19
tilde, it will be (by rectangular trigonometry,) R : MB : : S,BMK: BK,
i. e. As radius........ 10-00000
is to the distance.156 . 2-19312
so is the co-sine of the course..,..35° 40'. 9-90978
to the proper difference of tat...-127 .. 2-10290
equal to 2° 7'; and since the ship is-sailing from a north latitudetowards the south, therefore tire latitude come to will be 47° 53north. Hence the meridional difference of latitude will be iy3-4.
II. Produce BK to D, till BD be equal to 193-4; throughD draw DL parallel toMK, meeting DM produced in L; thenDL will be the difference of longitude: to find which by calcula-tion, it will be, (by rectangular trigonometry,) K : BD :: 1,
LBD : DL, i. e. as radius....!. 7 . 10 00000
is to the meridional difference of lat...193’4 . 2-28646
so is the tangent of the course.35° 40'. 9.85594
to minutes of difference of Ion .....,-,.138-8 . 2.14240
equal to 2° 18' 48 ', the difference of longitude the ship has madewesterlv.
Case LV. Given course and both latitudes, viz. the latitudesailed from, and the latitude-come to ; to find the distance sailed,and the difference of longitude. 0
Exam]}. Suppose a ship in latitude 54* 20' N. sails south 33 45'
E. until by observation she is found to be in lat. 51° 45' N. Re-quired the distance sailed, and the difference of longitude !
Geometrically. Draw AB, tig. 20, to represent the meridianof tire ship in the first latitude, and set off from A to B 155 theminutes of the proper difference of latitude: also AG equal to257 - 9 the minutes of the enlarged difference of latitude. 1 hroughB and G, draw the lines BC and GK perpendicular to AG; alsodraw AK, making with AG an angle of 33° 45', which will meetthe two former lines in the points C and K; so the case is con-structed, and AG and GK may be found from the line of equalparts: to find which, ....
By Calculation ; first, for the difference of longitude, it will be,(by rectangular trigonometry,) R : AG :: T, GAIv : GK,
i. e. As radius....!... 10-00000
is to the enlarged difference-of lat.257.9 .. 2.41145
so is the tangent of the course.33" 45'..... 9.84289
to minutes of difference of Ion. 172-3 .. 2.23634
equal to 2° 52' 18", the difference of longitude the ship has madeeasterly. This might also have been found, by- first finding thedeparture BC, by Case 2 of Plane Sailing, and then it would beAB : BC : : AG : GK, the difference of longitude required.Then for the direct distance AC , it will be, (by rectangular trigo-nometry,) R : AB :: Sec. A : AC ,
*. 0 . As radius......Ip'OOOOO
is to the proper difference of lat.,.155 . 2-19033
■so is the secant of the course.33° 45'.10-0801 j
to the direct distance.186-4.•. 2-27048
consequently the ship has sailed S. 33° 45' E. 186-4 miles, andhas differed her longitude 2° 52* 18” E.
Cask V. Both latitudes and distances sailed, given ; to find thedirect course, and difference of longitude ?
Examp. Suppose a ship from the latitude of 45° 26' IS. sails be-tween N. and E. 195 miles, and then by observation she is foundto be in lat. 48° 6' N. Required the direct course and differenceof longitude ?
Geometrically. Draw AB, fig. 21, equal to 160 the properdifference of latitude, and from the point B raise the perpendicularBD ; then take 195 in your compasses, and setting one foot ofthem in A, with tire other cross the line BD in D. Produce AB,till AC -be equal to 233-6 the enlarged difference of latitude.Through C draw CK parallel to BD, meeting AD produced in K:so the case is constructed; and the angle A may be measured bythe line of chords, and CK by the line of equal parts: to findwhich,
By Calculation: First, Eov the angle of the course BAD, it willbe, (by rectangular trigonometry,) AB : R :; AD : Sec. A.
