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3/10/2025 1:07:01 PM
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t <br />CURVE FABLES <br />Published by IKFUFFEL 8 ESSER CO. hf,�q E�--7'" <br />NOW TO USE CURVE TABLES"" <br />Table 1. contains Tangents and Externals to a 1.' curve. Tan. and f' <br />Ext. to any other radius may be found nearly enough, by dividing theTan. <br />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 />Divide Tan. opposite the iven Central Angle b the <br />To find Deg. of Curvve, having the Central Angleeand aExter nal: <br />Divide Ext. opposite the given Central Angle b the <br />To find Nat. Tan.. and Nat. Ex. Sec. for any angle by T bleeI : Tan. , <br />or L' xt. of twice the given angle divided by the radius of a I' curve will <br />be the Nat. Tan, or Nat. Ex. Sec. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about f2 ft. Angle <br />of Intersection or T. P. =23° 20' to the R. at Station <br />1 - 542+72. <br />p Lxt. in Tab. I opposite 230 20' =120.87 <br />120.87-12=10.07. Sa <br />Tan. in Tab. I opp. 23° 20'=11&3.1 E <br />1183.1 <br />10 =118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 g <br />�. 118.3.1-1-0,16=118.47—corrected Tangent- <br />,. 1. <br />(if corrected Ext. is required find in same tivav) <br />i Ang. 23'20'-23.33°-10=2.3333=L. C. <br />20 19y" =clef. for sta. <br />54` 1. <br />I. P. =sta. 542 } 72 <br />4' 492' <br />+50 Tan. = 1 .18.47 � <br />543 - <br />9° 4921* B. C. =sta. 541-x-53.53 <br />1 1 ° 40'= co cc 50 L. C. = <br />543 -}- 2 .33.33 <br />86.86 C. C.=Sta. 543$86.86 <br />100-53.53=16.47X3'(def. for 1 ft. of 10° Cur.) =139.41'= <br />2° 19',-' = def. for sta. 542. <br />i Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 5012' for a 10° Curve. <br />r s*A <br />x� <br />� l.P..kn 9.P3�20• <br />4� iB f <br />9� <br />- I' tOJCu�rO . <br />\W <br />Q�/ <br />
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