or more distances, is equal to the sum of the latitudes and departures of those distances respectively. Hence, if we have any number greater than 100, as 614, we have only to regard the last figure as a cipher, and recollect, that, 610+4=614, and that the latitude and departure of 610 are ten times as great, respectively, as the latitude and departure of 61, that is, equal to the latitude and departure of 61, multiplied by 10, or, with the decimal point removed one place to the right. Example 1.– To find the latitude and departure, the bearing being 291, and the distance 582. In this example, the latitude and departure answering to the course 291, and to the distance 58, were first taken from the table, and the decimal point moved one place to the right; then the latitude and departure answering to the same course, and the distance 5, were taken from the table and added. Excample 2.--To find the latitude and departure, the bearing being 62, and the distance 7855 chains. If the distances were expressed in whole numbers and decimals, the manner of finding the latitudes and departures would still be the same, except in pointing off the decimal places ; which, however, is not difficult, when it is remembered that, the column of distances in the tables may be regarded as decimals by removing the decimal point to the left 159. The manner of surveying with the compass, is to go entirely around the land, measuring the bounding lines with the chain, and taking all their bearings with the compass. In going round a piece of land, it is immaterial whether it be kept on the right hand or on the left; and all the rules deduced for one of the cases, are equally applicable to the other. To preserve, however, a uniformity in the language of the rules, we shall suppose it kept constantly on the left hand of the surveyor. Let ABCD (Pl. 6, Fig. 1) be the plan of a piece of land to be surveyed, NS the north and south line; the lines parallel to it, are meridians, and those at right angles, east and west lines. Let the work be commenced at the station A. On a sheet of paper, rule three columns, as below, and head them, stations, bearings, and distances. At station A, designated by 1, in the column stations, take the bearing to station B, which is the angle BAb, N 23° E, and enter it in the column bearings, opposite 1; then measure the distance AB=5 ch. 40 I., which insert in the column distances. The station at B, is station 2, in the field notes, opposite which, are written the bearing FBC, N 84° W, and the distance BC equal to 6 ch. and 601.; and similarly for the other bearings and distances. 160. In passing over the distance AB, the northing is Ab, and the departure, east, bB. For the distance BC, the northing is BF, and the departure, west, FC: Af, therefore, line CD, Cd is the southing, and Dd the departure easi: for the line DA, Dh is the southing, and hА the departure, east: hence we conclude that, CG is equal to the sum of the southings, and equal also to Af, the sum of the northings. If we consider also the departures, we see that bB, DD, and hA, are the departures east, and that their sum is equal to FC, the departure west ; and as the same may be shown for any other figure, we conclude that, when any survey is correctly made, the sum of the northings is equal to the sum of the southings, and the sum of the eastings to the sum of the westings. 161. It would appear plain, even without demonstration, that, after having gone entirely round a piece of land, the distance passed over, in the direction due north, must be equal to the distance in the direction due south; and the distance in the direction due east, equal to the distance in the direction due west. 162. In computing the area of a survey, it becomes necessary to know double the meridian distance of the middle of each line, from a given meridian. This meridian may be taken at pleasure, though it is generally most convenient to use the one that passes through the most easterly or westerly station. The manner of calculating these double distances, which are called, the double meridian distances of the lines, is now to be explained. Let all the departures in one direction, say west, be called plus ; and the departures east, minus : then, through whatever ; station of the survey the assumed meridian be taken, we shall have this general method. 1. The double meridian distance of the first line is equal to its departure. 2. The double meridian distance of the second line is equal to the double meridiun distance of the first line, plus its departure, plus the departure of the second line. 3. The double meridian distance of the third line, is equal to the double meridian distance of the second line, plus its departure, n 4. And, the double meridian distance of any line, is equal to the double meridian distance of the preceding line, plus its de parture, plus the departure of the line itself. It ought, perhaps, to be remarked, that plus is here used in its algebraic sense ; and that when the departures and double meridian distances are of different names, that is, one east and the other west, they have different algebraic signs, and are therefore to be subtracted. Demonstration. Let FL (Fig. 1) be the assumed meridian. Let n be the middle point of the distance BC; np is the meridian distance of BC, and 2 np is double the meridian distance of BC, and is plus. But BC is bisected at n, hence, np, equal to qF, is half of FC; hence the departure of the first line, is equal to the double of its meridian distance. Now, to 2np add 2Cq, equal to the departure FC; the sum is 2FC, and plus; to this sum add 2om, equal to the departure dD; this, being east, is minus, and the sum is 2ma, and plus. To 2 ma, add 2 mi, or dD, which being minus, the sum is 2Dv, to which add hA, or 2ts, the sum is 2sc. To 2sc add hA or 2rA ; the sum is 2Aw; to this add Aw or 2IP, the sum is 2PL. As the same may be shown for any meridian, and every figure, we may regard the demonstration as general. 163. Those lines, the middle points of which, lie on the east side of the assumed meridian, are said to have eastern meridian distances, and those, whose middle points lie on the west of it, to have western meridian distances. THEOREM, 164. Double the area of a piece of land, is equal to the difference between the sum of the products that arise from multiplying the double of each western meridian distance by its corresponding northing, and the double of each eastern meridian distance by its corresponding southing; and the sum of the products that arise from multiplying the double of each eastern meridian distance into its corresponding northing, and the double of each western meridian distance by its corresponding CASE I. When the assumed meridian passes entirely without the land. Let ABCD (Pl. 6, Fig. 2) represent a quadrangular piece of ground, NS the assumed meridian, from which the double distances are estimated; EF, IH, N'L, and PQ, the meridian distances of the lines AB, BC, CD, and DA respectively. These meridian distances are all east. Let NxW+S XE, designate the first set of products, that is, the products which arise from multiplying the northings by double the western meridian distances, and the southings by double the eastern meridian distances; and NxE+SxW designate the second set of products, or those which arise from multiplying the northings by double the eastern meridian distances, and the southings by double the western meridian distances. NxW+SXE NxE+SxW 2EF XAG=2 area AGB 2QPX AN=2 area ADN Sum=2AGB+2ADN 2HI X KG=2 area GBCK 2LNXKN=2 area KCDN Sum=2 area GBCDN The difference of these sums is twice the area of the figure ABCD. The line AB runs south, and its meridian distance is east; therefore the product 2FE X AG belongs in the first column. In a similar manner it might be shown that the other products are correctly placed. The product 2FE XAG is double the area of the triangle AGB, since 2EF=GB; and the product 21H XGK is double the area of the trapezoid GBCK. (178.*) CASE II. 165. Let ABC (Pl. 6, Fig. 3) represent a piece of ground, NS the assumed meridian, FE, PO, and ST, the meridian |