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j.- CURVE TABLES. Published by KEUFFEL & ESSER CO. I' HOW TO USE CURVE TABLES. Table I. contains Tangents and Externals to a 1°curve. Tan.and �' -- - ---- - -- --- - --- ---- _ - Ext.to any other radius maybe found nearly enough,bydividing the Tan. a 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. j --- --- ----- --- - - EXAMPLE. i - -------- ------ _ __ _ _. Wanted a Curve with an Ext. of about 12 ft. AnglL of Intersection or I. P.=23* 20' to the R. at Station —--- - - - --- -- 542+72. i Ext.in Tab. I opposite 230 20'=120.87 ---- ----- -- --- --- 120.87=12=10.07. Say a 10°Curve. Tan. iri 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. 27 191'=def.for sta. 542 I. P.=sta. 542+72 4°493 - " " " +50 Tan.= 1 .18.47 -- - - - --- - -- - - - _-- 70 19F= « n 543 B. C.=sta. 541+53.53 9°49,'= u « « +50 - --- - ----- -- 11040'= u K 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°19j'=def.for sta.542. Def.for 50 ft.=2°30'for a 10°Curve. Def.for 36.86 ft.=1°50+}'for a 10°Curve. II ' 1 —N I.P.An9.P'3/Q•PY>O ' 10•Creu --- —— 1 Y