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CURVE TABLES. = Publisfied by KELJFFEL 8&` ESSER CO.. IiOW 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, bydividing 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 dor 1: P.=23* 20' to the R: at Station M2+72. Ext. in •Tab. 1 opposite 23° 20' =120.87 120.87-12=10.07. Say a 10' Cui e. Tan. in Tab. I opp. 23' 20' =1183.1 1133.1=10 =118.31. Correction for A: 23' 20' -for a 10' Cur. =0.16 - 113.31� 0.15=,118.47=corrected Tangent. (If corrected F_xt. is required find in same way) Ang..23' 20' = 23.33' -10 = 2.3333 = L. C. 2' 191'= def. for sta: 542 1. P. =sta. 542 -72 40 492' _ cc " " +50 Tan. = 1 .18.47 701921'-- -B. C.=sta. 541-53.53 . 9049"— �,_ 1, « «" +50 L. C.= 2 .33.33 11040== •` . 543-}- . 86.86, E. C.=Sta. 543+86.86 100.-53.53=46.47X3'..(def..for 1 ft. of 10' Cur.)=139.41'= i 2' 19 -'=def. for sta. 542. II Def: for 50 ft. =2° 30' for a 10' Curve. Def. for 36.86 ft. =1' 50" for a 10' Curve. _ _ LP.An9.23°20 e°> ' 10• Gurve . 1 ` `` ' "44- Af ZOO,-, CURVE TABLES. = Publisfied by KELJFFEL 8&` ESSER CO.. IiOW 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, bydividing 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 dor 1: P.=23* 20' to the R: at Station M2+72. Ext. in •Tab. 1 opposite 23° 20' =120.87 120.87-12=10.07. Say a 10' Cui e. Tan. in Tab. I opp. 23' 20' =1183.1 1133.1=10 =118.31. Correction for A: 23' 20' -for a 10' Cur. =0.16 - 113.31� 0.15=,118.47=corrected Tangent. (If corrected F_xt. is required find in same way) Ang..23' 20' = 23.33' -10 = 2.3333 = L. C. 2' 191'= def. for sta: 542 1. P. =sta. 542 -72 40 492' _ cc " " +50 Tan. = 1 .18.47 701921'-- -B. C.=sta. 541-53.53 . 9049"— �,_ 1, « «" +50 L. C.= 2 .33.33 11040== •` . 543-}- . 86.86, E. C.=Sta. 543+86.86 100.-53.53=46.47X3'..(def..for 1 ft. of 10' Cur.)=139.41'= i 2' 19 -'=def. for sta. 542. II Def: for 50 ft. =2° 30' for a 10' Curve. Def. for 36.86 ft. =1' 50" for a 10' Curve. _ _ LP.An9.23°20 e°> ' 10• Gurve . 1 ` `` '