u PIIILADELrHIA .NAUTICAL COLLEGE, t South- West corner of Cliesnut and Eighth Sts. * \^ % In this long-established and well-known institution, the complete course of Naviga 2 tion and Nautical Astronomy is taught in the most satisfactory manner, and in | J the shortest possible time. f f The course includes all the diflfereut sailings: viz., Plane, Traverse, Parallel, ♦ # Middle Latitude, Mercator's, and Great Circle Sailing; f All the various methods of finding the ♦ i LATITUDE OF THE SHIP, I I by altitudes of th( ~ -■. — ..* ^ -r>-i^ o^.,--- ah ^v^g^arious methods | I IlIBRARY qP CONGRESS. I , ,^. ,J # and regulating a# --"- ^ '*^» ^^^ Fixed # I Stars; All the V f , \[^\^ ^ ^ I I by Dead Reck | ^^^^ Ehl I the use of Quad] f UNITED STATES OF AMERICA.! i and Steering | I Compasses, I @^^<^^^^^^^^^^^^^<^<^^^| the method of # ^1 Keeping a j0Ui.xic*x ^ _. , _ _ can at any time, # 4 with confidence and certainty, point out on the Chart ^ THE SIEilX^'S I'OSITIOlSr .if^T SE.A-. I 1?^ <•» # # # ^1 Acting Masters' I^Iates, Ensigns, and Masters for the XJ. S. Navy, and Revenue and ^ # Mercantile Marine Ofiicers, prepared for examination, and qualified for appointment. # *$* ^ I . ^ I t ENGINEERS FOR THE U. S. NAVY. I % Candidates for admission or promotion in either the volunteer or regular service are ^ ^ taught the course ordered by the Navy Department. I THE WEST POINT COURSE OF MATHEMATICS, S including Land and Topographical Surveying and Civil and Military % Engineering, with field practice in the use of Compass, Transit, Level, J Plain Table, &c., &c., taught in a course of special practical lessons adapted to the J % professional wants of oificers in the U. S. Army. I - ^r- GRADUATES ABE AWASDED A SPLENDID DIPLOMA. I Jf. B.— ''The United States Practical Navigator" will be sent by mail on the S receipt of $2.50. Ail communications must be addressed to the Authors, M. & G. W. EOCHE. ^rA r'oTO^ ^wC /r^^ K^. ^^ ^^ ^^ /^^ ^^^^ \ %^ '^ ^^ > \ \ ^i ^ v^ sS^ * . rsn WEMORY OF MAETIN KOCHE, PHILADELPHIA, FORMERLr Professor of Mathematics in the U, S. Navy, THK rOLLOWme pages, EXEMPUFYING his Km.ES FOR PRACTICAL NAVWATION, BEING THE SIMPLEST EVER DEVISED, ARE AFFECTIONATELY DEDICATED, BT HIS SONS, THE AUTHORS, (iii) PREFACE. Maiht persons wlio enter tlie maritime profession do not possess any knowledge of the ordinary rules of compntation ; otliers, again, begin tlieir career of following the sea, sufficiently prepared to perform any rule or method of calculation, but who from the nature of a seafaring life, do in the course of a few years, forget their early qualifications, and become altogether incompetent to solve even a question in Plane Sailing. The mariner, therefore, is in want of such practical methods as are most suitable to his calling, and most consistent with his limited acquirements. To supply the seaman's deficiency in this respect, and to facilitate the acquisition of knowledge so important as navigation, have been the Authors' aim and study for years, and it has been found that the best method by which the desired object can be accomplished, is to simplify the rules, so as to enable him without difficulty, to arrive at the greatest possible accuracy in the results. As the clearest and most convincing proof of the correctness and infallibility of these rules, they have worked out in full all of the examples in Bowditch's ISTavigator, and added Cases and Examples which that author has omitted to give in his "Epitome." (V) SIGNS USED IN THIS WORK. X {)■' V is the sign of Addition. *♦ " *' " Subtraction " " *' *' Multiplication, " " " " Division. " " " " the square. " " " the square root. " " " " equality. " " '* " a decimal, ** *' *' " is to, in proportion. " *' *' " so is, in proportion, " " " " to, in proportion. " " " " a degree. " " *' " a minute of a degree. " *' '* " a second of a degree. EXPLANATION. The reason why the answers in the foot-notes to Examples of Cases IV. and V. of Plane Sailing differ from those worked out by the Table of Miles is, that the table is extended to only five places of decimals, and the fifth is calculated to the nearest figure. This applies to the answers in the foot-notes of Cases IV., V., VL, and VII. of Middle Latitude Sailing, and all the Cases of Mercator's Sailing. (vii) ARITHMETIC DECIMAL FRACTIONS. When any quantity is considered as constituting a whole, it is called an integer, and when an integer is supposed to be divided into a certain number of equal parts, any number of these parts, considered in their relation to the whole, is^ called a fraction, which is expressed by two numbers placed one above another, with a line between them in the form of a common or vulgar fraction ; the lower of these, called the denominator, denotes the number of parts into which the integer or whole one is divided, and the upper, called the numerator, expresses the number of these parts which is contained in the fraction : for instance, suppose a mile divided into 6 equal parts, and 2 of these parts were to be considered as a fraction of the whole ; it would be written in numbers thus, |, where the figure under the line shows that the mile is divided into 6 equal parts, and the 2 above the line denotes the number ol" those parts contained in the fraction. Fractions are of two kinds. Vulgar and Decimal. Decimal fractions vary in the game proportion, and are managed by the same methods of operation, as whole num- bers. For this purpose every proper fraction is reducible to another whose denomi- nator shall be 10, 100, 1000, &c. : viz. unity with some number of ciphers annexed ; and fractions with such denominators are called decimal fractions, such as /j, yVs^ As the denominator of a decimal fraction is always 10, or 100, or 1000, &c., the said denominators need not be written, for the numerators may be made to express the true value of a decimal. For this purpose it is only required to write the nume- rator with a point, called the separatrix, before it, to distinguish it from a whole number, when it consists of as many figures as the denominator has ciphers annexed to unity ; so y^^ may be written .5 ; yV^ may be written .75 ; and j%%^j^ may be writ- ten .625. But if the numerator has not so many places as the denominator has ciphers, put as many ciphers before it, viz. to the left hand, as will supply the defi- ciency ; so write j(yf .05 ; y^^jf .005. And thus do these fractions receive the form of whole numbers. Therefore we may consider the separatrix as the fixed point whence whole num- bers proceed to the left, infinitely increasing, and decimals to the right, infinitely decrea-?ing towards 0, as the following table shows : — o 'S § 'S • «• '^ 'S § ^ -« .2 7654321 . 234567 In decimals, as well as in whole numbers, each figure takes its value by its dis- tance from the separatrix ; if it be in the first place after the separatrix, it signifies 2 ARITHMETIC. tenths ; if in the second, hundredths, &e., decreasing in each place to the right in a tenfold proportion. Consequently every single figure expressing a decimal has for its denominator 1, with as many ciphers as its place is distant from the separatrix. Thus 2 in the above table is /g, 3 is jgjj, and 4 is yjy'rtjj, &c. And if a decimal be expressed by several figures, the denominator is 1, with as many ciphers as the last figure is distant from the separatrix. So .234 signifies and must be read -^^^q. A cipher (or ciphers) placed at the right hand of a decimal fraction alters not its value, since every significant figure continues to occupy the same place ; so .5, .50, .oOO, &c., are all of the same value. But a cipher (or ciphers) placed to the left hand of a decimal does alter its value, every cipher decreasing it to j'jj of the value it had before, by removing every signi- ficant figure one place farther from the separatrix ; so .5, .05, .005 all express dif- ferent decimals, viz. .5, /^ ; .05, j§j ; .005, tuVs- "We may likewise observe the contrary efiect of ciphers being annexed to whole numbers and decimals ; every cipher to the right hand of a whole number increases its value ten times, but ciphers to the right hand of a decimal do not alter its value. Again, ciphers put to the left hand of a whole number do not alter its value, but every cipher put to the left hand of a decimal decreases its value to the j'j of what it would be without them, thus : — 5. .5 5. .5 50. .50 05. .05 500. .500 005. .005 5000. .5000 0005. .0005 It is also manifest from the above table that the places of decimals decrease in a tenfold proportion from the separatrix towards the right hand, consequently they increase in a tenfold proportion towards the left hand, as the places of whole num- bers do ; for ten-hundredths make one-tenth, and ten-tenths make a unit ; ten units make ten, ten tens make one hundred, &c. ; viz. j^q = j'^, |g = 1 ; 1 X 10=^ 10, 10 X 10 =100, &c., which proves that decimals are subject to the same law of nota- tion, and consequently of operation, as whole numbers are. Decimal fractions of unequal denominators are reduced to one common denomina- tor, when there are annexed to the right hand of those which have fewer places as many ciphers as make them equal in places with that which has most; so these decimals .5, .04, .125 may be reduced to the decimals .500, .040, and .125, which have all 1000 for their denominator. Of decimals that is the greater whose highest figure is greater ; whether they consist of an equal or unequal number of places : thus, .575 is greater than .395, and .5 greater than .395, for if it be reduced to the same denominator with .395 it will be .500, which is manifestly greater. A mixed number, viz. a whole number with a decimal annexed, is equal to an improper fraction, whose numerator is all the figures of the mixed number, taken as one whole number, and the denominator that of the decimal part ; so 32.405 is equal to YoVff » as is manifest from the method, given in all arithmetics, of reducing a mixed number to an improper fraction ; for 32, the integral part, being multiplied by 1000, the denominator of the fractional part, produces 32000, to which adding 405, the numerator of the fractional part, the sum 32405 is the numerator to 1000 for an improper fraction equal to the given mixed number. A pure whole number is one which has no decimal annexed to it. A pure decimal is one which has no whole number prefixed to it. A mixed decimal is one which has a whole number prefixed to it. ADDITION OF DECIMALS Is performed by writing down the given numbers to be added together so that the decimal points will be directly under each other (and if joined together would form a perpendicular line), and the figures will then be ranged in the proper order of their local values, viz. tenths under tenths, hundredths under hundredths, thousandths under thousandths, &c., and units under units, tens under tens, hundreds under hundreds, thousands under thousands, &c. ; then adding them up as if they were pure whole numbers, and in their sum placing the decimal point directly under the other decimal points in the given numbers. ARITHMETIC. d EXAMPLE I. EXAMPLE II. Add together 463., .422, and .914 463. .422 8.51 ^.43 '.914 8.51, .43, 7., Add together .000342, 267., .45, and 402. .000342 267. .826 3.5 .45 402. .826, 3.5, Answer 480.276 Answer 673.776342 SUBTRACTION OF DECIMALS Is performed by writing down the given numbers to be subtracted, the less from the greater, so that the decimal points will be directly under each other (and if joined together would form a perpendicular line), and the figures will then be ranged in the proper order of their local values ; then subtracting them as if they were pure whole numbers, and in their difference placing the decimal point directly under the other decimal points in the given numbers. EXAMPLE L From 716048. subtract .0000625 716048. .0000625 Answer 716047.9999375 EXAMPLE XL From 456321.23456789 subtract 42.3 456321.23456789 42.3 Answer 456278.934561 MULTIPLICATION OF DECIMALS Is performed by writing down the given numbers to be multiplied, as if they were pure whole numbers, and in their product placing the decimal point as many places from the right hand towards the left as there are decimal places in both the multi- plicand and multiplier together. N. B. — It sometimes happens that, in the multiplication of decimals, there will not be in the product as many figures as the rule requires decimal places, and in such cases the deficiency must be supplied by prefixing as many ciphers as there are places wanting. EXAMPLE L Multiply 34.26 by .542 34.26 .542 6852 13704 17130 18.56892 Answer. EXAMPLE II. Multiply .03526 by .00013 .03526 .00013 10578 3526 .0000045838 Answer. Note. — To multiply by 10, 100, 1000, &c., remove the decimal point in the multi- plicand to the right as many places as there are ciphers in the multiplier. EXAMPLE. .426 X 10 = 4.260 ) .426X100 = 42.600 I Answer. .426X1000 = 426.000] ARITHMETIC. DIVISION OF DECIMALS Is performed by writing down the given numbers to be divided, as if they were pure whole numbers, and in the quotient placing the decimal point as many places from the right hand towards the left as there are decimal places in the dividend exceed- ing those in the divisor. This is the general rule ; there are, however, several par- ticular cases, which we shall treat of separately in order. EXAMPLE I. Divide 5.73 into 258.0219 5.73)258.02' 19(45.03 Answer. 2292 2882 2865 1719 1719 EXAMPLE n. Divide 270.2 into 1275.6142 270.2)1275.6'142(4.721 Answer. 10808 19481 18914 5674 5404 2702 2702 CASE I. When the number of places of decimals in the divisor and dividend is equal, the quotient will be a pure whole number, provided the division terminates evenly ; but if the division does not terminate evenly, and it is deemed necessary to annex ciphers to the dividend, then the quotient will be either a mixed or pure decimal having as many places as ciphers were annexed. EXAMPLE I. Divide 52.475 into 18208.825 52.475)18208.825'(347. Answer. 157425 246632 209900 367325 367325 EXAMPLE IL Divide 23.8457 into 5365.3303 23.8457)5365.3303 '(225.002004-f Ans. 476914 596190 476914 1192763 1192285 478000 476914 1086000 953828 132172 N. B. — Five places of decimals will be suflficient for all calculations in practical navigation. Divide 53.497 into 42.968 EXAMPLE nL Answer .803184- CASE II. When the number of places of decimals in the dividend exceeds that in the divi- sor, the quotient will be either a mixed or pure decimal, the^ number of places in which will be equal to the excess, provided the division terminates evenly ; but if the division does not terminate evenly, and it is deemed necessary to annex ciphers to the dividend, then the quotient will have as many places more as ciphers were annexed. ARITHMETIC. EXAMPLE I. Divide 314.52 into 1339.8552 S14.52)1339.85'52(4.26 Answer 125808 81775 62904 188712 188712 EXAMPLE XL Divide .7402 into 19.73451 .7462) 19.7345 ' 1 (26.44667+ Answer. 14924 48105 44772 33331 29848 Divide 8.2467 into 3.521698 EXAMPLE III. Answer .42704-j- 34830 29848 ~49820 44772 50480 44772 57080 52234 4846 CASE III. When there are not as many places of decimals in the dividend as there are in the divisor, ciphers must be annexed first to make them equal, and the quotient will be a pure whole number, provided the division terminates evenly ; but if the divi- sion does not terminate evenly, and it is deemed necessary to annex more ciphers to tlie dividend, then the quotient will be either a mixed or pure decimal having as many places as more ciphers were annexed. EXAMPLE L Divide 26.82345 into 9119.973 26.82345)9119.973001(340. Answer. 8047035 10729380 10729380 EXAMPLE IL Divide 3456.782105 into 1451848.486 3456.782105)1451848.486000' (420.00000054+ Answer. 13827128420 6913564400 6913564210 Divide 584.29685 into 26.4953 19000000000 17283910525 17160894750 13827128420 3333766330 EXAMPLE III. Answer .04534+ CASE lY. When the divisor is a pure whole number, and the dividend a mixed or pure deci- mal, the quotient will be either a mixed or pure decimal having as many places as 6 ARITHMETIC. the dividend, provided the division terminates evenly ; but if the division does not terminate evenly, and it is deemed necessary to annex ciphers Uj the dividend, then the quotient will have as many places more as ciphers were annexed. EXAMPLE I. Divide 962. into 3354.625 962.)3354.625(3.48713+ Answer. 