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DIETZGF N'S RAILROAD CURVE ' ' f _ AND REDUCTION TABLES G ; 2 - CoDytight,1914•by Eugene Dietsgen Co.,New York Cr L` W� G ff PCMR C`.3 O CURVE FORMULAS f - - (1)Degree and sin.-- Radius=ft= . ree of CD iD � Tapgent=T=fttany (3)Length of Curve=L=100D(4) Middle ordinate=M=R(1—c6s. ,)(5) =Rvers s(6). External=E=Ttan4-(7)=R=cos.A—R(8)=Rexsec (9) Long Chord=C=2 R sin.z(10)o=Central Angle EXPLANATION AND USE OF TABLES �. Stations.—Given P. I.=Sta. 161.1-60.35 to find Sta, of P. C. and an P A==62* .49 =S* From From TablTable IV for 1° curve T— e V correction=.36 or T { 414.85 ft. P. C—Sta. P.I.—T=157+45.50, Also from (4) 1-- 746.00 and P.T.=Sta.P.C.+L=164+91.50. Offsets.-Tangent offsets vary (approximately) directl with y 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-458—Sta.P. C-54.50,hence offset--7.27 (54.50-100)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 o an6le=M D for 100 ft. Dfor 5 0 i ft. etc. For c ft:(in minutes).3 x C x D°oralefl.for 1 ft.from Table III x C. For Sta. 158 of above curve=.3 x 54.5 x 8%=136.2' or 2' 16.2,or--2.50 x 54.5=136.2'from Table III. For Sta. 159 deflec- tion angle=20 16.2 +8*28'+2==G°26.2',etc. Externals. May be found in similar manner to tangents. Thus E for curve above is 91.37. For from Table IV for 1°curve E=960.6 for 8° 20'---W.62-8 X1.2? 1/s and from Table Vcorrection=.10 or E----91.37 ft. Or suppose &==3r and E is measured and found to be 42 ft. What is D7 From Table IV E=230.9 and=42 .5 or D- 50 30r ' I s