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_ _ + CURVE TABLES. <br /> Published by KEUFFEL & ESSER CO. <br /> II HOW TO USE CURVE TABLES. <br /> 11 ' Table I. contains Tangents and Externals to a 10 curve. Tan.and <br /> Ext:to any other radius maybe found nearly enough,bydividing the Tan. <br /> - or Ext.opposite the given Central An le by the given degree of curve. <br /> / 1' To find Deg. of Curve, having the Central Angle and Tangent: <br /> y i Divide Tan. opposite the given Central An le 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.See.for any angle by Table I.:Tan. <br /> or Ext.of twice the given angle divided by the radius of a P 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.=23° 20' to the R. at Station <br /> - -- - -- -- 542+72. <br /> A Ext.in Tab. I opposite 23'20'=120.87 ` <br /> r,� } 120.87+12=10.07. Say a 10`Curve. <br /> — - - — - Tan.in Tab. I opp.23°20'=1183.1 <br /> 1183.1=10=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.16 <br /> g 118.31+0.16=118.47=corrected Tangent. <br /> (If corrected Ext.is required find in same way) <br /> Ang.23 20 =23.33 T 10=2.3333=L.C. <br /> 2°19j'=def.for sta. 542 I. P.=sta. 542+72 <br /> 404911= a a a +50 Tan.= 1 .18.47 <br /> 7°1911= a a a 543 <br /> 90 49,J'= (f a a +50 B. C.=stn. 541+53.53 <br /> 11°40'= a • a L.C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543+86.86 <br /> _y 100-53.53=46.47x3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 2°191'=def.for sta.542. <br /> r� Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°50j'for a 10°Curve. <br /> ---- <br /> ' d <br /> i - lo•Curve <br /> a <br />