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- <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a 1' curve. Tan. and <br />Ext. to any other radius may be found 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 />Divide Ext. opposite the given Central Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />or Ext. of twice the given angle divided by the radius of a 1° -curve will <br />be the Nat. Tan. or Nat. Ex. Sec.. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 it. Ang1_a <br />of Intersection or 1. 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'=1153.1 <br />11S3.1=10=118.31. <br />Correction: for A. 23' 20' for a 10' Cur. =0.16 <br />118.31.0.16=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' = dcf. f or sta. . 542 1. P. = sta. 542+72 <br />4' 49211= " It " +50 Tan. = 1 .13.47 <br />7' 192' _ " .543 <br />S. C.=sta. 511+53.53 <br />904912'= .:: " " +50 <br />11' 40' _ " 543 <br />L. C' — 2 •33.33 <br />86.86 E. C.=Sta. 543-1-86.86 <br />1 100-53.53=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' for a 10' Curve. <br />Def. for 36.86 ft.=1' 502' for a 10' -Curve. <br />s,4 <br />iz <br />IPnny.2i3j <br />*e0f <br />fV <br />DA <br />�9> <br />to' Curve <br />y <br />�a ydi <br />