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--- - s 4 "/1f. �, - - CURVE TABLES. ' <br /> _--_ f L A Published by KEUFFEL & ESSER <br /> HOW TO USE CURVE TABLES?`•7 9 <br /> Table I. contains Tangents and Externals to a 10 curve. Tan.and '71.37 <br /> - Ext.to any other radius may be found nearly enough,by dividing the Tan. -75- <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: o <br /> Divide Tan.opposite the given Central An le by the given Tangent. <br /> --�1 1 3. 7 - 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 1°curve will <br /> be the Nat:Tan.or Nat.Ex.Sec. -74,74 <br /> EXAMPLE. Lt.J-$ <br /> -- -- Wanted a Curve with an Ext.of about 12 ft. Angle 7�I � <br /> of Intersection or I. P.=23° 20' to the R. at Station <br /> 542-}-72'. <br /> Ext.in Tab. I opposite 23°20•=120.87 k, z <br /> ---- - <br /> 120.87+12=10.07. Say a 10°Curve. <br /> Tan.in Tab. I opp.23°20'=1183.1 (• s <br /> 1183.1=10=118.31. <br /> 6 4. 7 9 <br /> Correction for A.23°20'for a 10°Cur.=0.16 �g <br /> - - -- - <br /> 118.31+0.16=118.47=corrected Tangent. <br /> (If corrected Ext.is 4required find in same way) c 3 <br /> Ang.23°20'=23.33 -t-10=2.3333=L.C. ° <br /> 2°19Y=def.for sta. 542 I. P.=sta. 542+72 <br /> 4°491'= +50 Tan.= 1 .18.47 <br /> 7°19+)'= a u u 543 <br /> B.C.=sta. 541+53.53 7J.3� <br /> 90 49Ij'= a u u +50 - <br /> - 11°40'= u a rte+ L. C.= 2 .33.33 <br /> 86.86 1 E.C.=Sta. 543+86.86 <br /> 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= J°•/ <br /> 2°19V=def.for sta.542. <br /> --._— --- - -- - - - -- --- - - Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°501'for a 10°Curve. =_•�_ <br /> � I.P.An9.23°!075.3'L <br /> N 7 <br /> 10°Curve <br /> _ r <br />