2886 4686 3848 ~8382 7696 6865 6734 1310 962 3480 2886 ~594 EXAMPLE XL Divide 5263. into 17.4269 5263.)17.4269(.00331+ Answer. 15789 "16^579 15789 5900 5263 637 EXAMPLE TIL Divide 74693. into .76829 Answer .00001028-|- Note. — To divide by 10, 100, 1000, &c., remove the decimal point to the left as many places as there are ciphers in the divisor. EXAMPLE. Divide 10 into 4.26 Answer .426 " 100 " 42.6 " .426 " 1000 " 426 " .426 To divide by 60, remove the decimal point to the left one place, because there is one cipher, and proceed as in short division. EXAMPLE. Divide 60 into 8133.53992 6,0)813,3.53992 135.558998+ = 135.559 Ajiswer. This refers to Example of Case I. Plane Sailing, REDUCTION OF DECIMALS. The reduction of a vulgar fraction to a decimal is performed by annexing to the numerator as many ciphers as may be deemed necessary, and dividing the denomi- nator ; the quotient will be the decimal required, which must always consist of the same number of places as there were ciphers annexed. EXAMPLE II. To change .50041 of a degree to min- EXAMPLE L Reduce 30^ of a degree to a decimal. 30° 6,0)3,0. .5 Answer. 30^ = This refers to Example of Case I. Par- allel Sailing, under "Remark." utes. Rule.— Multiply by 60, and point off as many places in the product as in the given decimal. .50041 60 30.02460 Answer. This refers to Example of Case II. Par- allel Sailing, under " Remark." ARITHMETIC. INVOLUTION Is the finding of the po-svers of numbers. The first power is the root ; the second power or square, is the product of the first power or root, multiplied l)y itself; the third power or cube, is the product of the square multiplied by its root, &c. EXAMPLE I. What is the square of 244? 244 X 244 = 59536. Answer. This refers to Example of Case IV Plane Sailing. EXAMPLE II. What is the square of 136 ? 136 X 136 = 18496. Answer. This refers to Example of Case IV. Plane Sailing. EXAMPLE in. What is the square of 203 ? 203 X 203 = 41209. Answer. This refers to Example of Case V. Plane Sailing. EVOLUTION Is the finding of the roots of numbers, and of course, is the reverse of Involution. The root of a number is a factor, which being multiplied by itself a certain number of times, will produce the given number. Thus, if the given number is a square, the root is called the square root, if a cube, the root is called the cube root, &c. TO EXTRACT THE SQUARE ROOT. 1st. Separate the given number into periods of two places, beginning at the sepa- ratrix, going to the left in whole numbers, but to the right in decimals. If there should be an odd figure left in whole numbers, it must be used as a full period ; but if there should be an odd figure left in decimals, it must not be used until made a full period by the addition of a cipher. 2d. Find the greatest square in the left hand period or figure, and write its root down in the quotient, as in division ; square this root, and put it under the left- hand period or figure, and to the right of the remainder bring down the next period in order, for a dividend. 3d. Double the figure of the root already found, for a trial divisor, and find how often it is contained in the dividend exclusive of the right hand figure, and write the number in the quotient as the second figure of the root ; annex it also to the trial divisor, which makes the true divisor. Multiply the true divisor by the second figure of the root, and subtract the product from the dividend. To the remainder bring down the next period, for a new dividend. 4th. Double the figures of the root already found, for a trial divisor, and proceed as before until ail the periods have been brought down. N. B. — Every period of the given number must give a figure in the root. If the given number be a pure whole number, and the extraction of the root ter- minates evenly, the root will contain as many places of whole numbers as there were periods in the*^ given number; but if the extraction does not terminate evenly, and it be deemed necessary to annex periods of decimals, then the root will contain as many places of decimals as there were periods annexed. If the given number be a mixed decimal, and the extraction of the root terminates evenly, the root will contain as many places of whole numbers as there were periods in the given whole number, and as many places of decimals as there were periods in the given decimal ; but if the extraction does not terminate evenly, and it be deemed necessary to annex more periods of decimals, then the root will contain as many more places of decimals as there were periods annexed. If the given number be a pure decimal, and the extraction terminates evenly, the root will contain as many places as there were periods in the given decimal ; but if the extraction does not terminate evenly, and it be deemed necessary to annex more periods of decimals, then the root will contain as many more places of decimals as there were periods annexed. ARITHMETIC. EXAMPLE I. Extract the square root of 41040. 4|l0|40.(202.583+ Answer. 4 402)1040 804 4045)23600 20225 40508)337500 324064 405163)1343600 1215489 128111 This refers to Example of Case IV. Plane Sailing. EXAMPLE XL Extract the square root of 18327 l|83|27.(135.377+ Answer. 23 ]83 69 265)1427 1325 2703)10200 8109 27067)209100 189469 270747)1963100 1895229 67871 This refers to Example of Case V. Plane Sailing. EXAMPLE IIL Extract the square root of .17977 .17'97l70(.42399+ Answer. 16 82)197 164 843)3370 2529 8469)84100 76221 84789)787900 763101 "^4799 1 o .c rs 3-. — r^ OS M s i~ '3 "^ ffl «5 O us O .O pi .~r '"- i"- ■;: - s H^ r ■=■ ?: '.: ^ '^ CC M CO M •c .- o, a: oc =5 '- t * ^' -* -5 ?i c4 01 ^^■ oi oi m rn — ■ i-. rn ^ -t « ^ 35 » CO I - i is CO Ji r-; =5 C3 d d C3 2 O r-c ^( r3 TT O oooo = o o ri :^ .■'5 Tt io s §2S^^5 82?, .^^g g2§s^5 35 a ■* ^^ T-i ■>! — : — — ^ mm iiiill o o o 3; 35 rs g i5 ::: 2 g ?? So.3S8S 35 d 3: 00 3C od :;• ^ :^ 2 ;'' !(S I- i- d d d X) 3 S2^^^5 33 g2^^^5 825^55 g2Sg2g CO 1 c I"" 1, I, ,,: i~ -si d S =^. ^1 = =<> =^ ;0 ;i -.S •.-- 'C. o S g i £ 1 1=2 ^ ^ ^ p ^ ^. -f -f -f •* coco gg2Sd^ ri. c; r-l 35 iO CO 2 2 :i^' 2 2 s~sr^5s g2§g^g 82§§5g •-0 g2i^?g bi; 1 liiiiii S^ ^ ?5 ?) fl ?i R2^f KH 05 o5 oj o3 oi ?i S 55 s ?, 2 2 §2Sg^g S2§g^3 825^^5 s 8 ;::; oi c^o ? 5 o g2Sg5g ib liiiiii ?? ?i ?i ^ ?i ?5 iiiill iiiiii O -r j£ CO j-^ ^ ?, fj ol 05 o1 ^1 ^^^;^k^. "S 52§g^S §2g^^S 82^^55 82^^i5 8238^8 ^ ^ ;? ;? s M °3 >ii « 5 -* S ;:3 O ^ C-J CO '3' o i? So ?o ?i fc ?o O rt -M CO -C O O .-^ 0^ CO -T O S 2 i i g ^ 4il-fl-i- O i- O, CO -f o 8 S 2 5 8 S 8 8 c^ 8 5 8 5 ;^ s CO ?■ ii 3 liiii CC » 30 OC OC l^ CO CO ?0 CO CO CO ?:?^Sg2K t~ O ^ CO 05 O iiiiii Iiiiii ■4 is i2 15; ^' 1? ^ S2?,g5S 52^M^g g2§§^S g2^i^§ 05 820-58^8 CO •0 ii! 3 § S A ?i S 235;£s:^ CO CO 05 oi oi o4 S S5 i? :^ S 8 Iiiill ■^' -+ -^ ^ ^ ^! q o ^ S ^ q mm iiiiii 3 ° CM § 2 s -^ 5 s CO -1" §2S^^g 82^^^S 8288^8 bi . 5 aJ § :? s '^ liiiiii " 3C OC 2 =«' » ^ -2S^^S iiii S o ^ ^: o s ■i o ^? -^ ^ o 03 03 t- n o CO o .«■ O O O uO 2^.2§SS2 uO -l« -t -* -f -f -f ^T)-^ 3 §2Sg^§ S2§g^g g2S^^S oooooo Or-105 CO-* O O OOOOOO ' r-c 05 rC .0 ti c iifiiii o 2 2 ^ S "^ rf CO oq ^_ q S S85ggg=: CO 05 — O OCC l-H 35 OC I-- o o ..-: .o lO .« o ~, ■17 O »0 O >0 lO Or^ o o o § § S S o o iiiiii iSiiii 3 i2§S^S C5 g2§^5S g2Sg^g oooooo r- C^ CO ^ 05 i r ►3 S -o ^- S O uO S CC CO CC OC OC CO .o o UO O iC o q q =q co i, t^ t^ M ■* O, .-0 O q q o ir ^ t, i^ t-l i^ i~^ t-^ t~ UO .O O uo uO UO 2SS§sS CC uO CO t- 5 5 5S5S i S2SS5S CO g2Sg5S g2S^5S 82Sg§S 2 g2S8§g O uO O -f -t -I* o o -i< oc bi CO 2 ^ ;-2 ? 2 5 -* CO CO CO CO o? § s S g o S o5 ?5 ^ J-; ;2 iii ^ § ?^ ^1 ;j s §?5^§gg q q q q q q 35 35' 35 3C' 00' 00 s §2§g^S g2§^§§ 00 82gg^g 05 g2§g^g 52^852 1 ^ iiiiii iiisii iiiiii iSiSig gggggs °o §2§g^g gsss^s to 82§§§S g2§8^S . (9) o I ?3i o 9 ll !^ 1 S'! ?^' ?i www aaaa aaaa aaa b i? it is y.yyy. aaaa a 1 u r o i J= a •3 g ^•^'^'^ J = = ^ r.Kss s =^ =4 =^' CfcOJ CO a a a a ^■aaa a'a::^ '/. X X > > > >^ aixx X aaaa * M oi X o V2 o 2 K 5 3 W Miles in 1° of Longitude in Lutitude equal to the Coniple. uient of the Cour.so. d :n 'ooc i^ o ^ ?5 iiii 2gS^ ^ '- ==. -^ ■^; Iiii Ills ill 30 u-JM O i ^ 1 11 1 ^! 1 1? A oci^r,';L* t-S'i'i* ■^'szrj; o?"^"^* -* iiii q o. i^ cc F^ S a ^ ui O O >0 22gg ?3 -* Iflll- i '1 nH— let- 1» o O 1 >^ >5 crJxx -^. X X X ^- a; X X fe. '3 g ^ ^ si a a ^^aa-a aixx iXXX M . . . ^^■^^ X X X X aiaixai -aaa X 1 li = 1 3rS 1 5 3 1 3 ft ^ -J 1 i 52ZZ b >-. >-. >■• - a a a Jsaa ^2Z = c c c a s a s Zaas ^■^^^ zyyy. yy.yy. J22Z z o a z 1 1 1 Ed 1 ,.2 X C W c -s X ^ J; O 6 J ^ H CIO) ^U'S ^04r The Mariner's Compass is an artificial representation of the horizon of every pla<;e. It consists of a circular card divided into 32 equal parts, by lines drawn from the centre towards the circumference, called rhumb-lines, the extremities of which are called points ; the intervals are subdivided into half and quarter points. The points of the compass are counted from the meridian or the North and South points towards the East and West points of the horizon, and are named as follows: — Points. North-East Quarter. North-West Quarter. South-East Quarter. South-West Quarter. North North South South 1 N. by E. N. by W. S. by E. S.S.E. S. by W. 2 N.N.E. N.N.W. s.s.w. 3 N.E. by N. N.W. by N. S.E. by S. S.W. by S 4 N.E. N.W. S.E. s.w. 5 N.E. by E. E.N.E. N.W. by W. ' S.E. by E. S.W. by W. 6 W.N.W. E.S.E. w.s.w. 7 E. by N. W. by N. E. by S. W. by S. 8 East West East West ,1 (11) 12 MARINER'S COMPASS. The four principal points are called Cardinal Points, two of which, opposite each other, are called North and South points ; that which is on the right hand when you look towards the North is termed the East, and that which is on the left hand, or opposite, the West point. The names of the other points are compounded of these, according to their situation or position, but instead of the words the initials only are used. The North, South, East, and West are usually marked vrith a single letter, thus : N. S. E. and W. The North point is frequently marked with the Fleur de Lis. The North and South are markefl because they are on the meridian. The 4 points are expressed by two letters. The 2 and 6 points are expressed by three letters. The remaining 16 points are marked with the word bi/ in their names. The 3 and 7 points are always named and read backwards from the 4 and 8 points, but counted forward from the meridian, as all the others are. Under the card, along the North and South line, a small bar of steel is fixed, called the Needle, which being magnetized, or touched with a loadstone, acquires the peculiar property of pointing north and south, and consequently by the card, determines the direction of the other points of the horizon. The needle having a small socket in the centre, is supported, together with the card, on the point of a fine steel pin, on which it freely turns, and by the above-mentioned property, its points keep always in the same direction (the compass is here supposed to be true). These are confined in a circular brass box with a glass cover, the box being hung in brass hoops or gimbals, in order to counteract the motion of the ship. The whole of these are placed in a square wooden box with a moveable lid, which serves to support the gimbals and secure the compass from accident in removals. The compass is used to point out the direction that a ship sails at sea. For this purpose, it is so placed in the ship that the middle section of the wooden box, par- allel to its sides, may be parallel to the middle section of the ship along its keel. When it is thus fixed, that point of the card which coincides with a perpendicu- lar line marked on the inside of the circular box, and termed by seamen the '^Lubber's Faint/' will show the direction of the ship's head. To " box the compass,'' or, in other words, to name the various points of it accu- rately, is the first, the easiest and the simplest lesson in navigation. Many persons imagine this to be so puzzling, that they give it up at once and for ever, but the whole task should be accomplished in about half an hour. So begin with mastering one quarter of the compass, and the rest is easily accomplished, merely by a change of name, for the same arrangement applies to the other three quarters. We will now give two very simple rules by which you may learn the whole " mystery.'' KuLE I. — When the letters indicating two points are united, the point meant is halfway between the two; thus, N.E. is halfway between North and East; N.N.E. is half way between North and N.E. ; E.N.E. is half way between East and N.E. KuLE II. — When the letters are joined by the word by or the letter b, the point meant is the one which comes next after the first going towards the second ; thus, N. by E. is next to North going East ; N.E. by N. is next to N.E. going North ; N.E. by E. is next to N.E. going East; and E. by N. is next to East going North. ON THE TABLE OF MILES IN A DEGEEE OF LONGITUDE. CASE I. GiTEN the latitude to find the corresponding number of miles in a degree of lon- gitude. RULE. Find, in the Table of Miles, the next less and the next greater than the given latitude, and take out the corresponding miles in a degree of longitude. One-tenth of the difference between these will be the difference in miles for one minute in the latitude. Multiply this difference for one minute by the number of minutes the given latitude exceeds the next less, and subtract the product from the number of miles in a degree of longitude corresponding to the latitude next less than the given one, and you will have the number of miles in a degree of longitude in the given latitude. EXAMPLE L Required the number of miles in a de- gree of longitude in latitude 33° 52^ N. orS. By the Table of Miles in a degree of longitude, &c., on page 9, . . 33^50^ = 49.83966 Next less " greater 34 00 =49.74218 10 . 33° 50^ = -f 02 = . 33 52 = ). 09748 .009748 2 ^19496 Next less . . Given latitude . = 49.83966 = — .01950 = 49.82016 Next less . Given latitude This refers to Example of Case IV. Plane Sailing. EXAMPLE IL Required the number of miles in a de- gree of longitude in latitude 56° 18^ N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less . . . 56° 10^ = 33.40677 , 56 20 =33.26160 greater 10 ). 14517 .014517 8 .116136 56° 10^ = 33^0'677~ 4-08 = — .11614 56 18 = 33.29063 This refers to Example of Case V. Plane Sailing. (13) 14 ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. EXAMPLE IIL Required the number of miles in a de- gree of longitude iu latitude 28° iV N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less . . . 28° 10^ = 52.89465 " greater . . 28 20 =52.81207 Next less Given latitude 10 28° 10^ + 01 ). 08258 .00^^258 52.89465 — .00826 28 11 52.88639 This refers to Question IV. for exercise in Plane Sailing. EXAMPLE IV. Required the number of miles in a de- gree of longitude in latitude 25° 04^ N. orS. By the Table of Miles in a degree of longitude, &c., on page 9, . . 25° 00^ = 54.37850 Next less " greater 25 10 =54.30438 10 ).07412 .007412 4 Next less '* greater .029648 25° 00^ = 54.37850 + 04 = — .02965 Next less . 25 04 = 54.34885 Given latitude Next less Given latitude This refers to Question V. for exercise in Plane Sailing. EXAMPLE V. Required the number of miles in a de- gree of longitude in latitude 53° 36-' N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less . . " greater . . 53° 30^ . 53 40 = 35.68934 = 35.54886 Next less " greater 10 ). 14048 .014048 .084288 Next less . . . 53° 30^ + 06 = 35.68934 = — .08429 Next less . Given latitude . 53 36 = 35.60505 Given latitude This refers to Question II. for exercise in Parallel Sailing. EXAMPLE VL Required the numljer of miles in a de- gree of longitude in latitude 32° 17^ N. or S. By the Table of Miles in a degree of hjngitude, &c., on page 9, Next less " greater 32° 10^ = 50.79023 32 20 =50.69698 Next less Given latitude 10 32° 10^ + 07 ). 