BOOK 398 - Laserfiche WebLink

Home Browse Search

Ft LD o0 39 CURVE TABLES. Published by KEUFFEL 8s ESSER CO. HOW TO USE CURVE TABLES. �L �� Table I. contains Tangents and Externals to a V curve. Tan.and i --- -_- Ext.to any other radius may be found nearly enough,by dividing theTan. or Ext.opposite the given Central Angle by the given degree of curve. To find Deg. of Curve, having the Central Angle and Tangent: Divide Tan.opposite the given Central Angle by the given Tangent. To find Deg. of Curve, having the Central Angle and External: Divide Ext. opposite the given Central Angle by the given External. ,fes_ _ C-�✓1__ __o. �n __ _�c/ ___ To find Nat.Tan.and Nat.Ex.See.for any angle by Table I.:Tan. or Eat.of twice the given angle divided by the radius of a 1°curve will .o-O be the Nat.Tan.or Nat.Ex.Sec. EXAMPLE. - - ---- - -- -- -- Wanted a Curve with an Ext.of about 12 ft. Angle of Intersection or I. P.=23° 20' to the R. at Station 542+72. Ext.in Tab. I opposite 23°20'=120.87 --- -— - -- - - - ---- 120.87-- 12=10.07. Say a 10°Curve. Tan.in Tab. I opp.23°20'=1183.1 - - - -- ---- -- - --- -- --- _-_ 1183.1=10=118.31. Correction for A.23°20'for a 10°Cur.=0.16 118.31-}-0.16=118.47=corrected Tangent. (If corrected Ext.is required find in same way) Ang.23°20'=23.33°=10=2.3333=L. C. 2°191'=def.for sta. 542 I. P.=sta. 542+72 -C 3 L 40 49121= cc " 01 +50 Tan.= 1 .18.47 7'191'= « « 543 0 , u u u +50L.C.= 2 .33.33 a /•.rS 9°49 '- « u B. C.=ata. 541+53.53 11 40'= 543+ 86.86 E. C.=Sta. 543+86.86 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= 2°19F=def.for sta.542. Def.for 50 ft.=2°30'for a 10°Curve. Def.for 36.86 ft.=1°50}'for a 10°Curve s, -- ! LP.An9.Y3•YO• -- - - - '1 --- -� �f ---- - --- `` �� ------ - --- --- =s/ 10*Cur" O i � J s