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f3� �, , ��1,77 psi - . CURVE JABEES <br />published by KEUFFEL & ESSER CO. <br />p �, f ` . 72- HOW TO USE CURVE TABLES <br />I Table I. contains Tangents and Externals to a I' curve. .Tan. and <br />Ext. to any other radius may be found nearly enough, by dividing the Tan. <br />1 19 :. 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 />Di <br />vide Tan. opposite the given Central Angle by the given Tangent. <br />To find Deg, of Curve, having, the Central Angle. and External: <br />+yy � �t f Divide Ext. opposite the given Central Angle by the given External. <br />1 - y�%� Al, To find Nat. Tan. and Nat: Ex. Sec. for any angle by Table L: Tana <br />/ J or Ext. of twice the given angle divided by the radius of a 1' curve will <br />zsos',-- be the Nat. Tan. or Nat. Ex. Sec. <br />// t EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or 1. P.=28' 20' to the R. at Station <br />542+72. <br />% Ext. in Tab. I opposite 23' 20' =120.8+7 <br />„ 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 />118.31+0.16 —118.47 =corrected Tangent. <br />(If corrected Ext. is revired find is same way) <br />Ang. 23'20'=23.33 t10=2.3333=L. C. <br />2° 19�'=def. for sta. 542 1. P. =sta. 542+72 <br />4' 49"— " +50 Tan. = 1 .18.47 <br />7' 19J' = " " 543 <br />B. C.=sta. 541+53.53 <br />j 914 _ " " 548 50 <br />L. C. = 2 .33.33 <br />86.86 E. C. =Sta. 543+86.86 <br />f ` _ 1.00-53.53=46.47 X3'(def, for 1 ft. of 10' Cur.)=139.41'= <br />i' 2' 1912'=def. for sta. 542. <br />Def, for 50 ft. =2' 30' for a 10' Curve. <br />Def. for 36.86 ft, =1' 50 J' for a 10' Curve. <br />� o+a <br />]j x/ <br />• r <br />7 <br />- rangy <br />61` <br />,y , <br />