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PIETZGEN'S RAILROAD CURVE AND ��r-�� U �. I <�-:; � �•�� , 1�`� REDUCTION TABLES + Copyright,1914,by Eugene Dietzgen Ce.,Now York _ P.Gn7 - LLL _ - _ CURVE FORMULAS /'7Z. I G�, 4 L Radius=R=am50 f (1)Degree of Curve=D and sin.P_5*(2) « Tangent=T=Rtan� (3)Length of Curve=L=100 (4) Middle ordinate=llf=141—cos Y)M =Rvers g(6) - _ Extemal=E=Ttau--(7)=R-cos.4—R(8)=Rexsec {8) Long Chord=C=2 R sin.a(10)o=Central Angle G• _ - EXPLANATION AND USE: OF TABLES r' - - -_ Stations.—Given P. I.-eta. 161+60.35 to find Sta .of P. C. and P. T. A==62° 10' D---8° 20'. From Table IV for 1° curse T= - -- s ,- 3454.1 and—.8%-414.49 ft. From Table V correction=--36 or T= 414.85 ft. P. C.—Sta. P.L—T=157+45.50. Also from (4) IF-- 746.00 and P.T.=-Sta.P.C.+L=164+91.50. Offsets.—Tangent offsets vary (approximately). directl�r with 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=158-4ta.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_(2x 688.26)=2.16 ft. f Deflections.—Deflection angle--M D for 100 ft., 1/D for.50 ft., --- etc. For c ft.=(m minutes).3 x C x D`or—deft.for 1 ft.from Table III x C. For Sta. 158 of above curve=.3 x 54.5 x 8X-136.2' or 2°16.2',or=2.50 x 54.5136.2'from Table III. For Sta. 159 deflec- tion angle-2°16.2'+8°201-2==6°26.21,etc. ' Externals.—May be found in similar manner to tangents. Thus' • T for curve above is 91.37. For from Table IV for 1°curve E--9641.6 for 80 20' 60.6+8�i=91.27 and,from Table V correctionp-10 or E==01.37 ft. Or suppose p==32'and E is.measured and found to be 42 ft. What is D? From.Table IV E=230.9 and+42 9.5 or D� 5°30'.