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corrections the sum was found to be .16 only, instead of 17; this is ..04 was therefore placed for the correction, but on adding up the
* The first correction of departure is rather nearer .04 than 05;
In the above example the meridian distances are found by adding the eastings and subtracting the westings. By beginning at the 5th station, and adding the westings and subtracting the eastings, the same answer would be obtained. EXAMPLE 2. The bearings and distances of a survey are as follow, viz. Beginning at a corner stone, 1st. S 391 E, 24,25 ch.; 2d. S 1° W, 4. ch; 3d. S 54°1 W, 9 20 ch.; 4111. N 68° W, 3.35 ch.; 5th. N 7504 W, 3.95 ch; 6th S. 71° W, 2.20 ch.; 7th. S 650 W,9.90 ch. ; 8th. N 39°] W, 15.20 ch.; 9th. N 46° E, 27.25 ch. 10 the place of beginning. Required the area.
EXAMPLE 3. Required the area of a tract of land bounded as follows : 1st. N 75° E, 13.70 ch. ; 2nd, N 20°} E, 10.30 ch.; Srd. East, 16.20 ch.; 4th. S 33°7 W, 35.30 ch. i 5th. S 76o. W, 16.1. ; 6th. North, 9 ch.; 7th. S 840 W, 11 60 ch. ; 8:11. N 53°1 W, 11.60 ch ; 9ih, N 361 E., 19.36 ch.; 10th. N 22°1 E, 14 ch. ; 11th. S 76°3 E, 12 ch.; 12th. S 150 W. 10.85 ch. ; 13h. S 18° W, 10.62 ch. to the place of beginning:
81.38 1.19 11 52
99.83 1 1.61
65.76 2.73 11.70
56.86 50 46
10 N 221 E 14.00 12.93
1905 :| 57.57| 57.851 61.68, 61 9311 2& .2537.70 57.70161.8061 861 57.57
61 63 .28 Er. N. .25 Er. E.
Area. 110 A. 3 R. 6 P.
2286.1616 2)2215 7369 1107.86845 Sq. Ch.
1. Find the latitude and departure corresponding to each course and distance, and correct them as directed in the preceding rule.
2. Beginning at the first station of the survey, or any other convenient place, the first departure will be the first meridian distance, which place in the column of meridian distances, opposite the said departure, and mark it east or west according as the departure is east or west. To this meridian distance add the same departure, and the sum will be the second meridian distance, which place under the former in the same horizontal column, and mark it with the same name. Proceeding thus, find two meridian distances for each horizontal column, ob. serving that when the meridian distance and departure are both east, or both west, their sum is the next meridian distance, and of the same name; but when the meridian distance and departure are of different names, that is, one east and the other west, their difference is the next me. ridian distance of the same name with the greater. This done, the last meridian distance will be nothing, as in the preceding method.
3. Multiply the upper meridian distance in each hori- . zontal column by the corresponding latitude, and when the meridian distance is east, place the product in the column of north or south area, according as the latitude is north or south; but when the meridian distance is west, place the product in the contrary column, that is, in the column of south area if the latitude be north, and in the column of north area if the latitude be south. Half
the difference of the sums of the numbers contained in the columns of north and south area will be the area of the survey*,
* DEMONSTRATION. Let ABCDEFGHA, Fig. 79, represent the boundary of a survey, and let NS be a meridian passing through the first station A. In the line Ee takes Es equal to Ff, and draw st parallel to NS; then the courses, distances, latitudes, departures, &c. will be as in the following table.
That Ee-Ff and Ff+gp are the meridian distances and 2.seuv the area corresponding to the 5th station, may be shewn thus: From Ee+ so taking Fo =(Ff+fo) the remainder will be Ee-Ff, which is the first; but Ee-Ff-Ee-Es=es=ft; therefore from Fo taking Ee Ff =(ft) the remainder will be Ff+to=Ff+Es=Ff+Ff=Ff+gp, which last is the second meridian distance corresponding to the 5th station, and is west because the west departure Fo exceeds the east meridian distance Ee-Ff.
Now it is evident that the triangle Ffu is equal to the triangle Esv, and that the triangle Ftv is equal to the triangle Eeu: From the latter