image of which was made to coincide with each of the stations in measuring the horizontal angles. NOTE 4. In measuring the height of mountains at a distance, allowance must be made for the curvature of the earth, by adding to the calculated height 8 inches multiplied into the square of the distance in miles, or to twice the log. of the distance in feet add the constant log. 8-378641, and the result will be the log. of the correction in feet very nearly. EXERCISES. 1. A person 6 feet high, standing on the side of a river, observed that the top of a tower placed on the opposite side, subtended an angle of 59°, with a line drawn from his eye parallel to the horizon; receding backwards for 50 feet, he then found that it subtended an angle of 49°. Required the height of the tower, and the breadth of the river. Ans. Height of the tower 192.27 feet. Breadth of the river 111.92 feet. 2. To determine the altitude of a light-house, I observed the elevation of its top above the level sand on the seashore to be 15° 32′ 18′′, and measuring directly from it along the sand 638 yards, I then found its elevation to be 9° 56′ 26′′; required the height of the light-house. Ans. 302-463 yards. 3. It is required to find the height of Arthur's Seat, in the vicinity of Edinburgh, from the following observations, taken with a theodolite on Leith sands, about the medium height of the tide, the base being 1410-42 feet; at the west end of the base, being that nearest to Leith pier, the angle subtended by the top of Arthur's Seat, and the eastern extremity of the base, was 50° 44', and the elevation of its top 3° 59', whilst the angle subtended at the eastern station, by the western and the top of the hill, was 123° 29′, and the elevation of the top of the hill was 4° 17′ 30′′: Find the height of the hill above the medium level of the tide, allowing 5 feet for the height of the eye, and the necessary correction for curvature, by (Note 4). Ans. The observed altitude at the western station gives 821·174, that at the eastern gives 821-019 feet, the difference of the two being less than 2 inches. 4. Find the height of the top of the cross on the spire of Assembly Hall, Castle Hill, Edinburgh, from the following observations, the height of the eye being on a level with the sole of the entrance door; at the first station, the elevation of the top of the cross was 62° 36', angular bearing of the spire from the second station 35° 36', and the angular bearing of the spire and the first station, taken at the second, was 36° 22′, the distance between the stations being 195.6 feet. Ans. 235-309 feet. A 90. To find the distance of two inaccessible objects. Let the two objects be A and B; take any stations C and D, such that the objects and the other station can be distinctly seen from each, and that the distance CD can be accurately measured; then at the station C measure the angles ACB D B and BCD, and at the station D, measure the angles BDA, and ADC, then measure accurately the distance CD. SOLUTION. In each of the triangles ACD, BCD, we have given two angles, and the side lying between them, consequently (Art. 39) AC, AD, CB, and BD can be found, then in each of the triangles ACB and ADB we have two sides and the contained angle, consequently (Art. 40) the side AB can be found from either of these triangles, and if it be calculated from each, and the results be found the same, it is a proof of the correctness of the calculations. 91. This problem is of very extensive application in many departments, both of civil, military, and naval surveying, and may be solved in a different manner from the commonly adopted method which is given above. The method here referred to may be called surveying by Rectangular Co-ordinates, and consists in finding an expression for the perpendicular distances of the objects from the base line, and the distance of the middle of the base from the foot of the perpendiculars, in terms of the base and the angles at its extremities. 92. To find an expression for the perpendicular AD in terms of the base BC and the angles B and C, at its extremities. By (Art. 37) AC=BC sin. B sin. C A sin. (B+C) * sin. B+ log. sin. C+ log. BC—30}; or Add together the log. cosecant of the sum of the angles, the log. sines of each of the angles at the base, and the log. of the base; the sum, diminished by 30 in the index, will be the log. of the perpendicular. 93. To find an expression for ED, the distance of the Hence Log. AD={log. cosec. (B+C)+ log. foot of the perpendicular from the middle of the base, in terms of the base and the angles at its extremities. .. 