points give the diff. of longitude also 217; and which proves what is said in 94, that the mer. diff. of latitude is to the diff. of longitude as the proper diff. of latitude is to the departure.

CASE 8.

105. Given—one LATITUDE, the COURSE and the DIFFERENCE OF LONGITUDE, to find the DISTANCE and the DIFF.

OF LATITUDE.

Rule.-Set the index on the course, and the difference of longitude will give the meridional difference of latitude, by which the proper difference of latitude may be known, and which with the course will give the distance.

Example-A ship in lat. 42 deg. N. sails N. 38 deg. E. and her diff of longitude is 125 miles; required the distance, and diff. of latitude.

Here the course 38 deg. with the diff. lon. 125 give the mer. diff. of lat. 160; hence the proper diff. of lat. is 117, which with the course gives the distance 149 miles.

Lat. left 42 deg. equal 2782 mer. parts.
Add. mer. lat.

160

2942-43° 57′ the lat. in.

106. To find by the diagram the value of any given meridional parts in degrees of lat.-divide them by 60, which reduces them to degrees and minutes; then set the index on the arc of latitude, to the degrees found, and it will cut on the expanded arc, the true degrees of latitude answering to the meridional parts given.

Example.-What is the latitude for 2100 meridional parts? Here 2100 divided by 60 gives 35 deg. to which set the index on the proper arc of latitude, and it will cut on the expanded arc 33 deg. being the latitude for 2100 meridional. parts. It is seen that this

operation is the reverse of that by which the meridional parts are found for any given latitude, as shewn in No. 95, page 26.

CORRECTING THE DEAD RECKONING.

107. If the courses and distances which a vessel is supposed to sail were uniformly true, the difference of latitude by account would always agree with that from an observation of the heavenly bodies. But during a voyage it happens frequently that the latitude in by account differs from that by observation.

108. This difference arrises from an error in the course, or in the distance, and often in both.

109. When the course is more toward the meridian than the equator, the difference in the latitudes it is supposed arises from an error in the distance, rather than in the course.

110. But if the course is more toward the equator than the meridian, it is presumed the difference in the latitude arises from an error in the course rather than in the distance.

!

111. Various methods have been proposed to rectify these errors; but the most eminent mathematicians agree in this conclusion,-that, if after a careful re-examination of the incidents which affect the course and distance, there should still exist a difference between the latitude by account, and that by observation, the course and the distance should be corrected by the difference of latitude from observation, with the departure from account-leaving the departure by account unaltered.

112. Example.-A ship in lat. 44 deg. N. and lon. 60 deg. W. is supposed to sail N. 22 deg. 30 m. E. (N. N. E.) 264 miles, by which her diff. lat. would be 244 and her departure 101-and she would thereby be in lat. 48 deg. 04 m. N. and in lon. 57 deg. 35 m. W.; but by an observation of the sun, the ship is found to

be in lat. 47 deg. 40 m. N. what correction must be made to the course and distance? Here the diff. lat. by observation is 220, which with the departure 101, give the corrected course N. 24 deg. 39 m. E. and the distance 242 miles. As the mid. lat. is not materially changed, the longitude remains the same.

113. Example.-A ship in lat. 44 deg. N. and lon. 60 deg. W. is supposed to sail N. 67 deg. E. (E.N.E.) 264 miles, by which her diff. of lat. would be 101 and her departure 244-and thereby she would be in lat. 45 deg. 51 m. N. and lon. 56 deg. 10 m. W.; but by an observation of the 'sun the ship is found to be in lat. 46 deg. 15 m. N.; required the correction to be made to the course and distance. Here the diff. of lat. by observation is 135 miles, which with the departure 244 give the corrected course 61 deg. 03 m. and the distance 279; and the middle lat. not being much altered, the longitude remains the same.

QUESTIONS FOR PRACTICE.

114. What is the course and distance from Nantucket light-house in lat 41 deg. 22 m. N. lon. 70 deg. 00 m. W. and Cape Sable in lat. 43

65 deg. 32. W.?

