"in line." C then looks to A, and puts B in line at B'. Brepeats his operation from B', putting Cat C', and is then himself moved to B", and so they alternately "line" each other, continually approximating to the straight line between A and Z, till they at last find themselves both exactly in it, at B' and C'''. (181) A single person may put himself in line between two points, on the same principle, by laying a straight stick on some support, going to each end of it in turn, and making it point successively to each end of the line. The "Surveyor's Cross," Art. (104), is convenient for this purpose, when set up between the two given points, and moved again and again, until, by repeated trials, one of its slits sights to the given points when looked through in either direction. A Fig. 118. MAN (182) On Water. A simple instrument for the same object, is represented in the figure. AB and CD are two tubes, about 11⁄2 inches in diameter, connected by a smaller tube EF. A piece of looking-glass, GH, is placed in the lower part of the tube AB, and another, KL, in the tube CD. The planes of the two mirrors are at right angles to each other. The eye is placed at A, and the tube AB is directed to any distant object, as X, and any other object behind the observer, as Z, will be seen, apparently under the first object in the mirror GH, by reflection from the mirror KL, when the observer has succeeded in getting in line between the two objects. M, N, are screws by which the mirror KL may be adjusted. The distance between the two tubes will cause a small parallax, which will, however, be insensible except when the two objects are near together. Z--- (183) Through a Wood. When a wood intervenes between other, as in the figure, in which A and Z are the given points, proceed thus: Hold a rod at some point B' as nearly in the desired line from A as can be guessed at, and as far from A as possible. To approximate to the proper direction, an assistant may be sent to the other end of the line, and his shouts will indicate the direction; or a gun may be fired there; or, if very distant, a rocket may be sent up after dark. Then range out the "random line" AB', by the method given in Art. (169), noting also the distance from A to each point found, till you arrive at a point Z', opposite to the point Z, i. e. at that point of the line from which a perpendicular there erected would strike the point Z. Measure Z'Z. Then move each of the stakes, perpendicularly from the line AZ', a distance proportional to their distances from A. Thus, if AZ' be 1000 links, and Z'Z be 10 links, then a stake B', 200 links from A, should be moved 2 links to a point B, which will be in the desired straight line AZ; if C' be 400 links from A, it should be moved 4 links to C, and so with the rest. The line should then be cleared, and the accuracy of the position, of these stakes tested by ranging from A to Z. (184) To an invisible intersection. Let AB and CD be two Set stakes at the five given points, A, B, C, D, P. Set a sixth stake at E, in the alinements of AD and CP; and a seventh stake at F, in the alinements of BC and AP. at G, in the alinements of BE and DF. line. Then set an eighth stake Otherwise; Through P range out a parallel to the line BD. Note the points where this parallel meets AB and CD, and call these points Q and R. Then the distance from B, on the line BD, to a point which shall be in the required line running from P to the II. OBSTACLES TO MEASUREMENT. (185) The cases, in which the direct measurement of a line is prevented by various obstacles, may be reduced to three. A. When both ends of the line are accessible. B. When one end of it is inaccessible. C. When both ends of it are inaccessible. A. WHEN BOTH ENDS OF THE LINE ARE ACCESSIBLE. and measure past the obstacle; turn a third right angle at C; and measure to the original line at D. Then will the measured distance, BC, be equal to the desired distance, AD. If the direction of the line is also unknown, it will be most easily obtained by the additional perpendiculars shown in Fig. 109, of Art. (171). Fig. 122. A C B (188) By symmetrical triangles. Let AB be the distance required. Measure from A obliquely to some point C, past the obstacle. Measure onward, in the same line, till CD is as long as AC. Place stakes at C and D. From B measure to C, and from C measure onward, in the same line, till CE is equal to CB. be equal to AB, the distance required. CD and CE equal, respectively, to half of AC and CB; then will AB be equal to twice DE. E D Measure ED, and it will If more convenient, make Set a fourth stake, F, at the intersection of EA and CD. Measure AC AC, AF and FE. Then is AB = AF (FE-AF). As this expression is somewhat complicated, an example will be given: Let AC = 100, CE = 40, CD = 30, DE = 21, and CB = 80; then will AB = 51.7. B. WHEN ONE END OF THE LINE IS INACCESSIBLE. point A, set off AC, perpendicular to AB, and of A AC2 D AB = AD (192) Otherwise: At the point A, in Fig. Fig. 126. 126, set off a perpendicular, AC. At C set B If EC be made equal to AE, and D be set in the line of BE, and also in the perpendicular D from C, then will CD be equal to AB. If EC = AE, then CD=AB. (193) Otherwise: At A, in Fig. 127, measure a perpendicular, AC, to the line AB; and at any point, as D, in this line, set off a perpendicular to DB, and continue it to a point E, in the line of CB. Measure DE and also DA. Fig. 127. B A AC X AD Then is AB = DE - AC E D |