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629 <br /> F1ELSt3tjK . , <br /> - - + - - CURVE TABLES. <br /> Published by KEUFFEL & ESSER CO. <br /> 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,bydividing theTan. <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 /> 7 '�' �f- - - 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 /> 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 /> 2 � Sf 3.7 ��I' -T -- 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 /> 7- -- - � - -�`� Ext.in Tab. I opposite 23'20'=120.87 <br /> Al. 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=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 /> 4049211= a a a +50 Tan.= 1 .18.47 <br /> 7'1911= u a a 543---- -- ---- 9049}'= a a a +50 B.C.=sta. 541{53.53 <br /> 110400= a a 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°19F=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 10'Curve. <br /> Def,for 36.86 ft.=1°502'for a 10'Curve. <br /> tpx'� it <br /> LP.An9.23.201 <br /> lY <br /> I <br /> 10•Curve <br /> a= 4 <br /> • <br /> r . <br /> 1 <br />