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% -30 <br /> d <br /> y <br /> CURVE TABLES. <br /> Published by KEUFFEL Ss 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 maybe found nearly enough,bydividing theTan. <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 /> - - No 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 /> - 8 --- or Ext.of twice the given angle divided by the radius of a 1°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.=230 20' to the R. at Station <br /> ------------ - <br /> - 542+72. <br /> Ext.in Tab. I opposite 23°20'=120.87 <br /> -------- <br /> 120.87-. 12=10.07. Say a 10°Curve. <br /> Tan. in Tab. I opp.23°20'=1183.1 <br /> -- 1183.1=10=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.16 <br /> - --- <br /> -- - 118.31-}-0.16=118.47=corrected Tangent. � <br /> - <br /> (If corrected•Ext.is required find in same way) <br /> -------- Ang.23°20'=23.33°=10=2.3333=L.C. <br /> '-_ <br /> 2°19 def.for sta. - 542 I. P.=sta. 542+72 <br /> u j,— a a a _ <br /> 79°49 = « a n +50 Ba C=st. 541+53.53 <br /> FP,s -- - Z �e 11°�,_ « cr u 543+ L.C.= 2 .33.33 <br /> --- -- d8. 61_ <br /> Q-r�� o `� -�-_-_ 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°19Y=def.for sta.542. <br /> - Def,for 50 ft.=2*30'for a 10°Curve. ' <br /> -z - _ Def,for 36.86 ft.=1°50}'for a 10°Curve. <br /> -- -- - - "> <br /> ! I.P,An9.23'20' <br /> 10•Curve <br /> ------ ------ 4 <br /> • <br /> 1 <br /> r <br /> i e <br /> or, <br /> w <br /> ".( <br />