantities which enter into astronomical problems, is called the astro- nomical triangle. As three of its parts are sufficient to de- termine the rest, if three of the five quantities X, d, A, £, and Z are known, the other two may be found by the usual formulas of spherical trigonometry. These admit, however, of modifications which better adapt them for practical use. The following articles point out how X, c?, A, and t may be obtained. 117. The latitudes and longitudes of places on shore are given upon, charts, but more accurately in tables of geogra- phical positions, such as are found in books of sailing-direc- tions, and in Tab. LIV. (Bowd.) At sea it is sometimes necessary to assume them from the dead reckoning brought forward from preceding, or carried back from subsequent, determinations. (Bowd., p. 264.) NAUTICAL ASTRONOMY. 115 118. The altitude of an object may be directly measured at sea above the sea-horizon with a quadrant or sextant ; on shore, with a sextant and artificial horizon, or with an alti- tude circle. All measurements with instruments require cor- rection for the errors of the instrument. Observed altitudes require reduction for refraction and parallax ; for semidia- meter, when a limb of the object is observed ; and at sea, for the dip of the horizon. The reductions for dip and re- fraction are subtract iv e ; for parallax, additive. Strictly, the reductions should be made in the following order : for instrumental errors, dip, refraction, parallax, semidiameter. In ordinary nautical practice it is unnecessary to observe this order. Following it we should have, — 1st. The reading of the instrument with which an altitude is measured ; 2d. The corrected reading or observed altitude of a limb ; 3d. The apparent altitude of the limb ; 4th. When corrected for refraction and parallax, the true altitude of the limb ; 5th. The true altitude of the centre. Except with the sea-horizon, the observed and apparent altitudes are the same. For the fixed stars, and for the planets when their semidiameters are not taken into account, the altitudes of the limb and the centre are the same. Unless otherwise stated, the true altitude of the centre is the altitude which enters into the following problems, and is denoted by h. 119. The hour-angle of a body can be found, when the local time and longitude, or the Greenwich time, are given. (Probs. 34, 35.) For noting the time of an observation, a clock, chronometer, or watch is used ; at sea, only the last two ; but it will be necessary to know how much it is too fast or too slow" of the particular time required. 116 NAVIGATION. 120. The declination of a body can be found when the Greenwich time is known. (Prob. 21.) The polar distance of a heavenly body is the arc of the declination-circle between the body and the elevated pole of the place : that is, the north pole, when the place is in north latitude ; the south pole, when it is in south latitude. If PP' (Fig. 23) is the projec- tion of the declination-circle through an object, M ; P, the north pole ; P', the south pole ; E Q, the equator ; then the polar distances, P M = P Q-QM=90°-rf, P'M = P'Q + QM = 90°+ d. That is, the polar distance is 90°— d or 90° + J, according as the pole from which it is reckoned is N". or S. This, how- ever, is regarding declination, like the latitude, as positive when N., negative when S. To avoid, however, the double sign in the investigation of the formulas of Nautical Astronomy, we shall in most cases consider the declination, which is of the same name as the latitude, asfjositive, and that which is of a different name from the latitude, as negative ; hence the polar distance will be represented by jp=90°— d. When the declination is of a different name from the lati- tude, we have numerically p=90°+d. ALTITUDE AND AZIMUTH. 117 ALTITUDE AND AZIMUTH, 121. Problem 38. To find the altitude and azimuth of a heavenly body at a given place and time. Solution. Find the decimation of the body and its hour- angle at the given time. (Probs. 21, 34, and 35.) Then in the spherical triangle PMZ (Fig- 24), we have given PZ = : 90°- -L, PM = 90°- -d, ZPM = *, to find ZM = 90°- -A, PZM = Z. By Sph.Trig. (122), (123), if in the triangle ABC (Fig. 25), we have given b, c, and A to find a and B, we have tan (p = tan b cos A, cos (c — ', these become tan 0' == tan d sec t, sin h (cos (f) r — L) sin d sin r has the same name, or sign, as the declination, and is numerically in the same quadrant as t. 122. When t = 6 h , 0'= 90°, and the 3d of (97) assumes an indeterminate form. But from the 1st we have tan d cot t = tan (J)' sin t ' which, substituted, gives cot Z = sin (0' — L) tan 6? (98) sin 0'' sin £ which may be used when t is near 6 h . 123. A is the true altitude of M. If the apparent altitude is required, the parallax (Art. 54) must be subtracted, and the refraction (Art. 41) added. Z is the true bearing, or azimuth, of the body, reckoned from the N*. point of the horizon in north latitude, and from the S. point in south latitude. It is generally most conve- nient to reckon it as positive toward the east, which will re- quire in the above formulas — Z for Z, since t is positive when west. Restricting, however, Z numerically to 180°, it may be marked E. or W., like the hour-angle. 121. In Fig. 24, if 31 m be drawn perpendicular to the meridian, then Pr/i =

, its zenith distance, positive, or of the same name as the latitude, toward the equator. A convenient precept is to mark it N. or S., according as the zenith is 3\T. or S. of the point m. m falls on the same side of the zenith as the equator when Z > 90° ; at the zenith when Z = 90° ; and on the same side as the elevated pole when Z < 90°. It falls between P and Z only when t and Z are both less than 90°. 125. In the case of a Ursse Minoris {Polaris), whose polar distance is 1° 25', the more convenient formulas derived from (96) will be, since p and are small, (p z=. p cos £, (which gives within /; .5) sin h = sin (X + 0) — — , v YJ cos ^ ~ tan » sin £ cos 6 tan Z == -— -= — - ; cos (L + $) ' or approximately, h = X + 0, Z ' == p sin £ sec (X + ) . Z is a maximum, or the star is at its greatest elongation, when the angle ZMP (Fig. 24), or Z n P (Fig. 30), is 90°. We then have sin Z = sin p sec X, or nearly Z ' = p sec X. 126. Problem 39. To find the altitude of a heavenly body at a given place and time, when its azimuth is not required. Solution. The 1st and 2d of (96) or (97) maybe used; or, by Sph. Trig. (4), cos a = cos b cos c + sin b sin c cos A, we have sin h = sin L sin d+cos L cos d cos t; which, since cos t = 1 — 2 sin 2 -J- t, 120 NAVIGATION. cos (L—d) — 2 cos L cos d sin 2 it,) in t. \ (99) reduces to sin h or sin h = cos {L— d)— cos L cos c? versin (L—d) becomes numerically {L + d) when L and c? are of different names. Tab. XXVII. contains for the argument t in column P. M. the log sin % t in the column of sines ; which, doubled, is log sin 2 \t. It is well to note this, for mistakes are often made by regarding the logarithms in this table as log sin, log cos, &c, of t instead of ^ t. Tab. XXIII. contains for the argument t, log 2 sin 2 J t = log versin £, with the index increased by 5. It is sometimes necessary to compute the altitude of one, or both bodies, to use in connection with an observed lunar distance. The rules for this purpose on pp. 247, &c, Bowd., are derived from the above formulas. The result is evidently more accurate, the smaller the hour-angle £, especially if the altitude is near 90°. In these rules it is best to find the " sidereal time," or " right ascension of the meridian," from the mean local time, instead of the apparent (Art. 105). 127. Problem: 40. To find the azimuth of a heavenly body from its observed altitude at a given place. Solution. In this the Greenwich time of the observation must be known sufficiently near for finding the declination of the body. The observed altitude must be reduced to the true alti- tude. Then in the triangle PZM we have given the three sides to find the angle P Z M. In the triangle ABC, putting s = $-(a + b+c), we have //sin s sin (s — b)\ COS B a- sin a sin c For the triangle P Z M, ALTITUDE AND AZIMUTH. 121 B = Z, a — 90° — h, A being the true altitude, b — p, the polar distance, c = 90°— X, the co-latitude, 5= 90 o -i (i+A-p), 5-6= 90° —J- (Z + A+p), and the formula becomes cos^Z— yt fcos \ (L + h+27) cos | (Z+^ — p) cos Z cos A -P) or, if we put s r = ^ (Z-f-A+i?), -. r7 //COS «' COS (V- cos -J- Z = 4/ =- , z r \ cos L cos A which accords with Bowditch's rule, p. 160. In a similar way we may find from the formula • i t-» // sin (s — a) sin (s — V)\ sin | B = |/ ( - (100) sin a sin c sin | in which or, if we put cos \ (co Z+A+d) sin \ (co L-\-Ji—cl) cos Z cos /*. CO L — 90°— Z; sm s*=-| (co Z + A-fd), ^ ~_ /fcos s ff sin {s n —d) cos X cos A (101) (100) is preferred when Z> 90° ; (101), when Z< 90° If the body is in the visible horizon, then nearly h— — (33'+the dip). 128. If the bearing of the body is observed with a com- pass at the same time that its altitude is measured, or if the bearing is observed and the local time noted, the decimation, or variation, of the compass can be found. For, the true azimuth, or bearing, of the body can be found from its alti- tude (Prob. 40), or from the local time (Prob. 38) ; and the magnetic declination is simply the difference of the true and 122 NAVIGATION. magnetic bearings of the same object, determined simulta- neously if the object is in motion. It is marked JE. when the true bearing is to the right of the magnetic bearing, W. when the true bearing is to the left of the magnetic bearing. (Bowd., p. 161.) 129. The amplitude of a star when in the true horizon is its distance from the east or the west point, and is marked N. or S., according as it is north or south of that point. It is, therefore, the complement of the azimuth. Pkoblem 41. To find the amplitude of a heavenly body when in the horizon of a given place. Solution. Let the body be in the horizon at M (Fig. 27), A ==W M, its amplitude. The triangle P M N" is right angled at N, and there are given PN = Z, PM = 90°- ■* to find S NM = Z=.90°-i. . Fig . 2T . We have cos P M = cos P N" cos 1ST M, or sin d = cos L cos Z, whence cos Z — sin A — sin d sec X, (102) as in Bowditch, p. 159. By (102) A is N or S like the de- clination. As the equat'or intersects the horizon of any place in the east or west points, it is plain that the star Avill rise and set north or south of these points, according as its declination is N. or S. Tab. VII. (Bowd.) contains the amplitude, A, for each 1° of latitude up to 60°, and each 1° of declination to 23°. The convenience of this table, in the case of the sun, is the only ALTITUDE AKD AZIMUTH. 123 reason for introducing amplitudes. It is generally best to express the bearing of an object by its azimuth. In this problem the body is supposed to be in the true horizon, or about (33' + the dip) above the visible horizon. Hence the rule to " observe the bearing of the sun, when its centre is about one of its diameters above the visible horizon." (Bowd., p. 158.) Examples. (Probs. 38—41.) 1. 1865, Jan. 25, 2 h 33 m 13 s local mean time in lat. 49° 30' S. Ion o\ 102° 39' 15" E. ; required the sun's true altitude and azimuth. (97) h m s L. m. t. Jan. 25 2 33 13 (Jan. 25.) 0's dee, Lon£. Eq.oft, -6 50 37 18° 52' 48".7 S. - 37 ff .35 — 12 m 41 3 .67-0^561 ( 149 .4 (2^4 G. m. t. Jan. 24 19 42 36 +2 40 .3 -J 1 .5 +2 .40 ) .11 or Jan. 25 -4\39 18 55 29 S. ( 3 .4 - 12 39.3 ( 5 Eq'noft. —12 39.3 L. ap. t. 2 20 33.7 *= 3o° 8' 25" 1. sec 0.08738 1. cot 0.15251 d— 18 55 29 S. 1. tan 9.53512 1. sin 9.51098 ' ' — 22 44 50 S." 1. tan 9.62250 1. cosec 0.41266 1. sec 0.03516 L— 49 30 S. L—(p'= 26 45 10 X. h= 48 30 6 Z=S 124 44 23 W. 1. cos 9.95083 1. sin 9.65335 n 1. sin 9.87447 1. cot 9.84102 n \ The reduction for refraction and parallax of A=48°.5 is + 45" ; and the apparent altitude is h f =z-iS° 30' 51". If the compass bearing of the sun at the same instant had been ST. 34° 20' W. = S. 145° 40' W., the magnetic declination would have been 20° 56' W. 2. 1865, Sept. 27, 5 h 20 m 16 3 A.M. mean time in lat. 50° 15' K, long. 87° 3.0' W.;' required the altitude and azi- muth of Venus. (97) 124 NAVIGATION. L. m. t. Sept. 26 17 20 16 Long. + 5 50 12 21 12.4 G. m. t. Sept. 26 23 10 16 =Sept. 27— 0\829 Red. for long. Red. for L. m. t. + 57.5 (Sept. 27) KA. h m s s + 2 50.9 10 4 14.13 +11.645 Dec. of Yenus. o / // // + 12 32 25.0 —55.42 L. sid. t. 9.32 5 45 16.8 —9.65 .23 44.3 +45.9 1.1 R. A. of $ 10 4 4.5 . . . -JW + 12 33 11 - 5 H. A. of $ —4 18 47.7 *= 64 c 41' 55" E. l.sec 0.36919 1. cot 9.67461 d = 12 33 11 N. 1. tan 9.34766 1. sin 9.33714 f = 27 31 11 N. 1. tan 9.71685 1. cosec 0.33530 1. sec 0.05215 L— 50 15 N. f— L— 22 43 49 S. 1. cos 9.96489 1. sin 9.58703 n h= 25 42 42 1. sin 9.63733 Z=N101 38 17 E. 1. cot 9.31379 n 3. 1865, July 20, 5 h 58 m 20 s A.M., mean time in lat. 38° 19' 20" N., long. 150° 15' 30" E. ; required the sun's azi- muth. (98) L. m. t. July 19 17 58 20 Loner. —10 1 2 (July 19) O's dec. Eq'n oft. o / // H m s 8 + 20 49 8.7 —27.60 —5 58.34 —0.165 r 193.2 { 1.155 G. m. t. July 19 7 57 18 = 7 h .955 -3 39.5^ 24.8 1.4 -1.31 i .150 ( 9 Ec l'n of t. —5 59.6 + 20 45 29 I .1 — 5 59 65 L. ap. t. July 19 17 52 20.4 t = 91° 54' 54" E. 1. cosee 1.47604 n 1. cosec 0.00024 d-= 20 45 29 N. 1. tan 9.57868 1. tan 9.57868 ¥ = 95 2 17 N. 1. tan 1.05472 n 1. cosec 0.00168 £ = 38 19 20 N. 1. sin 9.92219 *' — L = 56 42 57 N. N 72 20 43 E. 1. cot 9.50279 4. Required the apparent altitudes of the sun and moon, Nov. 9, 1865, about 9 A.M., in lat. 18° 25' S., long. 84° 6' W. ; time by chro. 2 h 25 m 10 s ; chro. slow of G. m. t. 10 m 15 s . (99) ALTITUDE AND AZIMUTH. 125 T. by chro. 2 25 10 (Nov. 9) O's dec. Eq'n oft o / // ii m s s Chro. cor. +10 15 16 57 42.5 S. + 42.67 +16 0.76 — 0.230 G. m. t Xov. 9 2 35 25 = 2 h .590 — Long. —5 36 24 L. m. t. Nov. 8 20 59 1 Eq'noft. +16 0.2 L. ap.t. Nov. 8 21 15 1.2 L> d = 1° 25' 27" S. O's true alt.= 50 46 Ref. and par. +40 O's ap. alt. = 50 46 40 + 1 50.5 ^ = 16 59 33 S. Zr=18 25 S. t= — 2 h 44 m 58 s .8 ( 85.34 ( .460 \ 21.33 — .60 \ .115 ( 3.84 +16 0.16 ( 2i 1. cos 9.98061 1. cos 9.97717 cos .99969 -.22511 sin .77458 1. shr 2 1 t 9.09360 log 2 log 0.30103 9.35241 i* * 1. versin t = 9.39463 L. m. t. Nov. /ied. for % Red of * m L. sid. t. D's R. A. Z-d = X)'s true alt. Par. and ref. j)'s app. alt. Par. and ref. D's app. alt. (Nov. 9 2 h ) tfsdec. D'sB.A. 20 59 1 15 10 44.2 13 4 38.9 N. —7.420 ! 42 28.70 + 2.128C + 55.3 + 3 26.8 12 14 7.3 8 43 44.1 rf=13 16 3 30 23.2 Z=18 25 —4 22.8 63.84 + 1 15.36*! 10.64 8 43 44.1 31° 25' 16" S. cos .85336 —.36292 29 22 10 sin .49044 -49 Tab. XXIX. 1. cos 9.98871 1. cos 9.97717 1. sin 2 i t 9.29290 log 2 0.301 loo; 9.55981 90 U 03 ) * 1. versin t = 9.59393 28 33 (approx.) D's H. pa'x 56' 28".3 — 2".10 -47 47 28 34 23 ■ K K (4.2 -°-° i l.s tt=56 22 .8 log 3.5293 1. sin 9.9437 loo; 3.4730 7t'rz:28 33' p=4& 32" ref. = — 145 Par. and ref. =47 47 By Tab. XIX. 47' 48" 5. Find the altitude and azimuth of Polaris, 1865, Sept. 25 8 h 15 m P.M., in lat. 49° 16' N"., long. 85° 16' W., (Art. 125). 126 NAVIGATION. h m s L. m. t. 8 1.5 Long. 5 h 41™ 4 3 S 12 IT 16 p = 124 27 log 3.7048 1. cos 9.99987 Red. for long. + 56 t = 69 8 1. cos 9.5517 Red. of L. m. t. + 121 6 h , and consequently that 2 tj or the time that the body is above the true horizon, > 12 h ; and when the latitude and decimation are of different names, t < 6 h and 2 t < 12\ 2 t is an interval of sidereal time for a fixed star, of ap- parent time for the sun. In the case of the sun, t would be the apparent time of sunset, were the refraction and dip nothing, and (24 h — t) would be the apparent time of sunrise. Tab. IX. (Bovvd.) contains t for different values of L and d. 132. Problem 43. To find the hour-angle of a heavenly body at a given place, and thence the local time, when the altitude of the body and the Greenwich time are known. Solution. Find the declination of the body for the Greenwich time, and reduce the observed al- titude to the true altitude. Then in the triangle P Z M (Fig. 28) are given PZ = 90°-Z, Z M = 90° -A, to find ZPM = ^. For the triangle ABC (Fig. 29), we have j^ __ /fern (s — d) sin (a— c) sin-l- sin J) sin c in which, putting A — t a = 90° -A, b=p 9 c = 90° — Z, HOUR-ANGLE AND LOCAL TIME. 129 we have s—b = 90°— |- (Z+p + h), s—c = \ {L+p—li), cos \ (L + p -f h) sin ^ (Z +p — 7i) and sin | t or, if we put / cos L sin p s' = i (Z+p + h), sin -J t //cos «' sin (5' — Tij v (104) cos L sin ^? which is Bo wd itch's rule, p. 209. From Tab. XXVII. (Bowd.) we may take t directly from column P. M., corresponding to the log sin \ t. t is.— when the body is east of the meridian. When the object is the sun west of the meridian, t is the apparent solar time; when the sun east of the meridian, (24 h — i) is numerically the apparent time. "When the object is the moon, a planet, or a star, we have (Prob. 37), denoting its R. A. by a, the sidereal time = a 4- £, and the mean time = a—S' + t 9 in which S f is the "right ascension of the mean sun." (Art. 93.) Or the sidereal time may be converted into mean time by one of the other methods of Problem 32. 133. By the formula - . //sins sin (s — a)\ cos i A = A/[ — ^-tA , * V \ sin sin c y we may obtain for the triangle P Z M (z being the zenith distance), - M / [sm% (co L+p + z) sin b (co L+p — z)\ COS t t = A/ [ ^—. z V \ cos L sin p or putting s = i (co Z+p + z), /sin s sin (s — s)\ f (1 05) COS \ t = JU cos L sin p which is the rule in Bowditch's 2d Method, p. 210. 180 NAVIGATION. (105) is preferable to (104) when t considerably exceeds 6 h , which may be the case in high latitudes. If L =90°, the horizon and equator coincide, and p + h — 90° and p — z; so that both (104) and (105) be- come indeterminate. In very high latitudes, then, these equations approach the indeterminate form, and it is imprac- ticable to find with precision the local time from an observed altitude. So also if d = 90°, the star is at the pole and L = h; and the problem is indeterminate. A great declination is therefore unfavorable. 134. If the object is in the visible horizon (rising or set- ting), h = — (33' + dip) nearly. With the sun, the instants when its upper and lower limbs are in the horizon may be noted, and the mean of the two times taken as the time of rising or setting of its centre. The irregularities of refrac- tion would affect nearly alike the dip and the apparent posi- tion of the sun. 135. If the time at which the altitude is observed is noted by a watch, clock, or chronometer, we may readily find how much the watch or chronometer is fast or slow of the local time. (Prob. 50.) For, let C be the time noted, T, the local time deduced from the observation : c = T— G will be tjie correction of the watch or chronom- eter to reduce it to apparent time, when T is the local apparent time ; to mean time, when T is the local mean time ; or to sidereal time, when T is the local sidereal time. 136. The observed altitude is affected by errors of ob- servation, errors of the instrument, and errors arising from the circumstances in which the observation is made ; such as irregularities of refraction affecting both the position of the body and the dip of the horizon. Errors of the first HOUR-ANGLE AND LOCAL TIME. 131 class are diminished by taking a number of observations. Thus several altitudes may be observed, and the time of each noted ; and the mean of the altitudes taken as correspond- ing to the mean of the times, so far as the rate at which the body is rising or falling can be regarded as uniform during the period of observation. This period should then be brief 137. We may easily find how much a supposed error of V in the altitude will affect the resulting hour-angle, by di- viding the difference of two of the noted times by the dif- ference in minutes of the two corresponding altitudes. The effect will evidently be least when the body is rising or falling most rapidly. This will be the case wmen its diurnal circle makes the smallest angle with the vertical cir- cle. An inspection of the diagram (Fig. 30) shows that this is the case when the object is nearest the prime vertical, or bears most nearly east or west. Thus Z n being tangent to the diurnal circle n n\ the angle which it makes with it is ; and is there- fore less than the angle which any other vertical circle, as Z n\ makes with n 7i\ The diurnal circle m mil makes a smaller angle with Z m, the prime vertical, than with any other verti- cal circle, as Z m\ The diurnal circle o o' makes a smaller angle with Z o than with Z o'. The diurnal circles make right angles with the meridian ; so that at the instant of transit, the change of altitude is 0. 132 NAVIGATION. 138. At sea, and to a less extent on the land, the latitude is uncertain. To ascertain the effect of an error of 1' in the assumed latitude, the hour-angles may be found for two lati- tudes separately, differing, say, 10' ; and the difference of these hour-angles divided by 10. This is an essential feature of Sumner's method, which will be explained hereafter. This method will also show that an error in latitude least affects the deduced hour-angle when the body is nearest the prime vertical. Examples. (Prob. 43.) 1. At sea, 1865, March 20, 10 h 15 m 20 s G. mean time, in lat. 41° 15' S., long. 86° 45' W. (by account) ; observed P. M. altitude of the sun's lower limb 18° 20'; index, cor. of sextant —8' 20"; height of eye 18 feet; required the local mean time. (104.) G. m. t. h m s Mar. 20 10 15 20 G 's dec. Ecfn of t. o 1 li li m s a 10.256 —0 2 3.7 + 59.23 +7 33.82 —0.754 ( 592,3 ( 7.54 + 10 7.4 \ 11.8 + 7.73 \ .15 (3.3 (4 + o_ 8 4 +7 26.1 O 18° 20' 0" ( S. diam + 16' 5 In. cor.— 8' 20" + 52 \ par. + 8" dip. — 4 11 7i= 18 20 52 ( ref. —2 50 £=41-15 1. sec 0.12387 jp— 90 8 4 1. cosec 2 s = 149 43 56 S= 74 51 58 1. cos 9.41677 S—h= 56 31 6 1. sin 9.92120 / 19.46184 L. ap. t. Mar. 20 4 h 20 ra 28 3 1. sin i 9.73092 Eq. of t + 7 26 L. m. t. Mar. 20 4 27 54 Subtracting the local mean time from the G. mean time gives the long. + 5 h 47 m 2G S == 80° 51f W. If we take HOUR-ANGLE AND LOCAL TIME. 133 Z = 41° 25' S., we shall find the local ap. time 4 h 20 m 12 s ; so that for AZ= 10' S, A t = -16 s . 2. 1865, Jan. 1, 21 h at the Navy-Yard, Havana, in lat. 23° 8' 39" N"., long. 5 h 29 m 27 s W., the following altitudes of the sun were observed with an artificial horizon; re- quired the local mean time. T. by Chro, 2 © h m s 3 33 57.5 5310 Chro. fast of G. m s m. t. 42 37.7 34 29.3 20 Bar. 30.43 35 2.3 30 Ther. 75° 35 33.3 36 47 36 37.0 40 50 60 ^i ** nm f offthearc + 32 / lS\3 © sdiam -jonthearc-32 35.0 Index cor. — 8. 3 T. by Chro. 3 35 17.35 58 35 ©'« dec. Chro. cor. —42 37.7 h' = 29 17 30 i £ In. cor. — 4" — 22 53'42.2" + 14 ff .04 G. m. t. Jan. 2 3 52 39.6 = ! 3h.S78 + 14 42-J S. diam. + 16'18 ref. — 1 40 [42.12 + 54.4 ^l.