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J/_ 0 ,�, CURVE TABLES. <br /> 4� 2,9 -- , Published by KEUFFEL & ESSER CO. <br /> e 77- . HOW TO USE CURVE TABLES. <br /> �7"� •�, ;i Table I. contains Tangents and Externals to a 1°curve. Tan.and <br /> _ Ext.to any other radius rhay be found nearly enough,by dividing the Tan. <br /> G f` or Ext.opposite the given Central An le by the given degree of curve. <br /> • To find Deg, of Curve, having the Central Angle and Tangent: <br /> r - Divide Tan.opposite the given Central Angle by the given Tangent. <br /> To find Deg. of Curve, having the Central Angle apd 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 /> ---- 1 � be the Nat.Tan.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;� 542 -72. r. <br /> Ext.in Tab. I opposite 23°20'=120.87 <br /> { f u 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 /> def. 50 1.n.= <br /> .for sta. 542 1. 541+72 <br /> 4°491'= <br /> !!! 18.47 <br /> 7°1911= « « a 543 ' <br /> -"` <br /> ! B. sta.« « . 541- 53.53 <br /> i <br /> �T 11°40'= a a a 5 + L. C.= 2 .33.33 <br /> I d 86.86 E.C.=Sta. 543+86.86 <br /> 100-53.53=46.47X31(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 2°19 '=def.for sta.542. <br /> _ Def.for 50 ft.=2°30'fora 10°Curve, Z <br /> d 1 •j'o Def.for 36.86 ft.=1°50j'for a 10°Curve. :�-'`t <br /> I.P.An9.23.20' <br /> N <br /> 6.01 <br /> 'B4 <br /> NN <br /> 10.Curve <br /> db <br /> S r'^ <br /> FIELD B OK 08 <br /> f <br /> f. <br />