26

NAVIGATION.

which make equal to 120 ; then will A be the place the ship capedat. From A draw AB parallel to the WAN line CI), equal to 40,the motion of the current in twenty hours, and join CB ; then Bwill be the ship’s true place at the end of twenty hours, CB hertrue distance, and the angle SCB her true course. The calcu-lation wi.l be easy from this construction.

Examp. 3. Suppose a ship, coming out from sea in the evening,has a view of Scilly light, bearing NEAN distance four leagues, itbeing then flood-tide setting EN E two miles an hour, and the shiprunning after the rate of five miles an hour: Required upon whatcourse' and how far she must sail to hit the Lizard, which bearsfrom Scilly E-JS distance 17 leagues ?

Geomet. Having drawn the compass NESW, fig. 12, let Are-present the ship’s place at sea, and draw the NEAN line AS,which make equal to 12 miles; so S will represent Scilly. FromS draw SL equal to 51 miles, and parallel to the E J S line; then.L will represent the Lizard. From L draw LC parallel to the ENEline, equal to two miles, and from C draw CD equal to five milesmeeting AL in D ; then from A draw AB parallel to CD meetingLC produced in B; and AB will be the required distance, andSAB the true course. Hence the calculation may be easily per-formed.

Examp. 4. A ship from a certain headland in the latitude of34° 0’north, sails, SEAS 12 miles in five hours, in a current thatsets between N. and E. and then the same headland is found tobear WNW, and ship to be in the latitude of 33°■52' N. Re-quired the setting and drift of the current ?

Geomet. Having drawn the compass NESVV, fig. 13, let Arepresent the place of the ship, and draw the SEAS line AB, equalto two miles ; also the ESE line AC. Set off from A upon themeridian AD, equal to eight miles, the difference of latitude, andthrough D draw DC parallel to the east and west line WE, meeting AC in C. Join C and B with the right line BC ; (hen C willbe the ship’s place, the angle ABC the setting of the current fromthe SEAS line, and the line BC will he the drift of the current inthree hours. To find which, by calculation, there will be no dif-ficulty.

Of the Variation of the Compass , and how to find it

from the True and Observed Amplitudes or Azimuths

of the Sun.

T. The variation of the compass is how far the north or southpoint of the needle stands from the true south or north point of thehorizon towards the east or west; or it is an arch of the horizonintercepted between the meridian of the place of observation andthe magnetic meridian.

If. It is absolutely necessary to know the variation of the com-pass at sea, in order to correct the ship’s course ; for since the ship’scourse is directed by the compass, it is evident that if the compassbe wrong, the true course will differ from the observed.

III. The sun’s true amplitude is an arch of the horizon com-prehended between the true east or west point thereof, and thecentre of the sun at rising or setting; or it is the number of degrees,&c. that the centre of the sun is distant from the true east or westpoint of the horizon, towards the S. or N.

IV. The sun’s magnetic amplitude is the number of degrees thatthe centre of the sun is from the east or west point of the compass,towards the S. or N. point of the same at rising or setting.

V. Having the declination of the sun, together with the latitudeof the place of observation, we may from thence find the sun’strue amplitude, by the following astronomic proposition, viz.

As the co-sine of the latitudeis to the radius,

So is the sine of the sun’s declinationto the sine of the sun’s true amplitude,

which will be N. or S. according as the sun’s declination is N.or S.

Examp. Required the sun’s true amplitude in the latitude of 41°50' N. on the .23d day of April 1731. First,. I find (from thetables of the sun’s declination) that the sun’s declination the 23d ofApril was 15° 54' N. then for the true amplitude, it was by theformer analogy,

As the co-sine of the lat.41° 50'. 9-87221

is to radius.10-00000

So is the sine of the decl.15° 54',. 9 43769

fc) the sineoftheamplit..,..21° 35',.. 9-56548

which was N because the declination was N at that time; and con-sequently in lat. 41° 50' N. the sun rose on the 23d of April 21°35' from the E. part of the horizon towards the N. and set so muchfrom the W. the same way.

