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Jam_ <br />5 6 �r l A �� "I �� c. f -� = 1 ► w-� w a- d tide l.� w i t - <br />l� S <br />CURVE TABLES. <br />4..3 1n t 0 , . Published by KEUFFEL & ESSEI2 CO. <br />TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve. Tan, and <br />Ext. to any other radius maybe found nearly enough, by dividing the Tan. <br />«� or Ext. opposite the given Central Angie, by the given degree of curse. <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 />I „+, Divide Est. 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 i r or Ext: of twice the given angle divided by the radius of. a V curve will <br />be the Nat. Tan. or Nat. Ex. See. » <br />q <br />EXAMPLE. <br />, — <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or L P.=230 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 />+ S <br />Tan. in Tab. I opp. 23° 20' =1183.1 <br />' <br />1183.1=10=118.31. <br />L <br />Correction for A. 23° 20' for a 10° *Cur.= 0.16 <br />118.31-10.16 =118.47 =corrected Tangent. <br />4 <br />(If corrected Ext. is required find in same way) <br />Ang. 23° 20'=23,33'-- 10=2.3333 =L. C. <br />— <br />2"19,'=def. for sta. 542 f.P.=sta. 542+72 <br />4049 ca « r. +J0 Tan.= 1 .18.47 <br />°7o E9"= is cc cc 543 r <br />'= u o B. C.=sta. 541.+53.53 <br />Ja 491; +50 <br />11° 40' _ °° « Cf543 C J L2 33.33 <br />86-8e) E. C.=Sta. 543-x-86.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° C ur.) = 139.41'= <br />. <br />+ 'a <br />lin J <br />- <br />�'� <br />r 2' 19' = def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50.1' for a 10° Curve., <br />Sq <br />aK <br />s� <br />LP.An 9.2S°ZO• <br />