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"S <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />H01f TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve. Tan. 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. See. 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'ft. Angle <br />.of Intersection or I. P. =23' 20' to the R. at Station <br />542-x-72. <br />Est. in Tab. I opposite 23' 20'=120.87 <br />120.87-112=10.07. Say a 10' Curve. <br />Tan. in Tab. I opp. 23' 20'=1183.1 <br />1183.1=10=119.31. <br />Correction for A. 23' 20' for a 10' Cur. =0.16 <br />118.31+0.1.6 =118.47 =corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />An. 23'20'=23.33'-- 10=2.3333=L. C. <br />2' 19.1' def. for sta. 542 I. P: =sta. 542+72 <br />4049 2' _ +50 Tan. = 1 .18.47 <br />7' 192'= a « :: 543 B. C.=stn. 541-53.53 <br />9049 11 = {: « « +50 2 .33.33 <br />11° 40' = « 543 L. C. _ <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' 191'=def. for sta. 542. <br />Def. for 50 ft. =2' 30' for a 10' Curve. <br />Def. for 36.86 ft. =1' 50" for a 10' Curve. <br />-'A <br />