In the figure, draw L K parallel to G V, and passing through centre of first branch. = = Then MK= (R + r) X cos. B. NL - (R+r) X cos. C. NL – MK= D. A= (90° — C) — (90° — B)=B — C. Example. log. 3.352568 log. 9.916859 (R + r) cos. B= 1,860 3.269427 Add D 94 . XLV. TO PASS A CURVE THROUGH A FIXED POINT, THE ANGLE OF INTERSECTION BEING GIVEN. B To Y T R Given the intersection angle, A, of two tangents, to find the radius, R, of a curve which shall pass through a point, C; the position of said point, with reference to the tangents or the point of intersection, being known. 1. By what data soever point C is located, they may be commuted by simple processes to the form shown in the figure; namely, the ordinate BC and the distance IC to apex. Call the angle BIC a, and complete the triangle I CO. A In this triangle, x= a= 90° = (A + a). 2 Also, CO:10:: sin. x : sin. 2. R R But CO=R; IO= ..R: :: sin. X: sin. 2. cos. | A 'cos. A Hence sin. 2= The triangle ICO may then be cos. } A sin. X Escample. A = 400, BC= 32 feet, I B = 80 feet. Then BC = IB = 32 • 80 0.4 nat. tan. 21° 49'; and IC=BC : nat. sin. 21° 49' : :32 .372 = 86 feet. Also, x = 90° (1 A + a) 900 (41° 49') = 48° 11'. Or, since the sine of any angle is equal to the sine of its supplement, the supplement in this case, 52° 29', may be taken directly from the logarithmic table, from which supplement deducting x, or 48° 11', the remainder is the angle y= 4° 18'. 2. In the case of a rectangular intersection, the solution is more simple. It is quite plain, from the figure, that R2 = (R — a)2 + (R — b)?, from which equation, R=a+b+V2 ab. 3. Cases of this kind are disposed of with great ease in the field by means of the curve-protractor. TO FIND THE RADIUS OF A TURNOUT CURVE, THE FROG ANGLES, AND THE DISTANCES FROM THE TOE OF SWITCH TO THE FROG POINTS. 1. Draw the figure as in margin, C being the centre of the turnout curve, CK parallel to main track, and OK, IE, LM, perpendicular to it. Call the angle of the frogs at 0, F; that of the intermediate frog at I, 2 F'; the throw of the switch-rail for single turnout, D; its angle with main track, S; the gauge of the track, G; and radius of outer rail, R. 2. Usually the length and throw of switch-rail and the angles of the frogs at O are given. In that case, to find R, F', = 3. The angle HNW, between the line of the switch-rail prolonged and a tangent to turnout curve at frog point 0, NOP NII W = F – S. The angle NOL or NLO, between chord and tangent, half the intersection angle HNW } (F – S). The angle NOB=NOL+LOB. But NOL is seen to be # (F S), and NOB - F; then LOB= NOB - NOL=F - ] (F – S)=} (F+S). The distance LO, from toe of switch to point of main frog, LB • sin. LOB= (G–D) • sin. (F+S). 4. Again: the angle LCY N L0 => (F– S); LY= LO= }(G – D) • sin. † (F ·S). LY • sin. LCY= LC; i.e., ((G D) • sin. $ (F + S)] • sin. * (F – S) = R. 5. R may be found otherwise, as follows: OK= 0 C cos. KOC=R cos. F; LM=LC cos. CLM = Rcos. S; LM – OK= LB; i.e., R (cos. S cos. F) = (G D). Hence R (G-D) ; (nat. cos. S nat. cos. F). 6. If R be known, to find F. This equation gives nat. cos. F nat. cos. S - |(G-D) • R). 7. To find the angle, 2 F', of the middle frog at I. ΙΕ IP + PE or O K; i.e., R cos. F=G + A cos. F. Hence nat. cos. F' nat. cos. F +(IG : R). 8. The angle LIV, by similar reasoning to that used in relation to LOB, is found to be =} (F' + S). The distance LI, from toe of switch to point of middle frog, =LV • sin. LIV (1 G - D) = sin. I (F' + S). RULES FOR FROGS AND SWITCHES. 9. To find the Angle of Switch-Rail with Main Track. Divide its throw, in decimals, by its length: the quotient will be the natural sine of the angle sought. 10. To find the Distance from Toe of Switch to Point of Main Frog. Subtract the throw of switch-rail from the gauge of track, both in decimals; call the remainder a. Add together the angle of switch-rail with main track and the angle of the main frog; find the natural sine of half this sum, and call it b. Divide a by b: the quotient will be the distance |