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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 maybe 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 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. See. <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 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=318.47='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° 10F= def. for sta. 542 I. P. =sta. 542 { 72 <br />4° 491' = +50 Tan. = 1 .18.47 . <br />70 .191'= « " 543 <br />90 491' = " +50 B. C. =sta.. 541-}-53.53 <br />110 40'= ", " . 543+ L. <br />-1 <br />. = 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />.100-53.53=46.47X3'(def. for I 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' for a 10° Curve. <br />