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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 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. bf 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. f t. Angle <br />of Intersection or I. 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° 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 />. 2° 1921'=def. for sta. 542 I. P. =sta. 542+72 <br />4° 4921'= " " +50 Tan. = 1 .18.47 <br />7° 1921' « u « .543 <br />9° 4921' = <br />114 " " 543++50 <br />0'_ " " " B. C. = sta. 541 153.53 <br />L. C.= 2 .33.33 <br />86.86 E. C. =Sta. 543+86.86 <br />100-53.53=46.47X31(def. for I ft. of 10° Cur.)=139.41'= <br />2° 1921'=def. for sta. 542. <br />a Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50;' for a 10° Curve. <br />,) <br />I./,3 <br />aa/ <br />�4D( <br />5 <br />A <br />� <br />3f3 <br />00/ <br />6s <br />i <br />oS <br />IO <br />s <br />4. <br />/ <br />pQi <br />/o ° <br />I u <br />p0$ <br />03 <br />30, <br />J e <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 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. bf 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. f t. Angle <br />of Intersection or I. 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° 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 />. 2° 1921'=def. for sta. 542 I. P. =sta. 542+72 <br />4° 4921'= " " +50 Tan. = 1 .18.47 <br />7° 1921' « u « .543 <br />9° 4921' = <br />114 " " 543++50 <br />0'_ " " " B. C. = sta. 541 153.53 <br />L. C.= 2 .33.33 <br />86.86 E. C. =Sta. 543+86.86 <br />100-53.53=46.47X31(def. for I ft. of 10° Cur.)=139.41'= <br />2° 1921'=def. for sta. 542. <br />a Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50;' for a 10° Curve. <br />