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URVE TABLES.��P k_KL by KEUFFEL & ESSER Col. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents <br />and Externals to a 1° curve. Tan. and' <br />T <br />t. to any other radius may be found nearly enough, by dividing theTan. <br />Ext. opposite the given'Central Angle by the given degree of curve. <br />To find fide Tan. opposite t e givenhaving <br />ral Angle by he given Tangle and ent. nt <br />To find Deg. of Curve, .having the Central Angle and External: <br />Vide Ext. opposite the given Central Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. See. for any angle by Table I.: Tan. <br />Ext. of twice the given angle divided by the radius of a 1' curve will <br />the Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE, <br />Wanted a Curve with an Ext. of about 12 ft. Angle 8.4 <br />Of' Intersection or 1. P.--23' 20' to the R. at Sitation � <br />542+72. ,f <br />Ext. in Tab. I opposite 23' 20' =120.87 �� ,�°� -� <br />120.87-12=10.07'. Say a 10' Curve. `v <br />Tan. in Tab. I opp. 23' 20'= 1183.1 Z". <br />1183.1=10=118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 <br />118.31+0.16=118.47 =corrected Tangent. _ Z1,7s- <br />(If corrected Ext. is required find in same way)~ �— <br />Ang.23°20'=23.33° - 10=2.3333=L. C. - <br />2° 19,' =def. for sta. 542 1. P. =sta. , 542+72 <br />40 4912'= [: « +50 Tan. = 1 .18.47 � r <br />°tt « a <br />7 191'=. 543 B. C.=sta. 541+53.53 <br />} 90 4911 = +50 L. C.= 2..33.33 <br />e 110 40' = " " is 543+ <br />86.86 E. C. = Sta. 543+86 e6 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) =139.41'= <br />2' 19,'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10' Curve. ! <br />'�a c� <br />Def. for 36.86 f t. =1° 50,' for a 10' Curve. , 3393 <br />I.P.An9.23°20' <br />?) <br />- <br />10, Curve , <br />qlb <br />