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v /3 <br />¢� G 30/ <br />Olt <br />7 27 <br />`t33�fz= /3 Z33 = 3. ��4 <br />r y <br />y <br />1 ' <br />Published by KEUFFEL & ESSER CO.. <br />Q HOW TO USE CURVE TABLES <br />5' Table I. contains Tangents and Externals to a-1° curve: Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing the Tan. " <br />S_� 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 />S —— Divide Ext. -opposite the given Central Angle by the glven'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 />12 f Ann le .. <br />Wanted a Curve with an Ext, of about t. g <br />of Intersection 'or I., P.'=23° 20' to: the -R., at' Station <br />542+.72. <br />Ext. in -Tab. I opposite 23° 20' =120.87120.87.—.- 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 for. A. 23° 20' for a 10° Cur. =0.16 <br />118.31+0.16=118.47•=corrected Tangent.' <br />(If corrected Ext. is required find in swine way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C.; <br />2°192'= def. for sta. 542 '1: P. sta. 542+72,.+ <br />4° 492'= Tan. = 1 .18.:47 <br />7° :192' _ "- " 543 B. C. = sta. 541+•53.53 - <br />9°�491"= '« ° « x-50 ' 2 .33.33 <br />11° 40'= « « 543 ♦ . L. C. _ _ r . <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°• 192.' ='def. for sta. 542: <br />Def.- for 50 ft. =2` 30' for a 10° Curve. t <br />Def. for 36.86 ft. 1' 502' for a 10° Curve. <br />I.P.An9.23.20' <br />`\0 P <br />�o° Curve <br />G <br />0 <br />