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w CURVE TABLES. <br /> Published by KEU"EL 8s ESBER CO. <br /> ------ <br /> HOW TO USE CURVE TABLES. <br /> z p�P✓ Table I. contains Tangents and Externals to a 1°curve. Ta.and <br /> Ext.to any ol ethe her radius <br /> maybe found Angle nearly <br /> he enough, <br /> iven degree of curve. <br /> Lo� ` or Ext. <br /> Too <br /> 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 /> N opposite given Central Angle by the given External. <br /> Divide Ext. o d <br /> AAI" To find Nat.Ta,an Nat.Ex.Sec.for an angle by Table L:Tan. <br /> - or Ext.of twice the given angle divided by the radius of a 1°curve will <br /> SvQ IV)y/s/ani be the Nat.Tan.or Nat.Ex.Sec. <br /> Wanted a Curve with an Ext. of about 12 ft. Angle <br /> � e 7-V A,,j2 r15- { °542+7n72rsection or I. P.=23° 20' to the R. at Station <br /> - -f <br /> L.O f Zo P 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 /> 6 <br /> = .444 ZSCorrection 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°19Y=def.for sta. 542 I. P.=sta. 542+72 <br /> 40 491'= K K K +50 Tan.= 1 .18.47 <br /> ° II_ K K K <br /> 7 19 z = 543 <br /> o49;v B.C.=sta. 541x-53.53 <br /> K K K <br /> 9 +50 <br /> - - 11°40'= K .K " 543+ L.C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543 F86.86 <br /> - 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> v. 2°19;'=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> N ,— — Def.for 36.86 ft.=1°50}'for a 10°Curve. <br /> . p a w.A"9.za^z01 <br /> 1 <br /> T °rf°nv 10•Curve <br /> - �jo �° <br />