Method second. With the chord of 60° describe the circle NESW, Fig. 76, and draw the diameter NS. Take the several bearings from the line of chords and lay them off on the circumference from N or S according as the bearing is northerly or southerly, and towards E or W according as it is easterly or westerly, and number them 1, 2, 3, 4, &c. as in the figure. From A the centre of the circle, to 1 draw A 1, on which lay the first distance from A to B; parallel to A 2 draw BC on which lay the second distance from B to C; parallel to A 3 draw CD on which lay the third distance from C to D; Proceed in the same manner with the other bearings and distances. To find the area. By drawing lines as in Fig. 75, the survey is divided into two trapeziums AGFE, AEDB, and a triangle BDC. Measure the several bases and perpendiculars, on the same scale that was used in the protraction, and find the double areas of the triangle and trapeziums by probs. 2 and 6; the sum of these will be the double area of the survey. Bases. Perpens. EG 16.68 R ŞFa 7.502 = 203.6628 = 2 AGFE Ab 4.71 S EB 19.17 x ŞAc 5.85 267.4215 = 2 AEDB Dd 8.10) BD 19.23 X Ce 3.16 99.2268 = 2 BDC 2)570.3111ch.=2 ABCDEFG 285.15555ch.=28 A. 2R. 2P. EXAMPLE 2. The following field-notes are given to protract the survey and find the area. Ch. Ans. 46A. 2R. 9P. EXAMPLE 3. It is required to protract the survey and find the area from the following field-notes. Ch. ' W. SECTION 2. Containing three different rules for finding the areas of right-lined figures generally, when the bearings and distances of the boundaries are given. DEFINITIONS. 1. Meridians are north and south lines, which are supposed to pass through every station of the survey. 2. The difference of latitude, or the northing or southing of any stationary line, is the distance that one end of the line is north or south from the other end ; or it is the distance which is intercepted on the meridian, between the beginning of the stationary line and a line drawn from the other end, perpendicular to that meridian. Thus, if NS, Fig. 77, be a meridian passing through the point A of the line AB, then is Ab the difference of latitude, or southing of that line. 3. The departure of any stationary line, is the nearest distance from one end of the line to a meridian pass. ing through the other end. Thus Bb, Fig. 77, is the departure or easting of the line AB. But if ns be a meridian, and the measure of the stationary line be taken from B to A, then is BC the difference of latitude, or northing, and AC the departure or westing of the line AB. 4. The meridian distance of any station, is the distance thereof from a meridian passing through the first, or some other particular station of the survey. 5. The Traverse Table is a table containing the difference of latitude and departure corresponding to dif To find the difference of latitude and departure corresponding to any given course and distance, by means of the annexed Traverse Table. When the distance is any number of whole chains or perches not exceeding 100. Find the given bearing at the top or bottom of the table according as it is less or more than 15°. Then against the given distance, found in the column of distances at the side of the table, and under the bearing, if at the top, or over it if at the bottom, is the correspond. ing difference of latitude and departure. The difference of latitude and departure must be taken as marked at the top of the table when the bearing is at the top, but as marked at the bottom, when the bearing is at the bottom. Thus if the bearing and distance be S. 35° 15' E, dist. 79Ch., the diff. of lat. will be 64.51 Ch. S. and the dep. 45.59Ch. E.: but if the bearing and distance be S. 54° 75' E. dist. 79Ch. the diff. of lat. will be 45.59Ch. S. and the dep. 64.51 Ch. E. When the distance is expressed by any whole number of chains or perches exceeding 100. Divide the given distance into parts that shall not ex. ceed 100 each, and find as before the difference of lati. tude and departure corresponding to the given bearing and to each of those parts; the sums of the latitudes* and departures thus found will be the latitude and de. parture required. * For the sake of conciseness in the expression the word latitude only is some. EXAMPLES. 1. A line bears N. 201° E. dist. 117Ch. required the corresponding latitude and departure. Ch. Ch. Ch. Dist. 100 corresp. Lat. 93.67 and Dep. 35.02 17 15.92 5.95 Whole Dist. 119 Lat. 109.59 N. Dep. 40.97 E. 2. What is the difference of latitude and departure of a line bearing N.781° W. dist. 243 perches? Per. Per. Per. 97.90 8.76 42.10 Whole Dist. 243 Lat. 19.48 N. Dep. 237.90W. When the distance is expressed by chains or perches and decimals of a chain or perch. Find as above the latitude and departure corresponding to the given bearing and to the whole chains or perches. Then considering the decimals as a whole number, find the latitude and departure corresponding to it, removing the decimal point in each, two figures to the left hand if there be two decimals, or one figure to the left if there be but one decimal; these added to the former will give the difference of latitude and departure required. EXAMPLES. 1. If a line bear S. 41%" W. dist. 57.36 Ch. what will be the corresponding difference of latitude and depar |