The contents of the field=(2106.3-711.6)=697.35 sq. ch. Or 69.735 acres. In this example, the meridian distance of the first station A being nothing, cipher is placed at the head of the column of meridian distances. (Art. 117.) The first side AB being directly east and west, has no difference of latitude, and therefore the number in the column of areas against this course is wanting, as it is against the fifth course, which is directly north. (Art. 119.) The number against the fourth course in the column of multipliers, is only the length of the line DM; the figure DME being a triangle, instead of a trapezoid. Example III. Find the contents of a field bounded by the following lines; 121. When a field is correctly surveyed, and the departures and differences of latitude accurately calculated; it is evident the sum of the northings must be equal to the sum of the southings, and the sum of the eastings equal to the sum of the westings. If upon adding up the numbers in the departure and latitude columns, the northings are not found to agree nearly with the southings, and the eastings with the westings, there must be an error, either in the survey or in the calculation, which requires that one or both should be revised. But if the difference be small, and if there be no particular reason for supposing it to be occasioned by one part of the survey rather than another; it may be apportioned among the several departures or differences of latitude, according to the different lengths of the sides of the field, by the following rule; As the whole perimeter of the field, To the whole error in departure or latitude; So is the length of one of the sides, To the correction in the corresponding departure or latitude. This correction, if applied to the column in which the sum of the numbers is too small, is to be added; but if to the other column, it is to be subtracted.* See the example on the next page. * See the fourth number of the Analyst, published at Philadelphia. In this example, the whole perimeter of the field is 100 chains, the whole error in latitude .34, the whole error in departure .42, and the length of the first side 18. To find the corresponding errors, .34 .06 the error in latitude, 100: 18: 12:08 the error in departure. The error in latitude is to be added to 10.26 making it 10.32, as in the column of corrected northings; and the error in departure is to be added to 14.79 making it 14.87, as in the column of corrected eastings. After the corrections are made for each of the courses, the remaining part of the calculation is the same as in the preceding examples. 122. If the length and direction of each of the sides of a field except one be given, the remaining side may be easily found by calculation. For the difference between the sum of the northings and the sum of the southings of the given sides, is evidently equal to the northing or southing of the remaining side; and the difference between the sum of the eastings and the sum of the westings of the given sides, is equal to the easting or westing of the remaining side. Having then the difference of latitude and departure for the side required, its length and direction may be found, in the same manner as in the sixth case of plane sailing. (Art. 49.) Example V. What is the area of a field of six sides, of which five are given, viz. 123. Plotting by departure and difference of latitude.— A survey may be easily plotted from the northings and southings, eastings and westings. For this purpose, the column of Meridian Distances is used. It will be convenient to add also another column, containing the distance of each station from a given parallel of latitude, and formed by adding the northings and subtracting the southings, or adding the southings and subtracting the northings. Let AT (Fig. 33.) be a parallel of latitude passing through the first station of the field. Then the southing TB or LM is the distance of B, the second station, from the given parallel. To this adding the southing BH, we have LO the distance of CO from LT. Proceeding in this manner for each of the sides of the field, and copying the 7th column in the table, p. 65, we have the following differences of latitude and meridian distances. To plot the field, draw the meridian NS, and perpendicular to this, the parallel of latitude LT. From L set off the differences of latitude LM, LO, LR, and LP. Through L, M, O, R, and P, draw lines parallel to LT; and set off the meridian distances AL, BM, CO, DR, and EP. The points A, B, C, D, and E, will then be given. 124. When a field is a regular figure, as a parallelogram, triangle, circle, &c. the contents may be found by the rules in Mensuration, Sec. I, and II. 125. The area of a field which has been plotted, is sometimes found by reducing the whole to a TRIANGLE of the same area. This is done by changing the figure in such a manner as, at each step, to make the number of sides one less, till they are reduced to three. Let the side AB (Fig. 35.) be extended indefinitely both ways. To reduce the two sides BC and CD to one, draw a line from D to B, and another parallel to this from C, to intersect AB continued. Draw also a line from D to the point of intersection G. Then the triangles DBC and DBG are |