| 52 48 [ f x I par. +8—22 7i= 29 3212' Eq.oft. Z= 23 8 29 1. sec 0.0364304 +4 27.S1 +1.165 p= 11252 43 1. cosec 0.4102711 f 3 '^ 2 s = 165 33 29 +4.51 J-» s = 82 46 45 1. cos 9.0993144 + 4 32.32 I 1 s — h= 5314 33 1. sin 9.9037276 19.4497435 it=— 32 3 16.7 1. sin 9.724S718 L. ap. t. Jan. h m 8 1 19 41 53.7 *=— 64 6 33.4 = +295° 53' 26". 6 Eq. of t. + 4 32.3 L. m. t. Jan 1 19 46 26.0 We have also by subtracting the chro. time from the local mean time, Chro. cor. (L. m. t.) — 6 h 12 m 4 S .8 Long. - +5 29 27 .0 Chro. cor. (G. m. t.) —0 42 37 .8 As the Chro. is fast, the correction is subtractive. 134 NAVIGATION - . By comparing the first and last altitudes and the corre- sponding times, we find that for 2 Ah = + 50', A t = + 2 m 39 9 .5 ; or, for 2 A h— + 1', A t=z +f3M9; that is, an error of 1' in the double altitude will produce an error of 3 s in the resulting time. 3. At sea, 1865, Sept. 7, 8 h 4 m 16 s , G. mean time, in lat. 46° 16' N., long. 153° 0' E., the observed altitude of the moon's upper limb, W. of the meridian, was 21° 19'; index cor. of octant, —3'; height of eye 20 feet; required the local meau time. h m o / G. m. t. Sept. 7 8 4 16 "J 21 19 D's dec. In. cor. — 3 +6° 59' 34" + 1T\2 dip. — 4 +48 44.8 S. diam. — 17 + 7 22 2 .8 K — 20 55 S. diam. 16' 36" + 4" par.&ref. + 54 H. par. 60 49 h = 21 49 L = 46 16 1. sec 0.16033 p — 83 1. cosec 0.00325 b m s s 2 5 = 151 5 D's R. A. 11 12.2 + 2.40 s = 75 32£ 1. cos 9.39738 + 10.2 ~9j6 s—p == 53 43£ 1. sin 9.90644 1 1 22 .6 19.46740 D'sH. A. 4 22 21 1. sin j- 9.73370 L. sid. t. 5 23 43 _£ o —ii 6 18 -Red.for G.m.t. — 1 20 L. m. t. Sept. 7 18 16 5 Long. — 10 11 49 = 152° 57' E. 4. 1865, Sept. 30, in lat 30° 27' N"., the Chro. time of the setting of the sun's centre was ll h 16 m 6 s ; the Chro. cor., -f!5 m 25 s ; height of eye 16 feet; required the local time. HOUR-ANGLE AND LOCAL TIME. 135 h m a T. by Chro. 11 16 6 Chro. cor. + 15 25 O's. dec. O 1 II 2 54 31.9 Eq'n of t. — 58.30 —10 4.40- -0.801 G. m. t. Sept. 30 11 31 31 — 11 11.9s r 641.3 ( 29.1 -9.23-< 8.81 .40 11.525 A = - 3 5 44 ( - 37 k 1.5 -10 13.6 2 Lz= 30 27 1. sec 0.06446 P = 93 6 1. cosec 0.00064 2s = 122 56 s = 61 28 1. cos 9.67913 s — h = 62 5 1. sin 9.94627 h m s L. ap. t. Sept. 30 5 55 34 19.69050 1. sin \ 9.84525 Eq. of t. — 10 14 L. m. t. Sept. 30 5 45 20 Long + 5 h 30 ra 46 s = 82° 4l'i W. 139. Problem 44. To find the hour-angle of a heavenly body when nearest to, or on, the prime vertical of a given 'place. Solution. If d > X, and with the same name, as for the body whose diurnal path is n n' (Fig. 30), P Z n will be greatest, or nearest to 90°, when Z n is tangent to n n\ and consequently Znp = 90°. We then have cos t = -—4 = i r ( 106 ) cot L tan a v J If £?A = 1\0 ; t- 15 m _ 7* m Ah = is T x] 5"= 56". The uncertainty of altitudes at sea makes such a correc- tion of little practical importance; but it is generally ne- glected by those navigators who work out their latitudes to seconds, supposing that they have attained that degree of accuracy. In the above example, the maximum altitude of the sun would have been greater than the meridian altitude, and the latitude obtained from it in error, by nearly 1'. The sun would not have sensibly dipped until 9 or 10 minutes after noon. 143. A difficulty occurs at sea in measuring the meridian altitude of the sun when it passes near the zenith, on account of its very rapid change of azimuth ; the change being made from east to west, 180°, in a very few minutes. What is wanted is the angular distance of the sun from the'N". or S. points of the horizon. One of these points may be sufficiently fixed by means of the compass, and then the angular distance from this point observed within l m or 2 m of 144 NAVIGATION.. the meridian passage as determined by a watch regulated to apparent time. 144. From (108) we have z — Z — d, (111) by which the zenith distance maybe found when the latitude and declination are given. Also d = Z — 2, which may be used at sea for estimating the declination of a bright star from its estimated meridian altitude. If the time when it is near the meridian be also noted, and converted into sidereal time, we have the right ascension and declination of the star sufficiently near for de- termining what star it is. Example. July 16, 8 h 45 m , in lat. 11° N"., a bright star is seen near the meridian S., at an estimated altitude of 55°. L. m. t. July 16 8 h 45 ln L = 11° N. S Q V 37 2 = 35 N. L. sid. t. 16 22 d=24 S. The R. A. of a Scorpii {Antares) is 16 h 21 m , and its declina- tion 26° r S. 145. Problem 46. To find the latitude from an altitude of a heavenly body observed at any time, the local time of the observation and the longitude of the place being given. 1st Solution. Reduce the observed altitude to the true, altitude, and from the local time and longitude find the declina- tion and hour-angle of the body. (Probs. 21, 34, 35.) Then in the triangle P Z M (Fig. 32) there are given ZPM = £, PM = 90°-e?, ZM = 90°— A, to find PZ = 90°-Z. LATITUDE. 145 By Sph. Trig. (146), if in the triangle ABC (Fig. 33) are given a, 5, and A, we find c by the formulas tan (ft = tan b cos A, ^ . . cos cos a cos (ft = z — t — ■ r COS c = (ft ± 0' ; which, applied to the triangle P Z M, give Fig. 33. tan (ft = COS <£'- cot c? COS £, " cos-0 sin A (112) sin d 90°— Z = 0=b0'. These may be changed into a more convenient form for practice, if we put = 90°— 0" ; then tan r/ = tan d sec £, sin 0* sin h COS 0'= (113) sin a Here, observing that -f- and -— may be rendered by X. and S. respectively, we mark (ft" N". or S. like the declina- tion, and (ft' either X. or S. ; then the sum of (ft" and (ft' when of the same name, their difference when of different names, is the latitude, of the same name as the greater. There are two values of L corresponding to the same altitude and hour-angle, but which, imless (ft' is very small, will differ largely from each other ; so that we may take that value which agrees best with the supposed latitude (at sea the latitude by account). When t > 6 h , (ft" > 90°, as in (97). 146. In Fig. 32, if Mm be drawn perpendicular to the meridian, we shall have eft = P m, the polar distance of m, '=Z m, the zenith distance " 146 NAVIGATION. When ' is very small, (that is, when M m nearly coin- cides with M Z), (/>' cannot be found with precision from its cosine. If not greater than 12°, it can be found only to the nearest minute with 5-place tables ; if only 2°, it can be found only within 3'. The more nearly, then, that M m co- incides with Z m, or, in other words, the nearer the body is to the prime vertical, the less accurate is the determination of the latitude. If the body is on the prime vertical, cos '=. 1, and f ; it will also be numer- ically the change of latitude, as (f> n does not depend on h; Then from the 2d of (113) or, since A h and generally A

are so small that we may take cos A h—l, sin A h =A h sin 1", cos A 0'= l, sin A r =A 0' sin 1", coscb'—Ad)'. sin 6' sin V~ . ., (sin h + A h. cos h sin 1"). T T T sin a v ' Subtracting this from the second of (113), and reducing, we have or, since . ,, sin

' cot A. (114) But in the triangle M Z m, lrr7 t» r r^ t^ tan m Z cos M Z ra = — cos M Z P = z — tf~ ; tan M Z 7 that is, Z being the azimuth, LATITUDE. 147 — cos Z = — — t-, or sec Z = — cot 0' cot A, cot A and therefore J '= J A. sec Z. (115) If the body is on the meridian, Z— or 180°, and numer- ically A (p f =A h. The nearer Z is to 90°, the greater is A 0'. If Z = 90°, . or the body is on the prime vertical, sec Z = oo , and J 0' is incalculable. If Zis near 90°, (115) is inaccurate; since A r becomes too large for the assumptions cos A 0'=1, sin A '=A (/>'. sin 1" ; so, also, in (114) if X, sin {L—d) == sin z is negative ; that is, z will have a different name or sign from L (Art. 140). Properly A, 7i , and A h would also become negative to correspond. Still, however, we shall have numerically 7i = h + A h. We may therefore disregard the sign of L—d in (121) and consider h and h as always positive. If the star is observed at its lower culmination, then t will be the hour-angle from the lower branch of the meri- dian, and for d we may use 180°— d (Art. 140). A Q h and A h are then numerically subtractive. LATITUDE. 151 Examples. (Prob. 46.) 1. At sea, 1865, July 17 l h P. M., in lat. 36° 38' S., long. 105° 18' E., by account; time by Ckro., 5 h 47 m 14 s ; O., 30° 15'; N". W'y; index cor. + 2' 30"; height of eye, 17 feet ; Chro. cor. (G. m. t.) + I4 m 3 s ; required the latitude. By (113) h m s T. by Chro. + 12k, 17 47 14 Chro. cor. +14 3 G*s dec. Eq'n of t G / // // m 3 s + 21 20 29 —25.19 —5 43.8 —0.230 G. m. t. July 16 18 1 11 = 18.021 -7 34 ( 252 -4.1 j 2.3 + 21 12 55)202 _ 3 47.9 { 1-8 —Long. + 7 1 12 July 17 1 2 29 — 5 48 L. m. t. Eq. of t. _0 30° 15' ( In. cor. + 2'. 5 dip. — 4'.0 + 13 ( S.diam + 15 .8 ref.&par. — 1 .S L. ap. t. 56 41 h = 30 28 1. sin 9.70504 o / // ** = 14 10 15 1. sec 0.01342 d = 21 12.9 X. 1. tan 9.58903 1. cosec 9.44145 f = 21 49.1 N. 1. tan 9.60245 1. sin 9.57016 $' - 58 37.0 S. L - 36 48 S. 1. cos 9.71665 If we suppose an uncertainty of 3' in the altitude and 20' in the longitude, by (115) and (118) Z=S. 164° 40' W. A h = + 3' AL--ZW L cot (— h) 0.2304 a 1. cos L 9.903 Lcotf 9.7853 —A t= — 20' log 1.301 n 1. sec Z 0.0157 n 1. tan Z 9.438 n log 0.477 jZ=+4'.4 1og 0.642 log 0.493 n That is, an increase of 3' in the altitude will numerically decrease the latitude 3'.1 ; and a numerical increase of 20' in the assumed longitude will increase the latitude 4'. 4. This may be conveniently expressed in the following way : * Instead of changing t into arc, we may enter col. P. M. of Tab. XXVII. with 2 t = l h 53 m 22 9 . 152 NAVIGATION. Long. 105° 18' 20' E.; O, 30° 15' =fc 3' L = 36° 48' ± 4 '.4 zp 3'.1 S. By (119) t = h 56 m 41 a 2 1. sin £ 8.18227 ) ^or i. versin d— 21° 12'.9 N. 1. cos 9.96953 8.45283 1st L — 36 38 S. 1. cos 9.90443 log 8.35726 .02276 h = 30 28 sin .50704 Z = 58 .5 S. cos .52980 2d L — 36 47 .6 s. 1. cos 9.90345 log 8.35628 0.2271 Z =:58 0.7 s. cos .52975 3d i = 36 47 .8 s. 2. At sea, 1865, Jan. 5, 6 h P. M., in lat. 50° 36' K, long. 135° 25' W. (by account), time by Chro. 3 h 10 m 15 s ; Chro. cor. (G. m. t,)— 18 m 56*; Obs'd alt. of Mars, 45° 18'; S. E'y; index cor.- — 3'; height of eye, 19 feet; required the latitude. (113) T. by Chro. + 12\ 15 10 15 Chro. cor. —18 56 G. m. t. Jan. 5 14 51 19 = 14\855 S 19 22.1 Red. for G. m. t. 4- 2 26.5 G. sid. t. 9 54 7.6 —Long. —9 1 40 L. sid. t 52 27.6 ff = 45° 18' Mara 1 R. A. 3 55 25.6 — 8 /=- 3 258 h =45 10 or 45° 44' 30" d = 23 37 N. ' 0.4527 Z— X. 110° 46' E. 1. sec Z 0.4502 n J h=+ 5' log 0.699 Ji- - 14'.1 log 1.149 n 1. cos L 9.801 —J *= + 15', log 1.176 1. tan Z 0.421 n A £= — 25'.0, log 1.39S n 3. 1865, Feb. 17, near noon, at the light-house, TV. end of St. George's Island, Apalachicola Bay, long. 85° o lb" W. ; 5 observations with sextant No. 1, art. hor'n No. 3, A end toward observer : T. by Chro. h 16 m 21 8 .6; 2© 98° 14' 44", (S.) ; in. cor. +2' 30"; Chro. cor. (L. m. t.) — 18 m 30 s A ; Bar. 30.48, Ther. 43°. T. by chro. h 16 m 21 s .6 Chro. cor. —18 30 .4 L. m.t.Feb. 16 23 57 51 .2 Long. +5 40 21 G. m.t.Feb. 17 5 38 12 = Eq. oft. —14 14 .1 L. ap. t. ( 23 43 37 .1 or —0 16 22 .9 i— 4° 5' 43".5 d = — 11 46 55 .7 (j)"= — 11 48 41 .2 (j>'= + 41 26 10 .2 L= + 29 37 29 By (113) O's dec. — 11° 51' 52".8 +4 57 .1 —11 46 55 .7 = 5*637 ©48 Eq.oft. -52".71-14' 15".22 + s .2O4 ( FT020 + 1.15-1 .122 ( 6 -14 14 .07 7' 22" f \ In. cor. + V 15" j S. diam. +16 30 + 16 57 1 Ref. — 54 [Par. + 6 A=48 24 19 1. sec 0.0011104 1. tan 9.3192842 n 1. tan 9.3203946 n I. sin h 9.8738198 1. cosec 0.6899636 n 1. sin 9.3111004 n 1. cos 9.8748838 By (121) and (122) l s .96349 L= + 29° 37 d = — ll 47 L — d— +41 24 AJi=- 2".527 * = — 16 m .382 A h ' = + 678".l log 0.2930 1. cos 9.9392 1. cos 9.9908 1. cosec 0.1796 log 0.4026 2 log 2.4287 lo* 2.8313 h= 48° 24' 19" A Q h= + 11 18 h =z 48 35 37 Z Q = + 41 24 23 d= — ll 46 56 L= + 29 37 27 151 NAVIGATION. LATITUDE BY CIRCUM-MERIDIAN ALTITUDES. 151. Problem 47. To find the latitude from a number of altitudes observed very near the meridian, the local times being k?ioio?i. Solution. By (122) we see that very near the meridian the altitude of a body varies very nearly as the square of its hour-angle. Hence we cannot regard the mean of several altitudes as corresponding to the mean of the times, since this is assuming that the altitude varies as the hour-angle, Let /*!, A 2 , A 3 , &c, be the several altitudes ; t u t 2 , £ 3 , &c, the corresponding hour-angles expressed in minutes; and we have as the reduction of each altitude to the merid- ian, and the deduced meridian altitude, A l h = t\. A h h — h x + ^A) A 2 h — tl.A Q h h = h 2 + A 2 h > &c. (123) 4 h — tl.A h &c. h = h 3 + 4 h ) Thus the meridian altitude may be derived from each alti- tude, and the mean of all these meridian altitudes taken as the correct meridian altitude. But the following is a more expeditious method : — If n is the number of observations, the mean value of h Q will be 7 h 1 + h i + h 9 + ...h n , AJi-\-AJi + AJi-\-...A n h n — _ — u n n or, A I + A a + ^ + . : j, g ± j + q + ...g ( n n v J Whence the rule : — Take the mean of the squares of the hour-angles in minutes (Tab. XXXIII., Bowd.) ; multiply it by the change of alti- tude in l m from the meridian (Tab. XXXII.) ; and add the LATITUDE BY CIRCUM-MERIDIAN ALTITUDES. 155 product to the mean of the altitudes. The result is the mean meridian altitude required. (Bowd., p. 201.) From the meridian altitude thus found deduce the latitude as from any other meridian altitude. (Prob. 45.) Strictly, however, the declination to be used is that which corresponds to the mean of the times, and the hour-angles, £, are intervals of apparent time for the sud, and of sidereal time for a fixed star. , 152. It is unnecessary to reduce each observed altitude separately to a true altitude ; as the reductions, excepting slight changes of refraction and parallax, are the same for all, and may be computed for the mean of the observed altitudes, and applied to this mean with the reduction to the meridian. 153. Should it be desirable to compare the several obser-* rations with each other, and test their agreement, it will be sufficient to compute the several reductions to the meridian, A y A, J 2 hj 4* A, &c, and apply them separately to the read- ings of the instrument ; or to the half-readings when the altitudes are observed with an artificial horizon : applying, also, the semidiameter when both limbs of the body are observed. 154. If the altitudes are taken on both sides of the merid- ian, and at nearly corresponding intervals, a small error in the local time will but slightly affect the result ; for such error will make the estimated hour- angles and the corre- sponding reductions on one side of the meridian too large, and on the other side too small. (Bowd., p. 203.) 155. This method is rarely used at sea, as a single altitude on or near the meridian suffices. N"o increase of the number of observations will diminish at all the error of the dip, which affects alike each observation and the mean of all.* But on * Such an error is called constant ; those which affect the several obser* vations differently are called variable. 156 NAVIGATION. land it is preferable to measure a number of altitudes at the same culmination of the body, and thus diminish the " error of observation." Altitudes of the sun are used, but the best determinations are from the altitudes of a bright star. To facilitate the operations, and avoid mistaking one star for another, it is well to compute the altitude approximately be- forehand. (Art. 144.) If an artificial horizon is employed, the error of the roof is partially eliminated by making two sets of observations with the roof in reversed positions. 156. If two stars are observed which culminate at nearly the same altitude, one north, the other south of the zenith, the error of the instrument is nearly eliminated ; for such error (except accidental error of graduation) will make the latitude from one of the stars too great, and that from the 'other too small by very nearly the same amount ; the more nearly, the less the difference of the altitudes. The error peculiar to the observer is also eliminated. If the observations are made with an artificial horizon, the error of the roof is eliminated, if the same end is toward the observer in both sets of observations. 157. Bowditch's Tab. XXXII. extends only to d = 24° If a star is used whose declination is beyond this limit, or if greater precision than the table affords is required, A Q h may be computed for the star and place by (121). a l ;/ .9635 cos L cos d J ° /l " lin (L-d~)~ 158. If the observations are made at the lower culmina- tion of the star, we have only to use in the formulas I80°—d instead of d. (Art. 140.) 159. The altitudes observed at the same culmination are very nearly the same. To render the measurements inde- pendent, after each observation move slightly the tangent screw of the instrument. With the sextant, it is best to LATITUDE BY CIRCUM-MERIDIAN ALTITUDES. 157 make the final motion of the tangent screw at each observa- tion always in the same direction, for example, in advance. Examples. (Prob. 47.) 1. 1865, Jan. 10 h . Circum-meridian altitudes of Q ob- served at the Custom-House, Key West, Florida, 45" ]ST. of Light-House: lat. 24° 34' N., long. 81° 48' 31" W. T. by Watch. Sextant No. 1. Art. hor. No. 1. m 5 2 o 87 41 30 A end of Hor. h ^TTT . 5 30 41 30 Chro. 1843 4 58 24 6 26 41 5 50 41 35 Watch 11 39 20 1 7 30 7 8 4140-5. C— W +5 19 4 5J_9 11 7 30 41 50 Chro. cor. (L. m. t.) — 5 h 18^~34M) 7 55 41 55 9 2 O 86 36 25 _ r off arc ^ 4r>0 9 25 36 40 U S aiam * | on arc 32 6 .5 9 45 36 30 In. cor. + 20 .2 12 2 35 20 A. " 12 25 34 50 Bar - s 0.21 12 45 3 4 40 T k ei \ 78° h m s o / // T. by W. 8 41.2 2 "o 87 41 40 W. cor (L. ap. t.) -7 30.1 2 O. 86 35 44 L.ap. t. Jan. 10 1 11.1 2 O 87 8 42 Long. +5 27 14.5 G. ap. t. Jan. 10 5 28 25.6 (At h 9 m ) C— W. . 5* 19 m 6 3 .3 5.474 Chro. cor. (L. m. t.) —5 18 34 .0 W. cor. (L. m. t.) + 32 .3 Eq'n of t. +7 m 56 s .96 +0 S .998 t f — 2 m 30 9 6.2 2 4.0 1 40 2.8 22 0.1 0.0 + 25 0.2 1 30 2.2 1 55 3.7 j 4 .99 4-5.46 1 .47 + 8 2.42 Watch t. of ap. h h 7 m 30 s . 1 Hourly ch. — 8 .7 0's dec. —21° 54' 43".3 + 22".90 + 2 5 .3 —21 52 38 2 15 5.1 4 32 20.6 4 55 24.2 5 15 27.6 158 NAVIGATION. O 43 34 21 Ciln. cor. + 10".l < Ref. —58 .5 — 52 ( Par. + 6 .3 1".9635 log 0.2930 96.7 h = 43 33 29 Z= + 24° 34' 1. cos 9.9588 A h= 8.06 x 2".28 = +18 =—21 52 1. cos 9.9657 h = 43 33 47 X— fc+46 27 1. cosec 0.1398 z =+46 26 13 J h= 2".28 log 0.3573 d—— 21 52 38 Custom-House, X= + 24 33 35 Light-House, X=+ 24° 32' 50" 2. 1843, January 31 (civil date). Circum-meridian alti- tudes of © observed at E. Base station, Mullet Key in Tampa Bay; lat. 27° 37' 1ST., long 5 h 30 m 50 s W. T. by Chron. Sextant JVb. 2. Art. hor. No. 1 h m s O i II 5 30 2© 90 20 20 A end toward obser. 6 10 21 30 6 45 22 20 8 42 2 89 19 50 9 12 20 10 9 38 20 40 11 00 21 36 B end toward obser. 11 30 22 10 11 52 22 30 Index cor. + 1 13 15 2 © 90 27 50 13 45 28 10 Chron. cor. — 14 17 28 20 15 25 28 30 Bar. 29.95 15 50 28 40 16 18 28 40 Ther. 72° 17 53 2 89 23 18 20 22 50 18 40 22 50 20 50 21 A end toward obser. 21 12 20 40 LATITUDE BY CIRCUM-MERIDIAN ALTITUDES. 159 T by Chron. h . m a 21 37 Sextant No. 2. Art. hor. No. 1 o / // 89 20 30 22 50 2 © 90 24 10 23 20 23 40 23 40 23 30 Bq. of Time + 13 m 44 s 19+0 S .374 1".9635 log 0.2930 + 2 9 . 07 r 1.87 X = + 27° 37' 1. cos 9.9475 + 13 m 46 s . 3 •] .19 d =—17° 25' 1. cos 9.9796 — Chron. cor. + l m 39 s . 3 ( .01 L— d = +45° 2' 1. cosec 0.1503 Chr. T. ap. Noon + 15 m 26 8 . A Q h = 2" 346 lo g 0.3704 t m s —9 56 98.7 (k) O I II 45 10 10 2".346 t* i a + 3 51 S. diam / // —16 16 (K) o ; // 44 57 45 9 16 85.9 10 45 3 21 50 8 41 75.4 11 10 2 57 51 6 44 45.3 44 39 55 1 46 + 16 16 57 6 14 38.9 40 5 1 31 52 5 48 33.6 40 20 1 19 55 4 26 19.7 40 45 46 47 3 56 15.5 44 41 5 36 57 3 34 12.7 41 15 30 61 2 11 4.8 45 13 55 11 —16 16 50 1 41 2.8 14 5 7 56 1 9 1.3 14 10 2 56 —0 1 0.0 14 15 59 +0 24 0.2 14 20 64 52 0.8 14 20 2 66 2 27 6.0 44 41 30 14 + 16 16 60 2 54 8.4 41 25 20 61 3 14 10.5 41 25 25 66 5 24 29.2 40 30 1 8 54 5 46 33.3 40 20 1 18 54 6 11 '38.2 40 15 1 30 61 7 24 54.8 45 12 5 2 9 —16 16 58 7 54 62.4 11 50 2 26 70 + 8 14 67.8 11 45 +2 39 68 Mean —0 32 Mean 0's U. L. 44 57 58 Long. 5 h 30 50 O'sL. L 57 G. ap. T. 5 30 18 A B 44 57 57 58 160 NAVIGATION". 0\s Dec. — 1°7 28 26.0 + 41.55 h' = 44 57 57.4 + 3 48.7^207.75 -Jin. cor. +32.5 d-__17 24 37.3 j 20.71 Ref. —56.2 z Q = +45 2 20 ( - 21 Par. +6.2 Lat. - + 27 37 43 h = 44 57 40 3. 1865, May 22, 9 h , circum-meridian altitudes of a Vir- ginis (Sjiica) at Light-House on St. George's Island, Apala- chicola Bay, Florida, lat. 29° 37' K, long. 85° 5' 15" W. In. cor. — 3' 0" Bar. 30.04, Ther. 73° Chro. cor. (L.m.t.)+5 h 35 m 32 9 .9 Long. +5 40 21 T. by Chro. Sextant No. 1. Art. Hor. No. 1 h m s O 1 II 3 28 56 2 alt. 99 43 50 A. end 31 24 47 50 33 36 51 34 56 53 40 37 8 53 40 38 58 55 30 B. end 42 45 55 30 44 33 52 10 48 21 46 50 51 25 43 50 99 50 23 7i'= 49 55 12 ( i In. cor. — 1' 30" _2i7( Ref. - 47 7i = 49 52 55 h m 3 *'sR. A. 13 18 7.8 So —4 30.0 — Red for % -55.9 Sid. int. from h 9 16 41.9 Red. -1 31.2 L. m. t. of transit 9 15 10.7 — Chro. cor. — 5 35 32.9 Chro. t. of transit 3 39 38 {Mean) (Sid.) -10 42 — 10 44 115.2 1".9635 log 0.2930 8 14 8 15 68.1 Z = + 29°37' 1. cos 9.9391 6 2 6 3 36.6 d=r — 10 28 1. COS 9.9927 4 42 4 43 22.2 Z-eJrrr+40 5 l.cosec 0.1912 2 30 2 30 6.2 AJi == 2".606 log 0.4160 40 40 0.4 P= 49.86 log 1.6979 +3 7 + 3 7 9.7 Ah - + 2' 10" log 2.1139 4 55 4 56 24.3 h — 49° 52 55 8 43 8 44 76.3 h Q = 49 55 5 11 47 11 49 139.6 49.86 * =+40 4 tf=-.10 27 X=r + 29 37 55 34 21 LATITUDE BY CIRCUM-MERIDIAN" ALTITUDES. 161 160. Problem 48. To find the latitude from an observed altitude of Polaris or the North Pole-star. Solution. The formulas (112) of Prob. 46, tan ■==. cot d cos t cos sin A cos (/)'- sin d 90°— Z = ±4>' can be greatly simplified in the case of the Pole-star, since its polar distance is only 1° 25'. Putting we have d— 90°— p and 0'= 90°— cj>% tan