"VI. The sun’s true azimuth is the arch of the horizon interceptedbetween the meridian and the vertical circle passing through thecentre of the sun at the time of observation.

VII. 'fhe sun’s magnetic azimuth is the arch of the horizon, in-tercepted between the magnetic meridian ami the vertical, passingthrough the sun.

VII. Having the latitude of the place of observation, togetherwith the sun’s declination and altitude at the time of observation,we may find his true azimuth after the following method, viz.Make it, As the tangent of half the complement of the latitude is tothe tangent of half the sum of the distance of the sun from the poleand complement of the latitude ; so is the tangent of half the dif-ference between the distance of tiie sun from the pole and comple-ment of the altitude, to the tangent of a fourth arch, which fourtharch added to half the complement of the altitude will give a fiftharch, and this fifth arch lessened by the complement of the latitudewill give a sixth arch. Then make it, As the radius is to the tangentof the altitude ; so is the tangent of the sixth arch to the co-sine ofthe sun’s azimuth, which is to be counted from the S. or N. to theE. or VV. according as the sun is situated with respect to the placeof observation. It the latitude of the place and declination of thesun be both N. or both S. then the decimation taken from 90° willgive the sun’s distance from the pole; but if the latitude and decli-nation be on contrary sides of the equator, then the declination,added to 90° will give the sun’s distance from the nearest pole tothe place of observation.

Examp. In lat. 51° 32' N. the sun having 10° 39' N. declination,his altitude was found by observation to be 38° IS' : Required theazimuth? By the first of the foregoing analogies, it will beAs the tangent of £ the com-;

19° 14'.,

.9-54269

01° 1'.10-25655

9° 19'.

.7-51499

plement of the latitude \is to the tangent of 4 the sum iot the distance of the sun ffrom the pole and com- fplement of the altitude )

So is tl e tangent of half their )difference )'

to the tang, of a fourth arcli.40° 20'.9-92S85-

which fourth arch 40° 20', added to 19° 14', half the complementof the latitude,, gives a fifth arch 59° 34'; and this fifth arch lessenedby 38° 28', the complement of the latitude, gives the sixth arch 21°6'; then for the azimuth, it will be, by the second of the preced-ing analogies,

As radius.10-00000

is to Ihe tang, of the altitude.38° 18'. 3-89749

so is the tang, of the sixth arch....29° 6'. 9*58644.

to the co-sine of the azimuth.72° 14'. 9-48393

which because the latitude is N. and the sun S. of the place of ob-servation, must be counted from the S. towards the E. or \V. andconsequently, if the altitude of the sun was taken in the morning,the azimuth will be S 72° 15' E, or ESE 4° 45' E; but if the alti-tude was taken in the afternoon, the azimuth will be S. 72° 15 ; \V.W.S.W. 4° 45' W.

IX. Having found the sun’s true amplitude or azimuth bv thepreceding analogies, and his magnetic amplitude or azimuth by ob-servation, it is evident, if they agree, there is no variation ; but ifthey disagree, then if the true and observed amplitudes at the.rising or setting of the sun be both of the same name, i. e. eitherboth N. or both S. their difference is the variation ; but if they beof different names, i. c. one north and the other south, their sum isthe variation. Again, if the true and observed azimuth be botli ofthe same name, i. <-'■ either both E. or both W. their dilfevence isthe variation: but if they be ol different names, their sum is thevariation. And to know whether the variation is easterly, observethis general rule, viz. Let the observer’s face be aimed to the sun ;thru if the true amplitude or azimuth be to the right hand of theobserved, the variation is easterly ; but if it be to th‘e leit, westerly-To explain which, Let N ESYV,. fig. 14, represent a compass, andsuppose the sun-is really EAS at the time of observation, but the ob'server sees him off the east point of the compass, and so the trueamplitude or azimuth of the sun is to the right of the magnetic orobserved; here it is evident that the EAS point of the' compass

ought