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F ELD OOK 457 <br /> - CURVE TABLES. <br /> Publishmd by KEUMML & ESMIR CO. <br /> - How TO USE CMVB T"UN. <br /> tl Table I. contains Tangents and Externals to a 1°curve. Tan.and <br /> — 'Ext.to any other radius maybe found nearly enough,by dividing the Tan. ' <br /> .or Ext.opposite the given Central ale by'the given degree of curve. <br /> To find Deg. of Curve,having the CentraAngle*Ad Tangent: <br /> Divide Tan.opposite the given Central ale by the given Tangent. <br /> P — To find Deg. of Curve,.having the Central Angle and External: <br /> Divide Ext.o posite the given Central Angle by the given External. <br /> - { _, To find Nat.Tan.and Nat.Ex.Sec.for any an by Table L:Tan. <br /> or Ext.of twice the given angle divided by the radius of a V curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> RX"PLE. <br /> Wanted a Crave with an Ext.of about 12 ft. Angle <br /> of Intersection or I. P.=230 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 /> 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°19F=def.for sta. 542 1.P.—sta. 542+72 <br /> 40 491'= a a a +50 Tan.= 1 .18.47 <br /> — <br /> 70 10 '= a a a <br /> j 9°49j'= a a a 50 543 B.C.=sta. 541+53.53 � <br /> - 110 40'= a a a 543 L.C.— 2 .33.33 <br /> ` d 86.86 E.C.=Sta. 543+86.86 <br /> 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> u _ 2°19}'=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> f - Def.for 36.86 ft.=1°50}'for a 10°Curve. <br /> s. <br /> �x <br /> .P.An9.Y3•YO• <br /> 10•C%U" <br /> 47 <br /> y, <br />