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� I <br /> JP y� CURVE TABLES. <br /> { <br /> Published by KEUFFEL & ESSER CO. <br /> Tab1eHLO tall USE CURVE TABLES. j <br /> f f - - `°" s Tangents and Externals to a 1°curve. Tan.and <br /> - t 7 - Ext.to any other radius may be found nearly enough,bydividing the Tan. <br /> y y Y o or Ext.opposite the given Central Angle by the given degree of curve. <br /> 4-st';�j 1 , To find Deg. of Curve, having the Central Angle and Tangent: <br /> 4. s _ - Divide Tan.opposite the given Central Angle by the given Tangent. <br /> { 9 S , To find Deg. of Curve, having the Central Angle and External; <br /> Divide Ext. opposite the given Central Angle by the given External. <br /> 7 t b To find Nat.Tan.and Nat.Ex.Sec.for any angle by Table I.:Tan. <br /> � µ or Eat.of twice the Niven angle divided by the radius of a P curve will <br /> be the Nat.Tan.or at.Ex.Sec. <br /> � s z s EXAMPLE. <br /> 3 v37 / .,. /� Wanted a Curve with an Ext.of about 12 ft. Angle <br /> • �F Y D y6 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 /> 3, 120.87-12=10.07. Say a 10°Curve. <br /> — G f <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 /> f - (If corrected Ext.is r uired find in same way) <br /> - Ang.23°20'=23.33=10=2.3333=L.C. <br /> r' � rsy z — " 4°49 '-def.forsfa. 542 I. P.=sta. 542+72 <br /> 7°19#'- +50 an.= 1 .18.47 <br /> q 1r a u B. C.=st <br /> 9o 49 3 = a ,.{543 a. 541 50 I x3.53 <br /> 11°40'= a a a 543+ L. C.= 2 .33.33 <br /> 86.86 E. C.=Sta. 543+86.86 <br /> 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.411= <br /> "v 2°19y"=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 /> p'1 V <br /> 10•Curvy � <br /> `p <br />