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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 1°curve. Tan.and <br /> -- r- Ext.to any other radius maybe found nearly enough,by dividing 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 /> _ -- _-- -_ 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 V curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> ---- -- -—- S -- -" - --- <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 /> - 542+72. <br /> 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.23°20'=1183.1 <br /> 11831+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 /> (If corrected Ext.is required find in same way) <br /> - - --- - -- --- - - ._ Ang.23°20'=23.33°=10=2.3333=L.C. <br /> 2°19Y=def.for sta. 542 I. P.=sta. 542+72 <br /> -- -- - - - __ -- 4° 191,- a a a +50 Tan.= 1 .18.47 <br /> 70 19j'= " " " 543 <br /> -- - - 9°493'-_ a a a +50 B.C.=sta. 541+53.53 <br /> 11°40'= " a a 5,3+ L.C.= 2 .33.33 <br /> 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 /> Def.for 50 ft.=2°30:for a 10°Curve. <br /> 2"✓ o y 2 Def,for 36.86 ft.=10 50}'for a 10°Curve. <br /> --- - <br /> ♦t <br /> 7. 30 �o.�� px�� <br /> 13 <br /> - - - - - - _—_ 10•Curve <br /> � \X <br /> 4 el <br /> r <br />