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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 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 Cufve, 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 L. Tan. <br />or Est. of twice the given angle divided by the radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. Seca <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 />642+72. <br />_ Ext. in Tab. I opposite 23° 20r =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. 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° 191'=def. for sta. 542 I. Pa=sta. 542+72 <br />40 491' =. " " f/ +50, Tan. _ 1 .18.47 ` <br />70 191'= h " ca ;. 543 <br />9° 491' _ cc " " +50 B. C. = sta. -E41+53.53 <br />53.53 <br />110 40' _ 11' " 1154,3+ L. C. = 2 .33.33 <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. =20 30' for a 10° Curve. <br />" <br />Def. -for 36.86 ft. I!, 50;' for a 10',Curv& <br />.... -. - <br />x� <br />[i <br />