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- CURVE TABLES. <br />_ Published by KEUFFEL • & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table.I. contains Tangents and Externals to -a 10 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 />T -. 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. 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. Sec. <br />-EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or 1. P.=230 20' to _the R. at Station <br />542+72. <br />Ext. in Tab:•I opposite 23°.20' =120.87 <br />i 120.87 =12 =10.07:. Say a 10P Curve. <br />1 Tan. in Tab. I Topp. 23' 20' =1183.1 <br />1183.1=10=118.31." <br />Correction for A. 23° 20' for a 10P Cur. =0.16. <br />118.31,+0.16 =118.47 =corrected Tangent. <br />(If;corrected Ext. is,revired find in same'way) - <br />�Ang. 23°20'=23.33 =10=2.3333=L. C. <br />20 39;'='def: for sta. 542 I. P. =sta. 542-1-72 <br />40491 _ !' +50 Tan. = 1' .18.47 <br />7° 191'= u " « . 543. <br />90 492,'= " " +50 B. C. = sta. • 541 _}-53.53 <br />- 110 40'_; " 543+ L. C. _ .. 2 .33.33 <br />_ 86.86 E. C.'=Sta.' . " 543+86.86 <br />100-53:53=46.47X3'(def. for'1 ft. of 10° Cur.)=139.41'= <br />:.. 2°49;'=def. for sta. 542. .. <br />-- •- 'Def. for 50 ft. =2° 30' for a 10' Curve. - - <br />Def. for 36.86 ft. =1° 50j' for a 10° Curve. <br />8 I.P-An9.23*20!I <br />