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CURVE TABLES. <br /> ----- r i Published by KEUFFEL ds ESSER CO. <br /> - - <br /> 480 _-__ HOW TO USE CURVE TABLES. <br /> Table I. contains Tangents and Externals to a V 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. 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.and Nat.Ex.Sec.for any angle by Table I.:Tan. <br /> -- --- qt Ext.of twice the giaen angle <br /> the radius of a 1 curve will <br /> o <br /> be the Nat.Tan.or Nat.Ex.Sec. j <br /> - -- ------ -- Wanted a Curve with an Ext. of about 12 ft. Angle <br /> 4SS9 of Intersection or I. P.=230 20 to the R. at Station <br /> 542+72.Ext.in Tab. I opposite 23°20'=120.87 <br /> o4z e VE 4i 120.87=12=10.07. Say a 10°Curve. <br /> Tan.in Tab. I opp.23°20'=1183.1 <br /> � . 1 <br /> Correction for A.23*23 .20 for8a 10°Cur.=0.16 8.31-{-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°191'=def.for sta. 542 I. P.=sta. 542+72 <br /> - - - _ 40 4912'= a a u +50 Tan.= 1 .18.47 <br /> 7°1911— u u u <br /> _ - 3 — 543 B.C.=sta. 541+53.53 <br /> 9°4912'= u u ss ,+50 <br /> 110 40'= u « a 543-1- 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°1912'=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°5012'for a 10°Curve. <br /> �x <br /> _-- ^ —LP.An9.23240° <br /> -Y <br /> 10*Curve ' I <br /> \a <br /> t <br />