. Railroad construction. Theory and practice. A textbook for the use of students in colleges and technical schools . curve, but there is a consid-erable portion which is fitted tothe cast-iron filler, and this por-tion is always straight. Call thelength of this straight portionback from the frog-point / ( =FH,Fig. 138). Then we haver + hg = {g-f sin F) -^vers F =-^-fcotiF vers F / FH=/ \ F t B Fig. 138. g \eTs F -2fn. (82) BF^L = {g-f sin F) cot ^F+f cos F=2gn-f sin F cot iF-^f eos F=2firn-/(l -f cos F) +/ cos F-=2gn-f (83) Since r—\g = {L—f sec F) cot F^ andr + ig = (L—f cos F) cosec F,r=iL(c
. Railroad construction. Theory and practice. A textbook for the use of students in colleges and technical schools . curve, but there is a consid-erable portion which is fitted tothe cast-iron filler, and this por-tion is always straight. Call thelength of this straight portionback from the frog-point / ( =FH,Fig. 138). Then we haver + hg = {g-f sin F) -^vers F =-^-fcotiF vers F / FH=/ \ F t B Fig. 138. g \eTs F -2fn. (82) BF^L = {g-f sin F) cot ^F+f cos F=2gn-f sin F cot iF-^f eos F=2firn-/(l -f cos F) +/ cos F-=2gn-f (83) Since r—\g = {L—f sec F) cot F^ andr + ig = (L—f cos F) cosec F,r=iL(cot i^^-hcosec F)—y sec F cot F — y cos F cosec F ^ \ sm/ / r=L?i —5/cotii^ =^Ln-fn, Then from (83)r=2sfn2-2/n. ..... (84) 264, Effect of straight point-rails. The point switches,now so generally used, have straight switch-rails. This requires 300 KAILROAD CONSTRUCTION. §264. an angle in the alignment rather than turning off by a tangentialcurve. The angle is, however, very small (between 1° and 3°),and the disadvantages of this angle are small compared withthe very great advantages of the ^^=shr«i^ + a) r+i^=^-r FM 2smh(F—a)g-k 2smi{F + a)smi(F-a)g-k 1 cos a —cos F BF^L= FM cos ^{F + a) + DN^(g-k)cotKF-ha)+DN. (85) (86) 265. Combined effect of straight frog-rails and straight point-rails. It becomes necessary in this case to find a curve whichshall be tangent to both the point-rail and the frog-rail. Thecurve therefore begins at M, its tangent making an angle of a(varying from 0° 52 to 2° 36) with the main rail, and runs to H. § 266. SWITCHES AND CROSSINGS. 301 The central angle of the curve is therefore (F—d). The angleof the chord HM with the main rails is therefore i(F-a)+a=iiF + a);sinK-^ + a) r+i9 = HM 2sini{F-a) g—f sin F—k 2 sin i(F + a) sin i(F--a)_g—f sin F—k^ cos a—cos i^ ST=2r sin i(F-a). (87) (88) BF=L=HM cos i(F + a)-{-f cos F+DN = (g-f sin F-k) cot i(F + a) +/ cos F^DN. . (89) It may be more simple, if (r +Jg) has alread
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