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/y <br /> II <br /> Tf-,v,z CURVE TABLES. <br /> " Published <br /> — by KEUFFEL 86 F.SSER Co. <br /> TO USE CURVE TABLES <br /> a - sow , <br /> C `iso __ 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 /> ;y-,.a, 9r Ext.opposite the given Central Angle by the <br /> geven degree of curve. <br /> To find Deg. of Curve having the Central Angle and Tangent: <br /> S 7�3�,w; /y Divide Tan.opposite the given Central Angle b the iv <br /> 7 - To find DDivide Ext.oppeg. of Curgv�e, having the Central-Angleeaand External: <br /> osi <br /> To find Nat.TdhUs thand Nat.Ex.Se for anygle yan le by given <br /> Table In Tan. <br /> or Ext.of twice the given angle divided by the,radius of a 1°curve will <br /> bethe Nat.Tan.or Nat.Ex..Sec. <br /> y Ex"PLE. <br /> Wanted a Curve with an Ext,of about 12 ft. Anile <br /> �+Intersection or I. P.—230 20' to the R: at Statwn <br /> ` Ext.in Tab. I opposite 230 201-120.87 + <br /> 120.87+12=10.07. Say a 10 Curve. <br /> -- Tan.in Tab. I opp.23°20'=1183.1 <br /> 1183.1-f 10=118.31. <br /> j Correction for A.23"20'for a 10°Cur.—0.16 <br /> 118.31. 0.16—118.47—corrected Tangent. I <br /> (If corrected Ext.is required find in same way) <br /> Aug.23°20'=23.33 +10=2.3333-L.C. <br /> I I� <br /> 20 19j'=def.for sta. 542 I. P.=sta. 542}-72 <br /> 4"49 j'= • a a +50 Tan.= 1 .18.47 <br /> 7° 543 <br /> 9°491'= a a a +50B.C. sta. 541 -53.53 <br /> 110 40'= a • a 543+ L.C.= 2 .33.33 <br /> I 86.86 E.C.=Sta. 543+86.86 <br /> _ 100-53.53-46.47X3'(def.for 1 ft.of 10°Cur.)—139.41' <br /> _... 2°191'=def.for sta.542. <br /> ii Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1'50}'for a 10°Curve. <br /> 1F1ELCP <br /> r <br /> w <br />