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CURVE TABLES <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a 1' curve. Tan. and <br />-ht, to any other radius may be foun=d nearly enough, by dividing theTan. <br />r 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 />divide Tan.'opposite the given Central Angle by the given Tangent. <br />To find, Deg. of Curve, having the Central Angle and External: <br />?=vide Ext. opposite 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 />Ext, of twice the given angle divided by the radius of a 1° curve will <br />�e the Nat. Tan. or Nat. Ex. Sec. <br />i - 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 />542+72. <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 />1183.1110 =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 />i -- Aug. 23°20'=23.33 = 10=2.3333=L. C. <br />2° 19j"= def. for sta. . 542 1. P. =sta. 542-72 <br />4' 49j'= " " x-50 Tan. = 1 . x8.47 <br />7° 19j'= " 543 B. C.=sta. 541+53.53 <br />90 49j'= G . G +50 <br />543+ L. C. = 2 .33.33 <br />86.86 E. C. = Sta. 543+86.86 <br />100-53.53'•=46.47X3'(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. =1' 50V for a 10' Curve. <br />Cy <br />p' I.P.An9,23.20• <br />.r, <br />l� N , <br />ta4j <br />i0• Gitlwp ' <br />�' O <br />\; *a kAf J <br />� K <br />CI <br />p <br />®N11 <br />