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�. , CURVE TABLES. <br /> Published by KEUFFEL & ESSER CO. <br /> HOW TO USE CURVE TABLES. <br /> Table I. contains Tangents and Externals to a 10 curve. Tan.and <br /> Ext.to any other radius may befound nearly enough,bydividing 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 10 curve will <br /> be the Nat.Tarr.or Nat.Ex.Sec. <br /> - t - EXAMPLE. <br /> +f "= 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 /> 7 542 -72. <br /> t! 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.230 20'=1183.1 <br /> 1183.1=10=118.31. <br /> Correction for A.230 20'for a 100 Cur.=0.16 <br /> i <br /> - - -' -_ 118.31 x-0.16=118.47=corrected Tangent. <br /> 0 0 • -� (If corrected Ext.is required find in same way) <br /> i <br /> - �Zt- Ang.23020'=23.33°=10=2.3333=L. C. <br /> 20 191'=def.for sta. 542 I. P.=sta. 542+72 <br /> 40 49j'= « u u +50 Tan.= 1 .18.47 <br /> l <br /> 21 70 19j'= « a �r 543 -- B.C.=sta_ 541 53.53 <br /> - 4 i 90 49,= e �a a -f-50 <br /> 0 110 40'= « '1 543-f- L. C.= 2 .33.33 <br /> 86.86 F E.C.=Sta. 543+86.86 <br /> 4ti <br /> 100-53.53=46.47 X3'(def.for 1 ft.of 100 Cur.)=139.41'= <br /> Cr i� a T. ow L7 . 20 19j'=def.for sta.542. <br /> Def. for 50 ft.=20 30'for a 100 Curve. <br /> Def.for 36.86 ft.=1*50"for a 100 Curve. <br /> l� <br /> IAAn9.23.201 <br /> 1 lo•N Curve <br /> 4� <br /> i� - -- <br /> 0 <br /> A/ i <br /> F ELS OAC 541 e ; <br /> • <br /> r t <br />