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. 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. B 12.30 FIG. 4 D N 12.30 次 River 615 E Frequently offsets are made Also, in passing small lakes, bends of 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. 66 66 18, 26, multiply the base by 3. 66 21, 41, 66 multiply the base by 2·5. 26, 34, multiply the base by 2. SHORT METHOD OF FINDING THE AREA OF A MULTANGULAR FIELD. 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 EH. 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 H, 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. a b 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, 42-6= 36' 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 the number of minutes of a degree required for the correction of the course. Example. = Number of links to be corrected, 70 ÷ 2 = 35 — 5 = 30' answer. RULE FOR THREE MILES. Divide the whole number of links to be corrected by seven; the quotient will be the number of minutes of a degree required for the correction of the course. Example. Number of links to be corrected, 297 ÷ 7 = 423′ an swer. RULE FOR SIX MILES. Divide one-half of the number of links to be corrected by seven; the quotient will be the number of minutes required for the correction of the course. Example. Number of links to be corrected, 370+2=185÷7= 26% answer.* The distances given for corrections in the above examples, are those for which corrections are generally made in the survey of the public lands, and the calculation for the course of the corrected line can generally be mentally made by the surveyor, while he is occupied in adjusting his instrument. *The above rules are close approximations. |