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CURVE TABLES. Published by KEUFFEL & ESSER ' CO. a, TI0,M7 TO USE. CURVE TABLES. Table I: contains Tangents and Externals to a 1° curve. Tan. and t Est to any other radius maybe found nearly enough, bydividing theTan. or Est: opposite the given Central Angle by the given degree of curve. To find Deg. of Curve, having'tte Central Angie 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. =23° 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. 230 20' for a 10° Cur. =0.16 118.31=}-0.16 = 118.47 =corrected. Tangent. (If corrected Ext, is required find in sarne way) Ang.23°20'=23.33°=10=2.3333=L. C. 2° 191' = def. for sta. 542 I. P. = sta. 542.1-72 4° 49Y= ':cc+60 Tan. = 1 .1.8.47 70 191'= . J; " 543 9°491`= ': f 50 B. C.=sta: 541+53.53 - 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'= 2° 1.91'=def. for sta. 542. Def. for 50 ft. =2° 30' for a 10° Curve. Def. for 3.86 ft. =1° 50',-' for a 10° Curve. D -to o ;r 80 3•0 30 I ove 1.{3-E/� rr - -" . . ^o{ oo s- G,E 5 a . - Se c.. 3= L Oct (6,13 dao 148.7 t-5-+-•c7— Itfro • i. �e5 - ;i. z�- 321 c _ S . n CURVE TABLES. Published by KEUFFEL & ESSER ' CO. a, TI0,M7 TO USE. CURVE TABLES. Table I: contains Tangents and Externals to a 1° curve. Tan. and t Est to any other radius maybe found nearly enough, bydividing theTan. or Est: opposite the given Central Angle by the given degree of curve. To find Deg. of Curve, having'tte Central Angie 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. =23° 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. 230 20' for a 10° Cur. =0.16 118.31=}-0.16 = 118.47 =corrected. Tangent. (If corrected Ext, is required find in sarne way) Ang.23°20'=23.33°=10=2.3333=L. C. 2° 191' = def. for sta. 542 I. P. = sta. 542.1-72 4° 49Y= ':cc+60 Tan. = 1 .1.8.47 70 191'= . J; " 543 9°491`= ': f 50 B. C.=sta: 541+53.53 - 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'= 2° 1.91'=def. for sta. 542. Def. for 50 ft. =2° 30' for a 10° Curve. Def. for 3.86 ft. =1° 50',-' for a 10° Curve.