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Y <br /> DIETZGEN'S RAILROAD CURVE <br /> AND <br /> REDUCTION TABLES <br /> ODM7iat.1914,by Eugene Diatom Co..New York CUW <br /> PCA'0 <br /> je <br /> CURVE FORMULAS <br /> Radius=R=stems� (1) Degree of Curve=D and min.z=d0(� <br /> Tangent=T=Rtan& (3)Length of Curve=L=100E(4) ' <br /> Middle ordinate=M= R(1-cos. b(5) =Rvers�(6) <br /> External=E=Ttani-(7)=R+cos.#-R(8)-Rexmec#(9) <br /> Long Chord=C=2 R sin.s (10)o=Central Angle <br /> EXPLANATION AND USE OF TABLES <br /> Stations.--Given P. I.=.Sta. 161+60.35 to find Ste, of P. C. <br /> and P. T. 0==62° 10' Dom'20'. From Table IV for 1' curve T- <br /> 3454.1 and+8 --414.49 ft. From Table V correction=-.86 or. T <br /> - 414.85 ft. P. C—St& P.I--T�157+9,5.50. Also from (4) I,® <br /> 746.00 and P.T.=Sta.P.C.+L=164+91.50. <br /> Offsets.Tangent offsets vary (approximately) dhwg with <br /> D and with square of the distance. Thum tangent offset for Sta. <br /> 158 on above curve is 2.16 ft.found as follows. From Table III tangent, <br /> offset for 100 ft.-7.27 ft. Distance-158--ata.p. C.-5},�henceoffset-7.27 (54.50+100)a=2.16 ft. Also square of any distance <br /> divided by twice the radius equals(approximately)the distance from <br /> tangent to curve. Thum (54.50)*+(2 x 688.26)=2.16 ft. <br /> DeeecdowL-Deflection angle= D for 100 ft.; D for 50 ft. <br /> eta : For o ft.-(in minutes).3 z C x.A*or-- for I It.from Table <br /> III x C. For Sta. 158 of above curve*-.3 x 54.5 x 8�136.Z' or <br /> 2' 16.21,or-2.50 x 54.5136.2'from Table III. For Sta. 159 dedee- <br /> tion angle=-r 16.2'+8°20'+2==6'26.21,eta <br /> Externals.-Mry be found in similar manner to tangente. Thus <br /> E for.curve above is 91.37. For from Table IV for 1°curve E=980.6 <br /> for Be 20'=960.6+8%=-91.27 and from Table V correction-10 or <br /> i - E-41.37 ft. Or suppose A-32'and E is.measured and found to be <br /> 421t. What is D? From Table IV B-M.9 and+42-5.5 or,Dmm. <br /> r b°W. <br /> . e i • <br /> V <br />