Laserfiche WebLink
DIETZGEN'S RAILROAD CURVE <br />AND <br />REDUCTION TABLES <br />Copyright, 1914, by Eugene Dietigen Co., New York City <br />'MP <br />CURVE FORMULAS <br />Radius---R=---n-2(1) Degree of Curve=D and sin. D=R (2) <br />sin,Tangent=T=Rtan ° (3) Length of Curve=L=100D (4) <br />Middle ordinate=M=R(1—cos. a) (5)=Rvers 2 (6) <br />External=E=Tian 4 (7)=R=cos. s R(8)=Rexsec a (9) <br />Long Chord ---C=2 R sin. a (10) A=Central Angle <br />EXPLANATION AND USE' OF TABLES <br />Stations.—Given P. Leta. 161+60.35 to find Sta. of P. C. <br />and. P. T., p==�62° 10' D=8° 20'. From Table IV for 1° -curve T= <br />3454.1 and =8M=414.49 ft. From Table V correction=.36 or T= <br />414.85 ft. P.- C.=Sta: '.P.1.—T=157 +45.50. Also from (4) L= . <br />746.00 and P. T.----Sta. P. C.+L=164+91.50. <br />Offsets.—Tangent offsets vary (approximately) directly with <br />D and with square ofthe distance. Thus tangent offset for Sta. <br />158 on above curve is 2.16 ft. found as follows. From Table III tangent <br />offset for 1.00 ft. 7.27.ft. Distance=158—Sta. P. C.=54.50, hence <br />offset=7.27 (54.50-- 100)?==216 ft. Also square of any distance <br />divided by twice the radius equals (approximately) the distance from <br />tangent to curve. Thus (54.50).'=(2 x 688.26)=2.16 ft. <br />Deflections.—Deflection angle= MD for 100 ft., Y4 D for 50 ft. <br />etc. For c ft.=(in minutes) .3 x C x D. <br />or---deft. for 1 ft. from Table <br />III x C: For Sta. 158 of above curve=.3 x 54.5x 8y3=136.2' or <br />2° 16.2', or=2.50 x 54.5=136.2' from Table III. "For Sta. 159 deflec- <br />tion angle=20 16.2' +80 201 =2---6° 26.2', etc. <br />Externals.—May be found in similar manner to tangents. Thus <br />E for curve above is 115.37. For from Table IV for 1° curve E=960.6 <br />for 8° 20'=960.6=8!,J=115.27 and from Table V correction=.10 or <br />E=115.37 ft. Or suppose p=32° and E is measured and found to be <br />42 ft. What is D? From Table IV E=230.9 and _42=5.5 or D-- <br />5- 30'. <br />I <br />