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•4n/.Q� <br />"A <br />7 <br />6 Ao fl�_ <br />08 <br />00 <br />r� 55 <br />f dw�CJ i a <br />-9 4" � Y <br />CURVE :TABLES <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />Table I. contaias.Tangents and Externals to a 1° curve. Tan. and <br />Ixt. to any other radius maybe found nearly enough, by dividing the Tan. <br />oT r 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 />)ivide Tan. opposite the given Central Angle by the given --Tangent. <br />i` To find Deg. of Curve, having the Central Angle and External: <br />)ivide Ext. opposite'the given Central Angle by the given External. <br />B 9 A,.� : ; : I To find Nat. Tan. and Nat. Ex. Seca for any angle by Table i.: Tan. <br />-fr Ext. of twice'the given angle divided by the radius of a.1' curve will <br />he the Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE <br />}r�Wanted a Curve with an Ext. of about 12 ft. Angle <br />.�. 6 of Intersection or .l. P.=23° 20' to the R. at Station <br />i 542-72. <br />Ext. in Tab. 1 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 />1183.1-:=10=118.31: <br />Correction fog A. 23°.20' fora 10° Cur. <br />t _ 118.31I-0.16=118.47=corrected Tangent. <br />(1if corrected Ext. is required find in same way) <br />yAng.23'20'=23.33 =10=2.3333=L. C. <br />2° 19" def. for sta. 542 t: P. =sta. 542+72 <br />k 40 49,' _ « +50 Tan. = 1 .18.47 <br />7' 19j,'= « « « 543 B. C. =Eta. 541+53.53 <br />9° 4g, _ « ° « +50 <br />o �eL. C.= 2 .33.33 <br />..- A''<l' 11°40'= « « « 543-, <br />j `86.86 E. C.=Sta. 543 8fi.86 <br />521 A- 100-53.53=46:471 X3'(def. for 1a ft. of 101Cur.)-139.41'= <br />2° 191,' =def. for sta. 542. <br />Det. for 50 ft.= 2 30' fora 10° Curve. <br />Def. for 36.86 ft. 1* 501' for a 10° Curve. <br />IAAn9.23•201' <br />'�Q - <br />