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I � i - - - - CURVE TABLES. Published by KEUFFEL & ESSER CO. i -- --- - --- - _--4 HOW TO USE CURVE:TABLES. Table I. contains Tangents and Externals to a 1°curve. Tan.and - �- - Ext.to any other radius xray befound nearly enough,bydividing theTaa. br 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 An le 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 iiy Table L: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 \i,ith an Ext. of about 12 ft. Angle of Intersection or I. P.=230 20' to the R. at Station -- ----- - 542+72. I Ext.in Tab. I opposite 23°20'=120.87 -- 120.87=12=10.07. Say a 10°Curve. LL__ _ ,f 28 /� !a�/ -L� �-- ape�� Tan.in Tab. I opp.23°20'=1183.1 1183.1-10=118.31. ----- ---__- --✓rs� 1 q,72_ -_ _T_ - -- - - - ljlS Bi�o,E� Correction for A.23°20'for a 10°Cur.=0.16 118.31-}•0.16=118.47=corrected Tangent. Tp -�-_QJ -- - --_. -__ .11 (If corrected Ext.is required find in same way) Ang.23°20'=23.33°=10=2.3333=L. C. -- -- --- - - ----. _-___ 2°191'=def.for sta. 542 I. P.=sta. 542+72 - - p / - 4°491,_ (( « a +50 Tan.= 1 .18.47 -..W/ 0 -1,j_J_ Y - ---- - - - 7°19i?= :< « 543 9°491'= (f a B. C.=sta: 541+53.53 110 40'= « a u 54.8+ L. C.= 2 .33.,33 86.86 E. C.=Sta. 543+86.86• 100-53.53=46.47X3'(def.for I ft.of 10°Cur.)=139.41'= 2° 19,j'=def.for sta.542. r�7� 1 Def.for 50 ft.=2°30'for a 10°Curve. 7• +I`� 3`f-__-_ Def.for 36.86 ft.=1°501'fora 10°Curve. I. 1 IAAng.23°20' n1 7. 7 3 -z3 - ! zs - �- L 1o•Curn .01 "y - A 4