EB 6206 feet, the angle FEA = 40°. Required the radii of a c. c. to join A and B, and also the point of compound curvature. We observe the external secant E C is common to both curves. Now by construction of the tables we have: external secant a = tangent a tangenta, radius being unity. The angles EBC and EA C are measured by half their arcs CB and C A. 40° Call these angles x and y respectively. Then x + y = 2 20°; then 620-6 X tangent y =5057 x tangent x, or 620-6: 505-7 tangent y tangent x. Then by previous proposition 62065057: 620.6. - 505·7 :: sine (x + y = 20°) : sine (x — y) or, 11263 114.9 :: sine 20°: y-x. Neither of the radii being given, we will assume the condition, that the p. c. C shall be in line with the vertex E and the centres O and D. We have by logarithms: Now x + y = 20. Then 9° = 2° 11°; consequently COB = 18°, and A DC=22°. Now we have the length of tangent and curvature given, to find the radius. Let B be a point in a curve whose radius B F is given, and let D be another fixed tangent point. It is required to find point of c. c., the curve AB being produced, from which to start a curve to terminate in tangent at Ď, ́also the radius of last curve. Given the angles MDB, MB D, distance BD, and radius BF., Imagine the simple curve B CL to be run with a given radius BF till LN becomes parallel to D M. Now by the nature of a curve, upon whatever point on the curve the transit be placed, the difference between backsight on B and foresight on I, is always the × (38) 2 10 brods adt annet A same, namely, B+ D Now at the true point of c. curvature C, the difference between backsight on B and foresight on D is also therefore the transit reading the same on D as on L, CLD must be in the same straight line. Hence whenever the nature of the ground will admit of it, erect a flagstaff at D, curve round from B towards L until taking a backsight the foresight necessary to fall upon L should strike the flagstaff at D. The transit will then be at the point of c. curvature sought. Then measure CD, and make this proportion: sine HCL*: CD :: R:x=0 D. = Suppose HCL=8°, and the distance CD 600 feet. Then by substituting in the above proportion, we have by logarithms: When the ground will not admit of this method, ascertain by measurement or calculation the distance from B to D. 2 (B F) × B+ D = BL. Now angle LBD = B-D* The triangle will then have the two sides B D and BL, and the included angle B, to find the angle LDB = CD B.† Now in the triangle B CD we have the angle B CD to the sup plement of B+ D 2 also the angle at D, consequently the angle at B. These angles, together with base B D, determine the chord CD; from which, with the angle HCL, calculate R as before. HCL becomes known from the fact, that CBD gives CBMG CB, CBM being B-CBD. This taken from will give the angle HCL required. B+D PROPOSITION XVII. FIG. 16. Between two tangents to locate a curve passing through a given point. Suppose A B and CD to be the tangents permanently fixed with Fig.16 *Because N BL (isosceles) = exterior angles at N and M = - LBD which will make all the angles known. reference to some agreement between individuals; and let F be the given point at which it is necessary to keep a given distance from some building or other object. Suppose A B and CD produced to meet in E. The angle O ED, and consequently its half EBD, are known. The distance I E is also known. Let the angle OED= 60°, let IF = 17.5 feet. It is required to find the point B, so that the angle FBI shall = 30°. By natural sines: Now (35 + 17·5) × (35 — 17·5)* = √′ 52·5 × 17·5 = 30.3 = I B. Suppose IE measures 462 feet. Then B E will equal 462 + 30.3 = 492.3. By similar triangles FB: BE:: BI: BK, or Then BD =8524 and BH 852-4 35817.4. Now we have by geometry BH BF=BA, or /8173 x 35 169.1 BA. Hence A B+ BEAE, or 1691 + 4923 = 661·4. tangent 30° 0.57755 Now suppose it is inexpedient to produce the tangents to a vertex, the angle OED being known, find the point B as before, and turn off EBD = 40 E D, measure B D, and calculate by trigonometry the side ED BE, and also BA as before. = Again, suppose the angle at E is not known, neither is it practicable to measure a direct line between the two tangents, calculate by traverse the true course and distance between any two convenient points on the tangents by Proposition IX., from which calculate the position of E. Without ascertaining the distance to E, the radius A G can be calculated thus: Therefore commence at A, and run 800 feet of a 5° curve to C. PROPOSITION XVIII. FIG. 17. Given the length of a common tangent D G a, and the angles of intersection n and m, to determine the common radius CE CF = radius of a reversed curve to unite the tangents HD and B L. Now DC=R × tangent n, and CG Rx tangent m; we have therefore *The sum of two quantities multiplied by their difference is equal to the difference of their squares. Suppose it to be required to introduce 200 feet of tangent between the curves, that portion of the tangent DG taken by the two curves will be 600 feet. Then we have: 800 600: 3256'7 : 2442.5 radius: ON REVERSED CURVES, TURNOUTS, ETC. FIG. 18. AF 98 feet, A D = 102 feet, and D E = 102 feet. Let G gauge of track, and R radius of turnout, x = distance Fig.18. = |