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--- 13 aw CURVE TABLES. <br /> - 1 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 /> Ext.to any other radius maybe found nearly enough,by dividing theTan. <br /> u z 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 /> - 7s.s' r.e:"4P Divide Tan.opposite the given Central An le by the given Tangent. <br /> ----- - To find De of Curve havingthe Central Angle and External: <br /> f Divide Ext.opposite the given Cenal Angle by the given External. <br /> -- To find Nat.Tan.and Nat.Ex.Sec.for any angle by Table I.:Tan. <br /> " o �►�or $e �` _ 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 /> w l 9°° z ' - 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 /> � 0 120.87=12=10.07. Say a 10`Curve. <br /> µ N N Tan.in Tab. I opp.23 20'=1183.1 <br /> N 1183.1=10=118.31. - <br /> Correction for A.23°20'for a 10°Cur.-0.16 <br /> ~ 118.31-}-0.16=118.47=corrected Tangent. <br /> r (If corrected Ext.is required find in same way) <br /> Ang.230 20'=23.33 =10=2.3333=L.C. <br /> _ 2°19i'=def.for sta. 542 I. P.=sta. 542+72 <br /> t4 40 4921= a « « +50 Tan.= 1 .18.47 <br /> '\ oa a a <br /> So. .6T� d • 5 kS 14. - 9°4919j'=?2 -_ « a a +543 B. C.=sta. 541-1-53.53 <br /> 1 0 '= a K a +50L.C.= 2 .33.33 <br /> 11 40 543-1- ' <br /> 86.86 E.C.=Sta. 543+86.86 <br /> _ w100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 2°192'=def,for sta.542. <br /> 7- Def.for 50 ft.=2'30'for a 10°Curve. <br /> 788,7 - TX,tiJAP l Def.for 36.86 ft.=1°502'for a 10°Curve. <br /> I a*� <br /> x� <br /> j LP.An9.23.20' <br /> N <br /> 10•Cur" <br /> \?d 00 4A� <br /> 01 r - / <br /> -,A <br />