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J <br /> FIEL BO <br /> K 52.8 <br /> CURVE TABLES. <br /> Published by KEUFFEL 8s ESSER CO. <br /> HOW TO USE CURVE TABLES. <br /> " = Table I. contains Tangents and Externals to a 1°curve. Tan.and <br /> ---- --- - - - t.to any other radius maybe found nearly enough,bydividing the Tan. <br /> r Ext.opposite the given Central Angle by the given degree o 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 /> - -- - <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 /> ----------- <br /> - <br /> EXAMPLE. <br /> Wanted a Curve with an Ext.of about 12 ft. Angle <br /> i - of Intersection or I. P.=23° 20' to the R. at Station <br /> i <br /> -— - 542+72. <br /> Ext.in Tab. I opposite 23°20'=120.87 <br /> 120.872.12'=10.07. Say a 10°Curve. <br /> -- - Tab. 2 p.23* 0' <br /> Tan.in Tab. I o 23°20'-1183.1 <br /> 31. <br /> Correction for A.23°20'for a 10°Cur.=016 <br /> -- -- <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 2.10=2.3333=L.C. <br /> _ 2°19Y=def.for sta. 542 I. P.=sta. 542+72 1 <br /> - 4°491'= « « « -x-50 Tan.= 1 .18.4 <br /> 0 r= « « a543 <br /> - _ - 90 49/_ «« « •+ 50 B.C.=sta. 541�53.53 <br /> _ 11°40'= « « « 543-F L.C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543+86.86 <br /> - - - - <br /> i - - 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= ' <br /> 2°19Y=def.for sta.542. <br /> - -- — _ - ----- Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°50;'for a 10°Curve. <br /> e'A <br /> LP.An9.23°Y01 <br /> IO•Curva . <br /> 4 <br /> 40 q <br /> , <br /> X40/ <br /> 1 <br />