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CURVE <br />7 <br />_TABLES <br />Published by KEUFFEi & ESSER CO.. <br />HOW TO USE CURVE TABLES <br />Tan. <br />Table I. contains Tangents and Externals to a 1` curve. and <br />by dividing theTan. <br />t, <br />tt. to any other•radius maybe found nearly enough, <br />r <br />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 />`•=� <br />�%%,gj <br />�(� �� <br />__ _ <br />Tan. apposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />1 <br />�%Z <br />vide Ext. opposite the given Central Angle by the given External. <br />,3 <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 />9�, / t►d <br />�'C"/ �� <br />the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE <br />Wanted a Curve with an Ext, of about 12 ft. Angle <br />f <br />of Intersection or L P.=230 20' to the R.. at Station <br />542+72. <br />i <br />p 1 <br />f <br />h. <br />Ext. in Tab. I opposite 23° 20' =120.87 <br />120.87-12=10.47. Say a 14° Curve. <br />!� <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. <br />Tangent. <br />118.31--0.16=118.47=6orrected <br />:7rf, <br />(If corrected Ext. is required find in same way) <br />Aug, 23° 20'=23.33°=10=2.3333=L. C. <br />2° 191'= def. for sta. 542 1. P. =sta- 542+72 <br />}+ <br />4' 491' _ " +' •< x-50 Tan. = 1 .18.47 <br />v <br />7° 19j'_ " 543 B. C. =sta. 541+53.53 <br />2 <br />Cri <br />J <br />11° 40'= " 543+ L. C.= .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) <br />{ Ej <br />2° 191'=def, for sla. 542. <br />Def. for 50 ft. =2° 30' fora 10°. Curve. <br />Def. for 36.86 ft. =1° 501'f or a 10' Curve.Ile <br />I <br />O� <br />E <br />_ <br />.P.An9.23'PO� <br />{I( <br />N <br />@; <br />• 10Cur" <br />f <br />y <br />I <br />