Laserfiche WebLink
� . 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 /> or 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 Ext.of twice the given angle divided by the radius of a 1°curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> �EtAMPLE. <br /> _ Wanted a Curve with anExt.of about 12 ft. Angle G�J <br /> of Intersection or I. P.=230 <br /> 20 to the R. at Station <br /> 542+72. GGr <br /> Ext.in Tab. I opposite 23°20'=120.87 h 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 /> >S <br /> (If corrected Ext.is required find in same way) - <br /> , , <br /> — Ang.23°20'=23.33°=10:2.3333=L.C. � <br /> 2°19F=def.for sta. 542 I. P.=sta. 542+72 — <br /> 4°49"= " " " • +50 Tan.= 1 .18.47 <br /> 70 191 r= a ct a 543 <br /> o MY= u u « B.C.=sta. 541+53.53 <br /> 9 49-a +50 L.C.= 2 .33.33 <br /> 11°40'= « u u 543 <br /> 86.86 E.C.=Sta. 543+86.86 <br /> -- - - 100-53.53=46.47XT(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 2°191'=def,for sta.542. <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft. I*50j'for a 10°Curve. <br /> . 04 <br /> ?x <br /> >r <br /> LP.An9.23•PO1 <br /> 10•cur" <br /> P 9 <br /> \ / <br /> t' <br /> 1 <br />