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w <br />!I <br />CURVE TABLES" <br />Published by KEUFFEL & ESSER CO:, <br />HOW TO USE CURVE TABLES <br />x Table I. contains Tangents and Externals to a 1' curve. T. <br />—I <br />/ <br />Ext. to any other radius may be found nearly enough, by dividing t <br />or Ext. opposite the given Central Angle by the given degree of <br />_ <br />' <br />f <br />To. find Deg. of Curve, having the Central Angle and Ta <br />Divide Tan. opposite the given Central Angle by7 the given Tan€ <br />To find _Deg. of Curve, having the;. Central Angle and E: <br />Divide Ext. opposite the given Central Angle by'the given Exte <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table <br />or Ext. of twice the given angle divided by the radius of a 1° cu <br />txr <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 10° Curve. <br />Tan.'in Tab. I opp. 23° 20'= 1183.1 <br />I <br />1183.1=10 =118.31. <br />p Correction for A. 23° 20' for a 10° Cur. =0.16 <br />118.31-x-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 />i <br />2° 192"= def. for sta. 542 I. ;P. = sta. 542+72 <br />4° 491' _ 50 Tan. = 1 .18.47 <br />" <br />7° 1911= " " 543 B. C.=sta. 541+53.5c <br />90 491' = " " " +50 <br />11° 40' _ " ." " 543+ L. C.= 2 .33.33 <br />86.86 E. C. = Sta. 543 -{-86.8( <br />100-53.53=46.47XT(def. for 1 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° 50V for a 10° Curve. <br />4,4 <br />dx <br />' <br />I <br />>A <br />_ I.P.An9.23°20' ' <br />. <br />N / <br />• 10° Curve ' <br />in. and <br />he Tan. <br />curve. <br />lgent: <br />ent. <br />:ternal: <br />rnal. <br />Tan. <br />•ve will <br />