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i <br /> DIETZGEN <br /> PS RAILROAD CURVE <br /> AND <br /> REDUCTION TABLES <br /> CoWriaht.1914,by Eugene Dietspn Co..Now York City <br /> PC 'r <br /> + l <br /> c <br /> Ot <br /> a I <br /> T7 f CURVE FORMULAS <br /> Radius=R= (1)Degree of Curve=D and sin.L.'-L.0(2) <br /> a'T'Si3 <br /> 'L� ��•8g Tangent=T=RtanA(3)Length of Curve=L=100A(4) <br /> T 7Z Middle ordinate=M=R(1—cos. �y)(5)=Rvers (6) <br /> Externa1=E=Ttans-(7)=R=cos.#-R(8)=Rexsea#(9) <br /> M Long Chord=C=2 R sin.R(10)o=Central Angle <br /> EXPLANATION AND USE OF TABLES <br /> S -1`23.v s s e f� <br /> Stations.—Given P. I.—Sta. 161-i-60.36 to find Sta. of P. C. { <br /> and P.T. A-620 10' D—V 2('. From Table IV for le curve T- <br /> 3454.1 and+8X-414.49 ft. From Table V oorrection-636 or T= <br /> y /• Z 414.85 ft. P. 0—Sta. P.I.—T=167+45.50. Also from (4) L= <br /> 746.00 and P.T.—Sta.P.C.+L=164+91.50. <br /> otfee#a.—Tangent offsets vary (approximately) directly with <br /> D and with square of the distance. urs.tangent offset for Sta. <br /> 158 on abovecurve is 2.16 ft.found as follows. From Table III tangent <br /> offset for 100 ft.-7.27 ft. Distanae=158=Sta.P. C.==54.50,hence <br /> -offset-7,27 (64.50+100)9=4.16 ft. Also square of any distance <br /> 1 divided by twice the radiusuals(approximately)the distance from <br /> tangent to curve. Thus (54p+(Z x 688.26)-2.16 ft. <br /> Deflections.--Deflection angle—M D for 100 ft., Y/D for 50 ft., <br /> etc. For a ft..=(in minutes).3 x C z DP or--deft.for 1 ft.from Table <br /> III x C. For Ste. 158 of above curve—.3 x 54.5 x 84-436.2' or <br /> 211 16.2' or-1.50 x 54.5-136.2'from Table M. For Eta. 159 defleo- <br /> tion angle'16.21+91201+2-60 262',eta. <br /> Rztaawle.-May be found in.)Pular manner to tangents. Thus <br /> j E for awye above is 91.37. For from Table IV for r curve E=960.6 <br /> { for 8e *=QW.6+8%=91.27 and from Table V oorreodon=.10 or <br /> I Ee91.37 ft. Or w4nmee o—°and E is measured and found to;be <br /> 42 ft. What is D1 Front Table IV E=230.9 and+42-5.5 or D� . <br /> 5°80'. <br />