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tURVE TABLES Published -by KEUFFEL & ESSER CO.' .-HOW-40 USE CURVE TABLES Table 1. contains Tangents and Externals to a 1° curve. Tan. and Ext. to any other radius.inay 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 tlie.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.=23° 20' to the R. at Station 542+72. Eit. in Tab. I opposite 23° 20' =120.87 120.87=12=10.07. Say a 10° Curve. Tan. in Tab. I opp. 23° W= 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° 192' = def. for sta. 542 I. P. = sta. 542+72 4° 492` _ +50 Tan. = 1 .18.47 70 192' _ " " 543 9 49 1'— " " ` B. C.=sta. 541+53.53 2 — 11* 40'= " '° 543+ L. C. = 2 .33.33 86.86 E. C.=Sta. 543+86.86 100-53.53=46.47XX(def. for 1 ft. of 10° Cur.)=139.41'= 2° 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. s 4d - x> .. LP.An9.23°20� -.. .. ... 10° Curve I tURVE TABLES Published -by KEUFFEL & ESSER CO.' .-HOW-40 USE CURVE TABLES Table 1. contains Tangents and Externals to a 1° curve. Tan. and Ext. to any other radius.inay 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 tlie.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.=23° 20' to the R. at Station 542+72. Eit. in Tab. I opposite 23° 20' =120.87 120.87=12=10.07. Say a 10° Curve. Tan. in Tab. I opp. 23° W= 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° 192' = def. for sta. 542 I. P. = sta. 542+72 4° 492` _ +50 Tan. = 1 .18.47 70 192' _ " " 543 9 49 1'— " " ` B. C.=sta. 541+53.53 2 — 11* 40'= " '° 543+ L. C. = 2 .33.33 86.86 E. C.=Sta. 543+86.86 100-53.53=46.47XX(def. for 1 ft. of 10° Cur.)=139.41'= 2° 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. s 4d - x> .. LP.An9.23°20� -.. .. ... 10° Curve