COS sin "= sin A i r (125) (126) cosp ■0'-0, J the 2d value of i, or (180°— If t is more than 6 h or less than 18 h , cos t is negative, and we have numerically L = h + 0. Let $ represent the sidereal time, and a the right ascen- sion of the star, then t = S—a and =p cos (S —a). If we consider the right ascension and polar distance of the star to be constant, may be computed and tabulated for different sidereal times (right ascensions of the meridian), as in Bowditch, p. 206, and "Tab. I. for the Pole-star" in 162 NAVIGATION. the British Nautical Almanac. But owing to the change of right ascension and declination, such a table requires cor- rection for each year. It is better to take the apparent right ascension and declination from the Almanac, and compute t and may be found approximately in the traverse table (Tab. II.) in the Lat. col., by entering the table with t as a course, and p as a distance. 161. Formulas (126) maybe de- rived from Fig. 34, by regarding P M m as a plane triangle, and Z m = Z M. The first produces no error greater than 0".5. The error of the second is evidently greater the greater the altitude, or the latitude. This error, however, will not be more than 0'.5 in lati- tudes less than 20°, nor more than 2' in latitudes less than 60°. 162. We may use (125) with more exactness, but these formulas may be modified so as to facilitate computation. Put or log sin (h+A h) — log sin h = log seep — log sec 0. But A h being very small, representing by D ;/ the change of log sin h for 1", we have, with J A in seconds, log sin (h +Ah)— log sin h — A h X D M ; whence, by substituting in the preceding, we obtain log sec p — log sec (f> _ log cos — log cos p Jh = D m D. (127) LATITUDE BY CIRCUM-MERIDIAN ALTITUDES. 163 The difference of the log secants, or log cosines, of p and $ is readily taken from the table by inspection. D ;/ for log sin# is usually given in tables of 7 decimal places, and hence A h is readily found. We have then 0==pOOfl.t \ (12S) If D, is the change of log sin h for 1', then in minutes log sec p — log sec Ah = - D, (129) 163. The British Nautical Almanac contains three tables for the reduction of altitudes of Polaris, from which they may be found to the nearest second. 164. Altitudes of Polaris may often be observed at sea, with some degree of precision, during twilight, when the horizon is well defined, and the latitude found from them within 3' or 4'. Examples. (Prob. 48.) 1. At sea, 1865, March 31, 7 h 15 m 19 s , mean time in long. 160° 15' E. ; obs'd alt. of Polaris 38° 18' ; index cor. +3'; height of eye 17 feet : what is the latitude ? (128) h m s L. m. t. March 31 7 15 19 Long. h m s —10 41 3, 35 29 Red. for long. — 145 Red. of L. m. t. L. sid. t. *'sR. A. + 1 11 K— 38° 18' / 7 50 14 —2 - 1 9 14/ir: 38 16 ( In. cor. +3 Dip -4 Ref. —1 i- = 6 41 t— 100° 15' 1. cos 9.250 n — p=— 1 24.5 — ^r=+ 15.0 log 1.927/1 log 1.177 1. sec .00013 1. sec Jh=+ 0.8 h— 38 16 1. sec p — 1. sec d> 16 V, L=+ 38 32 164 NAVIGATION". 2. 1865, May 22, 9 h ; altitudes of Polaris, at light-house on St. George's Island, Apalachicola Bay: lat. 29° 37' N". ; long. 85° 5' 15" W. ; sextant No. 1, index cor. ~3'0"; Art. Hor. No. 1 ; Bar. 30.04, Ther. 73° ; Chro. cor..(L. m. t.) + 5 h 35 m 33 9 . KlyCkro. iA JXfhor.) T.ly Chro. ^dofHor.) 3 IT 25 56 32 20 4 2 18 56 34 19 26 32 30 6 3 34 20 21 39 32 50 8 27 34 50 30 35 32 80 10 15 34 50 3 22 16 T. by Chro. 14 40 4 .8 21 85 40 T. by Chro. 27^=56 32 32 27i'=56 34 44 Chro. cor. + 5 35 33 7*/= 28 16 16 Chrc . cor. + 5 35 33 7i/=28 17 22 L. m. t. 8 57 49 i-In. cor.-l 30 L. m . t. 9 43 54 Jin. cor.— 1 30 #0 4 30 Ref. -1 43 So 4 30 Ref. -1 43 Bed. for 2, + 56 h=2S 13 3 Red. for 2, + 56 7t=28 14 9 Red. for m. t. + 1 28 Red. for m. t. 1 36 L. sid. t. 13 43 L. sid. t. 13 46 56 * 's It. A. 1 9 82 ^'s R. A. 1 9 82 H 11 51 11 177° 47' 45" 1. cos 9.9996S n H 12 37 24 189 c 21' 0" I. cos 9.99419 n -p=- 1 24 44 log 3.70621 n — p=— 1 24 44 log 3.70621 n -0= + 1 24 40 log 3.70599 -$= + 1 23 37 log 3.70040 A*<= l.secp 1319 \ 1. sec<£ 1317 [• (Zf,=39.2) 2 J A*<= + 1 l.secp 1319 I I. sec 1284 y 7i = 28 13 3 h= 28 14 9 z= + 29 37 43 L= + 29 37 47 CZf„=39.2) 35 J 1. sec^> and 1. sec (f> are expressed in units of the 7th place of decimals. CHAPTER VIII. THE CHRONOMETER,— LONGITUDE. 165. Astronomically the longitude of a place is the dif- ference of the local and Greenwich times of the same instant. It is west or east, according as the Greenwich time is greater or less than the local time. (Art. 73.) The mean solar, the apparent, or the sidereal times of the two places may be thus compared. 166. A chronometer is simply a correct time-measurer, but the name is technically applied to instruments adapted to use on board ship. It is here used more generally, as in- cluding clocks which are compensated for changes of tem- perature. A mean time chronometer is one regulated to mean time ; that is, so as to gain or lose daily but a few seconds on mean time. A sidereal chronometer is one regulated to sidereal time. 167. A chronometer is said to be regulated to the local time of any place, when it is known how much it is too fast, or too slow, of that local time, and how much it gains or loses daily. The first is the error (on local time) ; the second is the daily rate. Both are + if the chronometer is fast and gaining. It is preferable, however, to use the correction of the chro- nometer, which is the quantity to be applied to the chrono- meter time to reduce it to the true time, and its daily change* Both are + when the chronometer is slow and losing. 166 NAVIGATION. They will be designated by c and A c. A chronometer is said to be regulated to Greenwich time, when its correction on Greenwich time and its daily change are known. If c is the chro. cor. to reduce to Greenwich time, and e, the chro. cor. to reduce to the time of a place whose longi- tude is X (-{- if west). .0 [ .15 +2 35.33 [ THE CHRONOMETER. 173 h 30 21 50 7 30 — 16 17.3 h = 49 51 12.fi L = 30 20 30 p = 68 1 20.0 L. ap. t. May 30 21 3 51.35 — Eq. of t, — 2 35.33 L. m. t. May 30 21 1 16.02 T. by Chro. 3 32 39.07 2 s = 148 13 2.7 0,Chro. cor.(L.m.t)— 6 31 23.05 s = 74 6 31.4 is— 7i = 24 15 18.7 O / // I * = 157 58 55.1* t= 315 57 50.2 \\n. cor. // a + 17.3 ref. — 51.6 S. diam. — 15' 48.4 par. +5.4 1. sec 0.0639749 1. cosec 0.0327661 1. cos 1. sin 9.4374537 9.6136320 19.1478267 9.5739134 T. by Chro. 3 35 5.18 O's dec. Eq'noft. oil, it m s* Chro. cor. — 42 26 +21 58 40.0 +21.04 +2 35.33 G. m. t. May 31 2 52 41 in h .041 + .8 2.878 + 21 58 40.8 — .02 + 2 35.31 —0.360 L. ap. t. May 30 21 6 17.89 — Eq. of t. — 2 35.31 L. m. t. May 30 21 3 42.58 T. by Chro. 3 35 5.18 0,Chro.cor.(L.m.t.) -6 31 3 2.60 O 50 7 30 + 15 19.5 j h = 50 22 49.5 I L — 30 20 30 p= 68 - 1 19.2 2 s = 148 44 38.7 s= 74 22 19.4 -h = 23 59 29.9 Mean —6 31 22.82 \ ' Red for 3 h .0 — .48 Chro.cor.(L.m.t.)-6 31 23.30 *■= 158 17 14.2 t = 316 34 28.4 i In. cor. // // + 17.3 ref.— 51.6 S. diam. + 15' 48.4 par. +5.4 1. sec 0.0639749 1. cosec 0.0327668 1. cos 1. sin 1. sin 9.4303807 9.6091709 19.1362933 9.5681467 May 31 h 2. Chronometer Correction. Pensacola Navy- Yard, 30° 20' 30 /; N"., 81° 15' 21 ;/ W. 1865, May 31 3 h . * In A. M. observations, \ t may be taken in the 2d quadrant ; or it may be taken in the 1st quadrant and marked — . 174 NAVIGATION. T. by Chro. Sextant No. 2. Art. Hor. No. 1. h m e o / m s 9 24 2.7 8 2© 100 40 A. end. Chro. cor. (G. m. t.) —42 27 ) 22.8 Daily change — 3.8 J 24 25.5 100 30 23.0 G's diam. off arc +32 12.5 ) 24 48.5 100 20 on arc —30 59.2 j 24.0 25 12.5 100 10 In. cor. + 36.6 22.3 25 34.8 100 23.4 25 58.2 99 50 28 33.5 2 97 40 B. end. Bar. 30.14 23.5 28 61 97 30 Ther. 76° 23.5 29 20.5 97 20 22.5 29 43 97 10 23.0 30 6 97 23.5 30 29.5 96 50 9 25 0.37 2 100 15 9 29 31.58 2 97 15 Computation. h m s T. by Chro. 9 25 0.37 O's dec. Eq'n oft. oil/ II ma e Chro. cor. — 42 21 +21 57 40.3 + 20.92 +2 36.35 —0.362 ( 167.36 ( 2.896 G. m. t. May 31 8 42 33 +3 2.2 \ 14.64 —3.15 \ .253 12.5 ( 8.709 +22 42.5 ( I 9 +2 33.20 ( '3 o / // ©50 7 30 filn.cor. u u -16 16.3 + 18 - 3 ref - 5LG S. diam. h— 49 51 13.7 [ +15' 48.4 par. +5.4 L— SO 20 30 1. sec 0.0639749 h m 8 p= 67 59 17.5 1. cosec 0.0328703 L. ap. t. May 31 2 56 12.04 2s = 148 11 1.2 — Eq. of t. —2 33.20 5= 74 5 30.6 1. cos 9.4379032 L. m. t. May 31 2 53 38.84 s-h= 24 14 16.9 1. sin 9.6133431 T. by Chro. 9 25 0.37 ~ ~~ 19.1480915 0,Chro.cor.(L.m.t.)- 6 31 21.53 \ t— 22 130.3 1. sin 9.5740458 t= 44 3 0.6 THE CHRONOMETER. 175 T. by Chro. 9 29 31.58 O '« dec. Eq'n of t Chro. cor. —42 27 +-22 42.5 +20.92 +2 33.20 — 0.352 G. m. t. May31 8 47 5 inO\076 +1.6 * — .03 8.785 +22 44.1 + 2 33.17 48 37 30 (i In. cor. „ „ ~ +15 17.8 J ,. +18-3 ref.-o-t.o I S. diam. h m g h— 48 52 47.8 L +15' 48.4 par. +5.6 L. ap. t. May 31 3 42.95 L — 30 20 30 1. sec 0.0639749 — Eq. of t. —2 33.17 p = 67 59 15.9 1. cosec 0.0328717 L. m. t. May 31 2 58 9.78 2s =147 12 33.7 T. by Chro. 9 29 31.88 s = 73 36 16.8 1. cos 9.4506544 0, Chro. cor.(L.Di.t.) -6 31 21.80 s-h = 24 43 29.0 L sin 9.6214453 19.1689463 (mean) —6 31 21.66 \t — 22 35 22.1 1. sin 9.5844732 Red for —3' . I +.48 t = 45 10 44.2 Chro.cor.(L.ni.t.)- 6 31 21.18, May SI O h . May 31 h Chro. cor. (L. m. t.) — 6 h 31 m 22 s .24 by A.M. and P.M. obs. Long. +5 49 1 .4 May 31 6 h Chro. cor. (G. m. t.) — 42 20 .84 3. Table of Chro. Corrections. Chro. 1876 ; fast of Greenwich mean time and gaining. Remarks. O , A.M. Key West Light-House. ©,A.M. " " " " , A.M. & P.M. Pensacola Navy-Yard. ©,A.M. &P.M. " " " G. m. t. Chro. cor. Daily Ch. h ' h m s 1865, May 1 3 — 40 20.5 -4.14 17 3 41 26.8 3.88 25 6 41 58.3 3.75 31 6 42 20.8 Long* of Key West Light-House, 81° 48' 40" W. Long, oi Pensacola Navy- Yard, 87 15 21 W. * The assumed longitudes of places, where the chronometer is rated, should be stated. 176 NAVIGATION. 4. Comparisons and Corrections of Chronometers. 1865, May 31, 6 h , G. mean time. Chro. 4375 Chro. 9163 Chro. 789 Cliro. 5165 hms hms hms hms Chro. 6 50 16.3 5 3 29.7 2 15 27.5 11 59 16.8 (1876) 6 30 6 31 6 32 10 6 33 30 (1876)— Chro. —0 20 16.3 +127 30.3 +4 16 42.5 —5 25 46.8 Cor. of (1876) —42 20.8 —42 20.8 —42 20.8 —42 20.8 Chro. cor. — 1 2 37.1 —0 45 9.5 +3 34 21.7 —6 18 7.6 or +5 41 52.4 175. 3d Method. (By equal altitudes.) A heavenly body, which does not change its declination, is at the same altitude east and west of the meridian at the same interval of time from its meridian passage. If, then, such equal altitudes are observed and the times noted by the chronometer, or by a watch and reduced to the chronometer (Art. 170), the mean of these times, or the middle time, is the chronometer time of the star's meridian transit. The corresponding sidereal time is the star's right ascen- sion, when the first observation is east of the meridian ; 12 h -f the right ascension when the first observation is west of the meridian. This, for a mean time chronometer, may be converted into local mean time (Prob. 32) ; and for a Greenwich chrono- meter into the corresponding Greenwich time. Subtracting the chronometer time, we have the correction of the chronometer. Example. 1865, Jan. 14, at Washington, in longitude 77° 2' 48' W., equal altitudes of a Canis Minoris were observed, and the times noted by a chronometer regulated to Greenwich mean time ; from which were obtained : THE CHRONOMETER. 177 Mean of chro. times ( % east) 2 h 16 n 35 3 .65 " "' " " (#- west) 7 59 16 .38 Chro. time of >fc's transit 5 7 56 .01 L. sid. t.= %'s R. A. 7 32 16 .26 Long. -ho 8 11 .2 G. sid. t. 12 40 27.46 ■JS (Jan. 14)- -19 35 51 11 Sid. int. from Jan. 14 h 17 4 36 .35 Red. to m. t. int. — 2 47 .86 G. mean time Jan. 14 . 17 1 48 .49 Chro. time 17 7 56 .01 Chro. cor. — 6 7 .52 176. If equal altitudes of the sun are observed in the fore- noon and afternoon of the same day, the mean of the noted times would be the chronometer time of apparent noon, were it not for the change of the sun's declination between the observations. Problem 50. In equal altitudes of the sun, to find the correction of the middle time for the change of the surfs declination in the interval between the observations. Solution. Let h = the sun's true altitude at each observation, t — half the elapsed apparent time between the observa- tions, y o — the mean of the chronometer times of the two ob- servations, or the middle chronometer time, AT = the correction of this mean to reduce to the chrono- meter time of apparent noon ; L = the latitude of the place, d = the sun's declination at local apparent noon, A d = the change of this declination in the time t ; then, when both observations are on the same day, t + A T Q will be numerically the hour-angle at the A. M. ob- servation, t—AT 01 the hour-angle at the P. M. observation, 178 NAVIGATION. d— A d) the declination* at the A. M. observation, d + A d, the declination* at the P. M. observation. By (116), Ave have for the two observations, sin A=sin L sin (d— A d)-\-cos L cos (d—Ad) cos (t+ AT ) ) / lq _\ sinh=smLsm(d+ Ad) + cos L cos (d+ Ad) cos (t—' A T Q ) ) ^ ' But sin (d ± A d) = sin d cos A d do cos d sin A d, cos (d ±A d) = cos J cos A d =F sin c? sin J J, cos (d ±AT )= cos £ cos A T =F sin £ sin J T^. Since J c?, and therefore ^ ^, are very small, we may put cos A d = 1, sin A d = A d. sin 1", cos A T = 1, sin J 7;= 15 AT . sin l" ; A d being expressed in seconds of arc, and AT Q in seconds of time ; we shall then have sin (d ± Ad) = sin d ± A d. sin V cos c?, cos (d ±A d) = cos d =F A d. sin l' 7 sin c?, cos ( £ ± zf T^) = cos £=F 15 AT . sin l" sin £. Substituting these in the two equations (135), subtracting the first from the second, and dividing by 2 sin 1", we shall have =:A d. sin L cos d — Ad. cos L sin d cos £ + 15 J T . cos i cos c? sin t. Transposing and dividing by the coefficient of AT^ we find the formula AT ,_ A d. tan L Ad. tan d . , Jio -~T5"7Tn"i" + ~T5la~nT' < 136 ' which is called the equation of equal altitudes. Let A h d = the hourly change of declination at the instant of apparent noon, and express £, which is half the elapsed apparent time, in hours, * Strictly, in the one case, A d should be the change of declination in the time t+ AT Q ; in the other, the change in the time t — AT Q . THE CHROMOMETER. 179 then A d = A h d. t, and (136) becomes A rn d h d. t tan L A h d. t tan d /-, oH \ 15 sin f T 15tan« v J If we put *=Tk7^1 ( 138 ) 15 sin £ 15 tan £ and » (7 = the chronometer time of apparent noon, we have A T — A. A h d. tan L + B. J h c?. tan J ) , _ * In these formulas, L and d are + when north, A d and zl h d are + when the sun is moving toward the north. The coefficient A is — , since t < 12 h , " " i? is + when t < 6 h , — when £ > 6 h . The computation of the .two parts of A T is facilitated by tables of log A and log B. Such tables are given in Chau- venet's " Method of finding the error and rate of a chrono- meter," in the American Ephemeris and Nautical Almanac for 1856, and reprinted in a pamphlet with his "New method of correcting Lunar distances." The argument of these tables is 2 t, or the elapsed time. The signs of A and B are given. Apply the two parts of A T , according to their signs, to the Middle Chronometer Time ; the result is the Chronome- ter Time of Apparent Noon. > Apply to this the equation of time (cidding, when the equation of time is additive, to mean time ; otherwise sub- tracting) ; the result is the Chronometer Time of Mean JVoon at the place. Applying to this the longitude (in time), subtracting if west, adding if east, gives the Chronometer Time of Mean JS T oon at Greenwich. 180 NAVIGATION. *12 h — Chro. T. at local Mean Noon, will be the Chro. corvee- lion if the chronometer is regulated to local time. *12 h — Chro. T. at Greenwich Mean Noon, will be the Chro. correction, if the chronometer is regulated to Greenwich time. 177. If a set of altitudes is observed in the afternoon of one day, and a set of equal altitudes in the forenoon of the next day, the middle time would correspond nearly to the instant of apparent midnight ; and half the elapsed time t, would be nearly the hour-angle from the lower branch of the meridian, or the supplement of the proper hour-angle. In this case 1 80° — (t + A T ) will be the hour-angle at the P.M. observation. 180°-(t-AT { j) " " " " " " A.M. " d— A d, the declination at the P.M. " d+Ad, " " " " A.M. " and we have for the two observations, as in (135) sin h=:sm L sin (d — A d)— cos L cos (d — A d) cos (t + AT ) ) , , ± v sin h=smL sin (d + A d) — cos L cos (d-\-A d) cos (t — A T ) ) ^ * ' Treating these in the same way as (135) we shall have = A d. sin L cos d+ A d. cos L sin d cos t — 15 A T cos L cos d sin t ; whence * This is better noted as h . \ These may be written — sin h= — sin L sin (d — A d)+cos L cos (d — A d) cos (t-\- A T Q ) — sin h = — sin L sin (d-\- J d)-hcos L cos (d -f- J d) cos {t—A T^). They differ from (130) in the signs of h and L and in reckoning the hour- angles from the lower, instead of the upper, branch of the meridian. This would be the case, if we suppose the observations to be referred to the lati- tude and meridian of the antipode. The only effect in (130) is to change the sign of tan Z, or of the first term in the equation of equal altitudes. THE CHRONOMETER. - 181 A d. tan L A d. tan d A T = 15 sin t 15 tan t or, putting as before A d = A h d. t . a=- *:■ ' b = 15 sin t 15 tan t A T ——A. J h d. tan Z + JB. A h d. tan Red.for Ah, 0«.055 x 0.5 = + .03 1st part of Eq. —2.89 0's dec— 21° 54' 43" J h <2 + 22\78 ch. in Id + 1 ".065 2d " " M —1.98 + 2 5 +.24 .213 Chro. t. of ap. noon 6 IT 3T.99 —21 52 33 +23.02 21 — Eq. of time — 8 2.41 1/ = +24°33 / .3 t. tan 9.6593 d =— 21° 52'. 6 1. tan 9.6037 m- Chro. t. of m. noon 6 9 35.58 J h 15 cos L sin Z which is a minimum, when Z = ±90°, and incalculable when Z= or 180°. f From (118) we find At ——- 15 cos L tan Z which is o, when Z= ±90°, and also incalculable when Z— or 180°. 194 NAVIGATION. forward to the time of observation from a preceding, or car- ried back from a subsequent determination. It may be very largely in error, especially in uncertain currents, or after running several days without observations. A small error may also result from the assumption that the mean of the instrumental readings corresponds to the mean of the noted times. The reduction of the mean of the altitudes to the mean of the times can be found,* but it can be avoided by limiting the series of observations, which are combined together, to so brief a period, that the error becomes insensible; or, when the body is near the meridian in azimuth, by reducing each observation by it- self. This last case, however, should be avoided in this problem. 198. At sea, it is usual to reduce longitudes obtained from day observations to noon by allowing for the run of the ship in the interval, and for currents when known. Those from night observations are recorded for the time of ob- servation ; or reduced to the commencement or end of the watch. 199. 2d Method. Altitudes in the forenoon and in the afternoon, or on different sides of the meridian, are prefer- able to single altitudes for finding the local time, for the reasons already stated in Article 174. The longitudes can be found from each set separately and then combined. At sea the longitudes derived from each can be reduced to noon, and the mean of the two taken as the true longi- tude ; or, if the difference can be regarded as due to cur- rents, the longitude at noon can be found by interpolating for the elapsed time. It is desirable that the observations should be made at nearly equal intervals from noon. Longitudes by A. M. and P. M. observations are enjoined Chauvenet's Astronomy, Vol. I., p. 214. LONGITUDE. 195 in the directions of the Navy Department whenever prac- ticable. Examples. (Prob. 51.) 1. At sea, May 17, 9 h 45 m A. M. ; 24° 50' K, 82° 18' W. by reckoning from preceding noon ; T. by Watch 9 h 30 m 15 s ; obs'd altitude of Q_ 58° 17'; Chro. — Watch + 5 h 12 m 26 s ; Chro. cor. + 25 m 15 s ; Index cor. of sextant -f- 3' 20"; height of eye 18 feet; re- quired the longitude. T. by W. 9 30 15 © ' 5 dec - E( l- °f *■ J o / // // m s s C—W +5 12 26 +19 23 40 +3^5 —3 51.1 +0.067 Chro. cor. +25 15 +1 45 j 10O5 +.2 ( ^20 G. m. t. May 17 3 7 56 = 3 d .13 + 19 25 25 I 4.4 3 50-9 ( 1 — Eq. of t. +3 51 G. ap. t. May 17 3 11 47 L. ap. t. May 16 21 45 46 Long. +5 26 1 1 / // O 58 17 \ In. cor.+ / // / // 3 20 dip. —4 3 + 14 36 ( S.diam. + 15 50 ref.&p.-31 h= 58 31 36 L = 24 50 1. sec 0.04214 p — 70 34 35 1. cosec 0.02545 2s= 153 56 11 s= 76 58 6 1. cos 9.35315 — h = 18 26 30 1. sin 9.50015 18.92089 81° 30' W. 1. sin -J t. 9.46045 May 17, noon, lat. by mer. alt. of O, 25° 8' N". ; run of the ship from 9^ A. M., E. N". E. (true) 18 miles. For E. N. E. 18', I = 6'.9 N"., p = 16'.6 E., D = 18'.4 E. At the time of the A. M. observations, then, the latitude carried back from noon was 25° 1' N". Using this in the computation of the time, we find the L. ap. t. May 16, 2ih 45m 29 s , and the long 81° 2 9 J' W. Applying D = 18'.4 E., we have for the longitude, May 17, noon, 81° 11' W., from observations at 9.45 A. JI. A 196 NAVIGATION. By P. M. observations, and reduced to noon, the longitude was found to be, May 17, noon, 80° 44' W. from observations at 3.45 P. M. As the position is in the Gulf Stream, where there is a strong easterly current, the difference of the two longitudes is attributed to that cause. We take, then, as the longi- tude at noon, 81° 11'- 2.2x27' = 81° l'-W. 2. At sea, 1865, Sept. 5, 4£ A. M., lat. 20° 16' S., long. 74° 20' W. T. by Chro. 10 h 36 ra 25^ ; Chro. cor. (G. m. t.) — l h 16 m 10 s ; obs'd alt. JL, 20° 16' 0", W. of meridian; Index, cor. + 2' 20"; height of eye 15 feet; required the longitude. h T. by Chro. 12 Chro. cor. h ra a + 10 36 25 — 1 16 10 D'si2.A h m s 22 40 41.9 + 2.370 ]) 's dec. o i n // —4 39 5 + 11.62 G-. m. t. Sept. 4 21 20 15 + 10 54 28.3 + 48.0 < 22 41 29.9*1 |47.4 t - 6 + 355J 232 —4 35 10 ( 3 Red. for G. m. t + 3 30.3 o / // J. 20 16 | [ In. cor. + 2 20 H. par. 60 39 G. sid. t. $>'si2.JL 8 18 13.6 22 41 29.9 + 15 14" h'= 20 31 4 ' Dip —3 49 [S.diam. + 16 33 +10" D 's G. h. ang. + 9 36 44 + 54 19 Par. and ref. h— 20 25 23 L— 20 16 1. sec 0.02776 p= 85 24 50 1. cosec 0.00139 2 5=126 6 13 8= 63 3 7 1. cos 9.65627 s -h— 42 37 44 1. sin 9.83074 19.51616 D's L. h. ang. +4 39 37 1. sin \ 9.75808 Long. +-4 57 7 or 74° 17' W. Note. — The examples under Problem 30 can be adapted to this by regard- ing the chronometer correction given, instead of the longitude. 200. 3d Method (Littrow's. By double altitudes of the sarne body.) LONGITUDE. 197 When two altitudes of a body have been observed, and the times noted by the chronometer or watch, the hour- angles and local times can be found from each separately ; and thence the longitude for each. But we may also com- bine them and find the hour-angle for the middle instant be- tween them. Problem 52. From two altitudes of a heavenly body, sup- posing the declination to be the same for both, to find the mean of the tioo hour-angles, the latitude of the place and the Greenwich time being given. Solution. Take the mean of the two noted times and re- duce it to Greenwich mean time ; and find for it the declina- tion of the body. Reduce the observed altitudes to true altitudes. Let h and h' be the two altitudes, T&nd T r , the corresponding hour-angles ; then we have, by (116), sin h = sin L sin d + cos L cos d cos T, sin h'= sin L sin d + cos L cos d cos T' / and by subtracting the first from the second, sin h ! — sin h = cos L cos d (cos T'— cos T). By PL Trig. (106) and (108), this reduces to sin i (h'— h) cos i (h' + h)=—cos L cos d sin J (T + T) sin -J (T'—T) ; whence sin i (T'+ T) = - S ^- %7 h) T < : 0Si f' +h ] - * x ' sin \ (T — T) cos L cos d Put H Q = \ (A' + A), the mean of the two altitudes, T = i (T+T), the middle hour-angle, t = (T'—T), the difference of the two hour-angles ; and we have . m sin \ (h—h r ) „ _ - sin T == . \ + — cos H sec L sec d. (146) sm \ t u v ' 198 NAVIGATION. t, for the sun is the elapsed apparent time ; for a star, the elapsed sidereal time ; and for the moon or a planet, the elapsed sidereal time — the increase of right ascension in the interval ; and can be found from the difference of the two chronometer times. Then, by (137), T can be found, and, as any other local hour-angle, subtracted from the corresponding Greenwich hour-angle, which in this case is to be derived from the mean of the noted times. T is + or — according as the second altitude is less or greater than the first ; so that it is on the same side of the meridian as the body at the time of its less altitude. , If (h — h!) is very small (146) becomes approximately sin To = Ha-aOrini 'coag. sin iK feoa I j ros a .