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75 <br />CURVE TABLES. '- <br />Pubhshed by KELIFFEL & 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. Sec: for any angle by Table I.: Tan. <br />i <br />or Ext. of twicelhe given angle divided' by. the radius of a P 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 230 20, =120.87 <br />120.87+12=10.07' Say a 1(Y' Curve. <br />Tan. in Tab. I:opp. 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 re find in same way) Ang.23'20'=23.33 ,,10=2.3333=L.•C. <br />2° 19Y= def. for sta. 542 I. P. = sta. 542-72 <br />4° 49i' _ ." " +50 Tan. = 1 .18.47 <br />7° 19;' = '� " a 543 <br />9°49 '= u �, �, +50B. C.=sta. 541+53.53 <br />110.40'—, '° " u 543 �- L. C. = 2 .33.33 <br />1 86.86 E. C.=Sta. - 543+86.86 <br />100-53:53=46.47XX(def. for 1 ft, of 10° Cur.)=139.41'= , <br />20 191`=def. for sta::542. <br />Def: for 50 ft. =2° 30' fora 10° Curve. - <br />Def. for 36.86 ft. =1° 50j' for a 10' Curve. <br />kv <br />r <br />