local attraction, when it arises, will first show itself in the difference between a true foresight and an erroneous back sight. When this difference appears, subtract the back sight from the foresight, and call the difference the correction for the next foresight. The correction will be positive when the foresight is the larger, and negative when it is less. Add this correction, with its proper sign, to the foresight of the next course, when the meridional and longitudinal letters of that course are both the same, or both dif ferent from the foresight of the previous course, and subtract it when one of the letters is the same and the other different the result will be the true bearing. The true bearing of any other course may be found by the same 1. S 85° 10' WN 85° 05′ ES 16° 15' WS 16° 20' W 2. N 16° 20' E S 18° 20' WN 15° 25′ W N 17° 25′ W N 27° 01' E 3. N 17° 25' WS 16° 10' E N 28° 16' E 4. S. 47° 18' EN 48° 10' WS 49° 15' WN 50° 07' W NOTE. If there be no course in the survey in which the foreward and back sights agree, take the one in which they agree the nearest, and add half the difference of the bearings to the least, and treat the result as the true bearing. 18. REMARK IV. In passing over the course AB, the northing is found to be IIB, and the departure, which is west, is represented by AH. Of the course BC, the northing is expressed by BG, and the departure, which is east, by GC. Of the course CD, the southing is expressed by CI, and the departure, which is east, by N P C B W K D E R ing is expressed by KA, and the departure, which is west, by DK. It is seen from the figure, that the sum of the northings is equal to HB+BG=HG; and that the sum of the southings is equal to CI+KA=PA=HG : hence, the sum of the northings is equal to the sum of the outhings. If we consider the departures, it is apparent that the sum of the eastings is equal to GC+ CF GF; and that the sum of the westings is equal to AH+DK=GF; hence also, the sum of the eastings is equal to the sum of the westings. We therefore conclude, that when any survey is correctly made, the sum of the northings will be equal to the sum of the southings, and the sum of the eastings to the sum of the weslings. It would indeed appear plain, even without a rigorous demonstration, that after having gone entirely round a piece of land, the distance passed over in the direction due north, must be equal to that passed over in the direction due south; and the distance passed over in the direction due east, equal to that passed over in the direction due west. Having now explained the necessary operations on the field, we shall proceed to show the manner of computing the contents of the ground. We shall first explain, THE TRAVERSE TABLE AND ITS USES. 19. This table shows the latitude and departure corresponding to bearings that are expressed in degrees and quarters of a degree from 0 to 90°, and for every course from 1 to 100, computed to two places of decimals. The following is the method of deducing the formulas for computing a traverse table; by means of these for mulas and a table of natural sines, the latitude and departure of a course may be computed to any desirable degree Let AD represent any course, and NAD ACB, expressed in degrees and minutes, be its bearing. unit of measure of the the radius of the table (Bk I., Sec. III., Art. 14). Let AC be the course, and also of natural sines Draw DE and CR parallel to NS, and AE perpen Idicular to AS. Then will DE be the latitude, and AE the departure of the course, and CB the co sine, and AB the sine of the bearing. From similar triangles we have these proportions, 1 : AC : CB :: AD: DE, or :: course : latitude, cos of the bearing 1 sin of the bearing :: : Whence, : AE, or course : departure. lat. course X cos of the bearing, dep. = course X sin of the bearing. We have then the following practical rule for compu ting the latitude and departure of any course. Look in a table of natural sines for the cosine and sine of the bearing. Multiply each by the length of the course, and the first product will be the latitude, and the second will be the departure of the given course. EXAMPLES. 1. The bearing is 65° 39', the course 69.41 chains: what is the latitude, and what the departure? Natural cosine of 65° 39' Length of the course .41231 69.41 Product, which is the Dif. of Latitude, 28.6184371. Natural sine of 65° 39' .91104 Length of the course. 69.41. 2. The bearing is 75° 47', the course 89.75 chains: what is the latitude, and what the departure? Product, which is the Dif. of Latitude, 22.0417025. 20. In this manner the traverse table given at the end of the book has been computed. When the bearing is given in degrees and quarters of a degree, and the difference of latitude and departure are required to only two places of decimals, they may be taken directly from the traverse table. If the bearing is less than 45°, the angle will be found at the top of the page; if greater, at the bottom. Then, if the distance is less than 50, it will be found in the column "distance," on the left hand page; if greater than 50, in the corresponding column of the right hand page. The latitudes or departures of courses of different lengths, but which have the N H B G E A S Therefore, when the distance is greater than 100, it may be divided by any number which will give an exact quotient, less than 100: then the latitude and departure of the quotient being found and multiplied by the divisor, the products will be the latitude and departure of the whole course. It is also plain, that the latitude or departure of two or more courses, having the same bearing, is equal to the sum of the latitudes or departures of the courses taken separately. Hence, if we have any number greater than 100, as 614, we have only to recollect that, 610+4= 614; and 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. EXAMPLES. 1. To find the latitude and departure for the bearing 2910, and the course 614. 4. 3.48 Departure for 4. Latitude for 610 . . 530.90 Departure for 610. In this example, the latitude and departure answering to the bearing 294°, and to the distance 61, are first taken from the table, and the decimal point removed one place to the right this gives the latitude and departure for the distance 610; the latitude and departure answering to the same bearing and the distance 4, are then taken from the table and added. 2. To find the latitude and departure for the bearing 6210, and the course 7855 chains. 55. Latitude for 7800. 3602.00 Departure for 7800 . 6919.00 REMARK. When the distances are expressed in whole numbers and decimals, the manner of finding the latitudes and departures is still the same, except in pointing off the places for decimals: but this is not difficult, when it is remembered that the column of distances in the table, may be regarded as decimals, by removing the decimal point to the left in the other columns. 3. To find the latitude and departure for the bearing 47, and the course 37.57. Latitude for 37.00 24.88 Departure for 37.00 .57 .38 Departure for |