■* v ' (148) sin \ t cos L cos d If T is very small, m _ sin \ Qi — li' ) cos H 15 sin 1" sin J t cos L cos d y and if both are small, T _ i(h — h')cosH 15 sin i t cos L cos d * (149) 201. To estimate the effect of small errors in the data of the problem, let A (Ji—h') be a small increment of the difference of the two altitudes, A H Q , of the mean of the altitudes, A L, of the assumed latitude, A t, of the elapsed time, and A T„ the corresponding change of the middle hour-angle ; the last two expressed in seconds of time, the rest in minutes of arc : LONGITUDE. 199 then, we have the formula,* Z, ^°-Ltani(A-^) 5 2tani*' cot ^ + cot ZJ tan V' li °° j in which each term may be computed separately for any supposed value of its numerator. The possible error in any case, on the suppositions made, would be the numerical sum of the several terms. As h—h\ however, is the change of altitude in the inter- val £, we may attribute all the error to h—h\ and regard t as exact ; or we may attribute all the error to t and regard h—hl as exact: so that one only of the first two terms in the second member of (150) is needed. When h—ti is so small that we may put cos-J- (h—h!) =1, we may use instead of (150) " T. = .\ J t {h - A ' ):C T ff \ , (151) sin -J- £ cos L cos a cos TJ v ' which is preferable to (150) and requires for computation four of the same logarithms as (147). The effects of errors are evidently least when T == ; that is, in the case of equal altitudes each side of the meri- dian. They increase rapidly as T increases; so that the method is especially adapted for altitudes nearly equal on both sides of the meridian, or for circum-meridian altitudes. But in the latter case, especially, a high altitude is neces- sary; for from (151) it appears that the effect of error in * Changing (146) into a logarithmic form, we have, 1. sin T = 1. sin -J- (A — h') — 1. sin -J- t + 1. cos H Q — 1. cos L — 1. cos d ; differentiating each term except the last, and reducing, we obtain, d T Q d(A — /*/) dt _dH Q dZ tan T 2 tan i (h — h') 2 tan \ t cot B Q cot L' This is reduced to (150) by multiplying A (h — h' ), A H^ and A L by 4, to reduce minutes of arc to seconds of time. 200 NAVIGATION. the difference of altitudes, h—h\ is least either when S is very near 90°, or t near 12 h ; so that, if t is small, H Q should be quite large. 202. The method presents no special advantages for ob- servations on shore, except in the case of two nearly equal altitudes of a fixed star on opposite sides of the meridian. In the case of the sun and planets, it is necessary to take the change of declination into consideration, to obtain pre- cise results. The special case for which the method provides is at sea, within the tropics, when the sun passes the meridian at a high altitude. In that case, when by reason of clouds ob- servations near noon only can be made, or it is desired to obtain the longitude as near noon as practicable, let a pair of altitudes, or several pairs, be measured and the times noted with all tire precision practicable. The altitudes should be reduced to true altitudes, and one of each pair for the run of the ship in the interval* by the method given in Prob. 58, and in Bowd., p. 183. From each pair the middle apparent time can be found by (146), and the mean of these times subtracted from the mean of the Greenwich apparent times for the longitude. If the altitude changes uniformly with the time, or nearly so, the mean of several altitudes observed in quick succes- sion can be taken for a single altitude, If the observations have been made with care, the errors of instrument, refraction, and dip will affect the two alti- tudes of each pair nearly alike ; and if the reduction for the run of the ship is carefully made, the difference of altitudes in comparison with the difference of times will be nearly exact. 203. This method was proposed by M. Littrow, Director * This may be avoided, if the course of the ship is at right angles to the bearing of the sun. LONGITUDE. 201 of the Vienna Observatory, and has been successfully used by Admiral Wullerstorf, of the Austrian navy, in 1857 and 1 858. It is highly commended by M. Faye in a full discus- sion of it in the Comptes Rendus of the French Academy, March 7, 1864. It should be used cautiously, and the errors to which the result is liable in any case carefully estimated. Altitudes greater than 80° and an interval of more than half an hour are recommended, but an intelligent navigator can readily determine by (150), when he can safely depart from these limits. This will be especially the case when the altitudes are on both sides of the meridian. Examples. 1. 1865, May 16, 11|- A. M., in lat. 25° 15' ST., long. 56° 20' W., by account; the ship running K". E. (true) 8 knots an hour. T. by Chro., 2 h 32 m 23 s Q's true alt., 81° 1' 0", " " " 2 53 11 " " " 83 40 30; Chronometer correction on G. mean time + 40 m 51 s ; re- quired the longitude. The distance sailed in the interval is 2'. 8. The sun's azimuth at the 1st observation is found to be N. 131° E., which differs 86° from the course. The reduction of the 1st altitude to the place of the 2d is (Prob. 58) 2'.8 x cos 86° — + 0'.2 = + 12". J st chro. t. 2 32 23 O's dec. Eq. of t 2d " " Elapsed chro. t. Mid. " " Chro. cor. 2 53 11 t - 20 48 2 42 47 + 40 51 O / " II + 19 10 4 +34.3 + 1 57 ( 103. + 19 12 1 1 14. o / // m s s —3 52.41 + 0.05 +.17 -3 52.2 G. m. t. May 16 3 23 38 h — 81 1 12 — Eq. of t. + 3 52 h' — 83 40 30 G. ap. t. 3 27 30 i (A— h')=- r 1 19 39 1. sin 8.3649 n L. ap. t. 23 41 44 H Q = 82 21 1. cos 9.1242 Long, at 2d obs. I + 3 45 46 56°26'.2W. L= 25 15 d— 19 12 1. sec 0.0436 1. sec 0.0249 Red. for h .l 0.6 E. t= h 20 m 48 s l.cosec-J-2 1.3433 Long, at noon 56 27 W. T Q =-0 18 16 1. sin 8.9009 n 202 NAVIGATION. By (150), if A (h-N)bz+l\ A 7; = +6».9; if AB = + 1\ A 7 7 = +2s.4; if JZ = + 10', A T Q = + 1K5. 2. At sea, 1865, June 29, lat. at noon by mer. alt. of O, 33° 25' N., long, by account 147° 10' E. ; near 11 A.M. T. by Chro., 1*55™ 54° ) , fltruealt . M o 2r 30 , ; " 1 P. M. " " " 3 45 ) Chro. cor. on G. m. t. — 36 m 28 s ; the ship run from 1st observation, to noon, N". 3 pts. W. 9'.9 ) " noon to 2d observation, N. 2 " W. 7 .2 ; ) required the longitude at noon. N. 3 W. N. 2 W. 9'.9 7.2, 8'.2 N. 6.6 5'.5 W. 2.8 JA=6'.6W. 3.4 N. 30° W. 17.0 14.8 8.3 10.0 The sun's azimuth was found to be N. 127° E. at the 1st observation ; N. 127° W. at the 2d observation. The difference of N. 30° W. and N. 127° E. is 157° ; the difference of S. 30° E. and N. 127° W, is 83°. It will be better, therefore, to reduce the second altitude to the position of the first. By Prob. 58, (or Bowd, p. 183,) this reduction is I7'.0xcos 83° = + 2'.1. The latitude at the time of the 1st observation was 33° 16'.8 N. h m s A. M. chro. t. 13 55 54 O's dec. Eq. of t. P. M. M " 15 45 +23° 16' 59"—- 7".3 +2 m 55 9 .6+0 8 .507 Elapsed " " t— 1 49 6 — 1 45 +7.2 ( 7 .10 + 23 15 14 +3 2 .8 ( .11 Mid. " " 14 50 27 — 36 28 Chro. cor. h = 74° 21' 30" 1. cosec j t 0.627 Mid.G.m.t. June 28 14 13 59 h' — 74 23 36 log T l T 8.824 — Eq. of t. __ 3 3 %(h—h')= —1 3 log 1.799 n Mid. G. ap. t. 14 10 56 H Q - 74 22 33 1. cos 9.430 " L. ap. t. 23 59 54 L = 33 17 1. sec 0.078 Long, at 1st obs ( —9 48 58 '( 147° 14'.5 E. d = 23 15 1. sec 0.037 T = — 6 9 .3 log 0.795* Red. to noon 6 .6 W, Long, at noon 147 7 .9 E. LONGITUDE. 203 By (151) if A (h—h ! ) = 1', A T = 3 s .l. It would require a change of l£° in i, or of 2-|° in jET , for either to change T one second of time. The accuracy of the result, there- fore, depends upon the accuracy with which the difference of altitudes has been found ; that is, in this case, mainly upon the course and distance made good. 204. 4th method. (By equal altitudes.) Let equal altitudes of a heavenly body be observed east and west of the meri- dian (Art. 175) and the times noted as in other observa- tions ; and the mean of the watch-times in each set, if a watch is used, reduced to chronometer time. If both sets have been observed at the same place, and the declination of the body has not changed, the mean of the two times will be the chronometer time of its meridian transit. If the declination has changed in the interval, as is ordi- narily the case with the sun, moon, or a planet, the correc- tion for such change, found by the methods of Problem 51, should be applied. Applying then the chronometer correction, we have the corresponding Greenwich time, which will be mean or side- real as the time to which the chronometer is regulated. Finding from this, by the method in Art. 189, the Green- wich hour-angle of the body (which in the case of the sun is the Greenwich apparent time), we have the longitude, if the first observation was east of the meridian, as the cor- responding local hour-angle is then 0. But if the first ob- servation was west of the meridian, the local hour-angle is 12 h and must be subtracted. This method should be used on shore, when practicable, in preference to either of the preceding. 205. Equal altitudes of the sun can be conveniently used at sea w T hen the sun passes the meridian near the zenith ; that is, when its declination and the latitude are nearly the same. Altitudes very near noon are then available for find- ing the time (Art. 196), and equal altitudes can be observed 201 NAViGvrio^. with only a short interval. In the example of Art. 196, an interval of eight minutes would have been sufficient. If the ship does not change her position in the interval, the middle time corresponds to apparent noon ; as the change of declination may be neglected, unless the interval between the observations is so great as to require it. 206. If the longitude only has changed, the middle time corresponds to apparent noon at the middle meridian, and will give the longitude of that meridian. This will be the longitude at noon, if the speed of the ship has been uniform. But if it has not. subtracting half the change of longitude, when the true course is icest, or adding it when the course is east, will give the longitude of the place where the first altitude w T as observed. This can then be reduced to noon by allowing for the run of the ship. If the change of longitude is west, the sun arrives at the corresponding altitude of the afternoon later than it would do if observed at the same place as in the forenoon ; if the change is east, it arrives earlier ; and the difference is the time of the sun's passing from the one meridian to the other, that is, the difference of longitude expressed in time. If, then, 2 t is the elapsed apparent time, A A, the change of longitude (+ when west), the hour-angle of the sun at each observation is t — \ A X ; and (137) becomes a rp_ A h d. t tan L A h d. t tan d °~~ 15 sin (t—i JA) + 15 tan (t~i A X)' ^ Z ' But even when the elapsed time is so great that it is thought necessary to correct for the change of declination, A X is never large enough to produce a change of I s . If the latitude only has changed, the middle time requires correction for such a change, which can be deduced in a similar way to that for a change of declination in Prob. 51. But, as in the fundamental formula, LONGITUDE. 205 sin h = sin JL sin d + cos L cos d cos t, L and c? enter with the same functions, they are interchange- able. If, then, J h X is the hourly change of latitude (+ toward the north and expressed in seconds), and A ' T^ the required correction, we have from (137) and (139), d'T =- A f- tUm f d + *£?*;* (153) lo sin t lo tan t v ' and A'T Q —A A h Z. tan d+B J h Z. tan X, (154) for which Chauvenet's tables can be used. If both latitude and longitude have changed, for t in the denominators of (153), we may substitute t — \ AX\ but this at sea is a needless refinement. The restriction of this method to a short interval between the observations, depends upon the uncertainty of the run of the ship and consequent imperfect determination of ^ h i, the mean hourly change of latitude in the interval. If its error is supposed to be - A h L, the consequent error in A' T is - A r T Q . When equal altitudes near noon are practicable, a merid- ian altitude of the sun can ordinarily be taken for latitude, so that L will be sufficiently exact. Moreover, the latitude and longitude are both found for noon. Examples. 1. At sea, 1865, March 17, noon, rat. by mer. alt. of the sun 3° 16' S., long, by account 84° 58' W. ; equal altitudes of the sun were observed at 5 h 34 m 18 s and 6 h 3 m 24 s G. mean time ; the ship running S. S. E. (true) 10 knots an hour ; required the longitude. For S. S. E., 10', J h i = - 9'.2, J h X = — 3'.8 206 NAVIGATION. h 1st G. m. t. March 17 5 34 18 0'* dec. Eq'n oft. 2dG. m. t. 6 3 24 — 1° 13' 10" + 59".25 +8 m 27 9 .5-0 9 .736 Elapsed time . 29 6 + 5 44 j 296. — 4 .3 ( 3 .7 Mid. G. m. t. March 17 5 48 51 - 1 7 26 I 48. + 8 23.2 ( -6 — Eq. oft. —8 23 j h i = - 552" log 2.742 w log 2.742n Mid. G. ap. t. March 17 5 40 28 L=— 3° 16' l.tan 8.756 n Red. for A L +5 rf = -l Utan 8.290 n j G. ap. t. of noon 5 40 33 • log A 9.406 n log£9.405 or long. 85° 8'W. { — 2 S .7 log 0.438 n log 0.903 ( — 2 S .7 log CM ( +8.0 In this example the sun's azimuth was 120°, and in l m the altitude changed 13'. An inequality of 30" in the altitudes would therefore affect the result only T \ of l m , or l s .2. An error of 1' in the hourly change of latitude would affect the result — , or S .6. 2. At sea, 1865, June 16, lat. at noon by mer. alt. of O, 22° 50' 1ST., long, by account 35° 59' W. ; equal altitudes of the sun were observed at 2 h 16 m 18 s and 2 h 31 m 42 s G. mean time ; the ship running S. (true) 14' an hour. O's dec. +23° 22'.4 Elapsed time h 15 m 24 8 Mid. G. m. t. June 16 2 24 — Eq. of t. —22.5 Mid. G. ap. t. 2 23 37.5 Red. for A L + 2.5 G. ap. time of noon 2 23 40 Long. 35° 45' W. The sun's azimuth was 72°; the change of altitude in l m was 13 '.2, so that an inequality of l' in the altitudes would affect the result ^\ of l m , or 2 S .3. An error of 1' in A h L would 2 s 5 affect the result —, or s . 2. 14: ' 3. At sea, 1865, June 29, h ; lat. by mer. alt. of 0, 33° 25' N., long, by account 147° 10' E. ; LONGITUDE. 207 Chro. cor. on G. m. t. — 36 m 28 s ; In. cor. of sex't +0' 50"; height of eye 18 feet. The ship run from 11 A. M. to noon N. 3 p'ts W. 11' from noon to 3 P. M. N. 2 " W. 8' required the longitude at noon. For N. 3 W. 11' jZ = +9'.l JX = + 7'.4 N. 2 W. 8 A£ = +1A zU = +3.7 whence A L = + 8 .25 = 495" A. M. chro. t. + 12 h h m a 13 55 54 Q'sdec. Eq'n of t. P. M. chro. t. 15 45 +23° 16' 59"-7".3 + 2 m 55 s .6+0 9 .51 Elapsed time 1 49 6 —1 45 + 7 .2 ( 7.14 + 2 2.8 ( .11 Mid. chro. t. 14 50 27 +23 15 14 Chro. cor. (G. m. t.) — 36 28 Mid. G. m. t. June 28 14 13 59 log J h L 2.695 log J h L 2.695 — Eq. of t. — 3 3 log A 9.410^1 log B 9.398 Mid. G. ap. t. 14 10 56 1. tan d 9.633 1. tan L 9.819 Red. for A L + 27 log 1.738 n log 1.912 G. ap. t. of noon 14 11 23 -54 s .7+81 s . 7= + 27 3 Middle long. \ ( or -9 48 37 147° 9'.2 E. Red. to noon 1.8 W. Long, at noon 147 7 .4 E. The sun's azimuth was 127° ; for A t = l m , A h = 10", and an inequality of 1' in the altitudes will effect the result ^ of l ra , or 3 s . An error of 1' in A h Z will affect the result 27* These observations reduced as single altitudes, give, as the longitude at noon, 147° 7'.8 E. ; reduced by Littrow's method (Ex. 2, p. 202), 147° 7'.9 E. 207. 5th Method (By transits.) Observe the transits of the sun or a star across the threads of a well-adjusted transit instrument, noting the times. Re- 208 NAVIGATION. duce the mean of the noted times for semi-diameter and errors of the instrument as in Art. 184 ; and thence find the Greenwich hour-angle of the body in the way described in Art. 189. This will be the longitude, if the upper culmina- tion has been observed, as the local hour-angle is 0. If the lower culmination has been observed, the local hour-angle is This method can be used only on shore. Example. 1865, May 17, I7 h 16 m 20 8 .5 G. mean time, the meridian transit of a Bootis (Arcturus) was observed ; required the longitude of the place of observation. G. mean time May 17 11 h 16 ra 20 8 .5 £» 3 40 47.20 Red. for G. m. t, + 2 50.24 G. sid. t. 20 59 57.94 *'sR.A. 14 9 32.81 * 's H. angle or Long. + 6 50 25.1 or 102° 36' 17" W. LONGITUDE. LUNAR DISTANCES. 208. Problem 53. To find the longitude by the distance of the moon from some other celestial object. Solutio7i. If we have given the local mean time and the true distance of the moon from some celestial object as seen from the centre of the earth, we may find, by interpolating the Nautical Almanac lunar distances (Prob. 28), the Green- wich mean time corresponding to this distance. The differ- ence of this from the local time is the longitude. The local time may be found for the instant of observation, either from an altitude of a celestial object observed at the same time, or by a chronometer regulated to the local time. At sea the correction of the chronometer on local time can LONGITUDE. — LUNAR DISTANCES. 209 be found from altitudes observed near the time of measuring the lunar distance, and reduced for the change of longitude in the interval by the formula (Art. 167), c r — c + AX, A X being in time and + when the change is west. In practice, the apparent distance of the moon's bright limb from the sun or a star is observed, and the true distance derived by calculation, as in the next problem. 209. Problem 54. Given the apparent distance of the moorts bright limb from a star, the centre of a planet, or the surfs nearest limb, to find the true distance of the moorfs centre from the star, or the centre of the planet or the sun. Solution. It is necessary that the altitudes of the two bodies should be known, either directly from observations at the same time, or from observations before and after, and interpolated to the time of observation (Bowd., p. 246) ; or computed from the local time (Prob. 38), (Bowd., pp. 247, &c). The Greenwich time is also supposed to be known ap- proximately, either from the local time and approximate longitude, or, as is preferable, from the time noted by a Greenwich chronometer. A complete record of the observations will include the ap- proximate latitude and longitude of the place, the local time and chronometer correction, the index corrections of the in- struments used, the height of the barometer and thermome- ter, and at sea, the height of the eye above the water, as well as the noted times of observation and the observed dis- tances and altitudes. Several observations may be made at brief intervals, and the means taken. 210. The preparation of the data embraces : 1. Finding the Greenwich mean time approximately from the chronometer time, or from the local time. 2. Taking from the Almanac for this time the semi-diame- 210 NAVIGATION. ter and horizontal parallax of the moon, and of the other body* when they are of sensible magnitude ; adding to the moon's semi-diameter its augmentation. (Art. 60.) At low altitudes the contractions produced by refractions should be subtracted from the semi-diameters of the sun and moon. Formulas for finding these are given in Art. 213. When the spheroidal form of the earth is taken into con- sideration, to the moon's equatorial horizontal parallax (Art. 57), as taken from the Almanac, should be added the aug- mentation to reduce to the latitude of the place, which is found in Tab. III. of Chauvenet's Method. The declinations of the two bodies to the nearest decree are reauired from the Almanac for this purpose. 3. Applying to the observed distance the index correction of the instrument, and, when the sun is used, adding the moon's augmented semi-diameter and the sun's semi-diame- ter ; when a planet or star is used, adding the moon's aug- mented semi-diameter if its nearest limb is observed, but subtracting it if the farthest limb is observed. 4. Applying to the observed altitude of each body the index correction, dip, and semi-diameter (when necessary), so as to find the apparent altitude of its centre. If the true altitude is computed, the parallax must be subtracted and the refraction added. In the following direct method it is necessary also to find the true altitudes. 211. To find the true distance, let D = the apparent distance of the centres, D'= the approximate true distance, h — the apparent altitude 1 n ^ , _, , A1 1 . _, y of S's centre, li = the true altitude ) _H= the apparent altitude ) r ^, , 1 ^ . ^ I of s centre, planet, or star. II— the true altitude J \ * The sun's horizontal parallax may be taken as 8".5. LONGITUDE.— LUNAR DISTANCES. 211 In Fig. 35, let m and S be the apparent places of the moon and other body ; m 1 and S', their true places. The true and apparent places of each are on the same ver- tical circle, Z m, Z S respectively, since they differ only by refraction and parallax, which act only in vertical circles, except so far as a small term of the moon's parallax is con- cerned, which will be subsequently considered. Z Then m S = -Z>, the apparent distance ; m r S'=jD', the true distance ; and in the triangle m Z S, h Uei Fig. 35. m S = JD Z m = 90°— h Y being given, ZS = 90°--Zn to find the angle Z, we have by Sph. Trig. (32),* 2 cos j Qi + #+ D) cos j (ft + E-L) 4 cos h cos i/ Then in the triangle m' Z S', Z m'= 90°- A' and Z S'= 90°-J7 ; being given, m! S' may be found by Sph. Trig. (I7),f sin 2 i D f = cos 2 1 {h' + H f ) - cos A' cos .fl 7 cos 2 | Z, or by substituting the value of cos 2 £ Z, and putting 5 = 1 (A + JST+Z>), (155) sin 8 \ D'=cos 2 \ (h'+ H') - cog h cosH cos s cos (s-D). To adapt this for logarithmic computation put . 