09325 .009325 7 .065275 o0T9'023" — .06528 32 1^ 50.72495 This refers to Question III. for exercise in Parallel Sailing. EXAMPLE VIL Required the number of miles in a de- gree of longitude in latitude 39° 3F N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, 39° 30^ = 46.29745 39 40 =46.18628 10 ). 11117 xTiTin 39° 30^ = 46.29745 + 01 = — .01112 59 31 46.28633 This refers to Example of Case I. Mid- dle Latitude Sailing. EXAMPLE VIIL Required the number of miles in a de- o-ree of longitude in latitude 48° 38'' N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, 48° 30^ = 39.75725 48 40 =39.62627 10 ). 13098 .013098 8 .104784 48° 30^ = + 08 = = 39.75725 = -.10478 48 38 = = 39.65247 This refers to Example of Case II. Mid- dle Latitude Sailing. ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. 15 EXAMPLE IX. Required the number of miles in a de- gree of longitude iu latitude -iC 26^ N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less '* greater 40° 20^ = 45.73747 40 30 := 45.62442 Next less Given latitude 10 ). 11305 .011305 6 ^07830 40° 20^ = 45.73747 4- 06 = — .06783 40 26 45.66964 This refers to Example of Case III. Mid- dle Latitude Sailing/. EXAMPLE X. Required the number of miles in a de- gree of longitude in latitude 48° oV N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less . . . 48° 50^ = 39.49509 " greater . . 49 00 =30.36355 10 ). 13154 .013154 Next less Given latitude 48° 50^ = 39.49509 + 01 = — .01315 48 51 39.48194 This refers to Example of Case IV. Mid- dle Latitude Sailinc/. EXAMPLE XL Required the number of miles in a de- gree of longitude in latitude 53° 08^ N. orS. By the Table of Miles in a degree of longitude, &c., on page 9, , 53° 00^ = 36.10975 53 10 =35.96933 Next less " greater 10 ). 14042 .014042 .112336 Next less Given latitude 53° 00^= 36.10975 + 08 = — .11234 53 08 35.99741 This refers to Example af Case Y. Mid- dle Latitude and Mercator's Sailing. EXAxMPLE XIL Required the number of miles in a de- gree of longitude in latitude 50° 43^ N. or o. By the Table of Miles in a degree of longitude, &c., on page 9, Next less " greater 50° 40^ = 38.02982 50 50 =37.89465 Next less Given latitude 10 50° 40^ + 03 ). 13517 .013517 3 .040551 38.0l982~ — .04055 50 43 .9^927 This refers to Example of Case VI. Mid- dle Latitude Sailing. EXAxMPLE XIIL Required the number of miles in a de- gree of longitude in latitude 50° 58^ N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less " greater Next less Given latitude 50° 50^ ==37.89465 51 00 =37.75922 10 ). 13543 1)13543 8 .108344 50° 50^ = 37y9"465~ -f 08 = — .10834 50 58 = 37.78631 This refers to Example of Case VII. Middle Latitude and Mercator^s Sailing. EXAMPLE XIV. Required the number of miles in a de- gree of longitude in latitude 48° 23^ N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, , 48° 20^ = 39.88771 , 48 30 =39.75725 Next less " greater Next less Given latitude 10 48° 20^ + 03 ). 13046 ^13046 3 .039138 39.88771 — .03914 48 23 = 39.84857 This refers to Example of Case VII. Middle Latitude Sailing. 16 ON THE TABLE OF MILES IN A DEGREE OF LONGITUDE. EXAxMPLE XV. Eequired the number of miles in a de- gree of longitude in latitude 35° H'' N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less . . . 35° 10^ = 49.04876 ** greater . . 35 20 =48.94798 10 ). 10078 .010078 .070546 Next less 35° 10^ = 49.04876 + 07 = — .07055 35 r 48.97821 Given latitude . This refers to Question I. for exercise in Middle Latitude Sailing. EXAMPLE XVL Required the number of miles in a de- gree of longitude in latitude 34° 47^ N. orS. By the Table of Miles in a degree of longitude, &c., on page 9, This refers to Question II. for exercise in Middle Latitude Sailing. EXAMPLE XVIL Required the number of miles in a de- gree of longitude in latitude 79° 56^ N. or S. By the Table of Miles in a degree of longitude, &c., on page 9, Next less . . . 79° 50^ = 10.59073 " greater . . 80 00 =10.41887 10 ).17186 .017186 6 .103116 EXAMPLE XVIIL Required the number of miles in a de- gree of lon";itude in latitude 10° 04^ N. or b. By the Table of Miles in a degree of longitude, &c., on page 9, d Course. • — ^ ■= 15.69254 the number of miles in the Departure = 15.7 Answer. TRAVERSE SAILING. 87 By the Table of Points opposite to the Compass, on page 10, 4th Course W.N.W. ... 6 points = 07° 30' gives 2'2.96101 the nnnilier of miles in a degree of longitude in latitude equal to it. Complement of the Course 2 " =22 30 " 55.43268 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. l^^. To find the Difference of Latitude on the 4th Course. 39 X 22.96101 60 1-1.92466 the number of miles in the Difference of Latitude = 14.9 Answer. 2(7. To find the Departure on the 4ih Course. • — -^ — "^ = 36.03124 the number of miles in the Departure = 3( Answer. By the Table of Points opposite to the Compass, on page 10, 5th Course S. by E. . . . 1 point = 11° 15' gives 58.84702 the number of miles in a degree of longitude in latitude equal to it. Complement of the Course 7 points = 78 45 " 11.70539 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. 1st. To find the Difference of Latitude on the 5ih Course. 40 X 58.84702 60 39.23135 the number of miles in the Difference of Latitude ■■ 2d. To find the Departure on the 5th Course. 40 X 11 70539 - o^orr, ^jjg number of miles in the Departure = 7.8 Answer. Having calculated all the Diflferences of Latitude and Departures corresponding to the several Courses and Distances sailed, arrange them, according to the Eule, in the form of a TRAVERSE TABLE. Course. Distance. Difference N. OF Latitude. S. Depai E. ITURE. W. S.S.E. S.E. W. by S. W.N.W. S. by E. 15 34 16 39 40 14.9 13.9 24.0 3.1 39.2 5.7 24.0 7.8 15.7 36.0 Differs 14.9 80.2 14.9 37.5 51.7 37.5 Remair mce of Latitu( ider 65.3 le made good. Remaic Departu der 14.2 re made good. 1 BY CASE VI. PLANE SAILING. 1st. To find the Distance made good. 65.3X60 |/(65.3)2 + (14.2)2 = |/4264.09 + 201.64 = -j/4465.73 = 66.826 ^^^ number of miles in the Distance made good = 66.8 Answer. 2d. To find the Course made good. = 58.62987 the number of miles in a degree of longitude in latitude equal to the Course made good, viz. S. 12° 16' W., because the DiiTerence of Latitude made good, is to the South- ward, and the Departure made good, is to the Westward, as per Example VII., page 20. Answer. Therefore, Block Island beai-s from the ship on the point of the compass di- rectly opposite, viz. N. 12"=' 16' E. The Latitude left (Block Island) ... 41° 10' N. Dif. of Latitude made good, 65.3 miles = 1 05 S. The Latitude the ship is in 40 05 N. Part III. of the Rule is not required in this Example. 38 TRAVERSE SAILING. EXAMPLE XL A ship from Mount Dcsort Rock, in the latitude of 43° 50^ N., sails for Cape Cod, in the latitude of 42° 03^ N., its Departure from the meridian of Muunt Desert Hock being supposed to be 84 miles to the Westward. She takes her departure from Mount Desert Kock bearing due North, distant by estimation 10 miles, Ijut by reason of contrary winds she is obliged to sail on the following Courses, viz. (1st, South 10 miles, the Departure reversed); 2d, W.S.W. 25 miles; 3d, S.W. 30 miles; and 4th, West 20 miles : required, 1st, the direct Distance and Course from Mount De- sert Rock to Cape Cod ; 2d, the Distance and Course made good by the ship, from Mount Desert Rock to where she has arrived ; and 3d, the direct Distance and Course from where she now is, to Cape Cod, the place bound to. Bemark.—The latitude of Mount Desert Rock lighthouse is 43° 58^ N. ; the lati- tude of Cape Cod (Highland) light is 42° 02^ N. The meridian of Cape Cod is only 83 miles to the Westward of the meridian of Mount Desert Rock. But as we want to work out all the Examples in Bowditch by our method, we must use his figures, although knowing them to be not strictly correct. Latitude of Mt. Desert Rock 43° 50' N. Cape Cod light . 42 03 N. Difference of Latitude . . 1 47 =107 miles. The Departure is 84 miles. Part I. To find the direct Distance and Course from the place left, to the port hound to. BY CASE VI. PLANE SAILING. \st. To find the Distance. -j/(107)a + (84)2 = i/ll449 + 7056 = |/lS505 = 136.033 the number of miles in the direct Distance from Mt. Desert Rock to Cape Cod. 2d. To find the Course. 107 X 60 ^ , oa (Yio = 47.19443 the number of miles in a degree of longitude in latitude equal to the direct Course, ido.ud Course. X 42.42637 — ■ = 21.21319 the number of miles in the Departure = 21.2 Answer. TRAVERSE SAILING. 39 On the 4th Course, which is due West, the ship makes no Difference of Latitude, and therefore the whole Distance sailed will be the Departure = 20 miles. Answer. Having calculated all the Differences of Latitude and Departures on the several Courses and Distances sailed, arrange them, according to the Rule, in the form of a TRAVERSE TABLE. Course. Distance. Difference of Latitude. N. S. Departure. E. W. South w.s.w s.w. West 10 25 30 20 10.0 9.6 21.2 40.8 Difference of Latitude made good 23.1 21.2 20.0 64.3 Departure made good BY case VI. plane SAILING. l5^. To find the Distance made good. l/(40.8)2 + (64.3)2 = -j/l664.&4 + 4134. -1/^5799.13 = 76.15202 the number of miles in the Distance made good = 76.2 Answer. 2d. To find the Course made good. 40.8 X 60 " ^ - = 32.14623 the number of miles in a degree of longitude in latitude equal to the Course made * '^•■^^^ good, viz. S. 57° 36' W., because the Difference of Latitude made good, is to the South- ward, and the Departure made good, is to the Westward, as per Example IX., page 21 . Answer. Tlierefore, Mount Desert Rock bears from the ship, on the point of the compass di- rectly opposite, viz. N. 57° 36' E. Part III. To find the direct Distance and Course from where the ship now is, to Cape Cod, the place bound to. Latitude of Mt. Desert Rock 43° 50' N. Cape Cod light . 42 03 N. Difference of Latitude . . 1 47 Difference of Latitude made good . 107 40.! miles S. Departure 84 miles W. Departure made good . 64.3 " W. Dtfiference of Latitude yet to make . . 66.2 S. Departure yet to make 19.7 W. |/(66.2)2 + (19.7)2 = |/4382.44 BY CASE VI. PLANE SAILING. l5^. To find the direct Distance yet to make. 388.09 = y^i 70.53 06902 the number of miles in the direct Dis- tance from the ship to Cape Cod light == 69.1 miles. Answer. 66.2 X ' 2d. To find the direct Course yet to make. 57.50771 the number of miles in a degree of longitude in latitude equal to the direct Course, viz. S. 16° 34' W., because the Difference of Latitude yet to make, is to the Southward, and the Departure yet to make, is to the Westward, as per Example X., page 21. Answer EXAMPLE in. A ship in the latitude of 37° 10'' N. is bound to a port in the latitude of 33° 00^ N., which lies 180 miles West of the meridian of the ship ; but by reason of contrary winds she sails on the following courses, viz. 1st, S.W. by W. 27 miles; 2d, W.S.W. ^ W. 30 miles; 3d, W. by S. 25 miles ; 4th, W. by N. 18 miles ; 5th, S.S.E. 32 miles; 6th, S.S.E.f B. 27 miles; 7th, S. by E..25 miles; 8th, South 31 miles; and 9th, S.S.E. 39 miles: required, 1st, the direct Distance and Course from the place left, to the port bound to ; 2d, the Distance and Course made good by the .ship, from the place sailed from, to where she has arrived ; and 3d, the direct Dis- tance and Course from where she now is, to the port bound to. 40 TRAVERSE SAILING. Latitude sailed from . . 37° WN. " of port bound to 33 00 N. Difference of Latitude . . 4 10 «= 250 miles. Departure 180 miles. Part I * Tojind the direct Distance and Course from the place lefty to the port hound to. BY CASE VI. PLANE SAILING. \st. To find the Distance. •j/(250)2 + (180)2 = |/62500 + 32400 = |/94900 = 308.05844 the number of miles in the direct Distance from the place left, to the port bound to. An?. 2d. To find the Course. 250 X 60 ' tine AtiQ ' ^ 48.69213 the number of miles in a degree of longitude in latitude equal to the direct Course, <}U».U{)» yi2. s. 350 45/ Yf ^ because the Difference of Latitude, is to the Southward, and the De- parture to the Westward, as per Example XI., page 21. Answer. Part II. To find the Distance and Course made good hy the ship, from the place sailed from, to where she has arrived. By the Table of Points opposite to the Compass, on page 10, 1st Course S.W. by W. . . 5 points = 56° 15' gives 33.33418 the number of miles in u degree of longitude in latitude equal to it. Complement of the Course 3 " =33 45 " 49.88810 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. \st. To find the Difference of Latitude on the \st Course. 27 y 33 33418 • —^ = 15.00038 the number of miles in the Difference of Latitude «= 15. Answer. oO 2d. To find the Departure on the \st Course. 27 V 49 88810 — ^ = 22.44965 the number of miles in the Departure = 22.4 Answer, By the Table of Points opposite to the Compass, on p^age 10, 2d Course W.S.W. i W. . . 6^ points = 73° 07-i-' gives 17.41706 the number of miles in a degree of longitude in latitude equal to it. Complement of the Course 1^ " = 16 52^ " 57.41635 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. 1st. To find the Difference of Latitude on the 2d Course. 30 X 17.417C 60 = 8.70853 the number of miles in the Difference of Latitude = 8.7 Answer. 2d. To find the Departure on the 2d Course. — ^ • = 28.70818 the number of miles in the Departure = 28.7 Answer. 60 By the Table of Points opposite to the Compass, on page 10, 3d Course W. by S. . . . 7 points = 78° 45' gives 11.70539 the number of miles in a degree of longitude in latitude equal to it. Complement of the Course 1 point = 11 15 " 58.84702 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. l5^. To find the Difference of Latitude on the 3tZ Course. 25 X 11 70539 — '. = 4.87725 the number of miles in the Difference of Latitude = 4.9 Answer. 60 2d. To find the Departure on the 2>d Course. 25 V 58 84702 — — '■ = 24.51959 the number of miles in the Departure = 24.5 Answer. * Bowditch has omitted to give this part of the Example. Why? TRAVERSE SAILING. 41 By the Table of Points opposite to the Compass, on page 10, 4tli Course W. by N. . . . 7 points = 78° 45' gives 11.70539 the mnnber of miles in a degree of longitude in latitude equal to it. Complement of the Course 1 point = 11 15 " 68.84702 tlio number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. 1.?^. Tojiiid the Difference of Latitude on the 4ih Course. 18 X 11 70539 . '- = 3.51162 the number of miles in the Difference of Latitude = 3.5 Answer. 60 2d. Tojind the Departure on the Ath Course. 18 y '^8 8470*2 . ^^-^ = 17.65411 the number of miles in the Departure •■= 17.7 Answer. By the Table of Points opposite to the Compass, on page 10, 6th Course S.S.E 2 points ^ 22° 30' gives 55.43268 the number of miles in a degree of longitude in latitude equal to it. Complement of the Coiirse 6 " =67 SO " 22.96101 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. \st. Tojind the Difference of Latitude on the 5t7i Course. 32 X 55.43268 60 = 29.56410 the number of miles in the Difference of Latitude = 29.6 Answer. 2d. To find the Departure on the 5th Course. 32 X 22.96101 • — = 12.24587 the number of miles in the Departure = 12.2 Answer. By the Table of Points opposite to the Compass, on page 10, 6th Course S.S.E. :J E. . . 2f points = 30° 56^' gives 51.45361 the number of miles in a degree of longitude in latitude equal to it. Complement of the Course 5^ " = 59 03J " 30.84612 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. 1st. To find the Difference of Latitude on the 6th Course. 27 X 51.45361 = 23.15412 the number of miles in the Difference of Latitude = 23.2 Answer. 2d. To find the Departure on the Qth Course. 27 X 30.84612 • — = 13.88075 the number of miles in the Departure = 13.9 Answer. By the Table of Points opposite to the Compass, on page 10, 7th Course S. by E. . . . 1 point ^ 11° 15' gives 58.84702 the number of miles in a degree of longitude in latitude equal to it. Complement of the Course 7 points = 78 45 " 11.70539 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. Jst. To find the Difference of Latitude on the 7th Course. 25 X 58.84702 = 24.51959 the number of miles in the Difference of Latitude = 24.5 Answer. 60 2d. To find the Departure on the 1th Course. 