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DIETZGEN'S RAILROAD CURVE AND REDUCTION TABLES CoWASht,1914,by Eugene Dietum Co..New York City 12 1 njel L31'4j . 11 77s` Po.T CURVE FORMULAS - - 1 86t. 1 qn a o { Radius=R=ems (1)Degree of Curve=D and sin.z= J ( Tangent=T=RtanQ� (3)Length of Curve=L=1008(4) Middle ordinate=M=R(1—ces. ',J(5) =Rvers (6) - External=E=Ttan`-(7)=R+cos.4—R(8)=Rexaec4(9) — _- Aug Chord=C=2 Rsin.s(10)A=Central Angle --:- EXPLANATION AND USE OF TABLES — Stttions.—Given P. I.==Sta. 161+60.35 to find Sta. of P. C. and P. T. 0==62° 10' D-8°20'. From Table IV for P curve T= i 8454.1 and+83 =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.—Sta.P.C.+1,=164+91.50. Offsets.—Tangent offsets vary (approximately) directly with D and with square of the distance. Thus tangent offset for Sta. { 158 on above curve is 2.16 ft.found as follows. From Table III tangent +! offset for 100 ft.7.27 ft. Distance=158—Sta.P. C.=54.50,hence j of(aet=7.27 (54.50=100p=2.16 ft. Also square of any distance f divided by twice the radius equals(approximately)the distance from tangent to curve. Thus (54.50p+(2x688.26}=2.16 ft. Deflections.—Deflection angle= D for 100 ft., �/D for SO ft., etc. For c ft:(in minutes).3 x C x D°or�lefl.for 1t.from Table III x C. For Sta. 158 of above curve=.3 x 54.5 x 8�§=1 .2' or 2° 16.2',of=2.50 x54.5=136.2'from Table III. For Sta. 15�de - on angle=2°16.2'+8°20'++2==60 26.21,etc. Hec- 3xt%1Ws.—May be found in similar manner to tangents. Thus E for curve above is 91.37. For from Table IV for V curve E-960.6 for 80 201=960.6+8%=91.27 and from Table V correction—jo or - E-91.37 ft. Or suppose &==32•and E is measured and found to be 42 ft. What is DY From T" IV E-330,9 and+42=5,5 or D= "�""LD BOO 080530'. 4