Ans.

deg. 26 m. N. and lon. course N. 57° 55′ E. distance 219 miles.

For the method of operation see Rule in Case 1, Middle Latitude Sailing, No. 83; and Case 1st, Mercator's Sailing No. 98.

115. A ship from Portland in lat. 43 deg. 39 min. N. lon. 70 deg. 13 min. W. sails S. E. 256 miles; what latitude and longitude is she in?

Ans. in latitude 40° 38′ N.

See No. 85, and No. 100.

long. 66° 09′ W.

116. A ship leaves Boon Island in lat. 43 deg. 6 m. N. and lon. 70 deg. 31 m:W. and sails E. S. E. till her departure is 145 miles; what distance has she sailed, and what latitude and longitude is she in?

See 88 and 103.

Ans. dist. sailed 157 miles.

lat. in 42° 06′ N.

lon. in 66 56 W.

117. A ship leaving Boston light house in lat. 42 deg. 20 m. N. and lon. 70 deg. 54 m. W. sails E. by SS. and by observation it is found she is in lat. 40 deg. 59 m. N. what distance has she sailed; and what longitude is she in? Ans. distance sailed 279 miles. lon. in 64° 57′ W.

See No. 86 and 101.

118. A ship from Isle of Holt in lat. 44 deg. north and lon. 68 deg. 31 m. W. sails southeasterly 224 miles and is in lat. 41 deg. 00 m. N. what course has she steered, and what lon. is she in?

See No. 87 and 102.

Ans. course steered S. E. & S. lon. in 65° 29′ W.

119. A ship bound to Philadelphia is in lat. 40 deg. 00 min. N. and in lon. 69 deg. W. what is her course and distance to Cape May in lat. 38 deg. 57 min. N. and lon. 74 deg. 57′ ̃ ̄W.?.

Ans. course S. 77° 04′ W. distance 281 miles.

120. A ship bound to Boston, and coming into the Bay, discovers Mount Desert Rock, which bears W.N.W. the true course, and distant 18 miles; Mount Desert Rock is in lat. 43 deg. 52 min, N. and lon. 68 deg. 09′ W. what lat. and lon. is the ship in, and what is her course and distance to Boston light-house which is in lat. 42 deg. 20 min. N. and lon. 70 deg. 54 min. W.? Ans. the ship is in lat. 43° 45′ N. long. 67° 45′ W.

the course is S. 58° 28′ W. distance 162 miles.

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121. Two ships leave the same port at the same time; one sails East at the rate of six miles per hour, the other sails E. N. E. Ten hours after sailing the latter bears from the former due north. How far are the ships apart, and what distance has each sailed.

Set the index on 6 points (E. N. E.) and the ship sailing east having run 60 miles in 10 hours; therefore 60 on the side marked equator will give on the index 65, which is the distance the other ship has sailed, from which the nearest line to the side marked meridian gives 25 miles the distance between the ships.

122. Two ships in lat. 44 deg. 30 m. north are 216 miles apart; they both sail due south at equal rates. When they are in lat. 32 deg. 18 m. N. what will be the distance between them?

Set the index on lat. 44 deg. 30 m. and 216 on the side marked equator will give on the index 303; then set the index on the other lat. 32 deg. 18 m. and 303 on the index will give on the side marked equator 256 miles, being the distance between the two ships in lat, 32 deg. 18 m. N.

CURRENTS.

123. A current is a stream in the ocean, by which all bodies on its surface, within the limits of the cur rent, are drifted toward that part of the horizon to which the current sets.

124. When a ship is affected by a current, the direction, or set of the current is entered in the traverse table as a course sailed; and the rate, or drift of the current is made up as a distance sailed.

125. On such course and distance the diff. lat. and departure are calculated, as on a course and distance sailed. Example-A ship sailing N. E. at the rate of 6 miles per hour for 10 hours, it is ascertained that a