21 cos h' cos H' . _. , ^ . sin f m = — — 7 ^ cos 5 cos (s— x>), (156) ^ cos A cos R v y ' v J cos 2 -J ^L _ sin -J- (a+6+c) sin J (6+c— a) sin 6 sin c f sin 2 i a = sin 2 •£■ (6-f-c) — sin 6 sin , and for the computation logarithms to four places ; as q and Q are required only within 30'. The contractions, A s and A jS, of the vertical semi-diame- 214 NAVIGATION. ters may eacli be found from the refraction table, by taking the difference of refractions for the limb and centre. Then, for the required corrections, we have the formulas,* A's = As cos 2 q, A f $ = AjS cos* Q. (161) This contraction for either body is less than l", if the alti- tude is greater than 40°. For a very low altitude, it is best to subtract it from the semi-diameter m the preparation of the data, so that D will be corrected for it. But, unless quite large, it will suffice to compute it subsequently, and subtract it from D' when the nearest limb is used, or add it to D' when the farthest limb is used. 214. Let A D = the reduction of the apparent distance to the true, or I) '= D + A B. A great variety of methods have been given for finding A Z>, requiring 4 or, at the most, 5-place logarithms ; but also needing special tables. Four such methods are con- tained in Bowditch's Navigator. They generally neglect to take into account the spheroidal form of the earth, the cor- rection of refraction for the barometer and thermometer, and the contraction of the semi-diameters of the sun and moon. These together, at very low altitudes and in extreme cases, may produce an error of 3 m in the calculated Greenwich time, and do actually, in the average of cases, produce errors from 10 s to l m . Prof. Chauvenet has given in the American Ephemeris for 1855,f a new form to the problem, with convenient tables, by which all these corrections are readily introduced. It is but little longer than the other approximative methods, in which they are neglected. * Chauvenet's Astronomy, Yol. I., p. 186. \ Reprinted in a pamphlet with his method of equal altitudes. LONGITUDE. — LUNAR DISTANCES. 215 215. The moon's mean change of longitude is 13°. 17640 in a clay (HerschePs Ast., p. 222), or 33" in l m of time. An error, then, of 33" in the distance will, in the average, produce an error of l m in the Greenwich time, or 15' in the longitude ; or an error of 10" in the distance will produce an error of about 20 s in the Greenwich time, or 5' in the longitude. We may, however, readily find the effect of an error of 1", and thence any number of seconds, in the distance, by taking the number corresponding in a table of common log- arithms to the " Prop. Log. of Diff." in the Almanac ; for this prop. log. is simply the logarithm of the change of time in seconds for a change of l" in the distance, (p. 95.) 216. Errors of observation are diminished by making a number of measurements of the distance. But even with a skilful observer a single set of distances is liable to a possi- ble error of 10" or even 20". Errors of the instrument are diminished by combining re- sults from distances of different magnitudes, especially from those measured on opposite sides of the moon. This cannot usually be done with longitudes at sea, but may be with de- terminations of the chronometer correction. The error pe- culiar to the observer, that is, in making the contacts always too close, or always too open, is not eliminated in this way, but will remain as a constant error of his results. The accuracy of the reductions of the observed to the true distance, depends more upon the precision w^ith which the differences of the apparent and true altitudes — that is, the parallax and refraction — have been introduced, than upon the accuracy of the altitudes themselves. 217. Lunar distances are used at the present day, not so much for finding the longitude, as for finding the Greenwich mean time, with which to compare the chronometer. They may thus serve as checks upon it, which in protracted 216 NAVIGATION. voyages may be much needed. If the chronometer cor- rection thus determined agrees with that derived from the original correction and rate, the chronometer has run well, and its rate is confirmed ; if otherwise, more or less doubt is thrown upon the chronometer, according to the degree of accuracy of the lunar observation itself. If the discordance is not more than 20 s , it is well still to trust the chronometer, as the best observed single set of distances may give a result in error to that extent. If it is large, then by repeated measurements of lunar distances, differing in magnitude, and especially on both sides of the moon, and carefully re- duced, the chronometer correction can be found quite satis- factorily. By taking the rate into consideration, observa- tions running through a number of days can be combined. Example. At sea, 1855, Sept. 7, about 6 h A. M., in lat. 35° 30' 1ST., long. 30° W. by account ; Time by chro. 8 h 29 m 57 9 .5 ; app. chro. cor. (G.m.t.) — 21 m l s .5 ; Observed distance of © and S> 43° 52' 30", index cor. — 20"; Observed altitude of JD 49° 31 ' 50", index cor. + 1' 0" ; Observed altitude of O 5° 21' 10", index cor. 0"; Bar. 29.10 inches ; ther. 75° ; height of eye 20 feet ; Required from these observations the chronometer correction on Greenwich time. Preparation. h m s / // // T.bychro. 12 h +8 29 57.5 D'sH.par. 54 19.4 D'sS.diani. 14 50.0 Chro. cor. -21 1.5 Aug. +3.6 Aug. +11.2 G. m. t. Sept. 6 20 8 56 J) >s Aug. H. par. 54 23.0 J) 's Aug. S. diam. 15 1.2 O's H. par. 8 ".5 0's S. diam. 15' 55". 1 V. cont. -21 .6 V. S. diam. 15 33 .5 _£ 49° 31' 50" 0. 5° 27' 10' Iu. cor. + 1 o In. cor. Dip —4 23 Dip -4 23 Aug. S. diam. + 15 1 V. S. diam. +15 34 LONGITUDE. — LUNAR DISTANCES. 217 H = 49° 43" 28' h = 5° 38" 21' Ref. —46 Ref. -8 12 Par. +35 10 Par. +8 #'=50 17 52 h'= 5 30 17 Obs'd dist, J) O 43° 52' 10' il= 49° 43' 3>'s Aug. S. diam. +15 1.2 h= 5 38 1. sec 0.002 _ *, ° . « «i D = 44 22 1. cosec 0.155 s S ' dlam ^" 15 5oJ 2 s = 99 43 Cont. A's = —21.6 s = 49 52 1. cos 9.809 D = 44 22 45 &-H= 9 1. sin 1418 Cont. Q's S. diam. (Ml). 17.384 2 log cos a 9.999 q = 2_J9 1. sin J 8.692 lo | 2 1.6 1.334 log A's 1.333 Computation of True Distance. (155-158) O I II O I II #=49 43 28 LsecjBT 0.1894554 h = 5 38 21 1. sec h 0.0021069 D = 44 22 45 #'= 50 17 52 1. cos H' 9.8053633 2 s = 99 44 34 h' = 5 30 17 L cos h' 9.9979925 s = 49 52 17 1. cos s 9.8092266 s — D= 5 29 32 o i a 1. cos (s-D) 9.998001 7 Compression. (159) 4(^+^=27 54 4.5 19.8021464 O'sdec :=+ 6°.3 L sin 9.040 * W= 52 46 39 ' 4 "-* 1 *" 1 9 ' 9Q1Q732 Z>' 1. cosec 0.150 s'= 80 40 44 «== 0.155 log 9.190 m=105 33 19 L cos s' 9.2094277 D'sdec.= +25°.3 1. sin 9.631 m— s'= 24 52 35 1. cos (m-s') 9.9577114 D' 1. cot 9.999 o t u 19.1671391 n'= .427 log ^630 j J)' = 22 32*23.? 1. sin 9.5835696 n-n' = - .272 log 9.435* .p = ^ ^^ Aiz log 1.363 n L 1. sin 9.764 Cor. f or Com P- ~ 4 Comp.— 3". 6 log 0.562 n D"— 45 4 43 true distance. Finding the Greenwich mean time and chronometer cor- rection. (82) True distance D"= 45° 4' 43" Distance at 18\ D = 46 3 17 P. L. 0.3433 Diff. +5 J)"— D Q = 58 34 log 3.5458 tz= 2 h 9 m 6 8 log 3.8891 218 NAVIGATION. G. m. t. of D 18 h m s Red. for 2d diff. -2 G. mean time, Sept. 6 20 9 4 T. by chro. 20 29 57 Chro. cor. — 20 53 by lunar. — 21 1 by previous cor. and rate. Difference + 8 This example is taken from the pamphlet of Prof. Chauvenet, where it is reduced by his method with far less labor of computation. The true distance by that method is 45° 4' 45" ; by Bowditch's 1st method, in which the small corrections are omitted, it is 45° 5' 44", differing very nearly 1' from the correct value. This would produce an error of 2 m 10 s in the Greenwich time. 218. Other lunar methods for finding the longitude, be- side that of lunar distances, are — 1. By moon culminations, or observing the meridian transits of the moon and several selected stars near its path, whose right ascensions are considered well determined. 2. By occidtations, or noting the instant that a star dis- appears by being eclipsed by the moon, or that it reappears from behind the moon. The first is called an immersion, the second an emersion. 3. By altitudes of the moon near the prime vertical. 4. By azimuths of the moon and stars observed near the meridian. These methods, except occasionally the second, are avail- able only on shore. They require good instruments, careful observations and determinations of the instrument correc- tions, and scrupulous exactness in the reductions, especially those which involve the moon's parallax. By each may be found the moon's right ascension, and thence, by inverse interpolation in the Almanac, the corre- sponding Greenwich mean time. Subtracting from it the local mean time, which must also be found from good ob- servations, gives the longitude. 219. If corresponding observations are made at two dif- LONGITUDE. — LUNARS. 219 ferent places, their difference of longitude can be found with much less dependence on the accuracy of the Ephemeris. When the two local times of the occupation of the same star have been noted, they can each be reduced to the in- stant of the geocentric conjunction of the moon's centre and the star in right ascension ; and the difference of the reduced times will be the longitude. By the other methods, the change of the right ascension of the moon, in passing from one meridian to the other, may be found. This, divided by the mean change in a unit of time, as l h or l m , computed from the Ephemeris, will give the difference of longitude in the same unit. CHAPTER IX. SUMNER'S METHOD: LATITUDE AND LONGITUDE BY DOUBLE ALTITUDES. CIRCLES OF EQUAL ALTITUDE. (SUMNER'S METHOD.) 219. Suppose that at a given in- stant the sun, or any other heavenly body, is in the zenith of the place M (Fig. 3 7) , on the earth ; arid let A A 7 A" be a small circle described from M as a pole. The zenith distance of the body will be the same at all places on this small circle, namely, the arc MA; for if the representation is transferred to the celestial sphere, or projected on the celestial sphere from the centre as the projecting point, M will be the place of the sun, or other body, and the circle A A' A" will pass through the zeniths of all places on the terrestrial circle, and M A, M A', &c, will be equal zenith distances. The altitude of the body will also be the same at all places on the terrestrial circle AAA"; hence such a circle is called a circle of equal altitude. It is evident that this circle will be smaller the greater the altitude of the body. 220. The latitude of M is equal to the declination of the body, and its longitude is the Greenwich hour-angle of the Fig. 37. CIRCLES OF EQUAL ALTITUDES. 221 body ; which, in the c&se of the sun, is the Greenwich appar- ent time, or 24 h — that apparent time, according as the time is less or greater than 12 h . This is evident from the dia- gram, in which, regarded as on the celestial sphere, P M is the celestial meridian of the place, whose zenith is M, and its co-latitude ; and also the declination circle, and co-declination, of the body M ; and if P G is the celestial meridian of Greenwich, 6PM is, at the same time, the longitude of the place, and the Green- wich hour-angle of the body. If, then, the Greenwich time is known, the position of M may be found and marked on an artificial globe. 221. If, moreover, the altitude of the body is measured, and a small circle is described on the globe about M as a pole, with the complement of the altitude as the polar radius, the position of the observer will be at some point of this cir- cle. His position, then, is just as well determined as if he knew his latitude alone, or his longitude alone ; since a know- ledge of only one of these elements simply determines his position to be on a particular circle, without fixing upon any point of that circle. As, however, he may be presumed to know his latitude and longitude approximately, he will know that his position is within a limited portion of this circle. Such portion only he need consider. It is commonly called a line of posi- tion.* 222. The direction of this line at any point is at right angles with the direction of the body ; for the polar radius M A is perpendicular to the circle A A' A" at A, A', A", and every other point of the circle. 223. Artificial globes are constructed on so small a scale that the projection of a circle of equal altitude on a globe * Inappropriately termed a line of bearing. 222 NAVIGATION. would give only a rough determination. But the projection of a limited portion maybe made upon a chart by finding as many points of the curve as may be necessary, and, having plotted them upon the chart, tracing the curve through them. The portion required is usually so limited that, when the altitude of the body is not very great, it may be regarded as a straight line ; and hence two points suffice. With high altitudes, three points, or if the body is very near the zenith, four may be necessary, and even the entire circle may be required. 224. Problem 55. From an altitude of a heavenly body to find the line of position of the observer ', the Greenwich time of the observation being known. Solution. From the given altitude, and assumed latitudes X n X 2 , X 3 , &c, differing but little from the supposed lati- tude, find the corresponding local times (Prob. 43), and thence, by the Greenwich time, the longitudes A x , A 2 , ^35 <&c. Thus we shall have the several points, whose positions are conveniently designated as (X 19 a x ,), (Z 2 , a 2 ,), (£ 3 , A 3 ,), &c. It facilitates the computation to assume latitudes differing 10' or 20', as the % sums and remainders differ 5' or 10', and only one of each need be written. Or, from the Greenwich time and assumed longitudes, ^1? ^2> ^3? &c., find the corresponding local times (Art. 77), and thence the hour-angles of the body (Probs. 34, 35). With these and the observed altitude, find the corresponding lati- tudes , i n i 2 , i 3 , &c. (Prob. 46). This is more convenient than the preceding method, when the body is near the meridian. In either mode the computation may be arranged so that the like quantities in the several sets shall be in the same line, and taken out at the same opening of the tables. The several points may then be plotted on a chart, each by its latitude and longitude, and a line traced through CIRCLES OF EQUAL ALTITUDES. 223 them, which, will be the required line of position. Two points connected by a straight line are sufficient, unless the altitude is very great, or the points widely distant. Thus in (Fig. 38), let A and B be two BjL_L such points plotted respectively on the parallels of latitude L 15 L 2 , and each in ,// its proper longitude ; A B is the line of // position, and the place of observation is ~r4r I " at some point of A B, or A B produced. Fig. 38. This is all which can be determined from an observed alti- tude, unless either the latitude, or the longitude, is definitely known. And as these are both uncertain at sea, except at the time when found directly by observation, the position of the ship found from a single altitude, or set of altitudes, is a line, of greater or less extent as the latitude, or the longitude, is more or less accurately known. In uncertain currents, or when no observations have been had for several days, the extent of this line may be very great. Yet, if it is parallel to the coast, it assures the na- vigator of his distance from land ; if directed toward some point of the coast, it gives the bearing of that point. 225. If there is uncertainty in the altitude, for instance of 3', the line of position having been computed and plotted, parallels to it on each side may be drawn at the distance of 3'. So, also, if there is uncertainty in the Greenwich time, parallels may be drawn at a distance in longitude equal to the amount of uncertainty. In either case, the position of the ship is within the in- closed belt. In Fig. 38, a b is such a parallel to the line of position A B, its perpendicular distance from it measuring a differ- ence of altitude; the distance A a on a parallel of latitude measuring a difference of longitude. 224: NAVIGATION. 226. Since the line of position is at right angles with the direction of the body (Art. 222), the nearer the body is to the meridian in azimuth, the more nearly the line of position coincides with a parallel of latitude ; and thus a position of the body near the meridian is favorable for finding the lati- tude from an observed altitude, and not the longitude. So also, the nearer the body is to the prime vertical, the more nearly the line of position coincides w^ith a meridian, and the less does any error in the assumed latitude affect the longitude computed from an observed altitude. So that, if the body is on the prime vertical, a very large error in the assumed latitude will not sensibly affect the result. Such a position of the body is, then, the most favorable for finding the longitude from an observed altitude. These conclusions have been previously stated, drawn from analytical considerations. 227. Two or more points of a line of position as (L Y A x ), (X 2 , A 2 ) etc., having been determined by Prob. 55, if the true latitude, X, be subsequently found, -p t> the corresponding longitude, A, may be obtained by interpolation. Or, the place of the ship may be found graphically upon the chart, by drawing a parallel in the latitude, i, -^ E Gr and taking its intersection P, with the line of position A B. So also, if the true longitude, /I, is subsequently found, the corresponding latitude, i, may be obtained by interpo- lation ; or, a meridian E F may be drawn in the longitude, A, which will intersect the line of position in P, the place of the ship. If there is uncertainty in either of these elements, two parallels of latitude (as in Fig. 38), or two meridians, may be drawn at a distance apart equal to the uncertainty. CIRCLES OF EQUAL ALTITUDE. 225 As altitudes, latitudes, and longitudes are never found at sea with much precision, and may under unfavorable circum- stances be largely in error, the position of the ship on the chart is not properly a point, but a belt, more or less limited according to the accuracy of the elements from which it has been formed. 228. In Fig. 39, if A is the position (i 19 A x ), B, the position (i 2 , A 3 ), both near P, the true position, whose latitude is i, and longitude is X ; the right triangles* A C B, A E P, being formed, CB = L 2 —L^ the difference of the two latitudes, AC = X 2 —X^ the difference of t?ie corresponding longi- tudes, E P = A L = L— i 15 the correction of X x , AE = JA = /l— A 1? the correction of X x ; then CB:EP = AC : AE or, (Z 2 — Z L ) : A L — (X 2 — A x ) : A A, whence we have, '*-«■£=£ } (162) and X — A x + A X ) as the formulas for finding A, the longitude of the true posi- tion, when its latitude, Z, is known. Or, we have ** = "tt\ (163) and L — L^AL ) as the formulas for finding Z, when X is given. They are the same formulas as for an interpolation. The several differences are most conveniently expressed in minutes of arc, or, in the case of longitudes, in seconds of time. The * This is different from the projection on a Mereator's chart, where G B and E P would be augmented differences of latitude. 226 NAVIGATION. local times may be used instead of the longitudes and in- terpolated in the same way. From the first of (162) we may readily determine how much a supposed error in an assumed latitude affects the resulting local time, or longitude. 229. Problem 56. To find from a line of position the azimuth of the body observed. Solution. We have the positions (X n A x ), (X 2 ^ ^2)5 or the latitudes and longitudes of two points, from which the azi- muth, or course of the line of position, can be found by middle latitude sailing. Adding or subtracting 90°, according as the azimuth of the body is greater or less, gives the azimuth required. Or, a perpendicular to the line of position may be drawn upon the chart, and the angle which it makes with a meri- dian may be measured with a protractor. The azimuth may thus be found to the nearest 1°. Example. At sea, 1 865, Nov. 23, 10 J A. M., by account in lat. 36° 50'N"., long. 65° 20' W. ; Greenwich mean time 2 h 40 m 47 s ; the sun's correct central altitude 29° 6' 25" ; to find the line of position. G. m. t. Nov. 23 2 L 40 m 47 s =2\68 's dec. Eq'n of t Eq. of t. + 13 18 —20° 25' 33"— 30".9 — 13 m 20 8 .4 + s .72 G. ap. t. 2 54 5 (61 .8 + 1 .9 ( 1 .4 —13 18 5 ( -5 -1 23 -J 18 ,1 — 20 26 56 ( 2 A 1. With assumed Latitudes. (Prob. 43.) h r= 29° 6' 25" L x = 36° 30', Z 2 = 36° 50', Z3 = 37° 10', £1 = 36 30 1. sec 0.09482 .09670 .09861 p— 110 26 56 1. cosec 0.02827 .02827 .02827 2s= 176 3 21 £-88 140 1. cos 8.53674 .49841 .45636 s-h — 58 55 15 1. sin 9.93271 .93346 .93422 G. ap. t. 2 h 54 m 5* 18.59254 .55684 .51746 CIRCLES OF EQUAL ALTITUDE. 227 f (1) 22 28 37 1. sin 9.29677 .27842 .25873 L. ap. t - I (2) 22 32 27 ( (3) 22 36 22 f Ai = 4 25 28 = 66°22'.0 W. — 57'.5 %z = 4 21 38 = 65 24.5 — 58'.8 Xs = 4 17 43 = 64 25.7 L x - 36° 30' N/ U — 36 50 Xg= 37 10 r For JZ = +40', AX—— 116'.3 ; or a change of 40' in latitude produces a change of —116' in longitude. From A L — + 40', J A = — 116'.3, we find, by middle lati- tude sailing, the dep. 93'.0, and then the bearing of the line of position, regarded as a rhumb line, which it nearly is, NT. 66°.7 E. ; the sun's azimuth therefore is ST. 156°.7 E. Suppose the correct latitude to be 36° 57' X., to find the corresponding longitude on the line of position, we have ; 36° 57' N. A-3 /u2 Lz — L2 Z 2 — 36° 50' N. h = 65° 24'.5 W. AL- +7' A% = — 2'.9x7= — 20'.6 58'.8 20 = -2\94 I =a 65° 4' W. 2. W%A assumed Longitudes. (Prob. 46.) G. ap. time 2 h 54 m 5 9 &, = 4 19 3* — 411 21 m 3 A 3 = 4 h 23 m 9 L. ap. time * = — 1 24 55 - 1 26 55 — 1 28 55 1. sec t 0.03052 .03201 .03354 d-_20° 26' 56" 1. tan d 9.57155 n .57155 .57l55?i 1. tanf 9.60207 w .60356?i .60509 71 f =— 21° 48'.1 -21° 52'.2 -21°56'.4 1. sin 9.56984 n .57114*1 .57244 ?? 1. cosec <# 0.45671 n .45671 n .45671 n h = 29° 6' 25" 1. sin A, 9.68703 .68703 .68703 1. cos 6' 9.71358 .71488 .71618 f=+ 58° 51'.7 + 58° 45'.5 + 58° 89'.2 S U = 37° 3'.6 IS T . Zs = 36° 53'.3 N. U = 36° 42'.8 K } 2,i = 64 45 W. A» = 65 15 W. A 3 = 65 45 W. 228 NAVIGATION". For A X = + 60', A Z = — 20'.8. From these, the bearing of the line of position is N. 66°.5 E. If the correct longitude is 65° 4' W., to find the corre- sponding longitude on the line of position, we have -k = 65° 4' W. Z 2 h = 64° 45' W. Z, = 37° 3'.6 N. Al= +19' JL=— 0\34xl9 = — 6'.5 Zi 10'.3 AT3^ = -ij X\ — A' 2 , or the difference of longitude on the first parallel is the greater. In this case m > 1, A Z > Z 2 —Z l and A X > X f 2 —X\. The point P is then, as in Fig. 40, in the same direction as B from A, and beyond B. But m will be — , when X\—X\ and X\—X' 2 have the same name, and X\—X" l < X\—X\, or the difference of longitude on the first parallel, is the less. A Z and A X will then have different names respectively from Z, 2 —Z l and X r 2 — X\. In this case P and B are in opposite directions from A. A negative value of m may be avoid- ed, so that P and B will fall always on the same side of A, or P and D always on the same side of C, (Fig. 40), if toe take as Z x the latitude of the parallel on ichich is the greatest difference of longitude. If the differences of longitude, X\—X\, X ,r 2 —X' 2 , on the two parallels have different names, their sum is taken numerically in finding m / in that case m is -f and less than ], Ji< Z 2 —Z L and A X < X r 2 — X\, with the same names CIRCLES OF EQUAL ALTITUDE. 