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1 <br /> DILTZGEN'S RAILROAD CURVE <br /> AND <br /> REDUCTION TABLES <br /> Cnv4sht,1914,by Eugene Dletftm Co..NevTork C+tr <br /> rr <br /> a <br /> . t ' <br /> CURVE FORMULAS <br /> t <br /> Radius=R=rte(1)Degree of Curve=D and sin. )=to <br /> t <br /> Tangent=T=Rtany(3)Length of Curve=L=100 (4) <br /> Middle ordinate=M=R(1—cos. �(5) =Rvers$(6) <br /> External=E=Ttan*-(7)=fit:-cos. —R(8)=Rexsec (9) <br /> Long Chord=C=2 R sin$(10)G_Central Angle <br /> - - - EXPLANATION AND USE OF TABLES <br /> _ - Stations.--Given VI.—Sta. 161+60-35 to find Ata. of P G. <br /> and P.T. n=-62* 10' D=8°20'. From Table IV for 1" curve <br /> $454.1'-and*8M"1t4V ft. From Table V oorreotion*%Wor T= <br /> 414.85 ft. P. C. P.I.—T=157+45.50. Also from:(4)-L <br /> 746.00 and P.T.—Sts.P.C.+L=184+91.50. <br /> Offsetar--Tangent offsets vary (approximately) direct l�yy with <br /> D and with=of the distance. Thus tangent offset f6r Ata. <br /> 188 on above cure is 2.16 ft.found as follows. From Table III tangent <br /> offset for 100 ft.-7.27 ft. Distance—IU—Sta.P. C.m.54.50t hence' <br /> offset=7.27 (54.50+100)2a2.16 ft. Also square of any distance <br /> divided by twice the radius equals(approximately)file distance from <br /> tangent to curve. Thus (84p+(2x688:26)=2.I8 ft. <br /> Deflections.—Deflection angle= D for 100 ft., Y D for 50 ft., <br /> etc. Fore ft.--on minutes).3 x C x D°or�de$:for*'l It.from Tattle <br /> III x C. For Ste. 158 of above curve=.3 x 54.5 x 81136.2' or <br /> 2' 18.212 or=4.50 x 54.5-136.2'ftom Table III. For Sta. 159 dedec- <br /> tloa:tinOe=2e 16.2'+8'20'+2=6°26.2,etc. <br /> 9*ernalse--May be found bi eh11aaK�meaner to tangents, Th. <br /> E for curve above is 91.87. For from Tattle IV for P curve 98Q <br /> for Se 20'-460.6+8%-*91.27 and from Table Vcorrection=.10.e!r <br /> - --- E�91.37 ft. Or suppose &—W and E is measured and found to be ,+ <br /> 42 ft. What is D? From Table IV E-230.9 and+42-5.6 or Dmm- <br /> - 5°30'. <br /> • <br /> e <br /> i <br />