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. CURVE TABLES <br />Published by KEUFFEL & 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 may be f ound nearly enough, bydividing <br />of curve. <br />or Ext. opposite the given Central Angle by the given <br />To find- Deg. of Curve, having the Central Angle given Tangent t. <br />Divide Tan. opposite the given Central Angle by <br />To find Deg. of Curve, having the Central h given le d External: <br />Divide Ext. opposite the given Central Angle by <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 1. 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 />Tan. in Tab. I opp. 23° 20'= 1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 23' 118.31+0.16=118.47 =' o .16 <br />cotrrecd Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C, <br />2°191'=def. for sta. .542 I. P.=sta. 542+.72 <br />40 4911= " " +50 .Tan. = 1 .18.47 <br />7° 191'= " " " 543. B. C.=sta. 541+53.53 <br />9-49121 = 'a it it+50 2 .33.33 <br />11° 40'= " _ " 543+ L. C. _ <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 />t 2° 191'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' fora 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curve. <br />i <br />i <br />r <br />. CURVE TABLES <br />Published by KEUFFEL & 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 may be f ound nearly enough, bydividing <br />of curve. <br />or Ext. opposite the given Central Angle by the given <br />To find- Deg. of Curve, having the Central Angle given Tangent t. <br />Divide Tan. opposite the given Central Angle by <br />To find Deg. of Curve, having the Central h given le d External: <br />Divide Ext. opposite the given Central Angle by <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 1. 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 />Tan. in Tab. I opp. 23° 20'= 1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 23' 118.31+0.16=118.47 =' o .16 <br />cotrrecd Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C, <br />2°191'=def. for sta. .542 I. P.=sta. 542+.72 <br />40 4911= " " +50 .Tan. = 1 .18.47 <br />7° 191'= " " " 543. B. C.=sta. 541+53.53 <br />9-49121 = 'a it it+50 2 .33.33 <br />11° 40'= " _ " 543+ L. C. _ <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 />t 2° 191'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' fora 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curve. <br />i <br />