231 respectively; and, as in Fig. 41, P is between A and B, and between C and D. When three or more latitudes are used in the computa- tions, those for which the differences of longitude are small- est should be taken as L x and X 2 . 232. The more nearly perpendicular the lines of position are to each other, the better is the determination of their intersection. Hence, the nearer the difference of azimuths of the body or bodies at the two observations is to 90°, the better is the determination of position from double altitudes. If the azimuths are the same, or differ 180°, the two lines of ]30sition coincide in direction, and there is no intersection. In this case the great circle joining the two bodies, or the two positions of the same body, is an azimuth circle, and passes through the zenith. An approach to this condition is generally to be avoided. (Bowd., pp. 181, 195, notes.) Still, however, if the two bodies, or positions of the same body, are near the meridian, the lines of position nearly coincide with a parallel of latitude. The latitude is then well determined, but not the longitude. If the two bodies, or positions of the same body, are near the prime vertical, the lines of position more nearly coincide with a meridian and the longitude is well determined, but not the latitude. When the difference of azimuths is small, the intersection of the two lines may be computed with tolerable accuracy, while it cannot be definitely found by the projection of the lines upon a chart. 233. The operations indicated in (164) are to subtract, 1. The first assumed latitude from the second, {JL^—L^) ; 2. for the first observation, the longitude corresponding to the first latitude from that corresponding to the second latitude, (A' 2 — A\) ; 232 NAVIGATION. 3. for each latitude, the longitude deduced from the first observation from the longitude deduced from the second, (A^ — X\) and (A" 2 — A' 2 ) ; 4. the difference of longitude for the second latitude from that from the first, \_{a\— X\) — (A" 2 — A' 2 )], (or add numerically these differences of longitude when they are of different names.) Then 5. To divide by this last result the difference of longi- tude, {h\— X\), for the first latitude, to obtain the coefficient m, (which will be — only when the dif- ference of longitude, (A /r 2 — A' 2 ), for the second lati- tude has the same name as and is greater than the difference of longitude, (A^ — A\), for the first lati- tude), 6. To multiply m by the difference, (X 2 — Xj), of the two assumed latitudes to obtain the correction of the first latitude L x \ and by the difference, A' 2 — A^), of the two longitudes derived from the first ob- servation, to obtain the correction of the first of these longitudes, X\. These corrections have the same name as the differences from which they are derived, when ni is + ; but contrary names when m is — ; and are to be applied accordingly. 234. The lines of position may be found from two as- sumed longitudes A L and A 2 , instead of two latitudes (Art. 224). The formulas for finding their intersection will differ from (164) only by an interchange of the letters L and A. We shall have, then, m {L\-L\)-(L\-L'^ AX — ni! (A. 2 — A x ), A = l x + A A A L = m' (i. 2 -X , 1 ), L = L\ + AL (165) CIRCLES OF EQUAL ALTITUDE. 233 Examples. 1. With Xi = 30° 28' N"., Z 2 = 30° 8' K Diff. 20' S. by 1st alt. X\ = 59 15 W., X 2 = 59 W. " 15 E. by 2d alt. W\ = 58 43 W., // 2 = 59 8 W. " 25 W. Differences, 32~ E. — 8 W. = 40 E. 32 m = — = .8 40 J Z = — 20' x .8 = — 16', L — 30° 28' — 16' = 30° 12' ]S T . A A=r — 15' x .8 = — 12', 1 = 59 15 — 12' = 59 3 W. The differences of longitude on the two parallels, 32' E. and 8' W., being in opposite directions, the intersection is between the two parallels, or L is between i x and X 2 . 2. With Li = 48° 10' S., L 2 = 48° 30' S. Diff. 20' S. by 1st alt. X'i = 88 16 E., a' 2 == 88 24 E. u 8 E. by 2d alt. ?."l = 88 30 E., 1\ = 88 55 E. " 25 E. Differences, 14 E., — 31 E. = 17 W. m = — • 4 = -82 17 AZ = ~ -.82x20'~ — 16', Z = 4S° 10' — 16' = 47° 54' S. A ?> = - ..82 x 8'=— 7', X = 88 16 — 7' = 88 9 E. In this example it is convenient to regard south latitudes and east longitudes as + . The differences of longitude on the two parallels, 14' E. and 31' E., being in the same direc- tion, the intersection is outside of the parallels and nearer the first, for which we have the smallest difference. 3. With A, = 165° 50' W., 2* - 186° 20' W. Diff. 30' W. by 1st alt. L\ — 36 16 S., X' 2 = 36 25 S. " 9 S. by 2d alt. L\ = 36 38 S., L\ — 36 29 S. " 9 N. Differences, 22 S, — "~T s. = 18 S. m' = 22 18 = 1.17 JZ=l.l7x 9' = + 10'.5, L— 36° 16' + 10'= 36° 26' S. A i— 1.17 x 30' = + 35'.1, \ = 165 50 + 35 y = 166 26 W. The differences of latitude on the two meridians, 22' S and 4' S., are in the same direction ; and the intersection is 234 NAVIGATION. outside of the meridians and nearer the second, on which the difference of latitude is least. 235. Problem 57 supposes the two altitudes observed at the same place. This at sea is rarely the case. Pkoblem 58. To reduce an observed altitude for a change of position of the observer. Solution. Let Z (Fig. 42) be the zenith of the place of observation ; h — 90° — Zm, the observed altitude; Z', the zenith of the new position ; h' = 90°— Z' m, the altitude reduced to the new position, Z f . d = Z Z', the distance of the two places, here referred to the celestial sphere ; C=FZ Z, the course ; Z = PZ m, the azimuth of m ; Z— C = m Z Z', the difference of the course and azimuth. Z Z', being small, may be regarded as a right line, Z 71 O as a plane right triangle, and O m, without material error, as equal to Z' m; so that we shall have Z O = Z Z' cos Z Z' m 71 m = Z m - Z O or, putting J /i = Z O, J A = c?cos(C r -Z)) , v A h = d cos (C—Z) is, then, the reduction of the ob- served altitude to the new position of the observer : it is additive when C—Z< 90° numerically ; subtractive when C—Z > 90°. (Bowd., p. 183.) It is smaller, and can, CIRCLES OF EQUAL ALTITUDE. 235 therefore, be more accurately computed the nearer C—Z approaches 90°. It is, therefore, better to reduce that alti- tude for which the difference of the course and azimuth is nearest 90°. If the second is the one reduced, then C is the opposite of the course. In practice Z Z' does not usually exceed 30', so that al- though an arc of a great circle of the celestial sphere, it may be regarded as representing the distance, c?, of the two places on the earth ; or, at sea, the distance run. The azi- muth, or bearing, of the body can be observed with a com- pass, or be computed to the nearest degree, or half-degree, from the altitude. The assumption, 71 m = O m, is more nearly correct, the greater Z' m or Z m, that is, the smaller the altitude. If we treat Z Z' m as a spherical triangle, d = Z Z' being ex- pressed in minutes and still very small, we shall find A h = d cos (C-Z)-i cP sin 1' tan h sin 2 {C—Z); (167) but the last term is inconsiderable unless d and h are both large. For instance, if d = 30', it will not exceed 1' unless h > 82°. Example. The two altitudes of the sun are 36° 16' 20", 58° 15' 20% the compass bearings of the sun respectively S. E. by E. -J E. and W. S. W. ; the ship's compass course, and distance made good in the interval N". N. W". \ W. 25 miles ; S. f| E. differs from N". 2\ W. 13 points, so that the re- duction of the 1st altitude to the position of the 2d is 25' X cos 13 pts. == — 25' cos 3 pts. = — 20'.8 = — 20' 48". S. 6 W. differs from S. 2| E. 8£ points, and the reduction of the 2d altitude to the position of the 1st is 236 NAVIGATION. 25' cos 8| pts. s= — 25' cos 7£ pts. = — 2' 30"; or — 2' 39", if the last term of (167) is included. 236. By (166) or (167) we may reduce one of the two alti- tudes for the change of the ship's position in the interval. But instead of this we may put down the line of position for each observation, and afterwards move one of them to a parallel position determined by the course and distance sailed in the interval. Thus in Fig. 43, let 2> A B be the line of position for the first observation, and A a represent in direction and length the course and distance sailed Fig 4a in the interval ; then a b, drawn parallel to A B, is the line of position which would have been found had the first altitude been observed at the place of the second. If the second observation is to be reduced to the place of the first, then A a in direction must be the opposite of the course. The perpendicular distance of A B and a b is the reduc- tion of the altitude for the change of position: for that dis- tance is A a X cos (B A a — 90°). LATITUDE BY TWO ALTITUDES. 237. Iii Sumner's method the latitude and longitude are both found by two altitudes, either of the same or different bodies. It is sometimes desirable to find the latitude only, or at least to make this the chief object of combining the two observations. 238. Problem 59. To find the latitude from tioo altitudes of the sun, or other body, supposing the declination to be the LATITUDE RY TWO ALTITUDES. 287 same at both observations, and the Greenwich time to be known approximately \ Solution. Let two altitudes, or sets of altitudes, be ob- served and the times noted by a chronometer, or a watch compared with it; reduce the altitudes to true altitudes, and at sea one of them for the change of the ship's position in the interval by Prob. 58. Find also the difference of the chronometer times of the two observations, and correct it for the rate in the interval. t Ac This correction is -^— (+ when the chronometer loses), t being the interval in hours of chro. time, and A c the daily change. The result is the elapsed mean time for a mean time chro- nometer ; the elapsed sidereal time for a sidereal chrono- meter. The Greenwich mean time of the greater altitude, or of the middle instant, should also be obtained from the chro- nometer times, sufficiently near for finding the declination of the body. In Fig. 44, let M and M' be the two positions of the body, h — 90° — Z M, the first altitude, ti= 90°— Z M', the second altitude, d = 90°- P M = 90° - P M', the common declination, and t = M P M', the difference of the hour-angles, Z P M and Z P M', of the body at the two observations : or, letting Tand T designate these hour-angles in the order of time, t = T f -T. 238 NAVIGATION. The method for finding t is different for different bodies. a. For a fixed star the angle M P M', or t, is the elapsed sidereal time. An elapsed mean time must therefore be re- duced to the equivalent sidereal interval. If, then, S and S' represent the sidereal times of the two observations, t m and s, the elapsed mean and sidereal times, we have, when the sidereal interval has been found, t=S*-S.= s; (168) when the mean time interval has been found, by (85), t = t m + . 00214 t m ; (169) or, with t m expressed in hours in the last term (87), t = t m + 9 S .8565 t m . (110) b. For the moon or a planet, if a and a -f A a represent the right ascensions of the body at the two times, we have (Art. Ill), T=S-a, T f =S'~ a -A a, and « = f-r=«-jc, (171) that is, the elapsed sidereal time diminished by the increase of the right ascension of the body in the interval. A h a, the change of right ascension in l h of mean time, may be obtained from the Almanac for the middle Green- wich time. The change in l h of sidereal time will be, by (86), (1— .00273) 4*j which can readily be found by regarding A h a as a sidereal interval, and reducing it to its equivalent mean time inter- val. Expressing t m and s in hours, when used as coefficients, Ave have Aa — t m . A h a = s. A h a (1— .00273) ; (172) and, for an elapsed sidereal time, t = 8—.s. A h a (1— .00273) ; (173) LATITUDE BY TWO ALTITUDES. 239 for an elapsed mean time, by (170), t = t m + t m (Qs.8565 -A h a). (174) c. For the sun, the angle MP IT, or t % is the elapsed ap- parent time. If, then, A h JS is the hourly change of the equa- tion of time (+ when the equation of time is additive to mean time and increasing, or sub tractive from mean time and decreasing ; that is, when apparent time is gaining on mean time), t = t m +t m . A h E, (175) by which t may be found from a mean time interval. If the sidereal interval is given, we have, as in (173), for a planet, t = s — s. A h a (1 — .00273). The reduction of the elapsed mean time to an apparent time interval, is commonly neglected by navigators; but on December 21, A h E= l s .25, and during a large part of the year exceeds s . 5. 239. We have given h = 90°- Z M, d = 90°- P M = 90°- P M', A'=90°-ZM', * = MPM', to find from the several triangles of Fig. 44, the latitude, Z=90°-PZ. Various solutions of this problem have been given, from which the following are selected. 240. (A.) 1st Method hi BowdiicNs Navigator. Let T (Fig. 44), be the middle point of M M' ; join P T and Z T ; and put 4 = MT = M'T=JM M', or half the distance of the two positions of the body, B — 90°— P T, the declination of T, j?=:90 o -Z T, the altitude of T, J = PTZ, the position angle of T. Fig. 44. 240 NAVIGATION. AsPTM and P T M' are equal right triangles, we have the angle PTM = PT W= 90°, 2 = 90°- Z T M = Z T M'— 90°, and J« = MPT = M'PT. 1. In the right triangle PTM, by Sph. Trig. (80), (84), and (82), sin A = cos d sin -J- Z, ) / ^ x sin B = sin d sec JLy ) or tan B == tan c? sec -J £/ 0-^) by which ^4 and .S, or M T and 90°— P T, can be found. 2. In the two triangles Z M T, Z M' T, by Sph. Trig. (4), sin h = sin H cos A + cos H sin A sin 97 sin h f = sin .Z? cos A— cos .Z? sin .4 sin q ; the half difference and half sum of which, by PI. Trig. (106) and (105), are sin \ {h—ti) cos \ (h-{-h f ) = cos J? sin A sin q, cos -J- (h—h') sin f (h + ti) = sin J? cos ^ty from which, COS J. v 7 sin a = sini(A-/Qcosi(A + ^) fl ^ ^ cos H sin J. ' v ' which determined and y, or 90°— Z T and the angle P T Z. 3. In the triangle PTZ, by Sph. Trig. (4), sin L = sin B sin II + cos B cos H cos #, To adapt this to computation by logarithms, put cos C sin Z= cos J? cos ^, | /180'i cos C cos Z = sin H y ) and then tan Z = cot iZ cos q, ) cos 6 7 = sin II sec Z, > (181) sin L = cos (7 sin (jS + Z), ) LATITUDE BY TWO ALTITUDES. 241 which determine (7, Z, and L. If, however, we add the squares of (180), we shall have cos 2 C = cos 2 H cos 2 q + sin 2 H, or 1 — sin 2 C — cos 2 H (1 — sin 2 q) + sin 2 H; whence sin (7= cos J? sin q. (182) Substituting this in (179), and the 2d of (180) in (178), we have . n sin \ (Ji — h') cos \ (h + h') Sill C — ; z « sin J. ~ cos i Qi—li') sin |- (h + h') COS ^ — ■: -pz , cos A cos 6 and (181) sin i = cos C sin {B + Z). (183) J£, being found from its cosine, may have two values numerically equal with contrary signs. Representing these, we have sin L = cos C sin (B±Z), which gives two values of X. The value which accords most nearly with the latitude by account may be taken. We shall see presently how the admissible value of Z may be selected. 241. To avoid using both the sine and cosecant of A and the cosine and secant of (7, we may take the reciprocals of (176) and the 2d of (183) ; we shall then have, as in the 1st method of Bowditch (p. 180), cosec A = sec d cosec* \ £, 1 cosec B = cosec d cos A, sin C = sin % (h—ti) cos $ (h + ti) cosec A, > (184) sec Z— sec \ (h—ti) cosec £ (h + ti) cos ^4 cos (7, sin L = cos C sin (J5±Z). J * log sec -J tf and log cosec -J ^ may be taken from Table XXVII. corre- sponding to t in the column P M (Art. 126). 242 NAVIGATION. It is unnecessary to find A and (7, as log cosine A can be taken from the tables corresponding to log cosecant A, and log cosine C corresponding to log sine C Indeed, we may dispense with A entirely by substituting the 1st of the pre- ceding equations in the 3d, and the 2d in the 4th, and em- ploying (177). We shall then have, for the solution of the problem,— tani?= :tan d sec* -J £, (185) sin C— sin \ (h—h f ) cos J (h + ti) sec d cosec* ^ £, sec Z=sec| {h—ti) cosec \ (]i + h!) sin d cosec B cos 6 r , sin i=cos G sin (B±Z). A, B, C, Z, and i, are each numerically less than 90°; A is in the 1st quadrant; C is + when the 1st altitude is the greater, — when it is the smaller ; B has the same sign, or name, as the declination ; and L the same as (B + Z) or (B— Z),from which it is obtained. 242. If Z O (Figs. 44 and 45), be drawn perpendicular to P T, we shall find from (182), C=±Z O, + when P T is west, — when it is east, of the meri- dian ; Z=T O, B + Z=90°-P O in Fig. 44, JB-Z=90°-P O in Fig. 45 ; Z, or T O, being -f or — according as P and Z are on the same side of M M', as in Fig. 44, or on opposite sides, as in Fig. 45. This may also be shown in an- other way : for, in the first case, the * log sec i t and log cosec i t may be taken from Table XXVII. corre- sponding to t in the column P M (Art. 126). LATITUDE BY TWO ALTITUDES. 243 angle q = P T Z < 90°, cos q is + ; and since, in (180), cos C and sin JI are -f , sin Z, and therefore Z, will also be + . In the second case, when M M' produced passes between P and Z, q = P Z Z > 90°, cos q and, consequently, sin Z and Z are — . Instead of marking Z + or — , we may use the symbols N" and S, as for c?, _Z>, and X. We shall then have the rule (Bowd,, p. 181) : — Mark Z north, or south, according as the zenith and north pole, or the zenith and south pole, are on the same side of the great circle, which joins the two positions of the body. By thus noting the position of this circle, the ambiguity of Z is removed. We may, however, remove the ambiguity by noting the azimuths of the two points M and M'. In Fig. 44, P Z M > P Z M'; in Fig. 45, P Z M < P Z M'; which would be the case also if one or both points were on the other side of the meridian. Hence we have the rule : — Z has the same name as the latitude when the azimuth of the body is numerically the greater at the greater altitude ; but a different name from the latitude when the azimuth at the greater altitude is the less. The azimuths are to be reckoned both east and west from to 180°, and from the N". point in north latitude ; but from the S. point in south latitude. 243. If Z is very small, it cannot be accurately found from its cosine, or secant ; its sign may be doubtful ; and the lati- tude cannot be determined with precision. This will be the case, when the altitudes are very great ; when M and M' are near the prime vertical ; or, in general, when M and M' are remote from the meridian and the difference of azimuths, 244 NAVIGATION. M Z M', is either very small, or near 1 80°. In each of these cases M M' intersects the meridian very near the zenith. It has been seen, with regard to lines of position derived from two altitudes (Art. 232), that the most favorable con- dition is when M Z M'=90° ; but that the latitude alone can be well determined when M and M' are quite near the meri- dian in azimuth and M Z M' quite small. Indeed, if both azimuths are 0, or 180°, the two altitudes become a meridian altitude. These conditions belong to all methods of finding the lati- tude from two altitudes. 244. The latitude having been found, we may proceed to find the hour-angle of the body from one of the altitudes (Prob. 43), if it is sufficiently near the prime vertical, and thence the longitude, if the times have been carefully noted by a Greenwich chronometer (Art. 188). Instead of this, by putting T = \ (T+T') = Z P T, the middle hour-angle, we have the formula (146), cos a cos L sin \ t v ' and from (185) • n — sni 2- (h—h') cos -J- (h + h') sin — j — : — 7— 1 x cos a sin -J- t whence S i n *^ = !^. (187) cosZ v y This could also have been obtained from the right triangle P O Z, (Figs. 44, 45) ; from which we have also 7DA tan Z tan ZrO — - — «a, sin P O 7 or * If we enter Table XXVII. (Bowd.) with log sin T 01 or log tan T„ we shall find 2 T Q corresponding in the P. M. column. LATITUDE BY TWO ALTITUDES. 245 Thus, by a brief additional computation,^ can be found by (186) or (187). We shall have also T=T Q -±t, T=T + it (189) for the hour-angles at the times of the observations. The longitude can be found from either T , T^ or T\ and the cor- responding chronometer time. (186) is the formula of Littrow's method* (Art. 200). The favorable conditions, as there stated, for finding 7J, are a small value of T and high altitudes near the meridian, or altitudes on each side of the meridian near the prime verti- cal. But such altitudes are unfavorable for finding the lati- tude. When both latitude and longitude are to be found from two altitudes, the nearer the difference of azimuths is to 90° the better will be the determination. The most favorable conditions for combining them will be equal azimuths of 45°, or 135°, on each side of the meridian. If one of the altitudes is very near the prime vertical, and the other very near the meridian, it will generally be better to find the time and longitude from the first by Prob. 43, and the latitude from the second by Prob. 46 or 47. 245. In this problem the declinations are supposed to be the same at both observations. This will be the case with the sun only at the solstices ; with a planet, or the moon, only when 90° from its node, and with the latter body for a very brief period. Navigators usually neglect the change of declination of the sun, or a planet, and use the mean de- clination, or that for the middle instant. It is better, how- ever, when the change is neglected, to employ the declina- tion at the time of the greater altitude,f except when the * The novelty of Littrow's method consists in finding T^ from very high altitudes near the meridian. (146), or (186), is by no means a new formula, f Chauvenet's Astronomy, Yol. L, pp. 276, 315. 246 NAVIGATION. hour-angle of this altitude is greater than the middle hour- angle. This can be the case only when the altitudes are on different sides of the meridian. When the middle declina- tion is used, we may, with little additional labor, find the correction of the computed latitude by the following formula from Chauvenet's Astronomy (Vol. I., p. 267). AZ = — ^: sin f, (190) cos L sin \ 1 v ' or, by substituting (187) JZ^-l^ZL (191) sin \ t v 7 in which A d is half the change of declination in the inter- val of the observations. Noting whether it is toward the north or the south, we can apply it with the same name to the computed latitude, when the lesser altitude was ob- served first / but with a different name when the lesser alti- tude was observed last. With this correction the preceding method can be em- ployed for altitudes of the moon at sea, when the elapsed time does not exceed l h . The correction of T^ the middle hour-angle, may also be found by the formula (Chauvenet's Astronomy, Vol. I., p. 268,) AT - Ad ( taU L C ° S To - i md ) • (1921 which differs from the equation of equal altitudes (136) only in the first term being multiplied by cos jT , and in chang- ing the signs, as it is here applied to the hour-angle instead of the chronometer time. (B.) Domoes^s Method ; JBowditcNs 2d Method: with an assumed latitude. 