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-- --- - -- CURVE TABLES. <br /> 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 maybe found nearly enough,by dividing the Tan. <br /> r 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 /> -- -- ---- or Est.of twice the given angle divided by the radius of a 1'curve will <br /> be the Nat.Tari.or Nat.Ex.Sec. <br /> 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 /> 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.23'2(Y=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 /> 2'19Y=def.for sta. 542 I. P.=sta. 542+72 <br /> - _ - 40 49j'= « u a +50 Tan.= 1 .18.47 <br /> -- B.C.=sta. 541+53.53 <br /> 900 91' +50L.C.= 2 .33.33 <br /> 11 40'=- a 543+ <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 /> 2'192'=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 /> a: <br /> ax <br /> // <br /> ot <br /> 10*Curve <br /> — 4o/ <br /> l <br /> Ap <br /> n - - <br /> p . <br /> ` 1 .as - <br /> 1 <br />