25 X n 70539 • — -^ = 4.87725 the number of miles in the Departure = 4.9 Answer. On the 8th Course, which is due South, the ship makes no Departure ; and therefore the whole Distance sailed, viz. 31 miles, will be the number of miles in the Difference of Latitude = 31. Answer. By the Table of Points opposite to the Compass, on page 10, 9th Course S.S.E 2 points = 22° 30' gives 55.43268 the number of miles in a degree of longitude in latitude equal to it. Complement of the Course 6 " «= 67 30 " 22.96101 the number of miles in a degree of longitude in latitude equal to it. BY CASE I. PLANE SAILING. Is^. To find the Difference of Latitude on the 9th Course. 39 X 55.43268 = 36.03124 the number of miles in the Difference of Latitude = 36. Answer. 2cL To find the Departure on the 9th Course. X 22.96101 = 14.92466 the number of miles in the Departure = 14.9 Answer. 60 42 TRAVERSE SAILING. Having calculated all the Differences of Latitude and Departures, on the several Courses aud Distances sailed, arrange them, according to the Kule, in the form of a TRAVERSE TABLE. 1 Difference of Latitude. Departure. Course. Distance. N. S. E. W. S.W. by W. 27 15.0 22.4 W.S.AV. i VV. 30 8.7 28.7 W. by S. 25 4.9 24.5 W. by N. 18 3.5 17.7 S.S.E. 32 29.6 12.2 S.S.E. f E. 27 23.2 13.9 S. by E. 25 24.5 4.9 South 31 31.0 S.S.E. 39 36.0 14.9 3.5 172.9 3.5 45.9 93.3 45.9 Kemainder 169.4 Remainder 47.4 Difference of Latitude made good. Departure made good. 1 BY CASE VI. PLANE SAILING. l5^. To find the Distance made good. |/(169.4)2 + (47.4)2 = |/28696.36 + 2246.7G = -1/30943.12 = 175.90657 the number of miles in the Distance made good = 175.9 Answer. 2d. To find the Course made good. 175.90657 57.78011 the number of miles in a degree of longitude in latitude equal to the Course made good, viz. S. 15° 38' W., because the Difference of Latitude made good, is to the Sonth- Avard, and the Departure made good, is to the WestAvard, as per Example XII., page 21. Answer. Therefore, the place sailed from, bears from the ship, on the point of the compass di- rectly opposite, viz. N. 15° 38' E. Part III. To find the direct Distance and Course from where the ship now is, to the port bound to. Latitude sailed from . . . 37° 10' N. " of port bound to . 33 00 N. Difference of Latitude . . 4 10 ^ Difference of Latitude made good . Difference of Latitude yet to make . 250 miles S. 169.4 " S. Departure 180 miles W. Departure made good . 47.4 " W. Departure yet to make 132.1 W. BY CASE YI. PLANE SAILING. 1st. To find the direct Distance yet to make. 1// (80.6)2 -j- (132.6)2 = i/ 6496.36 + 17582.76 = i/ 24079.12 = 155.17449 the number of miles in the direct Distance from the ship, to the port bound to = 155.2 Answer. 80.6 X 60 155.17449 '' 2d. To find the direct Course yet to make. 31.16492 the number of miles in a degree of longitude in latitude equal to the direct Course, viz. S. 58° 42' W., because the Difference of Latitude yet to make, is to the Southward, and the Departure yet to make, is to the Westward, as per Example XIII., page 22. Answer. »— The Latitude sailed from 37° 10' N. 169 Difference of Latitude made good is 169.4 miles = -— = 2 49 S. oO The Latitude the ship is in 34 21 N. Answer. PARALLEL SAILING. Tn the problems of Plane Sailing and Traverse Sailing, the Earth has been con- sidered as constituting an extended plane ; but in the solution of problems in which longitude is concerned, we must necessarily take into consideration the real spheri- cal form of the Earth, with the meridians meeting each other at the poles. The Departure made on any parallel of latitude will not give the same Difference of Lon- gitude that it would if made in a higher or lower latitude. Hence, the Difference of Longitude cannot be found by either of the two former sailings. The distance between any two meridians measured on a parallel of latitude is called the meri- dian distance, and decreases in proceeding from the Equator, where it is greatest, towards the poles, where it is nothing. When a ship sails due East or due West on the surface of the Earth, she keeps in the same latitude and describes a circle parallel to the Equator, and this is called parallel sailing. In Parallel Sailing, the distance sailed East or West, is the Departure made good by the ship, and is equal to the distance between the meridian sailed from and the meridian arrived at, in the latitude of the parallel sailed on. The Longitude of a place is its distance east or west from any given principal or first meridian, fixed at pleasure, such as that of Washington, Greenwich, Paris, &c., and is measured by that part of the Equator intercepted between the principal me- ridian and the meridian that passes through the given place, which is equal to the angle at the pole, between the two meridians. As Longitude is counted from the principal or first meridian. East and West, on the Equator, until it comes to the opposite meridian, therefore a ship's longitude cannot exceed 180° East or West. The Difference of Longitude between two places, if both East or both West, is found by subtracting the less longitude from the greater ; but if one longitude be East and the other West, the Difference of Longitude is found by adding both lon- gitudes together. Should their sum exceed 180°, subtract it from 360°, and the remainder will be the Difference of Longitude. In East longitude sailing to the Eastward, or in West longitude sailing to the Westward, the Difference of Longitude must be added to the Longitude left, to get the Longitude in, because the ship sails farther away from the principal or first meridian. But in East longitude sailing to the Westward, or in West longitude sailing to the Eastward, the Difference of Longitude must be siibtracted from the Longitude left, to get the Longitude in, because the ship sails nearer to the principal or first meridian. When the Longitude decreases and the Difference of Longitude is greater than the Longitude left, the Longitude left, must be subtracted from the Difference of Lon- gitude, to get the Longitude in, which will be of a different name from the Longitude left, because the ship in this case, crosses the principal or first meridian. This sailing is particularly useful in making small or low islands, in which case it is usual to run into the latitude and then steer due East or due West. (43) 44 PARALLEL SAILING. CASE I. Given the Difference of Longitude between two places in the same parallel of Latitude, to find the Distance between them. RULE. Multiply the Difference of Longitude in miles, by the number of miles in a degree of longitude in the latitude of the parallel on which the ship sailed, and divide the product by 60 ; the quotient will be the number of miles in the Distance. EXAMPLE. Suppose a ship, in the latitude of 49° 30^ North or South, sails directly East or West, until her Difference of Longitude be 3° 30^ (— 210 miles) ; required the Dis- tance sailed. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, Latitude 49° 30' gives 38.96685 the number of miles in a degree of longitude. 210 X 38.96685 — = 136.38398 the number of miles in the Distance sailed = 136.4 Answer. w CASE II. Given the Distance between two places on the same parallel of Latitude, to find their Difference of Longitude. RULE. Multiply the Distance in miles by 60, and divide the product by the number of miles in a degree of longitude in the latitude of the parallel on which the ship sailed; the quotient will be the number of miles in the Difference of Longitude, which divided by 60 gives the number of degrees and minutes to be added to, or subtracted from, the Longitude left, accordingly as we are increasing or lessening our longitude. EXAMPLE. Suppose a ship, in the latitude of 49° 30^ North or South and in longitude 36° 40^ West, sails directly West, the Distance 136.4 miles ; required the Difference of Lon- gitude and the Longitude in. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, Latitude 49° 30' gives 38.96685 the number of miles in a degree of longitude. 136.4 X 60 210 ■ o^^^V^. = 210.02467 the number of miles in the Difference of Longitude = -— = 3^= 30' W. 38.96685 ^ 60 Which added to the Longitude left 36 40 W. Gives the Longitude the ship is in 40 10 W. Answer. CASE III.* Given the Distance between two places on the same parallel and their Difference of Longitude, to find their Latitude. RULE. Multiply the Distance in miles by 60, and divide the product by the Difference of Longitude in miles ; the quotient will be the number of miles in a degree of longi- tude in the latitude of the parallel on which the places are. EXAMPLE. Suppose a ship having sailed due East or West the Distance 136.4 miles, found she had made a Difference of Longitude of 3° 30^ (= 210 miles) ; on what parallel of Latitude did she sail ? . — = 38.97143 the number of miles in a degree of longitude in the latitude of the parallel On which 210 tlie ship sailed, viz. 49° 30' N. or S., as per Example XIV., page 22. Answer. REMARK. The several Cases of Parallel Sailing may be solved by other rules, as follows : — Case I. Rule. — Multiply the number of miles in a degree of longitude in the latitude the ship sailed, by the number of degrees and minutes (turned into the deci- mal of a degree) in the Difi'erence of Longitude, and the product will be the miles in the Distance sailed. Example the same as before. 38.96685 X 3.5 = 136.38398, the number of miles in the Distance sailed = 136.4 Answer. * Bowditch has omitted to give this Case! Why? Question IT. for Exercise cannot be solved without it. PARALLEL SAILING. 45 Case II. Rule.— Divide the Distance in miles, by the number of miles in a de- gree of longitude in the latitude of the parallel on which the ship sailed, and the quotient will be the Diifference of Longitude in degrees and decimal of a degree. Example the same as befo-e. - = 3°.50041 = 3° 30' W. the Difference of Longitude. ^ 38.96685 "Which added to 36 40 W. the Longitude left. Gives .... 40 10 W. the Longitude the ship is in. Answer. Case III. Rule. — Divide the Distance in miles, by the number of degrees and minutes (turned into the decimal of a degree) in the Difference of Longitude be- tween the two places, and the quotient will be the miles in a degree of longitude, in the latitude of the parallel on which the places are. Example the same as before. — '- = 38.97143 the number of miles in a degree of longitude in the latitude S.5 of the parallel on which the ship sailed, viz. 49° 30' N. or S., as per Example XIV., page 22. Answer. QUESTIONS To exercise the Learner in the foregoing Rides. QUESTION L A ship in the latitude of 32° North or South, sails due East or "West, till her Differ- ence of Longitude is 384 miles ; required the Distance sailed. BY CASE I. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, Latitude 32° gives 50.88282 the number of miles in a degrea of longitude. 384 V 50 88282 '- = 325.65005 the number of miles in the Distance sailed = 325.7 Answer. 60 QUESTION IL A ship from the latitude of 53° 36^ North or South and longitude 10° 18^ East sails due West, the Distance 236 miles ; required the Difference of Longitude and the Longitude in. BY case II. Latitude 53° 36' gives 35.60505 the number of miles in a degree of longitude, as per Example Y., page 14. OQfi v/ CQ 3Q8 - = 397.69639 the number of miles in the Difference of Longitude = -^ = 6° 38' W. 35.60505 60 Which subtracted from the Longitude left 10 18 E. Gives the Longitude the ship is in ... 3 40 E. Answer. QUESTION IIL If two ships in the latitude of 44° 30^ North or South, Distant 216 miles, should sail directly South or North, until they arrive in the latitude of 32° 17^ North or South, what Distance would they be from each other ? BY case II. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, Latitude 44° SO' gives 42.79500 the number of miles in a degree of longitude. 216 X 60 = 302.83912 the number of miles in the Difference of Longitude. 42.79500 BY CASE I. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, Latitude 32° 17' gives 50.72495 the number of miles in a degree of longitude, as per Example VT., page 14. Vl9 R'^Q12 V 50 72495 "^ ^ = 256.02499 the number of miles that the ships are Distant from each other = 256. 60 Answer. QUESTION IV. A ship having run due East or West for three days, at the rate of 5 knots an hour (sailed a Distance of 360 miles) finds she has made a Difference of Longitude of 8° 16'' (= 496 miles) ; on what parallel of Latitude did she sail ? BY CASE III. — — = 43.54839 the number of miles in a degree of longitude in the latitude of the parallel on which *96 tlie ahip sailed, viz. 43° 28' North or South, as per Example XV., page 22. Answer. MIDDLE LATITUDE SAILI^s'G. "When a ship sails due North or due South, she keeps on the same meridian, and therefore does not change her longitude, and her distance run is the difference of latitude : consequently her place is easily determined by the latitude left and the difference of latitude. Again, vrhen a ship sails due East or due West, her differ- ence of longitude is found by the latitude in, and departure or meridian distance, as already explained in Parallel Sailing ; but T^vhen she sails upon any other course, she changes both her latitude and longitude. Now the difference of longitude can- not be found either from the departure, considered as a meridian distance in the latitude left or that come to ; for in the greater latitude it would give the difference of longitude too much, and in the less latitude too little : the departure is therefore taken as a meridian distance in the mean or middle of the two latitudes, and then the difference of longitude can be found as in Parallel Sailing. Hence this method, it is evident, is compounded of Plane Sailing and Parallel Sailing. Middle Latitude Sailing, therefore, is that in which the Earth and Sea are consi- dered as constituting a sphere or globe ; the meridians as great oircles on its surface meeting each other at the poles ; the ship as sailing at any time along a curved line (called the " Loxodromic" or oblique curve) cutting the meridians in the same constant angle, while on the same tack ; and the departure in going from one place to another as equal to the distance between their meridians on the middle par- allel of latitude between the two places. The Middle Latitude is half the sum of the two latitudes when they are of the same name, that is, both North or both South ; but half their difference when of contrary names, that is, one North and the other South. This method of sailing, although not strictly accurate, especially in high latitudes, approaches sufficiently near the truth for a day's run or the distance a sliip may sail in twenty-four hours. It is used principally in low latitudes, or when the two latitudes do not differ much from each other, and the difference of longitude is small, because the Meridional Parts, in Table III. of the Epitome, are calculated to whole miles only, which renders them inapplicable to day's work.* * Bo-^ditch says, " Middle Latitude Sailing is only an approximation, but it is very mnch used in calculat- ing short runs and days' works; but in crJculating large distances across distant parallels, it is liable to error." Yet notwithstanding this, he gives examples in several instances, as if he did not know they were unsuitable or improper, and in opposition to the well-known fact above stated. See Example of Case I., Questions for Exercise T., TI., Y., YIII., and IX. These properly require to be solved by Mercator's Sailing, which is perfectly accurate. (46) MIDDLE LATITUDE SAILING. 47 CASE I. Given the Latitudes and Longitudes of two places, to find their Distance and Bear- ing (or Course). RULE. Ist. Tofnd the Departure. Multiply the Difi'erence of Longitude in miles, by the number of miles in a degree of longitude in the Middle Latitude, and divide the product by 60 ; the quotient will be the number of miles in the Departure. 2d. Tojind the Distance. Add the square of the Difi'erence of Latitude in mJles, to the square of the Depart- ure in miles, and from the sum, extract the square root ; which will be the number of miles in the Distance. Zd. Tojind the Course. Multiply the Difi'erence of Latitude in miles, by GO, and divide the product by the Distance in miles ; the quotient will be the number of miles in a degree of longi- tude in latitude equal to the Course. EXAMPLE. Required the Distance and Course between Cape Cod lighthouse, in the Latitude of 42° 03^ N., Longitude 70° 04^ W., and the Island of St. Marv (one of the Azores or Western Islands), in the Latitude of 36° 59^ N., and Longitude 25° 10^ W. Be7nark.