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CURVE TABLES. <br /> A v vp'v� A>� S'� le ` 9.C%r -- Published by KEUFFEL 8s 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 may be f ound nearly enough,bydividing the Tan. <br /> i or Ext.opposite the given Central An le by the given degree of curve. <br /> To find Deg. of Curve, having the Centrsl 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. o <br /> i polite 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 Est.of twice the given angle divided by the radius of a V 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 /> 642-b72. <br /> Ext.in Tab. I opposite 23°20'=120.87 <br /> 120.87=12=10.07. Say a 10°Curve. <br /> + Tan.in Tab. I opp.23°20'=1183.1 <br /> __._--- <br /> 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 /> (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+72 <br /> 4°491'= a n u <br /> -f-50 Tan.= 1 .18.47 <br /> 7°19j'= u u 543 <br /> 9°497'= a u +50 <br /> B. C.=sta. 541+53.53 <br /> 110 40'= a 543+ L. C.= 2 .33.33 <br /> 86.86 E. C.=Sta. 543+86.86 <br /> j - - - — 100-53.53=46.47.X3'(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 /> Def.for 36.86 ft.=10 501'for a 10°Curve. <br /> I've21-2-111 <br /> to <br /> � t0•Gurw _. <br /> _____ <br />