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I <br />CURVE TABLES <br />Published by KELIFFEL & ESSER CO. <br />HOW TO. USE CURVE 'SABLES <br />— Table I: -contains Tangents 'and Externals to a 1° curve. 'Tan* <br />and <br />Ext. to any other radius may be found nearly enough, by dividing the Tan. <br />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 />Divide Ext. opposite the given Central Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. Sec., for any angle by Tablej.: Tan. <br />or Ext. of twice the given, angle'divided' by the.radius of A. l° curve will <br />be the Nat. Tan. or Nat. Ex. Sec. <br />-EXAMPLE, <br />I Wanted a Curve with an Ext. of about 12 ft. Angle <br />! '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 />I Tan. in Tab. I-opp. 23° 20'=1183.1 <br />1183.1,=10=118.31. - <br />Correction for A. 239 20' for a 10° Cur. =0.16 <br />118.31-x-0.16 =118.47-= corrected Tangent. - <br />(If. corrected Ext: is_ required find in same way) ' <br />'Ang. 23° 20'=23.33°=10=2.3333=L. C.• <br />2°•192'=def. for sta. 542 I. P.=sta. 542+72 <br />4° 492' _ "' { 50 Tan. = 1"..18.47 <br />o, 2/ _ u uu - <br />g° 492/._ « :« 543' B. C. sta.. 541+53.53 <br />31 °, 40'= " • " 543+ L. C. _ 2 .33.33 <br />86.86 E. C.=Sta: - 543-86.86 <br />100-53.53=46.47X3'(def: for 1 ft: of 10° Cur.) 139.41'=; <br />2° 192'='def. for sta. 542: <br />Def. for 50 ft. =2° 30' fo°r a-10° Curve. <br />Def. for 36.86 ft. =1° 502' for a 10° Curve. <br />...� A LP.An9.23°201 - .. <br />N <br />• `fig Al Bp> <br />I -curve <br />• sy <br />x, <br />V:. _.. <br />