B MEASURING DISTANCES OVER LAKES AND RIVERS.* It frequently happens in surveying the public lands that lakes, rivers, and bayous, the distance across which cannot be measured with the chain, interrupt the public lines. The following illustrations will assist the inex perienced surveyor in passing such obstacles. They are given on the principle of reducing the base, whatever * The four following illustrations are taken from Burt's "Key to the Solar Compass." Mr. Burt was the inventor of this instrument, which is a great improvement on the magnetic compass, and has been adopted by the government to be used in the running of standard and township lines. 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 Over lake 1862 links. FIGURE 2. Distance required over lake from D to F, course westfrom 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. Multiplied by the base K E • Subtract distance from D to K. Distance from D to F 242867250 16191150 1861-982250 = 3.545732 (Nat. co-tan. Fx KE)—K D=D F. 24820124 ·(3·545732 × 707)=2507-257-2250. 24820124 2506-832524 257 2250 lks. nearly. FIGURE 3. Distance required over lake from G to I, course south. To obtain a base in this example, we run Distance from S to I. Add distance from G to S Distance over lake from G to I.. Distance from G to S. 992. . H to S 896 links. Co tangent of the angle 16° 38' at I. 3.347319 20083914 30125871 26778552 2999-197824 992 3991 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 N 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. B 12.30 D River 66 66 A Angle 11°, 18', multiply the base by 5. 66 14, 2, multiply the base by 4. 66 615 E 18, 26, multiply the base by 3. 21, 41, 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 Fig. 5. 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 |