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CURVE TABLES. <br />Published by KEUFFEL 8s 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 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 findDeg. of Curve, having the Centeal 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. See. <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 />542-72. <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'= 1V83.1 <br />1183.1=10 =118.31. <br />Correction for A. 230 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° 19y"= det. for sta. 542 I. P. =sta. 542+72 <br />4° 49,' _ « «y « +50 "ran. = 1 .18.47 <br />7" 191' — " 543 <br />9° 49;' = it(! as +50 P C. = s to . 541+53.53 <br />53.53 <br />110 40' _ " " " 543 L. C.= 2 .33.33 <br />86.86 E. C. =Sta. 543+86.86 <br />100-53.53=46.47XX(def. for 1 ft. of 10° Cur.) = 139.41'= <br />2° 10F=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.. 3 <br />r, <br />r <br />L <br />