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LD 00 � <br /> ST, [ _ CURVE TABLES. <br /> lP ? ;!•_' - _-__— Published by KEUFFEL & ESSER CO. <br /> HOW TO USE CURVE TABLES. <br /> QC f, may bTable I. contains Tangents and Externals to a V curve. Tan.and <br /> 1 — Ext.to any other radius e f ound nearly enough,by dividing the Tan. <br /> or Ext.opposite the given Central Angle by the given degree of curve. <br /> To find Deg. of Curve, having the Central Angle and Tangent: <br /> Divide Tan.opposite the given Central Angle by the given Tangent. <br /> --- 1 To find Deg. of Curve, having the Central Angle and External: <br /> 1Divide Ext. opposite the given Central Angle by the given External. <br /> - _ - To find Nat.Tan.and Nat.Ex.Sec.for any angle by Table I.:Tan. <br /> or Ext.Df twice the given angle divided by the radius of a 1°curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> - i - - EXAMPLE. <br /> - -- - -- - -- - Wanted a Curve with an Ext. of about 12 ft. Angle <br /> of Intersection or I. P.=230 20' to the R. at Station <br /> - ii - -- 542- 72. <br /> 1 Ext.in Tab. I opposite 23°20•=120.87 <br /> } - - <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 /> 118.31-}-0.16=118.47=corrected Tangent. <br /> ---- -- --- (If corrected Ext.is required find in same way) <br /> Ang.23°20'=23.33°=10=2.3333=L. C. <br /> 2°19Y=def.for sta. 542 I. P.=sta. 542+72 <br /> 40 493z'= 11 11 44 +50 Tan.= 1 .18.47- <br /> ! - <br /> 70 193,= « 543 <br /> 904911_ a ca a B: C.=sta. 541+53.53 <br /> - - - 110 40"= « u « 543 L. C.= 2 .33.33: <br /> I 86.86 E. C.=Sta. 543 -86.86, <br /> J <br /> 100-53.53--46.47 X3'(def.for 1 ft. of 10°Cur.)=139.41'= <br /> 2°19"=def.for sta.542. <br /> -- ---- <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> --- 4 - _-_ ___— Def,for 36.86 ft.=1°50"for a 10°Curve. <br /> � X42 . <br /> %> <br /> I.RAn9.23.20' <br /> 1 `gyp A^ V BQ� <br /> 10•curve <br /> 0Ai <br /> It / <br />