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I <br />• .CURVE . TABLES. <br />Published by KEUFFEL 8v 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 />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 />1Wanted a Curve with an Ext. of about 12 ft. AngL <br />of Intersection or I. R=23* 20' to the R: at Station <br />542+72. <br />Ext. in Tab. I opposite 23° 20' =120.8! <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 />An5.23°20'=23.33°=LO=2.3333=L. C. <br />2° 191 -'=clef. for sta. 542 1. P. =sta. 542-x-72 <br />40491.1— " " " -x-50 Tan. = 1 .18.47 <br />7° 191'= " " " 543 <br />90 49 Z' = " B. C. = sta. 541+53.53 <br />x-50 <br />110 40: _ " " " 543+ L. C.= 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° 191' =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 />M <br />