—The latitude of Cape Cod (Hio-hknd) light is 42° 02^ N., and the lon- gitude is 70° 04^ W. The latitude of the Island of St. Mary is 36° 50' N., and the longitude is 25° 16^ W. But as we want to work out all the"^ Examples in Bowditch by our method, we must use his figures, although knowing them to be not strictly correct. Cape Cod's Latitude... 42° 03' N 42° 0.3' Cape Cod's Longitude... 70° 04' W. St. Mary's " ...36 59 N 36 59 St. Blary's "" ... 25 10 W. Difference of Latitude 5 04 2)79 02 Difference of Longitude 44 54 60 60 39 31 Dif. of Lat. in miles... 304 Middle Latitude. Dif. of Long, in miles... 2(i94 Is^ Tojind the Departure. By the Table of Miles in a degree of Longitude in every Latitude, as per Example \l'i., page 14, The Middle Latitude 39° 31' gives 46.28633 the number of miles in a degree of longitude. 2694 X 46.28633 ~ = 2078.25622 the number of miles in the Departure = 2078. Answer. 2d. Tojind the Distance. F (304;2 + (2078.256)!" == |/92416 + 4319148.001536 = -,/4421564.001536 = 2102.75153 the number of miles in the Distance = 2102.8 Answer. 3c?. Tojind the Course. 304 X 60 2102.752 = ^'^"'^^^ t^®cioT/\^'* ""^ ™"^' '"^ \ '^^^4^ «f longitude, in latitude equal to the Course, viz. S. 81° 41' L., as per Example X\I., page 22. Answer. Because the Island of St. :\Iary is in a less North latitude than Cape Cnd. therefore St. Mary is to the Southward of Cape Cod ; and because the Island of St. Mary is in a less West longitude than Cape Cod, therefore St. Mary is to the Eastward of Care Cod Hence, Cape Cod lighthouse bears from the Island of St, Mary, on the point of the compass directly opposite, viz. N. 81° 41' W. » j> i-^ ^<' "^ 48 MIDDLE LATITUDE SAILING. CASE II. Given both Latitudes, one Lonj^itude, and the Departure from the meridian, to find the Distance, Course, and Difference of Longitude. RULE. Isi. Tojind the Distance. Add the square of the Difference of Latitude in miles, to the square of the Depart- ure in miles, and from the sum extract the square root, which will be the Distance in miles. 2d. To find the Course. Multiply the Difference of Latitude in miles, by 60, and divide the product by the Distance in miles ; the quotient will be the number of miles in a degree of longitude in latitude equal to the Course. Zd. To find the Difference of Longitude. Multiply the Departure in miles, by 60, and divide the product by the number of miles in a degree of longitude in the Middle Latitude ; the quotient will be the Dif- ference of Longitude in miles. EXAMPLE. A ship in the Latitude of 49° 57^ North, and Longitude 15° 16^ West, sails South- Westerly till her Departure is 194 miles and Latitude in, is 47° 18^ North. Required the Distance, Course, and Longitude in. Latitude left 49° 57' N 49° 57' in 47 18 N 47 18 Difference of Latitude 2 39 2)97 15 60 48 38 Middle Latitude. Dif. of Lat. in miles... 159 \st. To find the Distance. |/(159)2 + (194)3 = |/25281 + 37636 ==^ |/62917 = 250.83261 the number of miles in the Distance = 250.8 Answer. 2d. To find the Course. 169 X 60 _—-__- == 38.03327 the number of miles in a degree of longitude, in latitude equal to the Course, viz. 250-833 g_ 50O 40^ ^ff ^ ^s per Example XVII., page 23. Answer. Because the place bound to, is in a less North latitude than the place left and to the Westward of it. Zd. To find the Difference of Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example VIII., page 14. The Middle Latitude 48° 38' gives 39.65247 the number of miles in a degree of Longitude. 194 X 60 294 ■ = 293.54540 the number of miles in the Difference of Longitude = — = 4° 54' W. Which added to the Longitude left 15 16 W. Gives the Longitude in 20 10 W, Answer. MIDDLE LATITUDE SAILING. 49 CASE III. Given one Latitude, the Course, and Distance, to find the Difference of Latitude and Difference of Longitude. RULE. Isi. To find the Difference of Latitude. Multiply the Distance in miles, by the number of miles in a degree of longitude, in latitude equal to the Course, and divide the product by 60 ; the quotient will be the Difference of Latitude in miles. 2cZ. To find the Departure.* Multiply the Distance in miles, by the number of miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the product by 60 ; the quotient will be the Departure in miles. Zd. To find the Difference of Longitude. Multiply the Departure in miles, by 60, and divide the product by the number of miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- ference of Longitude in miles. EXAMPLE. A ship in the Latitude of 42° 30^ North, and Longitude 58° bV West, sails S.E. by S. 300 miles. Eequired the Latitude and Longitude in. \st. To find the Difference of Latitude. By the Table of Points opposite to the Compass, on page 10, The Course S.E. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in lati- tude equal to it. 300 X 49.88810 249 — = 249.44050 the number of miles in the Difference of Latitude = — = 4^ 09' S. Answer. W) 60 Latitude left 42° 30' N 42° 30^ Diflerence of Latitude 4 09 S. Latitude in 38 21 N 38 21 Answer. 2)80 61 40 26 Middle Latitude. 2d. To find the Departure. By the Table of Points opposite to the Compass, on page 10, The Complement of the Course, yiz. 5 points = 56° 15' gives 33.33418 the number of miTes fn a degree of Ion.- gitude in latitude equal to it. 300 X 33.33418 • — = 166.67090 the number of mil^ in the Departure = 16S.T Answer. 3(f. To find the Difference of Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example IX., page 15, The Middle Latitude 40° 26' gives 45.66964 the number of miles in a degree of longitude. = 218.96954 the number of miles in the Difference of Longitude = — - = 3"= 39^ EL Whicli subtracted from the Longitude left 58 51 W. Gives the Longitude in. 55 12 W» Answer. * The Departure may be found otherwise, by Case IT. of Plane Sailing, thus ; |/(Di8tan<-e)2 — (Dif. of Lat.)2 == -|/(300y^ — (249.441)2 = |/'90000 — 62220.812481 = |^/ 27779.187519 =- 166.67090 the number of miles in the Departure = 166,7 Answer. 4 50 MIDDLE LATITUDE SAILING. CASE IV. Given both Latitudes and the Course, to find the Distance, Departure, and Differ- ence of Longitude. RULE. 1st. To find the Distance. Multiply the Difference of Latitude in miles, by 60, and divide the product by the number of miles, in a degree of longitude in latitude equal to the Course ; the quo- tient will be the Distance in miles. 2d. To find the Departure.'^ Multiply the Distance in miles, by the number of miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the product by 60 ; the quotient vs^ill be the Departure in miles. Zd. To find the Difference of Longitude. Multiply the Departure in miles, by 60, and divide the product by the number of miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- ference of Longitude in miles. EXAMPLE. Suppose a ship sailing from a place in the Latitude of 40° 57^ North, and Longi- tude 30° West, makes a Course of S. 39° TV., and then, by observation, is in the Lat- itude o«f 47° 44^ North ; required the Distance run, Departure, Difference of Longi- tude, and the Longitude in. Latitude left 49° 57' N 49° 57' in 47 44 N 47 44 Difference of Latitude 2 13 2)97 41 60 48 61 Middle Latitude. Dif. of Lat. in miles... 133 \st. To find the Distance. By the TtCble of Miles in a degree of Longitude in every Latitude, on page 9, The Course S. 39° W. gives 46.62882 the number of miles in a degree of longitude. ^ = 171.13871 the number of miles in the Distance= 171.1 Answer. 46.62882 2d. To find the Departure. By the Table fu'Jf. By the Table of Miles in a degree of Longitude in every Latitude, as per Example X., page 15, The Middle Latitude 48° 51' gives 39.48194 the number of miles in a degree of longitude. 107 701 V 60 164 ^ = 163.67129 the number of miles in the Difference of Lonsritude = — - -= 2° 44' W. 39.48194 60 Which added to the Longitude left .30 00 W, Gives the LongitTide in 32 44 W. AnsAver. * The Departure may be found otherwise, by Case IV. of Plane Sailivg, thus :— T/(Distance)2 — (Dif. of Lat.)2 = i/''(171 .13871)2— (13S)2 = -j/292'!8,45^n604641 — 17689 == |/ 11599.4580604641 = 107.70078 the number of miles in the Departure ^ 107.7 Answer. MIDDLE LATITUDE SAILING. 51 CASE V. Given both Latitudes and the Distance, to find the Course, Departure, and Differ- ence of Longitude. RULE. 1st. Tojind the Course. Multiply the Difference of Latitude in miles, by 60, and divide the product by the Distance in miles ; the quotient will be the number of miles in a degree of longitude, in latitude equal to the Course. 2d. To find tlie Departure.* Multiply the Distance in miles, by the number of miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the product by 60 ; the quotient vrill be the Departure in miles. Zd. To find the Difference of Longitude. Multiply the Departure in miles, by 60, and divide the product by the number of miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- ference of Longitude in miles. EXAMPLE. Suppose a ship sails 300 miles North-Westerly from a place in the Latitude of 37° North, and Longitude 32° 16^ West, until she is in the Latitude of 41° North ; required her Course, Departure, Difference of Longitude,, and Longitude in. Latitude in 41° 00' N... 41° 00' left 37 00 N 37 00 240 X Difference of Latitude 4 00 2)78 00 60 39 00 Middle Latitude. Dif. of Lat. in miles... 240 l5^. To find tlie Course. 48.00000 the number of miles in a degree of longitude, in latitude equal to tlie Course, vie. N. 36° 52' W., as per Example XVIII., page 23, Answer. 2d. To find the Departure. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XI., page 15, The Complement of the Course 53° 08' gives 35.99741 the number of miles in a degree of longitude. 300 X 35 99741 • —-^ — - = 179.98705 the number of miles in the Departure = 180. Answer. 60 3c?. To find the Difference of Longitude. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, The Middle Latitude .39° 00' gives 46.62882 the number of miles in a degree of longitude. 179 987 X 60 232 — : — —-— = 231.59968 the number of miles in the Difference of Longitude = — - = 3° 52' W. 46.62882 60 Which added to the Longitude left 32 16 W, Gives the Longitude in 36 08 W. Answer. The Departure may be found otherwise, by Case IV. of Plane Sailing, thus ; |/(Distance)^ — (Dif. of Lat.): = •^/(300)2 — (240)2 = -j/gOOGO — 57600 = i/32400 = 180 the number of miles in the Departure. Answer. 52 MIDDLE LATITUDE SAILING. CASE VI. Given one Latitude, the Course, and Departure, to find the Distance, DijBTofence of Latitude, and Difference of Longitude. RULE. 1st. Tojind the Distance. Multiply the Departure in miles, by 60, and divide the product by the number of miles in a degree of longitude, in latitude equal to the Complement of the Course ; the quotient will be the Distance in miles. 2d. Tojind the Difference of Latitude* Multiply the Distance in miles, by the number of miles in a degree of longitude, in latitude equal to the <}ourse, and divide the product by 60 ; the quotient will be the Difference of Latitude in miles Zd. Tojind the Difference of longitude. Multiply the Departure in miles, by 60, and divide the product by the number of miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- ference of Longitude in miles. EXAMPLE. A ship in the Latitude of 50° 10^ South, and the Longitude of 30° 00^ East, sails E.S.E. until her Departure is 160 miles ; required her Distance sailed. Difference of Latitude, and Difference of Longitude. \st. Tojind the Distance. By the Table of Points opposite to the Compass, on page 10, Th9 Complement of the Course 2 points = 22° 30' gives 55.43268 the number of miles in a degree of longi- tude in latitude equal to it. -— -— — = 173.18304 the number of miles in the Distance = 173.2 Answer. 55.43268 2d. Tojind the Difference of Latitude. By the Table of Points opposite to the Compass, on page 10, The Course E.S.E. 6 points = 67° 30^ gives 22.96101 the number of miles in a degree of longitude in latitude equal to it. 17*^18^ V 22Q6101 66 1. o. o ^ . _ ^ 66.27428 the number of miles in the Difference of Latitude = 66.3 — - 1° 06' S. 60 Latitude left 50° 10' S 50° 10' Difference of Latitude 1 06 S. Latitude in 51 16 S 51 16 Answer. 2)101 26 50 43 Middle Latitude. 3c?. Tojind the Difference of Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XII., page 15, The Middle Latitude 50° 43' gives 37.98927 the number of miles in a degree of longitude. == 252.70293 the number of miles in the Difference of Longitude = ^ = 4° 13' E. 37.98927 60 Which added to the Longitude left 30 00 E. Gives the Longitude in 34 13 E. Answer. * The Difference of Latitude may be found otherwise, by Case V. of Plane Sailing, thus : — ^(Dl3tauce)2 — (Departure)^ = |/ (173.183)2— (160)2 = |/-29992,351489 — 25600 == |/ 4392.351489 = 66.27474 the number of miles in the Difference of Latitude = 66.3 Answer. MIDDLE LATITUDE SAILING. 53 CASE VII. Given one Latitude, the Distance sailed, and Departure from the meridian, to find the Course, Dillerence of Latitude, and DiflFerence of Longitude. RULE. Isi. Tojind th^ Course. Multiply the Departure in miles, by 60, and divide the product by the Distance in miles ; the quotient will be the number of miles in a degree of longitude, in latitude equal to the Complement of the Course. Subtract the Complement of the Couree from 90°, and the remainder will be the Course in degrees and minutes. 2d. Tojind the Difference of Latitude.^ Multiply the Distance in miles, by the number of miles in a degree of longitude, in latitude equal to the Course, and divide the product by GO ; the quotient will be the Diti'erence of Latitude in miles. 3c?. Tojind the Difference of Longitude. Multiply the Departure in miles, by 60, and divide the product by the number of miles in a degree of longitude, in the Middle Latitude ; the quotient will be the Dif- ference of Lon*d. To find the Middle Latitude. Multiply the Departure in miles, by 60, and divide the product by the Difference of Longitude in miles ; the quotient will be the number of miles in a degree of lon- gitude, in the Middle Latitude. 4ith. To find the two Latitudes. If in North Latitude sailing South, or in South Latitude sailing North, to the Middle Latitude (which is half the sum of the two latitudes) add half the Difference of Latitude in degrees and minutes ; and the sum will be the number of degrees and minutes in the Latitude left. And from the Middle Latitude subtract half the Difference of Latitude in degrees and minutes ; and the remainder will be the num- ber of degrees and minutes in the Latitude in. Again, if in North Latitude sailing North, or in South Latitude sailing South, to the ^Iiddle Latitude add half the Dif- erence of Latitude in degrees and minutes, and the sum will be the number of degrees and minutes in the Latitude in. And from the Middle Latitude subtract half the Difference of Latitude in degrees and minutes, and the remainder will be the number of degrees and minutes in the Latitude left. '^^ When the ship crosses the Equator, subtract twice the Middle Latitude, from the Difference of Latitude, in degrees, &c., between the two places, and one-half of the remainder will be the less of the two Latitudes. Subtract the less Latitude from the Difference of Latitude, in degrees, &c., between the two places, and the remainder will be the greater of the two Latitudes, but of a different name from that of the less Latitude. EXAMPLE. A ship in North Latitude, and Longitude 51° 26^ W., sailed between the South and East 648 miles, when it was found she had made a Departure of 543 miles, and arrived in Longitude 41° 14'' W. ; required the Latitude left, the Latitude in, and the Course steered. \st. To find the Difference of Latitude. |./(64S)2 — (543)2 = |/41990J: — 294849 = |/l25055 = 353.63399 the number of miles in the Difference of Li 2d. To find the Course. 354 Latitude = — = 5° 54' Answer. 