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..CURVE TABLES. <br />Published" by KEUFFEL Sa ESSER CO. <br />HOW TO USE CURVE .TABLES. <br />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 />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 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 />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P. =23° 20' to the R. at Station. <br />542+72. <br />Ext. in Tab. I opposite 23° 20' =120.87 <br />120.87 =12 =10.07. Say a 10P 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 IOP 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-,k'-- def. for sta. 542 I. P. =sta. 542-x-72 <br />40 491' _ " " +50 Tan. = 1 .18.47 <br />70 191'= t/ " 543 <br />90 491' = " (( " j 50 B C. = sta. 541 { 53.53 <br />110401-- " 54.3+ 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° 191'=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 Y0° Curve. <br />A <br />23`/ <br />o_ <br />s 7 2 J <br />7, <br />GD. <br />a 76" <br />B7. res <br />- /S, <br />8 3 , <br />eo. <br />G0.� <br />i <br />g 7 3 <br />1, <br />-� S' <br />..CURVE TABLES. <br />Published" by KEUFFEL Sa ESSER CO. <br />HOW TO USE CURVE .TABLES. <br />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 />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 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 />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P. =23° 20' to the R. at Station. <br />542+72. <br />Ext. in Tab. I opposite 23° 20' =120.87 <br />120.87 =12 =10.07. Say a 10P 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 IOP 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-,k'-- def. for sta. 542 I. P. =sta. 542-x-72 <br />40 491' _ " " +50 Tan. = 1 .18.47 <br />70 191'= t/ " 543 <br />90 491' = " (( " j 50 B C. = sta. 541 { 53.53 <br />110401-- " 54.3+ 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° 191'=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 Y0° Curve. <br />A <br />