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FI LD B OK 5 5 <br /> 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,bydividing the Tan. <br /> or Ext.Opposite the given Central Angle by thegiven 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 Ext.of twice the given angle divided by the radius of a V curve will <br /> 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 <br /> 542- 72. <br /> - -- Ext.in Tab. I opposite <br /> p07. 23°20'=120.87 7d <br /> 120.8712=10.07. Say a 10°Curve. <br /> f Tan.in Tab. I opp.23°20'=1183.1 <br /> --_—_- -_ 1183.1=10=118.31. 7 2 3 8 F <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'=23.3.3 +10=2.3333=L.C. <br /> 2°19Y=def,for sta. 542 I. P.=sta. 542+72 <br /> --- --------- 4°49i'= a a a <br /> -._ -- --- - x-50 Tan.= 1 .18.47 <br /> 7°191-,'= " a a 543 <br /> -. - - 90491'= a a a B.C.=sta. 541 {53.53 <br /> 11°40'= a a a 5 3+ L.C.= 2 .33.33 <br /> 86.86 E. C.=Sta. 543 -86.86 <br /> 3•�v" D,�°?_ _.. 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 2°19j'=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> g Def.for 36.86 ft.=1°50j'for a 10°Curve. <br /> _ 0 �` 7 7. <br /> _. -lrll f.'�3 `: .S1 3 34 /,L _ .�. k-� Y_ �2__ .z_.._ i--__s�-�a.----- x�� I.P.An9.23•P0 <br /> �s- �o �7 3Z fL i� L 1 yb <br /> A. <br /> 10•Curve <br /> 30 <br />