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J l <br />'CURVE TAELE9�7-- <br />�j Published, by KEUFFEL ESSER CO. <br />is SOW 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 found nearly.enough, bydividing the T an. <br />or.•E, xt. opposite the given Central Angle by the given degree of curve. <br />-To find Deg. of Curve, having the Central Angle and Tangent: <br />Divide Tap. opposite the given Central Angle by the given Tangent. <br />.To find Deg: of Curve, having the Central Angle and External: <br />— i Divide Ext. opposite the given Central Angle by the given External. <br />To,find Nat. Tan. and Nat. Ex. •See. f or any angle by Table L: Tan. <br />--=i 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 />—i{ EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angl^ <br />of Intersection or 1. P. =230'. 20' to:.the 'R.. at Station <br />al 542+72. <br />I:xt. in Tab.'' -I opposite 23° 20• =120.87 <br />120.87 -12 p 10.07. • Say a 10° Curve. <br />'Tan. in Tab. I opp. 23° 20`=1183.1 <br />1183.1=10=1,18.31: <br />f Correction for A. 230 20'' for a 100 Cur. =0.16 <br />+• 118.31-1-0.16 =118.47=corrected Tangent. <br />(If corrected Ext. is required find in Baine way) <br />An-. 23020'=23.33'=10=2.3333=L. C. <br />20191'=def. for sta. 542 I, P.=sta. 542+72 <br />40 492" = « " -1-'50 Tan. = 1 .18.47 <br />70 19'= " '" 543 13. C.—sts. 541-{-53.5.3• <br />� :. 9049x'= « «• « +50 <br />9'. 2 .33.33 <br />i( .110 40' _ u .< 543:+ <br />86.86 E.' C. = Sta. 543 -86.86 <br />100-53.53=40.47X3'(def. for 14t. of 100 Cur.) 139.41'= <br />2° IV=def. for std. 542. <br />' Def. for 50 ft. =2'80' for a 10° Curve. <br />Def. for 36.86 ft.= 1* 50'7' for a 100 Curve. <br />Sq <br />g <br />10' Curve <br />r �� i <br />