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CURVE TABLES Published by KEUFFEL & ESSER CO. HOW TO USE CURVE TABLES Table I. contains Tangents and Externals to a 1° curve. Tan. and Ext. to any other radius may be found nearly enough, by dividing. the Tan. .or Ext. opposite the given Central Angle by the given degree of curve. To find Deg. of Curve, having the Central, Angle and Tangent: Divide Tan. opposite the given Central Angle by the given Tangent. To find Deg. of Curve, having the Central Angle and External: Divide Ext. opposite the given Central Angle by the given External. To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. or Ext. of twice the given angle divided by the radius of a 1° curve will be the Nat. Tan. or Nat. Ex. Sec. EXAMPLE Wanted a Curve with an Ext. of about 12 ft. Angle of Intersection or I. P.=230 20' to the -R. at Station 542+72. Ext. in Tab. I opposite 23° 20' =120.87. . 120.87=12=10.07. Say a 10° Curve. Tan. in Tab. I opp. 23° 20'= 1183.1 1183.1=10 =118.31. Correction for A. 23° 20' for a 10° Cur. =0.16 118.31+0.16 =118.47 =corrected Tangent. (If corrected Ext. is required find in same way) Ang. 23°20'=23.33°=10=2.3333=L. C. 2° 19 def. for sta. 542 I. P. =sta. 542+72 40 49'2'= " " " +50 Tan. = 1 .18.47 7° 192' _ 543 " " 543 B. C.=sta. 541+53.53 9° 492'= 110 40' _ " " 543+ L. C. = 2 .33.33 86.86 E. C.=Sta. 543+86.86 100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) = 139.41'= 20.192'=def. for sta. 542. Def. for 50 ft.=2°-30' for a 10°. Curve. Def. for 36.86 ft. =1° 502' for a 10° Curve. g g7 sed S g, 313. Y ty`C. 7 ge� A—, 1 .67 i 7 �L i 1 CURVE TABLES Published by KEUFFEL & ESSER CO. HOW TO USE CURVE TABLES Table I. contains Tangents and Externals to a 1° curve. Tan. and Ext. to any other radius may be found nearly enough, by dividing. the Tan. .or Ext. opposite the given Central Angle by the given degree of curve. To find Deg. of Curve, having the Central, Angle and Tangent: Divide Tan. opposite the given Central Angle by the given Tangent. To find Deg. of Curve, having the Central Angle and External: Divide Ext. opposite the given Central Angle by the given External. To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. or Ext. of twice the given angle divided by the radius of a 1° curve will be the Nat. Tan. or Nat. Ex. Sec. EXAMPLE Wanted a Curve with an Ext. of about 12 ft. Angle of Intersection or I. P.=230 20' to the -R. at Station 542+72. Ext. in Tab. I opposite 23° 20' =120.87. . 120.87=12=10.07. Say a 10° Curve. Tan. in Tab. I opp. 23° 20'= 1183.1 1183.1=10 =118.31. Correction for A. 23° 20' for a 10° Cur. =0.16 118.31+0.16 =118.47 =corrected Tangent. (If corrected Ext. is required find in same way) Ang. 23°20'=23.33°=10=2.3333=L. C. 2° 19 def. for sta. 542 I. P. =sta. 542+72 40 49'2'= " " " +50 Tan. = 1 .18.47 7° 192' _ 543 " " 543 B. C.=sta. 541+53.53 9° 492'= 110 40' _ " " 543+ L. C. = 2 .33.33 86.86 E. C.=Sta. 543+86.86 100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) = 139.41'= 20.192'=def. for sta. 542. Def. for 50 ft.=2°-30' for a 10°. Curve. Def. for 36.86 ft. =1° 502' for a 10° Curve.