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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 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. Sm 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+72. . <br />:Ext. in Tab. I opposite 23° 20' =120.87 <br />120.87=12 10.07. Say a IOP 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. P. = sta. 542+72 <br />40 4911'.= it « " +50 Tan. = 1 .18.47 <br />70 19;' 543 <br />90 OF " " " +50 B. C.=sta. .541+53.53 <br />110 40' = 5 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° 1912'=def. for sta. 542. <br />-Def. for 50 ft. =2° 30' for a 100 Curve. <br />Def. for 36.86 ft. =1° 50j' for a 10° Curv& <br />