CASE I. 91. In the first case, let BA (Fig. 10) represent the oblique line passing through A and the mark left on the staff at B; BE is equal to the horizontal distance measured (72). Place the theodolite at A, and measure the angle of elevation FAD, as also the angle of depression BAF'; from these data, together with those before found, the vertical distances, DC and DF, can be determined. Suppose the LDBC=38°, the horizontal distance BE=600 yards, the ZDAF=46° and the LBAF=ABE=27° 30' 25". In the triangle ABE, to find In the triangle ABD to find AB. 90°-EBA=EAB=62° AD. 29 35". As sin. LADB, 8° 9.143555 As sin. LEAB, 62 To AB, 676.47 2.830250 29' 35" 9.947901 So sin. LABD, 10° To BE, 600 2.778151 29' 35" 9.260349 So is radius 10. To AD, 885.20 2.947044 In the triangle FAD, to find DF. 10.000000 TO BE, 600 2.778151 | Is to AD, 885.20 2.947044 So is sin. EBA, So is sin. A, 46° 9.856934 To DF, 636.76 2.803978 Add AE= 312.43 161 30 25" But DBC-- ABE Gives CD=949.19 =DBA= 10° 29' 35" Hence DAB+ DBA= 172° 00'00" CASE II. 92. Let BA (Fig. 11) represent the oblique line passing through the mark left on the staff at B, and the second station A; and EA the horizontal distance between the points B and A, equal to 600 yards. Suppose the LCBD=38° ZFAD= 46', and the angle of elevation EAB=27° 30' 25". 180°—(BAE+FAD)= BAD=106 29' 35". Now, EAB= ABC; ABC+DBC=ABD=650 30 25'. 180°---(BAD+ABD=ADB=8°. From these data, the perpendicular CD is found to be equal 2869.19 yards. CASE III. 93. In the third case, the angle ADB is equal to the difference between the angles of elevation at the stations B and A, and the vertical line CD is found, as in the other cases. 94. If the ground between B and A inclines regularly, so that the oblique line can be measured with the chain, the calculation will be shortened by measuring this line, instead of the horizontal distance. But if the ground be broken and undulating, this cannot be done, and in such case the horizontal distance must be used. PROBLEM JII. To find the perpendicular distance of an object below a given horizontal plane. 95. From any point of the plane measure a base line, and at its extremities, the horizontal angles included between it and the lines drawn to the given object; and also the angle of depression at that extremity of the base which is in the given plane. Then calculate the horizontal distance between this extremity and the given object; there will then be known the base and angle at the base of a right angled triangle, the perpendicular of which is the distance sought. By measuring the angle of depression at the other extremity of the base, and calculating the perpendicular distance of the object below this point, we obtain, by a comparison of the disLet C (Pl. I. Fig. 12) be the given object, AB the base line, equal 672 yards, the horizontal angle ABC=39° 20', the horizontal angle BAC=72° 29, the angle of depression CAC= 27° 49, and the angle of depression CBC"=19° 10': CC' is the vertical line drawn from the object C, and meeting the horizontal line AC' at C': CC" is perpendicular to the horizontal line BC". The horizontal angle ACB=180°—(A+B)=180°— 111°49 To find BC". TO AC, 458.792 2.661617 To BC", 690.277 2.839024 a To find CC. To find CC". As radius 10.000000 As radius 10.000000 Is to AC', 458.792 2.661617 Is to BC", 690.277 2.839024 So is tang. CAC, So is tang. C'BC, 9.541061 To CC, 242.065 2.383932 | To CC", 239.93 2.380085 Hence, CC-CC"=242.065—239.93=2.135 yards—the difference of the elevations of A and B above the object C. We might also, as in the second method of the last problem, if the ground admitted of it, measure the base line directly towards or directly from the object, and then measure the angle of depression at each station: these data would be sufficient, from which to calculate the vertical distance. 96. In measuring a base line, if great accuracy be required, the theodolite should be placed at one extremity, and directed to the other; and the alignment of the staves made, by means of the vertical spider's line of the upper telescope. If the highest degree of accuracy be necessary, the base line should be measured with rods, which admit of being adjusted to a EXAMPLES. Example 1..Wanting to know the distance between two inaccessible objects, which lie in a direct line from the bottom of a tower of 120 feet in height, the angles of depression are measured, and found to be, of the nearest, 57°; of the most remote, 25o 30. Required the distance between them. Ans. 179.656 feet. Example 2.-In order to find the distance between two trees, A and B, which could not be directly measured because of a pool, which occupied the intermediate space, the distances of a third object C from each of them were measured, viz. CA=588, CB=672, and also, the contained angle ACB=55° 40. Required the distance AB. Ans. 593.8. Example 3.—Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured the angle of elevation of the top of the hill 40', and of the top of the tower 51'; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45'. Required the height of the tower. Ans. 83.9983 feet. Example 4.-Wanting to know the horizontal distance between two inaccessible objects E and W, I measured a horizontal base line AB, of 536 yards, and at the extremities A and B, the horizontal angles BAW=40° 16', WAE=57° 40', ABE =42° 22', EBW=7107. Required the distance EW. Ans. 939.52 yards. Example 5.-Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which both of them could be seen, two points, C and D, were chosen, at a distance from each other equal to 200 yards, from the former of which A could be seen, and from the latter, B, and at each of the points C and D a staff was set up. From C a distance CF was measured, , not in the direction of DC, equal to 200, and from D, a distance DE=200, and the following angles taken, viz.: AFC=83, ACF=54° 31', ACD=53:30, BDC=156° 25', BDE=54° 30, and BED=88° 30'. Required AB. Ans. 345.5, Example 6.-From a station P there can be seen three ob a jects, A, B, and C, whose distances from each other are known, viz. AB=800, AC=600, and BC=400 yards. Now there are measured the horizontal angles APC=33° 45, BPC=22 30. It is required, from these data, to determine the three distances PA, PC, and PB. Ans PA=709.33, PC=1042.66, PB=934 yards. OF MEASUREMENTS WITH THE TAPE OR CHAIN ONLY. 97. As it may often happen, that instruments for the measurement of angles cannot easily be obtained, it seems not to be out of place to explain the best methods of determining distances by means of the chain, or tape only. PROBLEM. 98. To trace, on the ground, the direction of a right line, that shall be perpendicular, at a given point, to a given right line. Let AB (Pl. I. Fig. 13) be the direction of the given line, and D the given point. FIRST METHOD. Measure from D, on the line AB, two equal distances DA, DB, lying on different sides of the point D. Take a portion of the chain, or tape, greater than AB; mark the middle point of it, and fasten its extremities at A and B. Then, taking the chain by the middle point, stretch it tightly on either side of AB, and place a staff at C, or E. AB and EC are then at right angles to each other (55)* SECOND METHOD. 99. From the point D, measure the distance DB, equal to 8; with the point B as a centre, and a radius equal to 10, mark on the ground the arc PI: then with D as a centre, and a radius equal to 6, mark in like manner the arc cutting it; the line DC is at right angles to AB (186). Remark. Any three lines, having the ratio of 6, 8, and 10, form a right angled triangle, of which the side 10 is the hypo |