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DIETZGEN'S RAILROAD CURVE AND 2 9 JJ g i REDUCTION TABLES / 173 oa l83 CoWr3ahf,1914,by Eugene Dietrgen Co.,New York ab P,G T P l z � a / CURVE FORMULAS Sw � a Radius=R=gam (1)Degree of Curve=D and sin.z= p.D Tangent=T=Rtan4 (3)Length of Curve=L=100$(4) • ? G Middle ordinate=M=R(1—cos. ')(5) =Rvers (6) External=E=Tta4-(7)=R=cos. —R(8)=Rexaec#(9) Long Chord=C=2 R sin.s(10)o=Central Angle / 0 0 EXPLANATION AND USE OF TABLES Stations.—Given P. I.oSta. 161+60.35 to find Sta, of P. C. and P.T. G�2o10' D--8 20'. From Table IV for P.curve,T= 3454.1 and+8% 14.49 ft. From Table V correcti6n=.38 or T. J '{f3 va 3 74&00 and P.T.=Sta.P.C.+L-164+9 Also from 50. (4) Ire Offsets.—Tangent offsets vary (approximately)' d' with D and with square of the distance. Thus tangent offr Ste 158 on above curve is 2.16 ft.found as follows. From Table III tangent l 6 6 b i offset for 100 ft.=7.27 ft. Distance--158---Sta.P., C.=54.5Qx hence f offset—_7.27 (54.50+100)2=2.16 ft. Also square of any distance o divided by twice the radius equals(approximately)the distance from / 03 tangent to curve. Thus (54.50)2+(2x688.43)==2.16 ft. .� 6� 0 a Deflections.—Deflection angle—etc. for 100 ft., 1/D for 50 ft., u etc. For c ft:(in minutes).3 x C x"De or=deft.for 1 t.from Table G, , # 'IIur I x C. For Sta. 158 of -above c ve=.3 x.54.5 x 8�-138.2' or 2° 16.2,or--2.50 x 54.5=136.2'from Table III. ' For Sta. 159 defleo. tion angle==2°16.21+8*201+2=6*262',etc. Externals.—May be found in similar manner to tangents. Thus E for curve above is 91.37. For from Table IV fora,curve FP=OW 6 for 8e 20'=960.6+8%=91.27 and from Table V.00rrecWm=*..10 or 1691.37 ft. Or suppose p-=_2°and E is measured and found to be 42 ft. What is DY From Table IV E-Mg and+42=4.6 or Dm•. / E 50 30'.