246. This method differs from the preceding in first find- ing T^ the middle hour-angle, by using an approximate lati- LATITUDE BY TWO ALTITUDES. 247 tude, and then the latitude from the greater altitude and its computed hour-angle, as in Problem 46 for finding the lati- tude from a single altitude. N Letting U == the assumed latitude, and as before / /^7l P T o =i (T r + T) = ZFT 1 /°/J (Fig. 46) the f ' z middle hour- \^g. angle. *==} (f-r)=MPir, the difference of 5. 3 the hour-angles, Fig. 46. we maj r use the formulas of Art. 244, sin c = sin \ (h — cos A')cosi-o+^r d sin ^ t sin T - •*■ sin cosZ (193) T= = T - ~i t, T< = T + ht and of Art. 149, (194) cos z = sin h + 2 cos rf cos J7 sin 2 £ T 7 cos z Q = sin A'— 2 cos c? cos i' sin 2 -J- 7"" selecting that which contains the greater altitude and less hour-angle. z = 90°— h is the meridian zenith distance from which the latitude may be found as from an observed meridian altitude. (Prob. 45.) Should this differ from the assumed latitude, the computa- tions should be repeated, using this new value. The method can be used to most advantage when T Q is ' small, and the greater altitude is observed near the meri- dian. But it has the inconvenience of requiring several re- computations, when the computed latitude differs widely from that assumed. If the observations are unfavorable for 248 NAVIGATION. finding the latitude, successive recomputations will approxi- mate very slowly, or may not approximate at all, to a con- clusive result. When the less altitude is near the prime ver- tical, it is preferable to find from it the greater hour-angle, and then the less by adding or subtracting t. 247. Mr. Douwes's formulas, however, are somewhat dif- ferent. From the triangles Z P M, Z P M' we have by Sph. Trig. (4) sin h = sin d sin L + cos d cos X cos jP, sin h' = sin d sin i + cos d cos L cos T f / the difference of which is sin h — sin h! =s cos d cos L (cos T— cos T r ). This by PI. Trig. (130) becomes sin A— sin h'= 2 cos d cos L sin f (T+ T 7 ) sin £ {T'-T) ; whence, since T = i (r+ r ) and i r = | (r- T 7 ), 2 sin 2^ = (sin ti — sin A) sec d sec i cosec % £, (195) which gives T 0J provided for L we use the approximate la- titude L'. • We then have as before T=T-it, r= T + ±t (196) and for finding the meridian zenith distance from the greater altitude, (116) cos z = sin h + cos i' cos d versin T 7 , ) or, j. (197) cos Zn = sin A' + cos U cos c? versin T", ) Mr. Douwes prepared special tables (Tab. XXIII., Bowd.,) to facilitate the use of formulas (195) and (197) calling log cosec \ t, log of " half elapsed time," log (2 sin T ), log of "middle time," log versin t, " log rising " ; LATITUDE BY TWO ALTITUDES. 249 increasing the indices of the last two by 5, so that natural sines, etc., to 5 decimal places may be treated as whole numbers. (193) and (194) are preferable, as they require only tables, which are in common use. 248. When the declination is taken for the middle time, we have for the correction of the computed latitude (191) A d. sin T AZ = sin 4 t (198) in which A d is the change of declination in half the inter- val of time between the observations. Example. At sea, 1865, April 16, A. M. and P. M., in lat. 37° 20' S., long. 12° 3' W., by account, two altitudes of the sun were observed for finding the latitude, the corrected times and altitudes being as follows : G. m. t. April 15 23 h 16 m 25 3 " " 16 1 IS 26 Mid.G.m.t. " — Eq. of t. Mid. G. ap. t. Elapsed m. t. Ch. of eq. of t. Elapsed ap. t. 16 17 25.5 + 16.4 17 42 O's true alt. 38° 16' 25" " 42 15 30, ^(h + li)- 40 15 57 £(£— h!)=z— 1 59 32 S. 15£pts.E. S. 161 W. 2 1 + 1 2 2 O's dec, d = + l0° 13' 57 ff Ch. in 1* +53.04 Ch. in l h .0l7 J d = + 53.7 Computation by (185), (187) and (191). 1. sec \ t \. tan d 1. tan B 0.01558 9.25651 9.27209 . .5 = + 10° 35' 52' Z-—4.1 45 25 B+Z -—37 9 33 1. cosec i t 0.58001 1. sec d 0.00696 1. sin d 9.24955 1. cos i(h+h') 9.88256 1. cosec \ (Ji+h') 0.18954 1. sin i(h+h') 8.54113 n 1. sec|(/i— A') 0.00026 1. sin C 9.01066 ?! 1. cos C 9.99771 1. cos C 9.99771 1. cosec B 0.73539 1. sin (B+Z) 9.78106 n 1. sec Z 0.17245 250 NAVIGATION. Xi =-36 55 52 1. sinZ 9.77877 t* AL— +26 mid. G.ap.t. = h l7 m 42 3 L= 36 55 26 S. T ^-0 29 28 Long. + 47 10 or 11° 47' 30" W. 1. sec L 1. sin C 1. sin 1\ 1. cosec -J- 1 log J d 0.09726 9.01066m 9.10792n 0.580 1.730 log(-JZ) 1.418 n IS The azimuth at the greater altitude being the greater, Z — , or S., like the latitude. Computation by (193), (194) an^ (191), with assumed lat. 37° 20' 8. t = 2 h 2 E1 2 s L , cosec \ 0.58001 i(h + h')= 40° 15' 57 1. , cos 9.88256 £(A_A')=- 1 59 32 1. sin 8.54113n log 2 0.30103 d = + !0 13 57 1. I. sec tan 0.00696 1. cos d 1. 2 cos d 9.99304 C 9.01066 n "O29407 U =-37 20 1. sec 0.09957 1. cosX' 9.90043 2;=— h 29 ra 37 s 1. sin 9.11023 n 21. s'miT' 7.67078 i*=+l 1 1 T' =+0 31 24 733 log 7.86528 h' - 42° 15' 30" sin .67248 z =-47 10 25 cos .67981 d= 10 13 57 L x =-36 56 28 2d Approximation. c 1. tan 9.01066 n 1. 2 cos d 0.29407 L x =-36° 56' 28" 1. sec 0.09732 1. cos L" 9.90268 T Q =— h 29* 1 28 s 1. sin 9.10798 n it = + 1 1 1 sin/i' .67248 2 1. sin i T 7.67488 T =+ 31 33 744 log 7.87163 z ft =-47 9 46 cos .67992 1. sin T 9.108 n which substituted in (202) gives h = i (h + K) + [i t *+T >] A h. (203) The difference of the two equations of (201) gives h-h' = (T*-T 2 ). A h = 2 T 1. A Q h. Hence Substituting this in (203) we have h = HM ) + (^J h + ^p^l. (205) The reduction to the meridian, then, is effected " by add- ing to the mean of the two altitudes two corrections; 1st, the quantity (| t) 2 . A A, which is nothing more than the common reduction to the meridian (120) computed with the half-elapsed time as the hour-angle; 2d, the square of one fourth the difference of the altitudes divided by the first correction." Several pairs of altitudes can be thus com- bined, and the mean of the meridian altitudes taken, from which the latitude can be obtained as from an observed me- ridian altitude. LATITUDE BY TWO ALTITUDES. 253 251. The restriction of the method corresponds with that of circum-meridian altitudes (Art. 150).* Quite accurate results can be obtained with hour-angles limited to 5 m when the altitude is 80°, to 25 m when the altitude is only 10°. If the interval t, however, exceed 10 m , AJi should be computed to two or three places of decimals, as it is given in Table XXXII. (Bowd.) only to the nearest 0".l. The accuracy of the method depends mainly upon the ac- curacy of the 2d correction, and therefore upon the preci- sion with which the difference of altitudes has been obtained. The altitudes, then, should be observed with great care. Errors of the tabulated dip and refraction, and a constant error of the instrument will affect both altitudes nearly alike. * Xote to Art. 150 (omitted in its proper place). From (117) we have with more exactness, cos L cos d 2 sin 2 4- t Ah: or putting sin (L—d) ' sin 1" 2 sin 2 i t , . cos L cos d m = — : — —j— and A = — — jj — jt-, sm 1 sm (L—dy Ah = Am and h = h -\-A m. Delambre's formula, obtained by developing in series the preceding equa- tion, (116) is h = h -\-A m —B n, in which 2 sin 4 !-* ,_ _ n = — , , £ = A 2 tan (Z— d) Table Y., of Chauvenefs Astronomy, contains m and ra, and Table VI. contains log m and log n for different values of t from to 30 m . Table VII. A gives the hmiting hour-angle at which the error resulting from neglecting the 2d reduction, Bn, amounts to 1". It varies from in the zenith to 36 m in latitude 40°, or to 67 m in latitudes 0° and 80°, for an altitude of 10°. Table XXXII. (Bowd.) gives J h only to the nearest 0".l ; if, then, it is taken from this table, AJi. t 2 may be in error 1", if t > 4 m . If, however, AJi is computed to the nearest 0".001, the error of using J /i. f, instead of A m, will not exceed 1", unless t > 20 m and h > 60°. 254 NAVIGATION. If the altitudes are equal, this second correction becomes 0. The most favorable condition is, therefore, that of equal alti- tudes observed on each side of the meridian. At sea, the method is especially useful for altitudes of the sun observed with a clear, distinct horizon. A long interval between the observations is to be avoided on account of the uncertainty of the reduction of one of the altitudes for the run of the ship. 252. The hour-angle of either altitude may also be ob- tained approximately; for we have from (204), in minutes, (Ji-h') and (206) Example. 1865, March 14, near noon, in lat. 45° 30' S., long. 120° 10' E.. by account, two altitudes were observed for latitude, T. by Chro. 4 h 15™ 20 s ; Obs'd alt. 46° 45' 30", (North) " " " 4 26 16 " " " 46 54 40; Index cor. of sextant +5' 20"; height of eye 18 feet. The sun's declination at noon —2° 31' 57", H. ch. +59". By preliminary computation AJi — 2".02 ; log AJi — 0.306. t = 10 m 56 3 h—h' = — 9' 10" i t = 5 28 = 5 m .47 l(h—h')=— 137 (ity= 29.9 '$(h + h')= 46° 50' 5" Aji= 2 ".02 1st cor. 29.9 x2".02= + l 137 2 2 log i {h-h') 4.274 log (1st cor.) 1.781 log (2d cor.) 2.493 log i (h-h') 2.137 n ar. co. log -} t 9.262 ar. co. log A Q h 9.694 log T Q 1.093 T«=~ ■ 12 m .4 T = — ■17 .9 r=- ■ 6 .9 2d cor. -— — • 60.4 = +5 11 46 56 16 In. cor. + 5 20 S. diam. + 16 7 Dip — 4 11 Ref. and par. — 48 K = : 47 12 44 z o — : 42 47 16 S. (12 m before h )d= : 2 32 9 S. Ls= 45 19 25 S. LATITUDE BY TWO ALTITUDES. 255 {JE) PresteVs method* by the rate of change of altitude near the prime vertical. 253. In the note to Art. 197 we have, for a very brief in- terval of time, and a small change of altitude, A t - Alh 15 cos L sin Z' i or, using the notation of this problem, T'-T=t = „ h '~ A . „ ; 15 cos L sin z 7 whence cos L = ~ cosec Z; (207) in which A'— h is expressed in seconds of arc, and t in seconds of time, and, Z being + when east, — when west, cos L is always positive. If Z is near 90°, its cosecant varies slowly. When Z = 90°, we have, cosZ=4^. (207') If, then, two altitudes are carefully observed near the prime vertical, and the times noted with great precision, the interval not exceeding 8 or 10 minutes, an approximate lati- tude may be found by (207 ; ), when the altitudes are within 2° or 3° of the prime vertical; or by (207) when they are at a greater distance, and Z is approximately known. The time of passing the prime vertical can be found by (107). Z maybe roughly computed from the altitudes, or found within 2° from the bearing observed by a compass, which will suffice, if the observations are made within 10° of the prime vertical. As, near the prime vertical, the altitude changes uniformly with the time, several altitudes may be observed in quick succession and the mean taken as a single altitude. The larger h'—h and t, consistent with the supposition of uniformity of change and the condition by which they are * Chauvenet's Astronomy, I., pp. 303, 311. 256 NAVIGATION. substituted for their trigonometric functions, the more accu- rate in general will be the result. Still the method does not admit of much precision. It is entirely unavailable near the equator, and in latitude 45° may give a result in error from 5 to 10 minutes, even when the greatest care has been bestowed on the observa- tions. It may, however, be useful to the navigator in high latitudes, as it can be used for altitudes of the sun, when it is almost exactly east or west, and consequently when no other method is practicable. There are occasions at sea, when to find the latitude only within 10' is very desirable. Examples. 1. 1865, June 15, 7 h A. M., in lat. 60° N., Ion. 60° W. ; T. by Chro. ll h 13 m 25 s .3, obs'd alt. © 27° 0' 23" ) O's Az. " " 27 48 42 J N.88°E.; log 8.8239 log 3.4622 ar. co. log 7.4137 1. cosec 0.0003 1. cos 9.7001 If A (ti—h) = 10", A log (h f -h) = A 1. cos L = .0015, and JX = 6'. If the difference of altitudes can be depended on within 5", the latitude is correct within 3'. 2. 1865, July 13, 5 h P. M., in lat. 54° 20' N., long. 113° W., by account ; the altitude of the sun's lower limb was ob- served at h 23 m 34 s by the chronometer, which was slow of G. mean time 10 m 18 s ; and the sextant remaining clamped the upper limb arrived at the same altitude at h 27 m 8 s -5 ; the true altitude of both limbs was 27° 18' 20"; required the latitude. The sun's diameter, 31' 33", is the difference of altitudes in this case. The sun's azimuth computed from the altitude and supposed latitude is 1ST. 88|° W. required ; 11 19 51 .0, the latitude. h'—h = t — z- L = i 48' 19" 6 m 2o 8 .7 88° 59° 55' N. LATITUDE BY TWO ALTITUDES. 257 TT log 8.8239 h — h' = W 33" log 3.2^2 t = 3 m 34 s .5 ar. co. log 7.6686 Z— 88£° 1. cosec 0.0002 L = 53° 56' N. 1. cos 9.7699 If we suppose t to be in error I s , 1. cos L will be in error .0020 and Z, 11'. If the elapsed time can be depended on within s . 5, the latitude is correct within 6'. The longitude obtained from the same observations is 113° 5' W. This method of observing the successive contacts of the two limbs of the sun with the horizon with the sextant clamped is recommended. 254. Problem 60. To find the latitude from two altitudes of different bodies, or of the same body when the change of declination is considerable, the Greemoieh times being known. Solution. The observed altitudes should be reduced to true altitudes, watch times to chronometer times, and the difference of the two chronometer times for the rate in the interval, as in Problem 59 (Art. 238), and to a sidereal in- terval, when the altitudes of two different bodies have been observed. When the latitude only is to be found, the Greenwich mean times of the observations are wanted only with suffi- cient exactness for finding the right ascensions and declina- tions of the bodies. If the longitude is also to be found, it is necessary to note the times by a Greenwich chronometer, or a watch compared with it. It is well also to note the azimuth, or bearing, for each observation ; or, as is sufficient, the difference of the azi- muths. Let M and M' be two positions of the body, or bodies, 258 NAVIGATION. h — 90°— Z M, the true altitude of M, h! = 90°— Z M', the true altitude of M', d = 90°— P M, the decimation of M, d'=90°— P M', the declination of M', r=ZPM, the hour-angle of M, f = ZP M', " « ' " M ; , t == T-T= M P M', the differ- ence of the hour-angles. £ is positive in the direction of the diurnal rotation, and will be positive and less than 12 h if M P M', estimated from P M in that direction, is less than 12 h , or 180°. We shall, as in the diagram, designate the two positions as M and M' respectively, so as to satisfy that condition. It will be seen hereafter, however, that it will be sufficient to have t nu- merically less than 12 h , without regard to its sign. 255. The method of finding t varies with the objects ob- served. But in any case we are at liberty to add or to sub- tract 24 h , either to change a negative into a positive result, or to reduce it within the numerical limit of 12 h . A posi- tive result greater than 1 2 h , or a negative result less than 12 h indicates that PM' is in the negative direction from P M. a. When two bodies are observed at different times, if a and a' are their right ascensions, S and /S", the sidereal times of the observations, by Art. Ill, T=S -a, T' = 8'-a f and t = S'—S—a' + a; or, t = s-ha— a! (208) LATITUDE BY TWO ALTITUDES. 259 when M has been first observed ; t= — (s + a'—a) (208') when M' has been first observed. In either case, the right ascension of the body first ob- served is added to the sidereal interval, and the right ascen- sion of the other body subtracted. If M' has been observed first, the sign is to be changed. b. When two bodies are observed at the same time. 5 = and t = a—a\ (209) the difference of their right ascensions. c. When the sun, moon, or a planet is observed at two different times, we have, as in Art. 238, t — s—s. 4 a (1 — .00273) (210) for an elapsed sidereal time ; t = t m + t m (9S.8565- J h a) (211) for an elapsed mean time; in which A h a is the change of right ascension in l h of mean time; (1 — .00273) A h a, the change of right ascension in l h of sidereal time ; and t m and s as coefficients are expressed in hours. In the case of the sun the last expression becomes, as in (175) t=t m +t m . 4E; in which J h E is the hourly change of the equation of time, employing for E the sign of its application to mean time. If this result exceed 12 h , it should be subtracted from 24 h . M' in that case is the position at the first observation. 256. We have given (Fig. 47) h = 90°-Z M, xl = 90°-P M, h' = 90°-Z M ; d l - 90°-P M', and ^MP M', 260 NAVIGATION. to find from the triangles P M M', Z M M' and PM'Z, or PMZ, L = 90°-P Z. The following method is selected as the most common. 25 7. Fourth Method of Bote ditches Navigator. 1. In the triangle P M M', (Fig. 48) MPM'-«, P M = 90°- J, PI' = 90°-c?', are given, from which w T e may find M M' — _5, the distance of the two positions, and the angle, PM'M = P, By Sph. Trig. (4) and (10)* we have cos B = sin d! sin c?+cos d! cos d cos t, cos dJ tan d — sin d' cos t cot F : sin t the 2d of which, by multiplying both numerator and deno- minator by cos d, becomes ™ cos d f sin d — sin d' cos d cos t cot P f — v- v— . cos • (215) sin jLj — ■ ^77 • i cos N ! ) We may take JV numerically less than 90°, and give it the same sign as that of cos q r : and the latitude i, numeri- cally less than 90°, with the same sign as J¥' + d r . There Avill be two values of L derived from the two values of q'. Unless q f is small, we may select the value w r hich agrees best with the known approximate latitude. If Z n (Fig. 47) be drawn perpendicular to P M', we shall find M' n =N' and 90° — Y n =N'+ d\ the declination of n. 258. Thus, by (212), (213), (214), and (215), the solution is effected. We have seen in each how the proper quadrant of the unknown quantities can be 'determined (with the re- striction of t to positive values less than 12 h ), except that Q' may have two values. The same results would be obtained by following the usual trigonometric precepts. 259. We may, however, select the proper value of Q' and avoid the double solution, by means of the noted azimuths. For, if Z and Z are the azimuths of M and M' (Fig. 47), reckoned as positive toward the right, MZM'=Z'-Z, and in the triangle M Z M' LATITUDE BY TWO ALTITUDES. 263 sin Q'=$in (Z'-Z)~. As cos h and sin _Z? are positive, (?' will have the same sign as (Z'—Z) restricted numerically to 180°. Hence, as Q' is to be subtracted from P\ Ave shall have q'=_P f — Q\ when M' bears to the right of M, q'=P r + Q', when M' bears to the left of M. Figs. 47 and 48 illustrate these two cases, for, in the first, where M' is to the right of M, PM'Z = P M' M - Z M' M ; and in the second, where JVT is to the left of M, PM'Z=PM'M-ZM'M. If M M' be extended to the meridian, in the first case, P and Z are on the same side of the intersection, and in the other they are on opposite sides ; so that q f =jP f — Q' when the zenith and north pole are on the same side of the great circle, which joins the two positions ; q r =P'+ Q\ when the zenith and north pole are on different sides, or the zenith and south pole are on the same side, of that circle. To use this criterion it will be necessary to note where the circle, which connects the two positions observed, crosses the meridian. 264 NAVIGATION. The doubtful case with either of these criterions is when Z'—Z is near 0, or 180°, or the great circle, which joins the two positions, passes near the zenith. These two conditions are coincident, except when the two positions are near the meridian on the same side of the zenith. 260. The hour-angle of M' may also be found, and thence the longitude, if the times have been noted by a chronometer regulated to Greenwich time. For, in the triangle P M' Z, we have, by Sph. Trig. (10), , rrt , cos d' tan Ji r — sin d' cos a' COt T = : ; — — , sin q which, by multiplying numerator and denominater by cos h y becomes , rril cos d' sin h' — sin d' cos h r cos q f cot T — tt~ -, — - cos h sin q Putting, as before, n' sin JSF= cos h' cos q ! ?i f cos iV'= sin h\ and eliminating n\ we have tan i\r= cot h! cos q' ) cot r= '<*£ cos gT ±*1. \ sin Is ) Or, L having been found, we have also . m1 sin q' cos h f /^h\ sm T — — - — ^ — . (217) cos L x 7 By (217) sin T and sin q' have the same sign, which will be positive when M' is west of the meridian, negative when M r is east of the meridian. In (216), if JST r has been taken cos q' less than 90° with the same sign as that of cos q\ — — ^7 is positive, and cos T' and cos (iV'-f- d!) have the same sign. We may take T\ then, numerically in the same quadrant as iV r/ + d\ and give it the positive sign when q 1 '< 180°, the negative sign when q' > 180°, or is negative. If the proper value of Q ', and therefore q\ has not been (216) LATITUDE BY TWO ALTITUDES. 285 previously determined, we shall have two values of T\ but may ordinarily take that which agrees best with its known approximate value. 261. The preceding formulas employ the angles at M', and the triangle P M' Z. We may also use the angles at M and trie triangle PMZ; and, as M and M' are similarly situated with regard to the triangles, except that the angles at each are estimated in opposite directions, Ave shall obtain, by interchanging accented and unaccented letters in the pre- ceding formulas, a set similar in form, but with this differ- ence of interpretation, that t is posi- tive in the opposite direction of the diurnal rotation, and q is less than 180° east of the meridian and great- er than 180°, or negative, west of the meridian. This difference is shown in Fig. 49, in which the primitive position of the triangles is east of the meridian, instead of west as in Fig. 47. We have, then, tan M~ tan d' sec £, sin d' cos (M — d) cos B cot P = sin M cot t sin (M- •d) Bin f cos M fcos s sin (s — h') hQ = i/(- cos li' sin B tan iV= cot h cos q, ) sin L : cot T-- sin li sin (iV+ d) > ' coTiV^ ' ) cot q cos (JV+ d) sin JV" (212') (213') (214') (215') (216') 266 NAVIGATION. sin T = — sin q cos h (217') q — P—Q, when M bears to the left of M', q = P + Q, when M bears to the right of M', 262. Either set of formulas may be used ; but, in general, the latitude can be best found from the altitude, which is nearest the meridian ; the hour-angle, from the altitude which is nearest the prime vertical. The distinction made with regard to M and M' (Art. 254), is important only so far as it may aid in determining the hour-angles and selecting the proper value of q or q'. So that it is sufficient practically to find t numerically less than 12 h without regard to its sign. 263. The most favorable condition is, as stated in Article (243), when the difference of azimuths is 90°. But altitudes near the meridian will give a good determination of the lati- tude, and altitudes near the prime vertical, a good deter- mination of the hour-angles, when the difference of azimuths is small, or near 180°; especially if the altitudes have been carefully observed, and their difference is nearly exact. 