60 648 353.fi.34 X 60 ■■ 32.74389 the number of miles in a degree of longitude in latitude equal to the Course, viz. S. 56° 56' E., as per Example XXII., page 24. Answer. Zd. To find the Middle Latitude. Longitude left 51° 26' W. in 41 14 W. Difference of Longitude 10 12 == 612 miles. 543 X fiO — — — — = 53.23529 the number of miles in a degree of longitude in the Middle Latitude, viz. 27° 28', as per ^^^ Example XXIII., page 24. Answer. Ath. To find the two Latitudes. Half the sum of the two Latitudes (the Middle Latitude) 27° 28' 27° 28' Half the difference of the two Latitudes (the Difference of Latitude) -|- 2 57 — 2 57 The sumf gives the Latitude left 30 25 N. 24 31 N. Answer. The difference! givet^ the Lat- itude in. Answer. * This Case cannot be worked by Murcatnr's Sailing. Bowditch has omitted to give it ! Why? t See Roche's Kuclid, Book V., Prop. XXX. (by the late M.uitin Koche, formerly Professor of Mathematics iu the U. S. Navy). 56 MIDDLE LATITUDE SAILING. . QUESTIONS To exercise the Learner in the foregoing Rules. QUESTION L Required the Distance and Course between two places, one in the Latitude of 37° 55^ North, and the Longitude of 54° 23^ West ; the other in the Latitude of 32° 38^ North, and the Longitude of 17° 05^ West. BY CASE I. Latitude of one place 37° 6^ N 57° 55' Longitude of one place 54° 23' W. " " the other place...32 38 N 32 38 " " the other place 17 05 W. DifiFerence of Latitude 5 17 2)70 33 Difference of Longitude 37 18 60 60 35 17 Dif. of Latitude in milea 317 Middle Latitude. Dif. of Longitude in miles 2238 1st. To find the Departure. Bj the Table of miles in a degree of Longitude in every Latitude, as per Example XV., page 16. The Middle Latitude 35° 17' gives 48.97821 the number of miles in a degree of Longitude. 2238 V 48 97821 ^ — — = 1826.88723 the number of nules in the Departure = 1826.887 Answer. 60 2d. To find the Distance. l/(317)2 -f (1826.887)2 = -1/100489 + 3337616.110769 = t/ 3438005.1107 69 = 1854.18583 the number of miles ' ' ' in the Distance =« 1854 Answer. Zd. To find the Course. —^ = 10.25782 the number of miles in a degree of Longitude in Latitude equal to the Course, viz.: 1854.186 g, 80° 09' E., or N. 80° 09' W., as per Example XXIV., page 24. Answer. QUESTION XL Required the direct Distance and Course from a place in the Latitude of 36° 55'' South, and the Longitude of 20° 00^ East, to another place in the Latitude of 32° 38^ South, and the Longitude of 8° 54^ West. BY CASE I. Latitude of one place 36° 65' S 36° 55' Longitude of one place 20° 00' E. •' the other place 32 38 S 32 38 " the other place 8 64 W. Difference of Latitude 4 17 2)69 33 Difference of Longitude 28 54 60 60 34 47 Dif. of Latitude in miles 257 Middle Latitude, Dif. of Longitude in miles 1734 1st. To find the Departure. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XVI., page 16, The Middle Latitude 34° 47' gives 49.27884 the number of miles in a degree of Longitude. 1734 X 49.27884 ^ ^424.16848 the number of miles in the Departure = 1424.158 Answer. 60 2d. To find the Distance. 1 /(•757VJ 4. a424.168j2 == i/66049 + 2028226.008964 =-i/ 209427 5. 008964 = 1447.16102 the number of milea y ^ ^ ^ '^ ' V y in the Distance = 1447 Answer. Zd. To find the Course. ^^' ^ = 10.65535 the number of miles in a degree of Longitude in Latitude equal to the Course, viz. : 1447.161 N. 79° 46' W., because the place bound to, is in a less South Latitude than the place left, and to the Westward of it, as per Example XXV., page 25. Answer. QUESTION IIL A ship from the Latitude of 37° 30^ South and the Longitude of 60° 00^ East, sails North 79° 56^ West, 202 miles ; required the Latitude and Longitude in. MIDDLE LATITUDE SAILING. 57 BY CASE III. IsL Tojind the Difference of Latitude. By the Table of Allies in a ilogrce of Longitude in every Latitude, as per Example XYIT., page 16, The Course N. 79° 56' W. gives 10.48761 the number of miles in a degiee of longitude in latitude equal to it. '^-— '■ — — = 35.30829 the number of miles in the Difference of Latitude = — = 0"= 35' N. Answer CO 60 Latitude left 37° 30' S 37° 30' Difference of Latitude 35 N. Latitude in 36 55 S 36 55 A uswer. 2 )74 -25 37 13 Middle Latitude. Id. Tojind the Departure. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XYIIT., page 16, The Complement of the Course, viz. 10° 04, gives 59.07621 the number of miles in a degree of longitude in latitude equal to it. 202 X 59.07621 — -: = 198.88991 the number of miles in the Departure = 198.890. Answer. 60 Zd. Tojind the Difference of Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XIX., page 16, Tlie Middle Latitude 37° 13' gives 47.78069 the number of miles in a degree of Longitude. -1— • • = 249.75358 the number of miles in the Difference of Longitude = -— - = 4° 10' W. 47.7SU6y ° 60 Which subtracted from the Longitude left 60 00 E. Gives the Longitude in 55 50 E. Answer. QUESTION IV. A ship from the Latitude of 34° 35^ North and the Longitude of 45° 16^ West, sails South 83° 36^ East, 101 miles ; required her Latitude and Longitude. BY CASE III. \st. To find the Difference of Latitude. By the Table of miles in a degree of Longitude in every Latitude, as per Example XX., page 16, The Course S. 83° 36' E. gives 6.68813 the number of miles in a degree of Longitude in Latitude equal to it. '— — '- =^ 11.25835 the number of miles in the Difference of Latitude = = 0° 11' S. Answer. Latitude left 34° 35' N 34° 35' Difference of Latitude 11 S. Latitude in 34 24 N 34 24 Answer. 2)68 59 34 30 Middle Latitude. 2d. To find the Departure. By the Table of miles in a degree of Longitude in every Latitude, as per Example XXI., page 17, The Complement of the Course, viz. 6° 24' gives 59.62592 the number of miles in a degree of longitude in latitude equal to it. 101 X 59.62592 . = 100.37029 the number of miles in the Departure = 100.370. Answer. 60 Zd. To find the Differ enoe of Longitude. By the Table of miles in a degree of Longitude in every Latitude, on page 9, The Middle Latitude 34° 30' gives 49.44761 the number of miles in a degree of Longitude. 100 370 X 60 122 '■ = 121.78950 the number of miles in the Difference of Longitude = = 2° 02' E. 49.44761 ■^ 60 Which subtracted from the Longitude left 45 16 W. Giyes the Longitude in 43 14 W. Answer. 58 MIDDLE LATITUDE SAILING. QUESTION V. A ship in the Latitude of 49° 57^ North, and the Longitude of 15° W West, sails South-Wcsterly till her Departure is 789 miles, and Latitude in 39° 20^ North ; re- quired the Distance, Course and Longitude in. BY CASE II. Latitude left 49° 57' N 49° 57 in 39 20 N 39 20 Difference of Latitude 10 37 2)89 17 60 44 39 Middle Latitude. Dif. of Lat. in miles... 637 1st. To find the Distance. 1/(637)2 + (789)2 =i/405769 + 622521 = i/ 1028290 = 1014.04635 the number of miles in the Distance ^ y * = 1014, Answer. 2cZ. To find the Course. 637 V 6C -————- == 37.69058 the number of miles in a degree of longitude in latitude equal to the Course, viz., 1014.04635 g 510 05/ -^y^^ ^s per Example XXVI., page 25. Answer. Because the place bound to, is in a less North latitude than the place left and to the Westward of it. 3^. To find the Difference of Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXII., page 17, The Middle Latitude 44° 39' gives 42.68473 the number of miles in a degree of longitude. ^ — = 1109.06172 the number of miles in the Difference of Longitude = = 18° 29' W. 42.68473 60 Which added to the Longitude left 15 16 W. Gives the Longitude in 33 45 W. Answer. QUESTION VL A ship in the Latitude of 42° 30^ North, and the Longitude of 58° bV West, sails S.E. by S. 591 miles ; required the Latitude and Longitude in. BY CASE III. \st. To find the Difference of Latitude By the Table of Points opposite to the Compass, on page 10, The Course S.E. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in latitude equal to it. — '■ = 491.39778 the number of miles in the Difference of Latitude = ■ — =8° 11' S. Answer. 60 60 Latitude left 42° 30' N 42° 30' Difference of Latitude 8 11 S. Latitude in 34 19 N 34 19 Answer. 2)76 49 38 25 Middle Latitude. 2d. To find the Departtire. By the Table of Points opposite to the Compass, on page 10, The Complement of the Course, viz., 5 points = 56° 15' gives 33.33418 the number of miles in a degree of longitude in latitude equal to it. 591 V^ 33 33418 . '' ' = 328.34167 the number of miles in the Departure = 328.342 Answer. 60 Sd. To find the Difference of Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXIII., page 17, The Middle Latitude 38° 25' gives 47.01065 the number of miles in a degree of longitude. 328 342 V 60 419 419.06504 the number of miles in the Difference of Longitude = — -- = 6° 59' E. 47.01065 ■ 60 Which subtracted from the Longitude left 58 51 W. Gives the Longitude in 51 52 W. Answer. MIDDLE LATITUDE SAILING. 59 QUESTION VIL Suppose a ship sailing; from a pliiee in the Latitude of 49° 57^ North, and the Longitude of 30° 00^ West, makes a Course of S. 3*.)° W., and then, by observation, is iu the Latitude of 45° 31^ A'orth; required the Distance and Longitude in. BY CASE IV. Latitude left 49057' N 49° 57' in 45 31 N 45 31 Difiference of Latitude 4 26 2;9o 28 60 47 44 Middle Latitude. Dif. of Lat. in miles... 266 IsL Tojind the Distance. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, The Course S. 39° W. gives 46.62882 the number of miles in a degi'ee of longitude in latitude equal to it. '- ^^ = 342.27758 the number of miles in the Distance = 342.3 Answer. 46.62882 2d. Tojind the Departure. By the Table of Miles in a degree of Longitude in every Latitude, on page 9, The Complement of the Course, viz., 51° gives 37.75922 the number of miles in a degree of longitude in lati- tude equal to it. Q 1.7 o'Tfi y 37 75Q22 -^^ — '—^ — = 215.40251 the number of miles in the Departure = 215.403 Answer. 60 2>d. Tojind the Difference oj Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXIY., page 17, The Middle Latitude 47° 44' gives 40.35487 the number of miles in a degree of Longitude. " ^ 40 V 48 ^ = 320.26320 the number of miles in the Difference of Longitude = -^ = 5° 20' W. Which added to the Longitude left 30 00 W. Gives the Longitude in 35 20 W. Answer. QUESTION VIIL A ship in the Latitude of 50° 10^ South and the Longitude of 30° 00^ East, sails E.S.E. until her Departure is 957 miles. Required her Distance sailed, and Latitude and Longitude in. BY CASE VI. 1st. To jind the Distance. By the Table of Points opposite to the Compass, on page 10, The Complement of the Course, viz. 2 points = 22° 30' gives 55.43268 the number of miles in a degree of lon- gitude in latitude equal to it. 957 X 60 ,, ,„, ■ = 1035.85105 the number of miles in the Distance = 1036. Answer. 00.43268 2d. Tojind the Difference oj Latitude. By the Table of Points opposite to the Compass, on page 10, The Course E.S.E. 6 points = 67° 30' gives 22.96101 the number of miles in a degree of longitude in latitude equal to it. 1035.851 X 22.96101 396 -— = 396.40309 the number of miles in the Difference of Latitude = -— = 6° 36' S. ^ 60 Answer. Latitude left 80° 10' S 50° 10' Difference of Latitude 6 36 S. Latitude in 56 46 S 56 46 Answer. 2)106 56 53 28 Middle Latitude. 2>d. Tojind the Difference oj Longitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXV., page 17, The Middle Latitude 53° 28' gives 35.71737 the number of miles in a degree of Longitude. 957 X 60 1608 -^nrr^^ = 1607.62117 the number of miles in the Difference of Longitude = — — = 26° 48' E. oo.ilVoT 60 Which added to the Longitude left 30 00 E. Gives the Longitude in 56 48 E. Answer. 60 MIDDLE LATITUDE SAILING. QUESTION IX. A ship in the Latitude of 49° 30^ North, and the Longitude of 25° 00^ West, sails South-Easterly 645 miles, until her Departure from the meridian is 500 miles ; required the Course, and the Latitude and Longitude in. BY CASE VII. IsL Tojind the Course. fiOO X 60 ■ — = 46.51163 the number of miles in a degree of longitude, in latitude equal to the Complement **** of the Course, viz., 39° 11', as per Example XXVII., page 25. The Complement subtracted from 90° gives the Course S. 50° 49' E., as per Question Answer. 2d. To Jind the Difference of Latitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXVI., page 17, The Course S. 50° 49' E. gives 37.90817 the number of miles in a degree of longitude in latitude equal to it. 645 X 37.90817 408 — = 407.51283 the number of miles in the Difference of Latitude = -— - = 6° 48' S. Answer. dO 60 Latitude left 49° 30' N 49° 3^ Difference of Latitude 6 48 S. Latitudein 42 42 N 42 42 Answer. 2)92 12 46 06 Middle Latitude. Zd. To find the Difference of Longitude. By the Table of miles in a degree of Longitude in every Latitude, as per Example XXVII., page 18, The Middle Latitude 46° 06^ gives 41.60399 the number of miles in a degree of longitude. 500 X 60 721 = 721.08468 the number of miles in the Difference of Longitude = — =12° 01' E. 41.60399 *' 60 Which subtracted from the Longitude left 25 00 W. Gives the Longitude in 12 59 "W. Answer. QUESTION X.^ A ship in the Latitude of 3° 24^ South and Longitude 38° 22^ West, sailed between the South and East until she made a Departure of 591 milee, and arrived in Longitude 28° 28^ West. Required the Latitude in, Distance sailed, and Course steered. BY CASE VIII. Longitude left 38° 22' W. in 28 28 W. Difference of Longitude 9 54 = 594 miles. \st. To find the Middle Latitude. 591 X 60 — — • = 59.69697 the number of miles in a degree of longitude in the ''^^ Middle Latitude, viz. 5° 46' as per Example XX., page 23. The sum of the two Latitudes 11 32 Subtract the given Latitude, or Latitude left 3 24 S. And the remainder will be the other Latitude, or Latitude in 8 08 S., as per Example. Answer. From which subtract the given Latitude, or Latitude left 3 24 S. Gives the Difference of Latitude between the two places 4 44 = 284 miles. 2(7. To find the Distance. -|/(284)2 -f (591)2 = -i/'80656 + 349281 = -i/429937 = 655.69581 the number of miles in the Distance = ^ *^ *^ 655.696 Answer. 2>d. To find the Course. 284 X 60 -—-——- "= 25.98765 the number of miles in a degree of longitude in latitude equal to the Course, viz. 655.69b s. 640 20' E., as per Example XXXI., page 26. Answer. * This Question is not in Bowditch. See Example of Case yill. MIDDLE LATITUDE SAILING. 61 COMPOUND COURSES. Given the latitude and longitude of the place sailed from, and the different courses and distances sailed, to lind the difference of latitude made good, the depart- ure made good, the latitude in, the distance made good, the course made good, the difference of longitude made good, and the longitude in. RULE. Isi. Tojind the Difference of Latitude made good, and the Departure made good. Make a Traverse Table of the different Courses and Distances sailed, and by it find the Difference of Latitude made good, and the Departure made good, by Part II. of the Rule for Traverse Sailing, on page 35. 2(7. Tojind the Distance made good. Add the square of the Difference of Latitude in miles made good, to the square of the Departure in miles made good, and from the sum of these two squares extract the square root, vrhich will be the Distance in miles made good, by Part I. of the Rule for Case VI. Plane Sailing, on page 31. 3cZ. Tojind the Course made good. Multiply the Difference of Latitude in miles made good, by 60 and divide the pro- duct by the Distance in miles made good, the quotient will be the number of miles in a degree of longitude, in latitude equal to the Course made good, by Part I. of the Rule for Case V., Middle Latitude Sailing, on page 51. 4/7i. To Jind the Difference of Longitude made good. Multiply the Departure in miles made good, by 60 and divide the product by the number of miles in a degree of longitude in the Middle Latitude, the quotient will be the number of miles in the Difference of Longitude made good, by Part III. of the Rule for Case V., Middle Latitude Sailing, on page 51. EXAMPLE. A ship takes her departure from Cape Henlopen, in the Latitude of 38° 47^ North and the Longitude of 75° 05^ West, bearing W, by N., distant 20 miles, and then sails on the following true courses, viz., E.N.E; 15 miles, S.E. 26 miles South 16 miles, W.S.W. 6 miles, N.W. 10 miles, and East 30 miles ; required the Difference of Latitude made good, the Departure made good, the Latitude in, the Distance made good, the Course made good, the Difference of Longitude made good, and the Longitude in. \st. To find the Difference of Latitude made good, and the Departure made good Having calculated all the Differences of Latitude and Departures made on the several Courses and Distances sailed, by Case I., Plane Sailing, arrange them, ac- cording to the Rule, in the form of a TRAVERSE TABLE. Difference of Latitude. Departure. Course. Distance. N. S. E. W. E. by S. 20 3.9 19.6 E.N.E. 15 5.7 13.9 S.E. 26 18.4 18.4 South 16 16.0 • W.S.W. 6 2.3 5.5 1 N.W. 10 7.1 7.1 East 30 30.0 12.8 40.6 12.8 81.9 12.6 12.6 1 Remain der 27.8 69.3 Rer aainder 1 Differ< mce of Latitude made good. Departure made good. 62 MIDDLE LATITUDE SAILING. 