2ED=DC-BD= BC{sin. B cos. C—cos. B sin. C}. sin. (B+C) But (Art. 52), {sin. B cos. C-cos. B sin. C}=sin (B-C). By substituting this value, and dividing both sides by 2, we sin. (B-C). whence obtain, ED=BC sin. (B+C) Log. ED={Log. cosec. (B+C)+ log. sin. (B—C)+ log. 1⁄2 BC-203. Or, add together the log. cosecant of the sum of the angles at the base, the log. sine of their difference and the log. of half the base, the sum diminished by 20 in the index will be the log. of the distance of the foot of the perpendicular from the middle of the base. 94. The foot of the perpendicular falls always on that side of the middle of the base which is adjacent to the greater angle at the extremity of the base, and its distance from the middle of the base should be marked when on one side, and when on the other, (Art. 28). Also, if in an extensive survey some of the objects should lie on the one side, and others on the other side of the base, the perpendiculars should be marked + on the one side, and on the other. 95. The perpendiculars DG and CF being thus calculated, and also EG and EF, the distance of the two objects C and D can be found, for CI being drawn parallel to AB, the triangle DIC will be right angled, and ĎI is the difference of the perpendiculars, and IC is the difference of the distances of the perpendiculars from the middle of the base, since EF is -; and if one of the perpendiculars had been measured downwards, DI would still have been their difference, since the perpendicular measured downwards would be minus. Now Log. DI+10— Log. IC=Log. tan. DCI. But Log. sec. DCI+ Log. IC-10= Log. DC; hence we have the following rule for finding the distance of two objects:-From the log. of the difference of the perpendiculars, increased by 10, of an index, subtract the log. of the distance of the perpendicular, the remainder will be the log. tan. of an angle; to the log. secant of this angle add the log. of the distance of the perpendiculars, and the sum, rejecting 10, from the index, will be the log. of the distance sought. 96. If the distance between the objects A and B be given, to find the distance between the stations C and D, (Art. 90); measure the angles at C and D as before, and assume CD any length, as 100 or 1000, and from this assumed length find AB; then the calculated length of AB is to the assumed length of CD, as the true length of AB is to the true length of CD. EXERCISES. 1. Find the distance between the two objects A and B, (Fig. art. 90), on the supposition that CD is 300 yards, LACB 56°, LBCD=37°, ZADB=55°, and LADC=41°. Ans. AB-341.25 yards. 2. Being desirous of finding the distance between two objects A and B, I measured a base, CD, of 384 yards, on the same horizontal plane with the objects A and B. At C I found the angle BCD=48° 12′, and ACD=89°18'; at D the angle ADC was 46° 14', and BDC 87° 4'. It is required from the data to compute the distance between A Ans. AB-358.5 yards. and B. 3. Wanting to know the distance between two inaccessible objects A and B, (Diag. art. 90), I measured a base line CD of 360 yards: at C the horizontal angle ACB was observed with a sextant, and found to be 53° 30', and the angle BCD 38° 45'; at D the horizontal angle BDA was 67° 20′, and the angle ADC 44° 30'. Required the distance between A and B. Ans. 548-149 yards. 4. Wishing to know the distances and positions of a number of the principal objects in and about Edinburgh, I took a station on the top of Arthur's Seat, and with a theodolite measured the angular bearings of the following objects, with a line drawn through the top of Nelson's Monument, on the Calton Hill, viz. :-Spire of Assembly Hall (H), Castle Tower (T), Dome of St George's Church (G), Spire of St Andrew's Church (A), Melville's Monu ment (M), Spire of North Leith Church (L), Inchkeith Light-house (K), and Berwick Law (B). I then went to the top of Nelson's Monument, and took the angular bearings of the same objects with a line drawn through the former station on the top of Arthur's Seat. The angles were as given in the following table, where the single letters in the first column signify the objects after whose names they are placed above: The base of the above triangles, namely, the distance. between the top of Arthur's Seat and Nelson's Monument, was determined from the third triangle, of which a side, namely, the distance from Nelson's Monument and the top of the dome of St George's Church, was previously determined from a similar survey, in which the base was measured, and the distance so determined agreed to the tenth of a foot with that deducible from data given in Wallace's Theorems and Formule, and hence concluded to be very correct. The base thus determined was 5675 feet. The following is the method of calculating the perpendicular, and the distance of its foot from the middle of the base in the first triangle; and the others are exactly similar: For the perpendicular. For its distance from the middle of the base. Log. (5675) = 3.452936 Log. cosec. 126° 56′-10·097271 =9-654059 Log. sin. 73° 20' 9.981361 =9.993172 Log. 5675 3.498468 3400 73.531568 NOTE 1. The foot of the perpendicular falls always on that side of the middle of the base which is adjacent to the greater angle. All the perpendiculars in the above fall upon that side of the base which is adjacent to Nelson's Monument except the last, which is therefore |