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E <br />9 <br />CURVE •TABLES.' <br />4r <br />J Y . Published by KEUFFEL & E$SER CO. <br />!! HOW T®, FTSE CURVE TABLES. <br />Table 1. contains Tangents and Externals to'a V curve. Tan. and <br />Ext. to any other radius may be found nearly enough, bydividing theTan. <br />or Ext. opposite the given Central Angle by the given degree of curve. <br />To find Deg., of Curve, having the Central Angle. and Tangent: <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />Divide Ext: opposite the given Central Angle by "the given External. <br />To find Tan. and Nat. Ex. Seci for aor Ext of twi et the given angle id'vided' by th nyeradiusoe bf a 1' curve will <br />be the Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE. <br />Wanted a Curve- with an Ext. of about 12 ft. Angle <br />of Intersection or .I. P.=230 20' to the R. at Station <br />1 542-}-72. <br />Ext. in Tab. I opposite 23' 20, =120.87 <br />120.87'—*. 12 =10.07. Say a 10° Curve. <br />Tan. in Tab. I opp. 239 20'=1183.1 <br />1183.1+10=118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 <br />.118.31+0.16=118.47=corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang.23*20'=23.33°=10=2.3333=L. C.; <br />20 19'=def. for sta. 542 I. P. =sta. 542-72 <br />4' 49v"_— " a " ' -+50 Tan. _' 1 .18.47 <br />70 191,1= a 543 <br />gc 491,1= " " " +50 B. C. =sta. - " 541+53.53 <br />S 11° 40' = a a it 543+ L. C.= 2 .33.33 <br />86.86 E: C. = Sta. .543+86-86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 192 def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1' 501' for a 10° Curv& <br />,, <br />G <br />s <br />i <br />� e <br />b3 47 <br />y� <br />. <br />f <br />_ <br />/8 <br />�4 <br />�' S <br />95.3 <br />0 <br />9s" 30 <br />7i� <br />9 - <br />CURVE •TABLES.' <br />4r <br />J Y . Published by KEUFFEL & E$SER CO. <br />!! HOW T®, FTSE CURVE TABLES. <br />Table 1. contains Tangents and Externals to'a V curve. Tan. and <br />Ext. to any other radius may be found nearly enough, bydividing theTan. <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 Tan. and Nat. Ex. Seci for aor Ext of twi et the given angle id'vided' by th nyeradiusoe bf a 1' 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.=230 20' to the R. at Station <br />1 542-}-72. <br />Ext. in Tab. I opposite 23' 20, =120.87 <br />120.87'—*. 12 =10.07. Say a 10° Curve. <br />Tan. in Tab. I opp. 239 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 same way) <br />Ang.23*20'=23.33°=10=2.3333=L. C.; <br />20 19'=def. for sta. 542 I. P. =sta. 542-72 <br />4' 49v"_— " a " ' -+50 Tan. _' 1 .18.47 <br />70 191,1= a 543 <br />gc 491,1= " " " +50 B. C. =sta. - " 541+53.53 <br />S 11° 40' = a a it 543+ L. C.= 2 .33.33 <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. <br />Def. for 36.86 ft. =1' 501' for a 10° Curv& <br />