2il. To find the Distavce made good. -i/(-27.S)''2 + (69.;i)2 = W-,1±U + 4802.49 =1/ 5575.33 = 74.60813 the nunibor of miles iu the Distance = *'*'*' 74.7 Answer. 3cZ. To find the Course made good. 27 8 V 60 -—- • = 22.33885 the numljer of milos in a dojrroe of lonRitndo in latitude equal to the Pourso ma^lo '*-66813 goofj^ vjy_^ g_ 680 OS/ -y .^^ p^.p Example XXTITI., p. 25. Because the DifTerenc- of \mX\- tude made good is to the Southward, and the Departure made good is to the Eastward. Latitude of Cape Henlopen 38° 47' N 38° 47 Difference of Latitude 28 S. Latitude in 38 19 N 38 19 Answer. 2)77 06 38 33 Middle Latitude. Ath. To find the Difference of Longitude made good. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXYIIT., page 18, The Middle Latitude 38° 33' gives 46.92375 the number of miles in a degree of longitude. AQ ^ \/ fin 8Q — = 88.61184 the number of miles in the Difference of Longitude made good = — ~ = 1° 29' E 46.92375 60 Which subtracted from the Longitude of Cape Henlopen 75 05 W Gives, the Longitude in 73 36 W. Answer. Remark. — This example is the same as that given by Bowditch, at the end of Mercator's Sailing, in his Practical Navigator, where he has attempted . to exem- plify the working of compound courses, both by Middle Latitude and Mercator's Sailing. MERC A TOR'S SAILING. Mercator's Sailixg is called after Gerard Mercator. a Flemisli geographer, who invented and published in the year 156G his chart, on which the meridians are parallel straiglit lines, and consequently all the parallels of latitude are equal to the Equator. The degrees of latitude and the meridian distances are increased in the same proportion. Hence it follows that the Rhumbs which form equal angles with the meridians will be straight linos, which renders this projection of the Earth's surface much more easy and proper for the mariner's use than any other. Mercator's Chart m.ay be roughly explained as follows : — Suppose the globe to be surrounded by a hollow cylinder which touches it everywhere at the Equator, and that the degrees of latitude and longitude are expanded proportionally from the Equator towards the Poles, until it meets the concave of the cylinder, marking upoa it all the lines that are drawn on its surface. Then the cylinder being opened its whole length, and unrolled, will be Mercator's Chart, on which the places bear the same relation and proportion to each other, as do the corresp.mding places on the surface of the globe. The increased meridian between the two parallels is called the meridional differ- ence of latitude, in contradistinction to the actual difference, which is called the proper difference of latitude. The meridional difference of latitude is obtained by finding in Table III. of the Epitome the meridional parts of the two latitudes, and taking the difference between them, when the latitudes are of the same name, but by adding them together when the latitudes are of different names. In Table III. of the Epitome, the meridional parts are only calculated to the nearest mile or minute, and therefore Mercator's Sailing is not strictly accurate when the difference of latitude is small, consequently it is not much used on ship- board in calculating short runs and day's work. To find the Meridional Difference of Latitude. EXAMPLE I. What is the ^Meridional Difference of Latitude, between Cape Cod Light, and the Island of St. Mary, one of the Azores (or Western Islands) ? Latitude of Cape Cod light... 42° 0-3' N Meridional Parts 2786 as per Table III., of the Epitome, " St. Mary 36 59 N " " 2391" " " " " Proper Difference of Latitude 5 04 395 60 Meridional Difference of La^titude in miles. Answer. Proper Dif. of Lat. in miles... 304 EXAMPLE IL What is the Meridional Difference of Latitude, between two places, one in W^ 56'' North Latitude, and the other in 3° 24^ South Latitude. Latitude of one place 14° 56' N Meridional Parts 906, as per Table IIT., of the Epitome, " the other place... 3 24 S " " 204" " " " " Proper Difference of Latitude 18 20 1110 60 Meridional Difference of Latitude in miles. Answer, Proper Dif of T^t. in miles.... 1100 (03) 64 MERCATOll'S SAILING. CASE I. Given the Latitudes and Longitudes of two places, to find the Course (or their Bearing) and Distance. RULE. isi. To find a Distance correspnndinr/ to the Meridional Difference of Latitiide and the Difference of Longitude. Add the square of the Meridional Difference of Latitude in miles, to the square of the Difference of Longitude in miles, and from the sum of these two squares extract the square root; which will be a Distance in miles (but not the dis^a,nce between the two places). 2d. To find the Course from this Distance. Multiply the Meridional Difference of Latitude in miles, by 60, and divide the product by the distance obtained above, the quotient will be the number of miles in a degree of longitude in latitude equal to the Course. Zd. To find the Distance between the two 'places.* Multiply the Proper Difference of Latitude in miles, by 60, and divide the product by the miles in a degree of longitude in latitude equal to the Course ; the quotient will be the number of miles in the Distance between the two places. EXAMPLE. Required the Course and Distance between Cape Cod lighthouse, in the Latitude of 42° 03^ North, and the Longitude of 70° 04^ West ; and the Island of St. Mary (one of the Azores or Western Islands), in the Latitude of 36° 59^ North, and the Longitude of 25° 10^ West. 8i^^ See our remark on this question under Case I., Middle Latitude Sailing. Cape Cod's Latitude 42° 03' N Meridional Partsf 2786 Cape Cod's Longitude.... 70*^ 04' W. St. Mary's " 36 59 N " " 2391 St. Mary's " 25 10 W. Proper Dif. of Latitude... 5 04 Meridional Dif. of Latitude 395 Difference of Longitude 44 54 60 60 Proper Dif. of Lat. in miles 304 Dif. of Longitude in miles 2694 \st. To find a Distance corresponding to the Meridional Difference of Latitude and the Difference of Longitude. t/(395)2 -I- (2694)2 = i/l66025 + 7257636 = -1/7413661 = 2722.80389 the number of miles in a Distance corresponding to the Meridional Dif. of Lat. and the Dif of Long. 2d. To find the Course. 395 ^ 60 ■ = 8.70430 the number of miles in a degree of longitude, in latitude equal to the Course, 2722.804 yiz.^ g. gio 40/ e., as per Example XXIX., page 26. Because the Island of St. Mary is in a less North latitude than Cape Cod, therefore St. Mary is to the Soiithward of Cape Cod; and because the Island of St. Mary is in a less West longitude than Cape Cod, therefore St. Mary is to the Eastward of Cape Cod. Hence Cape Cod lighthouse bears from the Island of St. Mary, on the point of the compass directly opposite, viz., N. 81° 40' W. Answer. Zd. To find the Distance. ^04- V 60 Z> — == 2095.51601 the number of miles in the Distance = 2096.t Answer. 8.70430 * * The Distance may be found otherwise, by the Departure, thus : As Mer. Dif. of Lat. : Dif. of Long. : : Proper Dif. of Lat. : Departure. 395 : 2694 : : 304 : 2073.35696 1 /(Prop. Dif. of Lat.)2 + (Departure)^ =-|/ (304)2 + (2073.35696)2 == i/ 4391225.0835804416 = 2095.52501 = *' *^ *^ 2096. Answer. + The Meridional Parts are taken from Table TIT., of the Epitome. \ Bowditch's answer is not strictly correct. His course by logarithms proportioned to seconds is 81° 39' 31", the secant of which put in the 26. canon, gives 3.32129, the exact logarithm answering to the Distance 2095.524, which at most is only = 2096 and not 2098 miles, as given by him. MERCATOR'S SAILING. 65 CASE II. Given both Latitudes and the Departure to find the Distance, Course, and Differ- ence of Longitude. RULE. 1st. To find the Distance. Add the square of the Proper Difference of Latitude in miles, to the squnro of the Departure in miles, and from the sum of these two squares, extract the square root ; which will be the number of miles in the Distance. 2d. To find the Course. Multiply the Proper Difference of Latitude in miles, by 60, and divide the product by the Distance in miles ; the quotient will be the number of miles in a degree of longitude, in latitude equal to the Course. M. To find the Difference of Longitude.* Multiply the Meridional Difference of Latitude in miles, by 60, and divide the product by the number of miles in a degree of longitude, in Latitude eqiinl to the Course; the quotient will be a Distance in miles, corresponding to the Meridional Difference of Latitude. Multiply this Distance in miles, by the miles in a Je<^ree of longitude, in latitude equal to the Complement of the Course, and divide the pro- duct by 60 ; the quotient will be the number of miles in the Difference of Lon- gitude. EXAMPLE. A ship in the Latitude of 49° 57^ North and the Longitude of 15° 16^ West, sails South- Westerly until her Departure is 197 miles, and then by observation is in the Latitude of 47° 18^ North; required her Distance, Course, and Longitude in. Latitude left 49° 57' N Meridional Parts S470 as per Table III., of tlio Kpitnme, in 47 18 N " " 3229" " " « " Proper Difference of Latitude 2 39 241 60 Meridional Difiference of Latitude. Proper Dif. of Lat. in miles... 159 Ist. To find the Distance. 1/(159)2 + (197)2 ^t/25281 + 38809 = t/ 64090 = 253.16003 the number of miles in the Distance == 253.2 Answer. 2d. To find the Course. 159 X 60 - ocQiftA " == 37.68368 the number of miles in a degree of longitude in latitude equal to the Course, viz.: '^'*-^*^ S. 51° 06' W., as per Example XXX., page 26. Answer. 241 X Zd. To find the Difference of Longitude, = 383.72049 the number of miles in a Distance corresponding to the Meridional Dif. of Latitu'le. By the Table of miles in a degree of Longitude in every Latitude, as per Example XXIX., page !«. The Complement of the Course, viz. 38° 54', gives 46.69456 the number of miles in a degree of longitude in latitude equal to it. '■ ^ — '■ = 298.62766 the number of miles in the Dif. of Longitude = 298.6 = — = 4° 59' W. Which added to the Longitude left 15 Ifi W. Gives the Longitude in .'20 15 W. Answer. * The Difference of Longitude may be found otherwise, by the Departure, thus : As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 159 : 197 : : 241 : 298.59748 = 299 miles. Answer. 66 MERCATOR'S SAILING. CASE III. Given one Latitude, the Course and Distance, to find the Difference of Latitude and Difference of Longitude. RULE. l5^. To find the Proper Difference of Latitude. Multiply the Distance in miles, by the miles in a degree of longitude, in latitude equal to the Course, and divide the product by GO ; the quotient will be the number of miles in the Proper Difference of Latitude. 2d. To find the Difference of Longitude.* Multiply the Meridional Difference of Latitude in miles, by 60, and divide the pro- duct by the number of miles in a degree of longitude, in latitude equal to the Course ; the quotient will be a Distance in miles, corresponding to the Meridional Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the |)roduct by 60 ; the quotient will be the number of miles in the Difference of Longitude. EXAMPLE. A ship in the Latitude of 42° 30^ North, and the Longitude of 58° 51^ West, sails S.W. by S., 300 miles; required the Latitude and Longitude in. \st. To find the Proper Difference of Latitude. By the Table of Points opposite to the Compass, on page 10, The Coxirse S.W. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in latitude equal to it. 800X49.8S810 249 — ■ = 249.44050 the number of miles in the Proper Dif. of Lat. = — - = 4° 09' S., as per Example. Latitude left 42° 30^ N .Meridional Parts 2822, as per Table III., of the Epitome, Proper Difference of Latitude 4 09 S. Latitude in 38 21 N " " 2495" " " " " " Meridional Difference of Latitude 327 2d. To find the Difference of Longitude. 327 X 60 • -= 393.28016 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 49.^ool0 By the Table of Points opposite to the Compass, on page 10, Tlie Complement of the Course, viz, 5 Points = 56° 15' gives 33.33418 the number of miles in a degree of longitude in latitude equal to it. 393.28016 X 33.33418 218 — = 2l8.49453t the number of miles in the Dif of Long.= — =3° 38' W. as per Example. Which added to the Longitude left 58 61 W. Gives the Longitude In 62 29 W. Answer. * The Difference of Longitude may be found otherwise, by the Departure, thus : l/(Distance)2— (Prop. Dif. of Lat.)2 = -|/(300)2 _ (249.4405)2 = -i/goooO — 62220.56304025 == -|/27779.43695975 = 166.67164 Departure. As Proper Dif. of Lat. : Departure : : Meridional Dif of Lat. : Dif of Longitude. 249.4405 : 166.67164 : : 327 : 218.49550 = 218 mUes. Answer. f Eowdi roll's answer is not strictly correct, for by proportion his logarithm 2.33944 answers to 218.495 = 218 only, and not 219 miles, as given by him. MERCATOR'S SAILING. 67 CASE IV. Given both Latitudes and the Course, to find the Distance, and Difference of Lon- gitude. RULE. 1st. To find the Distance. Multiply the Proper Difference of Latitude in miles, by 60, and divide the product by the miles in a degree of longitude, in latitude equal to the Course ; the qliotieut will be the number of miles in the Distance. 2d. To find the Difference of Longitude.* Multiply the Meridional Difference of Latitude in miles, by 60, and divide the product by the number of miles in a degree of longitude in latitude equal to the Course ; the quotient vrill be a Distance in miles, corresponding to the Meridional Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the pro- duct by 60 ; the quotient will be the number of miles in the Difference of Longitude. EXAMPLE. A ship from the Latitude of 49° 57^ North and the Longitude of 30° We'st, sails South 39° West, till she arrives in the Latitude of 47° 44^ North ; required the Dis- tance run, and the Longitude in. Latitude left 49° 57' N Meridional Parts 3470, as per Table in. of the Epitome. in 47 44 " " 3268 " " Proper Dif. of Latitude 2 13 Merid. Dif. of Lat. 202 60 Prop. Dif. of Lat. in miles... 133 1st. To find the Distance, By the Table of Miles in a degree of Longitude in every Latitude, en page 9, The Course S. 39° W. gives 46t62882 the number of miles in a degree of longitude in latitude equal to it. -— — — r = 171.13879 the number of miles in the Distance = 171.1 Answer. 4o,o.^8b2 2d. To find the Difference of Longitude, oryo v/ AA — r~— ,„- -- = 259.92508 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 46.62882 By the Table of Miles in a degree of Longitude in every Latitude, on page 9, The Complement of the Course, viz., 51°, gives 37.75922 the number of miles in a degree of longitude in lati- tude equal to it. S'iQ Q2'i08 V ^7 7'iQ22 164 I—: ;^ = 163.57614 the number of miles in the Dif. of Long. = 163.6 = — - = 2° 44' W., per Ex. 60 60 "Which added to the Longitude left 30 00 W. Gives the Longitude in 32 44 W. Answer. * The Difference of Longitude may be found otherwise, by the Departure, thus : i^/(Di8tance)2 _ (Prop. Dif. of Lat.)3 = |/(171.13879)a — (133)2 = 1/29288.4854426641 — 17689 = -j/ll599.4854426641 = 107.70091 Departure. As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 133 : 107.70091 :: 202 163.57582 = 164 miles. Answer, 68 MERCATOR'S SAILING. CASE V. Givon both Latitudes and the Distance, to find the Course, and Difference of Lon- gitude. RULE. 1st. To find the Course. Multiply the Proper Difference of Latitude in miles, by 60, and divide the pro^ duct by the Distance in miles ; the quotient will be the number of miles in a degree of longitude, in latitude equal to the Course. 2d. To find the Difference of Longitude.