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nd <br />3n. <br />e. <br />t: <br />al: <br />CURVE TABLES <br />Published by KEUFFE18, ESSER CO. <br />HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a 1° curve. Tan. a <br />!Ext. to any other radius maybe found nearly enough, by dividing the T, <br />or Ext. opposite the given Central Angle by the given degree of cury <br />To find Deg. of Curve, having the Central Angle and Tangen <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and Extern <br />Divide Ext. opposite the given Central Angle.by the given External. <br />To find Nat. Tan. and Nat. Ex. See. for any angle by Table I.: T <br />r Ext. of twice the given angle divided by the radius of a 1° curve v <br />be the Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P.=230 20' to the R. at Station <br />542+72. <br />Ext, in Tab. I opposite 230 20' =120.87 <br />120.87-12=10.07. Say a 10° Curve. <br />Tan. in Tab. I app. 23° 20'=1183.1 <br />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 T 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° l9', -'=def. for sta. 542 1. P.=sta. 542-72 <br />4° 491'= " ' x-50 Tan. = 1 .18.47 <br />543 B. C.=sta. 541d 53.53 <br />54+3500 L. C. = 2 .33.33 <br />86.86 E. C. = Sta. 543+86.86 <br />100-53.53=46.47>(3'(def. for 1 ft. of 16° Cur.) =139.41'= <br />2° 194'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curve. <br />`° i.P.An9.23.20 <br />e "! <br />of <br />nd <br />3n. <br />e. <br />t: <br />al: <br />