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CURVE TABLES. <br /> a � 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 th6Tan. <br /> Ext.xt.opposite the given Central Angle by the g�'ven 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. j <br /> I To find Nat.Tan.and Nat.Ex.See.for any angle by Table I.:Tan. <br /> for Ext.of twice the given angle divided by the radius of a 1°curve will <br /> LIJ) 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 1. P.=230 20' to-the R. at Station <br /> 542+72. <br /> Ext.in Tab.I opposite 23°20•=120.87 . <br /> ------ - <br /> 120.87+12=10.07. Say a 10°Curve. <br /> Tan.in Tab. I opp.23°20'=1183.1 <br /> -- - 1183.110=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.16 <br /> 118.3L+0.16=118.47=corrected Tangent. <br /> (If corrected Ext.is required find in same way) <br /> Ang.23°20'=23.38°=10=2.3333=L.C. <br /> 2°191'=def.for sta. 542 1. P.=sta. 542-x-72 <br /> a } 40 492'_ « it « -!-50 Tan.= 1 .18.47 <br /> 7*19 = « « [{ 543 <br /> ' c 2 — B.C.=sta. 541+53.53 <br /> 90492'= « « a -F50 <br /> « «. « 543+ L.C. 2 .33:33 <br /> 11°40'= <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 /> p Ga 2°192'=def.for sta.542. <br /> f L o o Def.for 50 ft.-2 30 for a 10•Curve. <br /> c _. <br /> Def,for 36.86 ft.=1°50"for a 10°Curve. <br /> ,,P7 4,0 - ---- <br /> I � r <br /> .P.An 23°201 . <br /> v*�-7o - 9 <br /> Z � C y7 t CNN <br /> lo•Curve <br /> i <br /> 0 � 4 <br /> 4 <br /> 1 <br /> w <br />