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DIETZGEN'S RAILROAD CURVE -, _c AND REDUCTION TABLES (7ny®right,1914,by Eugene Dietsgen Co.,New York City PC T_ P r A CURVE FORMULAS • r - a a Radius=R= '50 (1) n �n'bi3 Degree of Curire=D and sin,2.a(2) 31 Tangent=T=Rtan a (3)Length of Curve=L=i00D(4) E Q U Middle ordinate=M= R(1—cos. Y)(5) =Rvers (6) Ezternal=E=Tian*-M=R=cos.A—R(8)=Rexsec (9) Long Chord=C=2 R sin.a(10)A=Central Angle 9 I — EXPLANATION AND USE OF TABLES Stations.—Given P. I.= Sta. 161+60.35 to find Sta, of P. C. and P. T. A=62° 10' D=8° 201. From Table IV for 10 curve T= 3454.1 and=8%=414.49 ft. From Table V correction=.36 or T= 414.85 ft. P. C.—Sta'. P.L—T=157+45.50. Also from (4) L= 746.00 and P.T.=S'ta.P.C.+1=164+91.50. Offsets.—Tangent offsets vary (approximately) directly with D and with square of the distance. Thus tangent offset for Star 158 on above curve is 2.16 ft.found as follows. From Table III tangent e offset for 100 ft=7.27 ft. Distance=158—Sta.P. C.=54.50,fence offset=7.27 (54.50.1-100)2=2.16 ft. Also square of any distance divided by twice the radius equals (approximately)the distance from tangent to curve. Thus (54.50)2+(2x688.26)--2.16 ft. Deflections.—Deflection angle--% D for 100 ft., %D for 50 ft., etc. For o ft.=(in minutes).3 x C x D*or==defl.for I ft.from Table III x C. For Sta. 158 of above curve=3 x 54.5 x 8X=1`36:2' or 2° 16.2,or=2.50 x 54.5=136.2'from Table III. For Sta. 150 de8w, tion augle==20 16.21+80 201+2=00 26,2',etc. ` External&—May be found in similar manner to tangents. !Thus, - E for curve above is 91.37. For from Table IV for r curvQ Fo=W.6 forcr�96o.e+81/y 127 p*d from Table V eorreaitip_.!0;p11t _ 7 ft. Or suppose A=ra i'end F is measured sad.found to e. 5°fit What is D7 From Table 13' 230.�r and+435.5 or I! _ , r •