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� 1 CURVE TABLES. Published by KEUFFEL 86 ESSER CO. HOW TO USE CURVE TABLES. Table I. contains Tangents and Externals to a 1°curve. Tan.and - - _- - - - -'- - - 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 ��' - -- ----- _ To find Nat.Tan.and Nat.Ex.Sec.for any angle by Table I.:Tan. or Ext.of twice the given angle divided by the radius of a 1°curve will 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.=230 20' to the R. at Stationk -- _---0 _------ __ ------. - --- 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--418.47=corrected Tangent. ---� --- - _a- - a_- _._ _ _- (If corrected Ext.is required find in same way) , Ang.23°20'=23.33 =10=2.3333=L.C. e 2°19F=def.for sta. 542 I. P.=sta. 542+72 - -- 4°492"_- ro a a +50 Tan.= 1 .18.47 - -_- me rz 7°19,'= �c a a 543 s 90 4911= ro a a 5 B. C.=sta. 541+53.53 --- - 110 40 - it a a 543+ L.C.= 2 .33.33 - -_ 86.86 E.C.=Sta. 543+86.86 - - 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= _ 2°191'=def.for sta.542. --- Def.for 50 ft.=2°30'for a 10°Curve. -- ---- Def,for 36.86 ft.=1°50j'for a 10°Curve. a, la curve -- ---_.------- ----- 4 \X VL % t ' I ` A -