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CURVE TABLES. <br />Published by KEUFFEL 8y ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve. Tan. and <br />Ext. to any other radius may bef ound nearly enough, by dividing theTan. <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 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 abouf'12 ft. Angle <br />sof Intersection or, I. P.=230 20' to the R. at Station <br />54.2+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 />1183.1=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°197'=def. for sta., 542 I. P.=sta. 542+72 <br />40 497'= 11 " " +50 Tan. = 1 .18.47 " <br />o z, _ t< <br />719, (' " 543 B. C.=sta. 541+53.53 <br />90 491 -' _ +50 L. C. = 2 .33.33 <br />11' 40' = " 543+ <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° 197' =def. for sta. 542. <br />Def -for 50 ft. =2°-30'.for a. 10° Curve. <br />Def. for 36.86 ft. =1° 507' for a 10° Curv& <br />