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ti - <br /> - - - -- --- - -- - CURVE TABLES. <br /> Published by KEURML & n=R CO. j <br /> HOW TO USE CURVE TABLES.. <br /> Table I. contains Tangents and Externals to a 1°curve. Tan.and <br /> -- Ext.to any other radius may befound nearly enough,bydividing theTan. <br /> or Ext.opposite the given Central Angle by the given degree of curve. <br /> To find eg <br /> T 8 . of Curve,D having the Central Angle and Tangent: <br /> � g <br /> Divide Tan.opposite the given Central Angle by the given Tangent. <br /> To find Deg. of Curve having the Central Angle and External: <br /> `Divide Ext.opposite the given Central Angle by the given External. <br /> To find Nat.Tan.andNat:Ex.Sec.for any angle by Table L:Tan. <br /> -'- 7 -- or Ext.of.twice the given angle divided by the radius of a 1°curve will <br /> 'be the Nat.Tan.or Nat.Ex.Sec. , <br /> EXAMPLE. <br /> - - - --- --- ------- - 1-= --- Wanted a Curve with an Est.of about 12 ft. Angle <br /> / G d of Intersection or 1. P.=23° 20' to the R. at Station <br /> _ -- -- - - - ---- -- _. 542+72. <br /> Ext.in Tab. I opposite 23°20•-120.87 <br /> - ' `' 120.87+12=10.07. Say a 10°Curve. <br /> Tan.in Tab.I opp.23°20'=1183.1 <br /> 1183.1+10=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.16 <br /> -- -"—' <br /> 118.31+0.16=11&47=corrected Tangent. <br /> (If corrected Ext.is required find in same way) <br /> - - <br /> -� Ang.23°20'=23.33 +10=2.3333=L.C. <br /> _ 2°19Y-def.for sta. 542 1. P.-sta. 542+72 <br /> 4-49j'= a a a +50 Tan.= 1 .18.47 <br /> _-_ 7 19} _ 543 <br /> -- _-- t o i_ a a s <br /> - - -- <br /> - 9°491'= a u a +� B.C.=sta. 541+53.53 <br /> - 110-40'= a a a 548+ L.C.= 2 .33.83 <br /> _ - <br /> 86„86 548+86.86 <br /> 100-53.53-46.47X3'(def.'for I ft.of 10°Cur.)®139.41'= <br /> 2°191'=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°50}'for a 10'Curve. <br /> •x <br /> S (G•Our" <br /> _ - -- --- <br /> �� ani <br /> 4',' <br /> .. . . . i. _ <br /> +' 0 - .. <br />