264. If we put in (212' &c.) M=-A, P = C, Q=Z, we shall have 2=:90°-#, tan A = — tan d' sec £, ~ sin d' cos (A + d) cos G — P=90°-P, cot F- — sin A tan t cos A sin | Z = j/7- sin {A + dy /cos s sin (s - -A') cos h' sin B G = F±Z, tan 1= cot h sin 6r, T _ sin h sin (d + I) -Lj — — SID COS / (218) LATITUDE BY TWO ALTITUDES. 267 the formulas of Bowditch's 4th method, if we take h r and^c?' as the greatest altitude and the corresponding declination. By attending to the signs of the quantities and their func- tions, the proper result can be obtained. We may give to Z the same sign, or name, as that of the latitude, when the zenith and elevated pole are on the same side of the great circle, which passes through the two j30sitions observed ; but a different sign, or name, from that of the latitude, when the zenith and elevated pole are on opposite sides of that circle. The precepts, which Bowditch gives (p. 194) are based on the consideration of the trigonometric functions, and possess the advantage that the sum of quantities of the same name, or the difference of quantities of different names, is taken, and the name of the greater given to the result. If the sum exceed 180°, it should be subtracted from 360°, and the name changed. Example. At sea, 1865, May 5, 7 h P. M., in lat. 36° 41' N"., long. 168° 57' W., by account; altitudes of a TJrsse Majoris and the moon were observed and the means noted as follows : required the latitude and longitude. T. by Chro.6 h 41 m 27 s : alt, of* 62° 18'30 r/ ; bearing N".byE.|E. " " " 6 53 8 ; " " JD44 56 50 ; " S.E.fS." Chro. at 18 h fast of G. m. t, 36 m 48 s , losing daily 10 s .2 ; Index cor. of sextant —4' 20"; height of eye 18 feet. Ship running NT. E. by X. (true), 10 knots an hour. From these data, reducing the second altitude to the posi- tion of the first, we find, a Ursce Maj. Moon, G. ra. t. May 5 18 h 4 m 39 3 .3 IS 11 16 m 20 s .4 Elap. in. t. ll m 41U R. A. 10 55 24.3 11 37 57.0 Red. +1.9 Dec. d'— + 62° 28' 45" d=—0° 53' 54" -Elap. sid t.-ll 43 .0 True alt. h'= 62 9 29 h= 45 39 54 Diff. ofR. A. 42 32.7 *=30 49.7 268 NAVIGATION. Computation by (212- -216). t- 7° 42' 26" 1. sec t 0.00394 d— — 53 54 1. tan d 8.19535 71 1. sine* 8.19530 n M' = — 54 24 1. tan M-' 8.1992971 1. cosec 31' 1.80077 n d'=+ 62 28 45 M-d' = - 63 23 9 lcos(M'-d') 9.65126 .5=: 63 38 39 1. cosec B 0.04766 1. cos B 9.64733 h= 45 39 54 1. cot t 0.86859 h'= 62 9 29 1. sec 7i' 0.33066 1. sec M 1 0.00005 2s= 171 28 2 Ism(M'-d') 9.95136 n s= 85 44 1 1. cos s 8.87153 1. cotP' 0.82000 n *—A = 40 4 7 1. sin (s—h) 9.80869 19.05854 i Q = 19 46 19 1. sin -J- Q' 9.52927 G'= 39 32 38 P' = 171 23 36 1. cot A' 9.72278 q'— 210 56 14 1. cos q' 9.93335 ^ 1. cot q' 0.22230 or -149 3 46 1. tan iV' 1. sin ti 9.65613 n 1. cosec N' 0.38441 n JST'z=— 24 22 20 9.94657 1. cos (N' + d') 9.89590 d'= + 62 28 45 1. sec iV" 0.04054 1. cot T' 0.50261?i N' + d'—+ 38 6 25 1. sin (N'+d') 9.79038 r'=r — l h 9 m 48 8 1. sin L 9.77749 ■X-'sR.A.= 10 55 24.3 Sid. time 9 45 36 .3 Lat. 36 c ' 48'.3 N. — S —2 53 28.5 Long. 168 52 .4 W. Red.forG.m.t. — 2 58.2 L. m. t. 6 49 9 .6 G. m. t. 18 4 39 .3 Long. 11 15 29.7 265. When equal altitudes have been observed, (213) and (213') reduce to the simple form, // CQg-ftJ?+»)\ , y 1 2 ^ i » ™° *' sin £ (?'= sin | # : cos £ P cos hj ' We have also from the isosceles triangle Z M M' (Fig. 47), cos Q'= cos Q = tan J- ^ tan A. (220) 266. When a lunar distance has been measured and re- duced to the true or geocentric distance (Prob. 55), we have in the triangle P M M' (Fig. 47), LATITUDE BY THBEE ALTITUDES. 269 PM = 90°-e?, 3PM'=-90°— : dr," and JM M'=j5, the geocentric distance, from which we may find P M' M = P' and P M M'= P. By Sph. Trig* (30), sm ^ -y [ sin b cos d' : ) which reduces to • i™, //cos j(B+d+d') sin j(B+d '-d)\ Sin ^r - y ^ sin i? cos # J or putting (221) "»*'=•( -•*££?' ) l So also we shall have • i r> //cos s' sin ($' — d')\ #««,>.% m *. Pas .r \ ^ ^ Id > < 221 ) These may be employed instead of (212) and (212') ; and if the altitudes of both bodies have been observed, the lati- tude and hour-angles can be found by the subsequent formu- las. From a lunar distance, then, and the two observed alti- tudes, the longitude, latitude, and local time may all be found. 267. PKOBLEMf 61. To find the latitude from three alti- tudes of the same body near the meridian, and the chrono- meter times of the observations. The Greenwich time, or the longitude, will be required only with sufficient exactness for taking the declination from the Ephemeris. * smH A = ^*(«+»7«)Bh *(«-»+«) (80) sin o sin c v ' f Chauvenet's Astronomy, Vol. L, p. 299. 270 NAVIGATION. Let A, h\ h\ be the true altitudes, h 0J the meridian altitude, Tj T\ T", the chronometer times, T^ the chronometer time of meridian passage, a, the change of altitude in I s of chronometer time from the meridian, then we have, from the three observations (120), the differ- ences of the times being expressed in seconds, h Q = h +a(T-T )*) h = h'+a(r-T o y\ (222) V=A f +a(Z*-JS) , J the differences of which are = h> -7i + a [(r -T o y-(T -T yi = £"-# + a [(T f '~Ty-(i«^To)% From these we obtain 7i — 7i T- h'- -T a (T'+T)-2a T oy (223) (224) TF — f7 = a(T»+T')-2aT , the difference of which is y_y h-h' _, Tlf T , rpn rpi rpi rp t(/ yJ. J. ]• If now we put h y b = T , T , the mean change of altitude in I s of the chro- nometer from the first to the second observa- tion, c = rpn T n tne mean change in I s from the second to the third observation, we shall have, from (224) and (223), _ c—b 2 o/ (225) LATITUDE BY THREE ALTITUDES. 271 from which a and T Q may be fonnd. h may then be found by either of the three equations (222) ; and thence the lati- tude as from an observed meridian altitude (Prob. 45). The correction and rate of the chronometer need not be known ; it is sufficient to have the rate uniform during the period of the observations. This method is restricted like other methods of circum- meridian altitudes (Arts. 150, 247), to hour-angles varying with the meridian zenith distance of the body. Its accuracy depends upon the precision with which a, &, and c are ob- tained : hence the altitudes should be carefully observed, so that their differences shall be nearly exact, and the intervals of time should be greater, the greater the distance of the middle observation from the meridian. 268. The computation is facilitated if the observations are made at intervals of exact minutes of time. For then, ex- pressing these intervals in minutes, and taking a, 5, and c as changes of altitude in l m of chronometer time, , 7i — ~h! ~h! — h" ^ _ c — h , , h c are easily found. - — and - — , however, will be in minutes Z co l co of time. Reducing them to seconds, we shall have, instead of (225), CO CO In (222) h = h +a(T-T o y\ h = h' + a(F-T o yy h Q = h n +a(T"-T Q y) we may now use, in computing the reductions, a table of * If the intervals are reduced by the methods of Art. 238 to intervals of hour-angle, and the declination has not changed, a will be the change of alti- tude in l m of hour-angle, as in (119) and Tab. XXXII. (Bawd.). 272 NAVIGATION. " squares of minutes and parts of a minute" as Tab. XXXIII. (Bowel.). 269. For the sun, as its declination usually changes in the interval, T is the chronometer time of the maximum alti- tude* (Art. 142) ; in which case the meridian declination is to be employed. If the latitude and longitude also have changed, as is usu- ally the case at sea, c and 5, being observed changes of alti- tude, are no longer due to the diurnal rotation alone, but are affected by the change of position. But with altitudes near the meridian, a change of latitude has the same effect as a change of declination in the opposite direction, while a change of longitude is equivalent to a change in the rate of the chronometer. If, then, the motion of the ship has been tolerably uniform in the interval of the observations, T is still the chronometer time of the maximum altitude. The method, then, can be used at sea when the sea is smooth and the horizon well defined, and meridian altitudes of the sun are prevented by passing clouds. But the altitudes should be very carefully observed, and on both sides of the meridians when practicable. The intervals should not be less than 10 m . 270. If the three altitudes are observed at equal intervals of time, the process of computation becomes much simpli- fied.! Let t be this common interval, T 7 , the time from the maximum altitude at which the second observation w r as made ; then we have h -h + a (T-ty h = 7i f +a T 2 h = h f '+ a (T+t) 9 . * Chauvenet's Astronomy, Vol. I., pp. 299 and 244. f Chauvenet's Astronomy, Vol. I., p. 309. LATITUDE BY THREE ALTITUDES. 273 Half the sum of the second and third equations is which, subtracted from the third, gives = h f - i (h + h") - a f; whence a f = h'- \ (h + h°). The difference of the first and third gives al - T? ' which, substituted in the second equation, gives h . If we put A— a f, we have, as the formulas for computa- tion, EXAMPLES. 1. At sea, 1865, Sept. 16, in lat. 40° 0' K, long. 60° 0' W., by account ; the following altitudes of the sun were observed near noon ; index cor. + 2' 10*; height of eye 20 feet. T. by Chro. 4 h 25 m 15 9 Biff. Obs'd alt. of O. 52° 24' 20" (S.) Biff. 32 37 15 15 5 20 10 14 50 4' 10' 5 20 T+ T') = 4* 28 re l 45 s . 250" = S5".1 log6 1.553 ^^1= -7 a 34 _ 320" 5 = 64 .0 log 30 1A11 T Q = 4: 21 11 _ 28".3 ~12~ = 2 .36 ar. co. log a 9.627 T—4: 25 15 454 s , 30 6 log— — a 2.657 T-T Q = 4 4 a(T-T Q f = 2".36 x 16.5 = 39" 274 NAVIGATION. 1st alt. of 52° 24' 20" ( In. cor. +2' 10" dip —4' 24" +*13 45 -J S. diam. +15 58 ref. & par. —38 Mer. alt. h = 62 38 5 (Red. +39 Mer. zen. dist. z = 37 21 55 N. Mer. dec. d — 2 28 48 N. Lat. 39 50 43 N. If the middle altitude had been 30" greater or less than 52° 20U0", the result would have been varied only 20" ; but if the middle altitude had been 1/ less, the latitude would have been 39° 40'. The interval is too short, un- less the differences of the altitudes can be relied on within 40". 2. At sea, 1865, May 8, in lat. 35° 50' S., long. 60° 0' E., by account ; the following altitudes of the sun were observed at equal intervals near noon : index cor. + 2' 0* ; height of eye 20 feet. Obs'd alt. of 0.36° 44' 20" (N.) T. by watch ll b 50 m 20 s 36 51 40 12 20 36 52 40 12 10 20 *(* + 70 = :36 48 30 h- -h" = -8' 20" A = 3 10 i (*-; h")=. - 125" 2 log 4.194 log A 2.279 h' = :36° 51' 40" t Red. + 1' 22" 1.915 + 13 42 -J In. cor. + 2 Dip -4' 24" Mer. alt. h = :37 5 22 ( S. diam . + 15 53 Ref. & par.- -1 9 Mer. zen. dist. h — 52 54 38 S. Mer. dec. d = ■11 1 19 N Lat 35 41 19 S. If either of the altitudes be changed 1', the reduction to the meridian will be changed less than 40" : so that, if the differences can be depended on within 1', the reduction is correct within 40". CHAPTER X. AZIMUTH OF A TERRESTRIAL OBJECT. 271. Ix conducting a trigonometric survey, it is necessary to find the azimuth, or true bearing, of one or more of its lines, or of one station from another. Thence, by means of the measured horizontal angles, the azimuths of other lines or stations can be found ; and, still further, a meridian line can be marked out upon the ground, or drawn upon the chart. For example, suppose at a station, A, the angles reckoned to the right are B to <7, 48° 15' 35"; C to D, 73° 37' 16"; D to E, 59° 45' 20"; and that the azimuth of D is N". 35° 16' 15" E. ; the azimuths of the several lines are A B, N\ 86° 36' 36" W. A Z>, N. 35° 16' 15" E. A (7, N. 38 21 1 W. A E, N". 95 1 35 E. If upon the chart a line be drawn, making with A B an angle of 86° 36' 36" to the right, or with iDan angle of 35° 16' 15" to the left, it will be a meridian line. Or, if a theodolite or compass be placed at A in the field, and its line of sight, through the telescope or sight-vanes, be directed to D, and the readings noted ; and then the line of sight be revolved to the left until the readings differ 35° 16' 15" from those noted,, it will be directed north. If a stake or mark be placed in that direction, it will be a meri- dian mark north from A. 276 NAVIGATION. 212. If the azimuth of a terrestrial object is known, it may be conveniently used in finding the magnetic declina- tion, or variation of the compass. For, let the bearing of the object be observed with the compass, — the difference of this magnetic bearing and the true bearing is the magnetic declination, or variation, required. It is east if the true bearing is to the right of the magnetic bearing ; but west if the true bearing is to the left of the magnetic bearing.* 273. Peoblem 62. To find the azimuth, or true bearing, of a terrestrial object. Solution. Let Z (Fig. 50) be the zenith, or place, of the observer ; O, the terrestrial object ; M, the apparent place of the sun, or some other celestial body ; Z — N Z O, the azimuth of O ; z = N Z M, the azimuth of M ; ^ v = Z-s = MZO, the azimuth an- gle between the two objects, or the difi&rence of azimuth of M and O. The problem requires that z and £ be found ; then we have Z=z + $. Or, numerically, * This has reference to the two readings. The actual direction of the object is the same ; but the true and magnetic meridians, from which the angles are estimated, are different. When the magnetic declination is east, the magnetic meridian is to the right of the true meridian ; when the mag- netic declination is tvest, the magnetic meridian is to the left of the true meridian. It is sometimes necessary to distinguish between the magnetic bearing and the compass bearing. The latter is affected by the errors of the instrument employed and by local disturbances ; the former is free from them. AZIMUTH OF A TERRESTRIAL OBJECT. 277 Z=2 + £, when the azimuth of the terrestrial object is greater than that of the celestial, Z — z — ^ when it is less. The sign of £ should be noted in the observations. 274. 2 = XZ M, the azimuth of the celestial body, may- be found from an observed altitude (Prob. 40), or from the local time (Prob. 38). In the first case, the most favorable position is on or nearest the prime vertical ; for then the azimuth changes most slowly with the altitude. In the latter, positions near the meridian may also be successfully used. 275. £ = M Z 0, the azimuth angle between the two ob- jects, may be found in one of the following ways : — 1st Method. (By direct measurement.) M Z O, being a horizontal angle, may be measured direct- ly by a theodolite or a compass, by directing the line of sight of the instrument first to one of the objects and reading the horizontal circle, then to the other and reading again. The difference of the two readings is the angle required. Or, the telescope or sight-vanes of a plane table may be directed successively to the objects, and lines drawn upon the paper along the edge of the ruler in its two positions, and the an- gle which they form measured by a protracter. At the instant when the observation is made of the celes- tial object, either its altitude should be measured, or the time noted, so as to find its azimuth simultaneously. The instrument should be carefully adjusted and levelled. With the compass or plane table, it is not well to observe objects whose altitudes are greater than 15°. A theodolite can be used with greater precision than the other instruments ; but the greater the altitude of the object, the more carefully must the cross-threads be adjusted to the axis of collimation, and the telescope be directed to the object. 278 NAVIGATION. The error of collimation is eliminated by making two ob- servations with the telescope reversed either in is Vs, or by rotation on its axis. Low altitudes are generally best. 276. If the sun is used, each limb may be observed alter- nately ; or a separate set of observations may be made for each. To find the azimuth reduction for semi-diameter, when but one limb is observed ; Let h— 90° — Z s (Fig. 51), the altitude of the sun, s = S s, its semi-diameter, s f = S Z s, the reduction of the azimuth for the semi-diameter. We have . sin S s sin- SZs = .., „ 3 sin Zs or, since s and s r are small, rig - 51 - s f — s sec A, (229) which is the reduction required. The sign, with which it is to be applied, depends upon the limb observed. 277. If the observations are made at night, and the ter- restrial object is invisible, a temporary station in a conve- nient position may be used, and its azimuth found. The horizontal angle between this and the terrestrial object may be measured by daylight, and added to, or subtracted from, this azimuth. A board, with a vertical slit and a light behind it, forms a convenient mark for night observations. The place of the theodolite should be marked, that the instrument may be replaced in the same position. But in doing this, ancl selecting the temporary station, it should be kept in mind that a change of the position of the instru- AZIMUTH OF A TERRESTRIAL OBJECT. 279 ment of g^Vs °f tne distance of the object may change the azimuth 1'; or of 2W000 of the distance may change the azimuth more than l" . 278. 2d Method. Finding the difference of azimuths of a celestial and a terrestrial object by a sextant y sometimes called an " astronomical bearing" Measure with a sextant the angular distance M O (Fig. 52) of the two objects, and either note the time by a watch regulated to local time, or measure simultaneously the alti- tude of the celestial object. Measure, also, the altitude of the terrestrial object (if it is not in the horizon), either with a theodolite which is furnished with a vertical circle, or w T ith a sextant above the water-line at the base of the ob- ject, when there is one. Correct the readings of the instru- ments for index errors, and when only one limb of the sun is observed, for semidiameter.* Observed altitudes of either object above the water-line are also to be corrected for the dip by (53) or Tab. XIII. (Bowd.), if the horizon is free; but by (55) or Tab. XVI. (Bowd.), if the horizon is obstructed. The altitude of the celestial object, when not observed simultaneously, may be interpolated from altitudes before and after, by means of the noted times. (Bowd, p. 246.) Or the true altitude may be computed for the local time (Prob. 38 or 39), and the refraction added and the parallax subtracted to obtain the apparent altitude. * It is best in measuring the distance of the sun from the terrestrial ob- ject to use each limb alternately. 280 NAVIGATION. Let h! = 90° — Z O (Fig. 50), the apparent altitude of O, H' — 90° — Z M, the apparent* altitude of M. D — M O, the corrected distance. We have then in the triangle M Z O, the three sides from which £ = M Z O, may be found by one of the follow- ing formulas : — 1. By Sph. Trig. (164) we have . 1V /sin i (Z> + 7T— h') sin HZ) — H' + h') sm | £ = j/- cqs ^ cqs v 1 or, letting d == H'—h\ sin i >• - / sin Hi) + ^) sin j (2> — ) 1* (231) COS J £ = i/ ^7^ TT 1 | r COS jH cos h' J (230) is preferable when £ < 90°; (231), when $ > 90°. 279. If O is in the true horizon, or its measured altitude above the water line equals the dip, h = 0, and the right triangle M m O gives cos $ = cos MO = cos D sec H' ; (232) or more accurately when £ is small (Sph. Trig., 105), tan \ $ = V (tan J {D + H f ) tan £ (D-H f ) ). (233) If the terrestrial object is in the water-line, A' is negative, and equals the dip. * The true altitude of M is used in finding g, its azimuth. AZIMUTH OF A TERRESTRIAL OBJECT. 281 280. If both objects are in the horizon, or H and h are equal and very small, we have simply $ = Z>. (234) In general the result is more reliable the smaller the in- clination of M O to the horizon. If M O is perpendicular to the horizon, the problem is indeterminate by this method. 281. If the terrestrial object presents a vertical line to which the sun's disk is made tangent, the reduction of the observed distance for semidiameter is s' = s sin MOZ (235) and not s, the semidiameter itself. This follows from the sun's diameter through the point of contact, O, being per- pendicular to the vertical circle Z O and not in the direc- tion of the distance O M. As the altitude of the terrestrial object is always very small, we may find MOZ by the formula ^ r ^ „ sin h' COS MOZ— - — — ; , sin D 7 D' being the unreduced distance. 282. When precision is requisite, the axis of the sextant with which the angular distance is measured must be placed at the station Z ; and if the object seen direct is sufficiently near, the parallactic correction must be added to the sex- tant reading. If J represent the distance of the object, cZ, the distance of the axis from the line of sight or axis of the telescope, this correction is p = — cosec V = 206265" — . (236) It is 1', when A = 3437.75 cl 283. If the distance of the terrestrial object and the dif- ference of level above or below the level of the instrument 282 NAVIGATION. are known, Ave may find its angle of elevation, nearly, by the formula tan h' = — , A A being the distance of the object, and E, its elevation above the horizontal plane of the instru- ment. If the object is below that plane, E and h! will have the negative sign. Note. — The horizontal angle between two terrestrial objects may also be found by measuring their angular distance with a sextant, and employing the same formulas (230 to 234) as for a celestial and terrestrial object ; H' and K representing their apparent angles of elevation. Each of these may be found by direct measurement, or from the known distance and the elevation, or depression, from the horizontal plane of the observer. If the two objects are on the same level as the observer, we have simply as in (234) Example. 1865, May 16, 5f A. M. in lat 38° 15' 1ST., long. 76° 16' W.; the angular distance of the sun's centre from the top of a light-house measured by a sextant (©to the right of L. H.), 75° 16' 25", index cor. —1' 15"; altitude of O above the sea-horizon observed at the same time, 10° 18' 20", index cor. +2' 10"; observed altitude of the top of light-house above the water-line, distant 7300 feet, 1° 15' 20", index cor., + 2' 10"; height of eye, 20 feet; required the true bearing of the light-house. From the data we find O'sap. alt. H'.= 10° 31' 57"; ap. alt. of L. H. fc'= 1° 1' 34" O'strue" E = 10 27 7; ang. disk D = 75 15 10 ©'s dec +19 9 30. AZIMUTH OF A TERRESTRIAL OBJEC3 283 Computation (100) and (230). H = 10° 27' 7" 1. sec 0.00726 H' - 10° 31' 57" 1. sec 0.00738 L = 38 15 l.sec 0.10496 V = 1 7 34 1. sec 0.00008 P = 70 50 30 d = 9 24 23 2s = 119 32 37 D =75 15 10 s = 59 46 18 1. cos 9.70196 i (£+d) = 42 19 46 L sin 9.82827 p — s = 11 4 12 1. cos 9.99184 £(£— e*) = 32 55 24 1. sin 9.73521 19.80602 19.57094 iZ = 36° 53'.0 1. cos 9.90301 if = 37° 36'.2 f= / 75 12.4 1. sin 9.78547 O 's azimuth Z=N. 73 46 .0E. True bearing of L. House (Z-J) = N. 1 26 .4 "W 4