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CURVE TABLES. <br /> r_ Published by KEUFFEL & 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 may be found nearly enough,by dividing the Tan. <br /> _ or Ext.opposite the given Central Angle by the given degree of curve. <br /> ---- -- may '' `5 ---- To find Deg. of Curve, having the Central Angle and Tan$ent: <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 /> - - - - or Ext.of twice the given angle divided by the radius of a 1°curve will <br /> be the Nat.Tan.or Nat.Ex.EXAMPLE. <br /> -- - -- - Wanted a Curve with an Ext. of about 12 ft. Angie <br /> — of Intersection or I. P.=230 20' to the R. at Station <br /> 542+M <br /> 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 /> 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°191'=def.for sta. 542 I. P.=sta. 542+72 <br /> 40 4911_ " " " +50 Tan.= 1 .18.47 <br /> -- 7'191'= " ti " 543 <br /> B.C.=sta. 541+53.53 <br /> 9°49"= « " cc x-50 <br /> 110 40'= " ca " 5u+ <br /> L.C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543+86.86 <br /> --- -- - - --- - -- - -----� <br /> 100-53.53=46.47 X3'(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°50j'for a 10°Curve. ' <br /> I <br /> N <br /> 10•Curve <br /> 46 YL <br />