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i <br /> CURVE TABLES. <br /> Published by KEUFFEL 8s 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 may be 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 /> 9 - - 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 /> - t2 <br /> 542+72. <br /> f Ext.in Tab. I opposite 230 20'=120.87 <br /> 120.87=12=10.07. Say a 10°Curve. <br /> Tan.in Tab. I opp.23°20'=1183.1 <br /> -----_—__-- _ 1183.13-10=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.16 <br /> --- - 118.31 x-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 /> _ � I <br /> 2°194'=def.for sta. 542 I. P.=sta. 542+72 <br /> _-- 40 491'_ " " " +50 Tan.= 471 <br /> -...i 7o 19j'= a a a - 543 <br /> 90 491?_ a a u +50B.C.=sta. '41+ <br /> 41•}-53.53 <br /> --- <br /> 11°40'= a a u + L.C.= 2 .33.33 <br /> 86.86 E.C.=Stn. 543-{-86.86 ' <br /> 100-53:53=46.47x3'(def.for 1 ft.of 100 Cur.)=139.41' <br /> �� ca Etc �- '•''''� - <br /> 2°19F=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 100 Curve. <br /> t G-' <br /> Def.for 36.86 ft.=10 50}'for a 10°Curve. <br /> e <br /> P.Ang.23•L-01 <br /> I <br /> 17 7 10•Curve <br /> 4� <br /> 3 <br /> I <br />