In the figure, draw L K parallel to G V, and passing through centre of first branch. .. (R+r) × cos. C― (R+r) × cos. B = D. A: Example. R=1,433, r= 819, B34° 20', D=94. Cos. C[(R+ r) cos. B + D] ÷ (R+r). XLV. TO PASS A CURVE THROUGH A FIXED POINT, THE ANGLE OF INTERSECTION BEING GIVEN. 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 ICO. IC=BC÷ nat. sin. 21° 49′ = 32 ÷ .372 = 86 feet. Also, x = 90° — († A + a) = 90° 32 80 = 0.4 nat. tan. 21° 49'; and 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)2, from which equation, R=a+b+√2 ab. Example. a = 40, b = 80. Then R40+80+ √6,400 — 200. 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 O, 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 HN W, between the line of the switch-rail prolonged and a tangent to turnout curve at frog point O, NOP NHW FS. The angle NOL or NLO, between chord and tangent, = half the intersection angle HNW (FS). The angle NOB=NOL+LOB. But NOL is seen to be = (F S), and NOB = F; then LOB = NOB-NOL=F (F — S) = (FS). LO, from toe of switch to point of main frog, LOB (G— D) ÷ sin. † (F + S). 4. Again: the angle LCY = NLO= = The distance D) sin. (F + S). LY÷sin. LCY= (G — D) ÷ sin. † (F + S)] ÷ sin. † (F — S) 5. R may. be found otherwise, as follows: OK = OC cos. KOC = R cos. F; LM = LC cos. CLM= R cos. S; LM-OK=LB; i.e., R (cos. S D). Hence R = (G — D) ÷ (nat. cos. S · cos. F) (G nat. cos. F). 6. If R be known, to find F. This equation gives nat. cos. F nat. cos. S 7. To find the angle, 2 F', of the middle frog at I. IPPE or OK; i.e., R cos. F' = G + R cos. F. Hence nat. cos. F = nat. cos. F + († G ÷ R). = 8. The angle LIV, by similar reasoning to that used in relation to LOB, is found to be (FS). The distance LI, from toe of switch to point of middle frog, - LV sin. LIV 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 |