may be its course or courses, to a right-angled base to the course of the line to be measured. This can be readily done if care be taken to run and measure the base, at such angles that their latitude and departure can be taken from the traverse table. FIGURE 1. Distance required over lake from A to C, course eastright-angled base-from A to B 690 links. Angle at C 20° 20.' Natural co-tangent of the angle at C = 2.698525 690 242867250 16191150 1861-982250 Over lake 1862 links. FIGURE 2. Distance required over lake from D to F, course west— from D to E, S. 20° E. 752 links--gives 707 links southing, which is the right-angled base K G, and 257 links easting from G to K. Angle at F 15°. Natural co-tangent of the angle at F . (Nat. co-tan. Fx KE)-KD=D F. =DF) = 3.545732 707 24820124 (3·545732 × 707)=2507-257-2250. 24820124 Distance required over lake from G to I, course south. To obtain a base in this example, we run Distance from G to S. 992. . H to S 896 links. Deputy surveyors will be allowed pay for the distance across lakes or ponds not meandered where they are required to continue the lines of the public surveys across them; but no offsets or lines run in triangulating will be paid for. Where the distance across a lake or other body of water is ascertained by offsetting, it is not enough to say in the field notes "8.65 over lake and set a meander corner," but the mode by which the distance is ascertained must be stated and described in full. DISTANCE OVER A RIVER BY "OFFSET." Example.-Fig. 4. In running a line north, intersect the right bank of a river at A (course N.N.E.) and erect an object, turn the compass sights to west, to an object at B, and pass over the river to it, then run and measure a line north to C, and "offset" east into line at D, the distance between A and D will be equal to the distance between B and C. Or, if a line be run and measured from A, N. 60° E., until an object in line at D bears N. 30° W., the distance A D will be twice that of A E, for the reason that the triangle thus formed is one-half of an equilateral triangle. FIG. 4 C D B 12.30 River N 12.30 615 A E Frequently offsets are made in passing small lakes, bends of rivers, etc.: sometimes the distances can be advantageously taken over such obstacles with the telescope and rod. Also, it often happens that a suitable angle can be taken, and the base to that angle measured afterwards; in such cases the distance can be taken from the traverse table; but if no traverse, or other proper tables are at hand, the following angles, on a right angle base, and the multiplier to it, will give the distance. These may be committed to memory. Angle 11°, 18', multiply the base by 5. 14, 2, multiply the base by 4. 18, 26, multiply the base by 3. 66 21, 41, multiply the base by 2.5. SHORT METHOD OF FINDING THE AREA OF A MULT- Example, showing how to reduce the plot of a multangular field to a field of equal area having only three or four sides, by which its contents may be readily found. To reduce such a field the only instruments required, after the meanders are properly laid down, are a good parallel-rule* and a fine protracting point. In the following figure, first extend the base E H to an * The triangle and the rule are the best. indefinite length; then placing the rule on the angles 1 and 3, move it parallel from the angles 1 and 3 to the angle 2, and mark the exact point of intersection at A, on the base E H. Now place the rule on A and the angle 4; then move it parallel to the angle 3, finding the point B on the base E H; place the rule on B and the angle 5, and move, parallel, to the angle 4, finding the point C on the base E H. Now place the rule on the point C, and the terminating point 6 on the line F G, and move the rule, parallel, to the angle 5, finding the point D on the base E HI, from which point draw a line to 6, the process then being complete. The line D 6 thus drawn, leaves the same area of lake to the left, that there is of land to the right. (Fig. 5.) Any figure may be calculated upon the same principle FIG. 6. i 1Ъ a h by drawing a base and erecting a perpendicular line from it, passing through the figure. Place the rule at a and c, then move, parallel, back to b, marking the point 1 on the base; then from 1 to d, and move forward to c and so on to the angle at i, leaving a triangle to the right of the perpendicular. Pro ceed in like manner with that portion of the figure to the left of the perpendicular line, throwing it into two triangles. (Fig. 6.) CONVENIENT RULES FOR CORRECTING THE COURSE OF RANDOM LINES, WHEN THE CORRECTION DOES NOT EXCEED 200 LINKS TO EACH MILE. RULE FOR HALF A MILE, OR FORTY CHAINS. From the number of links to be corrected in that distance, subtract one-seventh; the difference will be the number of minutes of a degree required for the correction of the course. Example. Number of links to be corrected, 42-636' answer. RULE FOR ONE MILE, OR EIGHTY CHAINS. From half of the number of links to be corrected in that distance, subtract one-seventh, the difference will be |