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CURVE. TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE -CURVE TABLES. <br />rve. Tan. and <br />Table I. contains Tangents and Externals to a 1b° div ding theTan. <br />Ext. to any other radius maybe found nearly enougiv�en degree of curve. <br />or Ext. opposite the given Central Angle by the given, <br />le and Tangent: <br />To find Deg. of Curve, having the Central Angle <br />Tangent. <br />'Divide Tan.. opposite the given Central Angle by the given <br />i To find Deg. of Curve, having the Centrath given External- <br />the <br />x ernal: <br />al. <br />;Divide Ext. opposite the given Central Angle by <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.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. <br />120.87+12-10-07- I Say°a 10° C rve. <br />I Tan. in Tab. I opp. 23° 20' =1183.1 <br />1183.1=10 =118.31. <br />Corr ection.for A. 230 20' for a 10° Cur. =0.16 <br />j <br />118..31+0.16 =118.47 =corrected Tangent.. <br />(If corrected Ext. is required find in same way) <br />Ang. 23'20'=23-330+10 =2.3333 =L. C. <br />} 72 <br />2° 19;' =def . for sta. 542 I. P. = sta. 1 :18,47 <br />4° 49,' = a a (-50 - Tan. <br />7° 19" a as ca . 543 B. C.=sta. 541+53.53 <br />9° 49;'— a « +3+ L. C. = 2 .33.33 <br />11°40',= a a 54 <br />86.86 E. C. = Sta. 543+96-gra.. <br />100_53.53=46.47X3'(def. for for . of 0* Cur.). 139.41' = <br />2° 191' = def. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. 1* 50V for a 10° Curve. <br />•. vie <br />-- x> <br />