* Multiply the Meridional Difference of Latitude in miles, by 60, and divide the pro- duct l)y the number of miles in a degree of longitude, in latitude equal to the Cfuirse ; the quotient will be a Distance in miles, corresponding to the Meridional Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the product by 60 ; the quotient will be the number of miles in the Difference of Longitude. EXAMPLE. A slup from the Latitude of 37° North and the Longitude of 32° 16^ West, sails 300 miles North-Westerly, until she is in the Latitude of 41° North ; required the Course steered and Longitude in. Latitude in 41° N Meridional Parts 2702, as per Table III. of the Epitome. left 37 N " » 2393 " " " " Proper Dif. of Latitude 4 Merid. Dif. of Lat. 309 60 Prop. Dif. of Lat. in miles... 240 \st. To find the Course. - — = 48.00000 the number of miles in a degree of longitude in latitude equal to the Course, viz., 300 N. 36° 52' W., as per Example "XVni., page 23, 2d. To find the Difference of Longitude. 309 V fiO _ = 386.25000 the number of miles in a Distance corresponding to the Meridional Dif of Latitude. 48.00000 By the Table of Miles in a degree of Longitude in every Latitude, as per Example XI., page 15, The Complement of the Course, viz. 53° 08', gives 35.99741 the number of miles in a degi-ee of longitude in latitude equal to it. ■ ' ^ -^''^"'^" ^^J = 231.73333 the number of miles in the Dif. of Long. = ^ = 3° 52' W., as per Example. (jO 60 Which added to the Longitude left 32 16 W. Gives the Longitude in 36 08 W. Answer. * The Difference of Longitude may be found otherwise, by the Departure, thus : •i/(Did. To find the Difference of Longitude. Multiply the Meridional Difference of Latitude in miles, by 60, and divide the product by the number of miles in a degree of longitude, in latitude equal to the Course ; the quotient will be a Distance in miles, corresponding to the Meridional Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the pro- duct by 60 ; the quotient will be the number of miles in the Difference of Longitude. EXAMPLE. A ship from the Latitude of 50° 10^ South and the Longitude of 30° East, sails E.S.E. until her Departure is 160 miles ; required the Distance sailed, and the Lati- tude and Longitude in. \st. To find the Distance. By the Table of Points opposite to the Compass, on page 10, The Complement of the Course, viz., 2 Points = 22° 30', gives 55.43268 the nnmber of miles in a degree of longitude in latitude equal to it. /^. _ = 173.18304 the nnmber of miles in the Distance = 173.2 Answer. 55.43268 2d. To find the Proper Difference of Latitude. By the Table of Points opposite to the Compass, on page 10, The Course E.S.E. 6 points = 67" 30^ gives 22.96101 the number of miles in a degree of longitude in latitude equal to it. 173.18304 X 22.96101 66 . .== 66.27429 the number of miles in the Pi-oper Dif. of Lat. = 66.3 = — = 1° 6' S., as per Ex. Latitude left 50° W S Meridional Parts 3490, as per Table III. of the Epitome. Proper Dif. of latitude 1 06 S. Latitude in 51 16 S " " 3594 " " « « Answer. Meridional Dif. of Lat. 104 ScZ. To find the Difference of Longitude. 104 X 60 ■ - = 271.76505 the number of miles in a DLstance corresponding to the Meridional Dif of Latitude. 271.76505X55.4.3268 „.,„„._ ,^ u . -, - , t... . x 251 . — = 2ol.O/ i 1 the number of miles m the Dif of Long. = — =4° 11' E., as per Example. dO 60 Which added to the Longitude left 30 00 E. Gives the Longitude in 34 11 B. A nswer, • The Difference of Longitude may be found otherwise, by the Departure, thus : As Proper Dif. of Lat. : Departure : : Meridional Dif. of Lat. : Dif. of Longitude. 66.27429 : 160 :: 104 : 251.07776 =2.51 miles. Answer. 70 MERCArOR'S SAILING. CASE VII. Given one Latitude, the Distance sailed, and Departure, to find the Course, Differ- ence of Latitude, and Difference of Longitude. RULE. 1st. Thjind the Course. Multiply the Departure in miles, by 6<3, and divide the product by the Distance in miles ; the quotient vf'iW be the number of miles in a degree of longitude, in lati- tud. To find the Difference of Longitude.^ Multiply the Meridional Difference of Latitude in miles, by 60, and divide the pro- duct by the number of miles in a degree of longitude, in latitude equal to the Course ; the quotient vrill be a Distance in miles, corresponding to the Meridional Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the product by 60 ; the quotient vrill be the number of miles in the Difference of Longitude. EXAMPLE. A ship in the Latitude of 49° 30^ North and the Longitude of 25° West, sails South-Easterly 215 miles, making 167 miles Departure ; required the Course steered, and the Latitude and Longitude in. \st. To find the Course, — -— — ^ 46.60465 the mimber of miles in a degree of longitude in latitude equal to the Complement '-^^ of the Course, viz., 39° 02', as per Example XIX., page 2-3. Answer. The Complement subtracted from 90° 00', gives the Coiarse S. 60° 58' E., as per Example. 2d. To find the Proper Difference of Latitude. By the Table of Miles in a degree of Longitude in every Latitude, as per Example XIII., page 15, The Course, S. 50° 58' E., gives 37.78631 the number of miles in a degree of longitude in latitude equal to it. 215 X 3<.er Difference of Latitude 10 37 Meridional Dif. of Latitude 899 60 Prop. Dif. of Lat. in miles 637 \st. To Jind the Distance. v/ (637)2 + (789)2 = t/405769 -f 622521 =|/l028290 = 1014.04635 the number of miles in the Distance = *^ V V jQj^^^ Answer. 2d. To Jind the Course. = 37.69058 the number of miles in a degree of longitude in latitude equal to the Course, viz.: lOU.046a5 51° 05' W., as per Example XXVI., page 25. Answer, . Zd. To Jind the Difference of Longitude. 8d9 y 60 „V-— — - = 1431.12682 the number of miles in a Distance corresponding to the Meridional Dif, of Latitude. 37.690.dS By the Table of Miles in a degree of Longitude in every Latitude, as per Example XXX., page 18, The Complement of the Course, viz., 38° 55 gives 46.68360 the number of miles in a degree of longitude in latitude equal to it. 1431.126S2 X 46.68360 1114 —■ =1113.50253 the number of miles in the Dif. of Longitude = -— = 18° 34' W., as per Q. Which added to the Longitude left 15 16 W. GiTBs the Longitude in _. 33 50 W. Answer. QUESTION XL A ship in the Latitude of 42° 10^ North and the Longitude of 58° 51^ West, sails South-West by South 591 miles ; required the Latitude and Longitude in. BY CASE III. \st. To Jind the Proper Difference of Latitude. i5y the Table of Points opposite to the Compass, on page 10, The Course S."W. by S. 3 points = 33° 45' gives 49.88810 the number of miles in a degree of longitude in latitude equal to it. 501 y 49.88810 491 — = 491.39779 the number of miles in the Proper Dif of Lat. = — — = 8° 11' S., as per Question. 60 60 Latitude left 42° .30' X Meridional Parts 2822 per Table lU., of the Epitome. Propor Dif. of Latitude 8 11 S. Latitudei.ii .,,.^. 34 19 X. " " 2194 « " « « " « Answer. Meridional Dif. of Latitude 628 Id. To find the Difference of Longitude. 62^ X fiO ■ , = 755.29034 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. By the Table of Poiats opposite to the Compass, on page 10, The Complement of the Course, viz. 5 points = 56° 15', gives 33.33418 the number of miles in a degree of longitude in latitude equal to it. 75'i 29034 V 3.3 '53418 420 " ' ' — , ' •' = 419.61640 the number of miles in the Dif of Long. = — - = 7° 00' W. as per Question. 60 60 Which added to the Longitude left 58 51 W. Gives the Longitude in 65 51 W. Answer. MERCATOR'S SAILING. 73 QUESTIONS To exercise the Learner in the foregoing Rules. QUESTION III. A sliip from the Latitude of 49° 57^ North, and the Longitude of 30° 00^ West, sails S. 39° W. till she arrives in the Latitude of 45° 31^ North ; required the Dis- tance run and Longitude in. BY CASE IV. Latitude left 49° 57' N Meridional Parts 3470, as per Table III. of the Epitome. in 45 31 N " " 3074 " " « " Proper Dif. of Latitude 4 26 Merid. Dif. of Lat. 60 Prop. Dif. of Lat. in miles 266 l5^. To find the Distance. By the Table of Miles in a degree of Longitiido in every Latitude, on page 9, The Course S. 39° "W. gives 46.62882 the number of miles in a degree of longitude in latitude equal to it. • = 342.27759 the number of miles in the Distance = 342.3 Answer. 46-62882 2d. To find the Difference of Longitude. ^ = 509.55611 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 46.62S(>2 By the Table of Miles in a degree of Longitude in every Latitude, on page 9, The Complement of the Course, viz., 51°, gives 37.75922 the number of miles in a degree of longitude in latitude equal to it. 609.55611 X 3/ .75922 ^ ggQ g^-^Qg ^-^^ number of miles in the Dif. of Long. = ^ = 5° 21' W. as per Question. 60 60 Which added to the Longitude left 30 00 W. Gives the Longitude in 35 21 W. Answer. QUESTION IV. A ship from the Latitude of 50° 10^ South and the Longitude of 30° 00^ East, sails E.S.E. until her Departure is 957 miles; required the Distance sailed, and the Lati- tude and Longitude in. BY CASE VI. 1st. To find the Distance. By the Table of Points opposite to the Compass, on page 10, The Complement of the Course, viz., 2 Points = 22° 30' gives 55.43268 the number of miles in a degree of longitude in latitude equal to it. 957 X 60 — — = 1035.85105 the number of miles in the Distance = 1036. Answer. 55.43268 2d. To find the Proper Difference of Latitude. By the Table of Points opposite to the Compass, on page 10, The Course E.S.E. 6 points = 67° 30' gives 22.96101 the number of miles in a degree of longitude in latitude equal to it. 1035.85105 X 22.9610 1 ^ ggg^Qg^j^Q ^^^ number of miles in the Proper Dif. of Lat. = — = 6° 36' S., as per Qu. 60 ^ 60 ' ^ Latitude left 50° 10' S Meridional Parts 3490, as per Table III. of the Epitome. Proper Dif. of Latitude 6 36 S. Latitude in 56 46 S « " 4157 Answer. Meridional Dif. of Lat. 667 2>d. To find the Difference of Longitude. cr,7 X 60 22.96101 1742.95469 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 1742.95409 X 55.4-3268 1610 ^ = 1610.27749 the number of miles in the Dif of Long. = — - = 26° 50' E., as per Qu. Which added to the Longitude left 30 00 E. Gives the Longitude in 56 50 E. Answer. 74 MERCATOR'S SAILING. QUESTIONS 7b exercise (Tie Learner in the foregoing Rules. QUESTION V. A ship in the Latitude of 49° 3(K North, and the Longitude of 25° 00^ West, sails South-Easterly 645 miles, making 500 miles Departure ; required the Course, and the Latitude and Longitude in. BY CASE YII. \st. To find the Course. 600 X 60 — ^jT — = 46.51163 the number of miles in a degree of longitude, in latitude equal to the Complement »**^ of the Course, viz., 39° 11', as per Example XXVII., page 25. Subtract the Complement from 90° OC, and the remainder will be the Course, S. 50° W E., as per Question. Ana. 2d. To find the Proper Difference of Latitude. Bt the Table of Miles in a degree of Longitude in every Latitude, as per Example XXTI., page 17, The Course, S. 50° 49' E., gives 37.90817 the number of miles in a degree of longitude in latitude equal to it. 645 X 37.90817 408 — == 407.51283 the number of miles in the Proper Dif. of Lat. = — - = 6° 48' S., as per Question. DO 60 Latitude left _ 49° BC N Meridional Parts 3428, as per Table in., of the Epitome, Proper Difference of Latitude 6 48 S. Latitude in 42 42 N " " 2839" " " " " " Answer. Meridional Dif of Latitude 5S9 Zd. To find the Difference of Longitude. 589 X 60 •————- = 932.25286 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. — 722.67667 the number of miles in the Dif. of Long. = — - = 12° 03' E., as per Qn. 60 60 932.25286X46.51163 „____, , , -, ■ ,^. r,-r rr '^3 imber of miles in the Dif. of Long. = — ^ 60 "Which subtracted from the Longitude left 25 00 W. Gives the Longitude in 12 57 W. Answer COMPOUND COURSES. Given the latitude and longitude of the place sailed from, and the different courses and distances sailed, to lind the difference of latitude made good, the depar- ture made good, the latitude in, the distance made good, the course made good, the difference of longitude made good, and the longitude in. RULE. Isi. To find the Difference of Latitude made good, and the Departure made good. Make a Traverse Table of the different Courses and Distances sailed, and by it find the Difference of Latitude made good, and the Departure made good, by Part II. of the Rule for Traverse Sailing, on page 35. 2d. To find the Distance made good. Add the square of the Difference of Latitude in miles made good, to the square of the Departure in miles made good, and from the sum of these tTvo squares extract the square root, which will be the Distance in miles made good, by Part I. of the Kule for Case YI. Plane Sailing, on page 31. M. To find the Course made good. Multiply the Proper Difference of Latitude in miles made good, by 60, and divide the product by the Distance in miles made good, the quotient will be the number of miles in a degree of longitude, in latitude equal to the Course made good, by Part I. of the Rule for Case Y., Mercator's Sailing, on page 51. Ath. To find the Difference of longitude made good. Multiply the Meridional Difference of Latitude in miles, by 60, and divide the product by the number of miles in a degree of longitude, in latitude equal to the Course ; the quotient will be a Distance in miles, corresponding to the Meridional Difference of Latitude. Multiply this Distance in miles, by the miles in a degree of longitude, in latitude equal to the Complement of the Course, and divide the pro- duct by 60 ; the quotient will be the number of miles in the Difference of Lrm- gitude'made good, by Part II. of the Rule for Case Y., Mercator^s Sailing, on p. 51. MERCATOR'S SAILING. 75 EXAMPLE. A ship takes her departure from Cape Henlopen, in the Latitude of 38° 47^ North and the Longitude of 75° 05^ West, bcarino; W. by N., distant 20 miles, and then Bails on the foUo^ving true courses, viz., E.N.E. 15 miles, S.E. 26 miles, South 16 miles, W.S.W. 6 miles, N.W. 10 miles, and East 30 miles ; required the Difference of Latitude made good, the Departure made good, the Latitude in, the Distance made good, the Course made good, the Difference of Longitude made good, and the Longitude in. Eemark.—See Compound Courses, in Middle Latitude Sailing, where this Ex- ample is worked out by that method. 1st. To find the Difference of Latitude made good, and the Departure made good. Having calculated all the Differences of Latitude and Departures made on the several Courses and Distances sailed, by Case I., Plane Sailing, arrange them, ac- cording to the Rule, in the form of a TRAVERSE TABLE. i Difference of Latitude. 1 Departure. Course. Distance. \ N. , " S. E. 1 W. E. by S. 20 3.9 19.6 j E.N.E. 15 5.7 13.9 S.E. 26 18.4 18.4 South 16 16.0 w.s.w. 6 2.3 5.5 N.W. 10 7.1 7.1 East 30 30.0 12.8 40.6 12.8 81.9 i 12.6 12.6 1 Remain der 27.8 69.3 Remainder i Difference of Latitude made good. Departure made good. 2d. To find the Distance made good. 1 / (27.8)2 -t- ita.3)2 = -I./772.84 + -tSO-'.-ta = -1/5575.33 = 74.66813 the immber of miles in the Distanco = *■ " * 74.7 Answer. 2>d. To find the Course made good. 27.S X ^0 -— = '22.."3S85 th-e nnmber of miles in a degree of londtude in latitude equal to the Course made ' 4.ti(.^ !3 „noA, viz., S. 68^^ OS' E.. as per Example XX VIII., p. 25. Answer. Because the Difference of Latitude made good is to the Southward, and the Departure made good is to the Eastward. r^atitude of Cane Henlopen 3S° 47' N Meridional Parts 252S, as per Table III., of the Epitome. Proper Differeiice of Latitude... 2S S. Latitude in. 38 19 X. Answer. » 2492 Merid. Dif. of Lat. 36 Ath. To find the Difference of Longitude made good. 90.69253 the number of miles in a Distance corresponding to the Meridional Dif. of Latitude. 36 X 60 22.33S85 By the Table of Miles in a degi-ee of Longitude in every Latitude, as per Example XXXI., page 18, The Complement of the Courst, viz. 21*^ 52', gives 55.68313 the number of miles in a degi-ee of longitude ()fl.K9253 X 55.68313 GO in latitude equal to it. 90 "60 89.73571 the number of miles in the Dif. of Longitude = -~ = 1^ 30' E., as per Ex. Which subtracted from the Longitude left 75 05 W. Gives the Longitude in „. 73 35 W. Answer l^ "i?N \ \