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iso <br />CURVE TABLES. <br />i Published by KEUFFEL & ESSER 'CO. <br />HOS!' TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a 1' curve. Tan. and <br />a Ext. to any other radius maybe f ound nearly enough, by dividing the Tan. <br />'i or Ext. opposite the given Central Angle by the given degree of curve. <br />To find Deg. of Curve, having the Central Angle and Tangent: <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />j 6 Divide Ext. opposite the given Central'Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. See. f or any angle by Table I.: Tan. <br />or Est. of twice the given angle divided by the radius of a V curve will <br />r; be the Nat. Tan. or Nat. Ex. Sec. <br />-EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />11 of Intersection or I. •P.=23° 20' to the R. at Station <br />.542+72. <br />Ext. in Tab. I opposite 23' 20'= 120.87 <br />120.87 =12 =10.07. Say a 10' Curve. <br />Tan. in Tab. I opp. 23' 20'=1183.1 <br />1 f 1183.1=10 =118:31. !" <br />Correction for A. 23' 20' for a 10' Cur: =0.16 <br />. 118.31-1-0.16 =118.47 =corrected Tangent. <br />i '(If corrected Ext. is required find in same way) <br />i Ang. 23'20'=23.33'=10=2.3333=L: C. <br />j 2° 191'= def. for sta. 543 I. P. =sta. 542-72 <br />' V 491'= " " " +50 Tan. = 1 .18.47 <br />7' 19;' = u u « 543 <br />' 9' 49v= " " " +50 B. C:'=sta. 541+53.53 <br />11° 40' _ ° " 543+. L. C.= 2 .33.33 <br />_ f 86.86 E. C.=Sta. '543+SG.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10' Cur.) =139.41'— <br />2' 19"=def. for sta. 542.. <br />Def. for 50 ft. =2' 30' for a 10° Curve. <br />Def. for 36.86 ft.=1' 501,' fora 10' Curve. <br />i� <br />