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CURVE TABLES. <br /> j Published by KEUFFEL & ESSER CO. <br /> C11,.ss <br /> HOW TO USE CURVE TABLES. <br /> Table I. contains Tangents and Externals to a 1° curve. Tan.and <br /> To find Deg. of Curve, having the Central and Tangent: <br /> -- - g <br /> - curve. <br /> or Ext.opposite the given Central Angle by the given degree f cue. <br /> Ext.to an other radius maybe found nearly enough,b dividing theTan. <br /> - - _- Divide Tan. opposite the given Central Angle by the givenTangent. <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. <br /> \ — - or Ex Nat. oe Nat.Ex. le divided b rthe radiuls of a 10 curve ll <br /> X g' g Y r <br /> EXAMPLE. <br /> 48 <br /> 9 _ Wantgd a Curve with an Ext.of about 12 ft. Angle <br /> of Intersection or I. P.=23* 20' to the R. at Station <br /> 542+72. <br /> X Ext.in Tab. I opposite 230 20'=120.87 <br /> 120.87--. 12=10.07. Sa a 10 Curve. <br /> - - - <br /> Tan.in Tab. I opp.23°20'=1183.1 <br /> y 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 /> -- <br /> (If corrected Ext.is required find in same way) <br /> - - -- -- Ang.23°20'=23.33°=10=2.3333=L. C. <br /> 2°191'=def.for sta. 542 I. P.=sta. 542-x-72 <br /> 4°49-1 = a a a +50 Tan.= 1 .18.47 <br /> 7°191'= a a a 543 <br /> ------ -_ B.C.=sta. 541 53:53 <br /> 90 493/= ra a a +50 <br /> 110 40,= a a a 543{ L.C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543+86.86 <br /> -- -- 100-53.53=46.47X31(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 /> �l TPDef.for 36.86 ft.=I'501'for a 10°Curve. <br /> _ a <br /> � •fix <br /> •� -� 1, 7�' q/ <br /> •�-Ct-x+43- s _ �!i -.Gl/'l.._- >A IAAn9.231201 <br /> --- - - - t0•Curve <br /> p d/ <br /> 0/ <br /> i <br /> J , . <br /> Y • <br /> r <br /> 1 <br />