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CURVE TABLES. <br /> y Published by KEUFFEL 8b ESSER CO. <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 maybe found 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.op�osite the given Central Ile by the given Tangent. <br /> To find Deg. of Curve, having the Central Angle and External: <br /> - ---- ----- - - z --- Divide 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.of twice the given angle divided by the radius of a V curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> ---- -- <br /> EXAMPLE. <br /> Wanted a Curve with an Ext. of about 12 ft. Angle <br /> of Intersection or I. 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°CurOe. <br /> Tan. in Tab. I opp.23°20'=1183.1 <br /> _ 1183.1=10=118.31. <br /> ij <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°19j'=def.for sta. 542 I. P.=sta. 542+72 <br /> 4°49j'= " " " +50 Tan.= 1 .18.47 <br /> P <br /> 7*o ra a a <br /> ---- _._ 49 j __'= u u a +43 B. C.=sta. 541+53.53 <br /> go-- - - 11°40'= a • 543+ L.C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543+86.86 <br /> 100-53.53=46.47X3'(def.for 1 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°507'for a 10°Curve. <br /> � eiex <br /> P.Ang.2 3-201 <br /> >r <br /> N <br /> - 10•curve <br /> a <br /> - - - % .dpi <br /> i <br /> -7 Mid <br /> • <br /> • <br />