£. e. As the proper difference of lat.l6u . 2.20412
is to radius...........10-00000
so is the distance... 195 . 2-29003
to the secant of the course.34° 52’. 10-08591
which, because the ship is sailing between N. and E. will be N.34° 52' E. os NEfcN 1° 7 easterly. Then for the difference of
(by rectangular trigonometry,)
AC :: T,.10-00009
A : CK.
i. e. As radius....
is to the meridional difference of lat.203-6 . 2-36847
so is the tangent of the course.34° 52'. 9-84307
to minutes of difference of Ion...162 8 . 2-2115-4
equal to 2° 42' 48", the difference of Ion . easterly.
Case VI. One latitude, course, and difference, of longitudegiven ; to find the other latitude and distance sailed ?
Exam]). Suppose a ship from lat. 48° 50' N. sails S. 34° 40' E.till her difference of Ion. is 2“ 42': required the latitude come to,and the distance sailed?
Geometrically, I. Draw AE, fig. 22, to represent the meridianof the ship in the first latitude, and make the angle EAC equal to34° 40', the angle of the course; then draw FC parallel to AE, atthe distance of 164 the minutes of difference of longitude, whichwill meet AC in the point C. From C let fall upon AE the per-pendicular CE; then AE will be the enlarged difference of lati-tude. To find which by calculation, it will be, (by-rectangulartrigonometry,) '1', A : R :: CE : AE,
i. e. As the tangent of the course.34° 40’..— 9-83984
is to the radius.—10-00009
so is minutes of difference of Ion.164 . 2-21484
to the enlarged difference of lat.237.2 . 2-37500
and because the ship is sailing from a N. lat. southerly, thereforeFrom the meridional parts ) ,.„o 3-mn.o
of the tat. sailed from \ . 48 J °. 3a6b 9
take the meridional difference of lat.
237-2
3129-7H ence
the properCE:
and there remains......
the meridional parts of the latitude come to, viz. 46° 9'.for the proper difference of latitude,
From the latitude sailed from.4S° 50' N.
take the latitude come to.....46 9 N.
and there remains........... - N.
equal to l6l , the minutes of difference of tat.
H. Set off upon AE the length AD equal todifference of latitude, and through I) draw Dll ,then AB will be the direct distance. To find which by calcula-tion, it will be,(by rectangular trigonometry,)R: AD :: See. A: AB.
i. c. As radius...10-00000
is to the proper difference of tat.161 . 2.206S3
so is the secant of the course.34° 40'.10-08488
to the direct distance... 195-8 ...... 2.29171
Case VII. One latitude, course, and departure, given; to findthe other latitude, distance sailed, and difference of longitude ?
Exump. Suppose a ship sails from 54*36' N, 42° 33' W. untilshe has made of departure 1 - 16 -miles: required the latitude she isin, her direct distance sailed, and how much she has altered herlongitude ?
Geometrically, L Having drawn the meridian AB, fig. 23,make the angle BAD equal to 42° 33'. Draw FD parallel to ABat the distance of 116 , which will meet AD in D. Let fall uponAB the perpendicular DB. Then AB will be the, proper differ-ence of latitude, and AD the direct distance: to find which bycalculation, first, for the distance AD, it will be, (by rectangulartrigonometry,) S, A : BD t: R : AD.
i. e. As the sine of the course.42° 33'..,.9-83-010
is to the departure..116 .. 2 06446
so is radius.10-00000
to the direct distance........171-5 . 2-23436
Then for the proper difference of latitude, it will be, (by rectan-gular trigonometry,) T, A : BD :: R : AB,
«. e. As the tangent of the course...42” 33'...... 9-96281
is to the departure..........116 ............... 2-Q6446
so is radius...10-00000
to the proper difference of tat.126-4 . 2-10165
equal to 2° 6': consequently the ship has come to tat. 52® ; 0'N.and so the meridional difference of latitude will be 212-2.
II. Produce AB to E, till AE be equal to 212-2; and throughE draw EC parallel to BD, meeting AD produced in C ; thenEC will be the difference of longitude; to find which by calcula-6° n > b "ill be, (by rectangular trigonometry,) R : AE .-: T,A : EC.
i. e. As