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3/10/2025 1:37:12 PM
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CURVE TABLES <br />Published by KEUFFEt & ESSER CO.' <br />HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a I' curve. Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing the Tan. <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 `l'angent- <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 />Divide Ext. opposite the given Central Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. See. for any angle by Table I.: Tan. <br />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 />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or 1. P.=23° 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 />1183.1=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 required find in same way) <br />Ang. 23° 20',=23.33'=10=2.3333=L. C. <br />2' 19;'=def. for sta_ 542 1. P. =sta. 542-72 <br />4° 491'— " " " +50 Tan. = 1 .18.47' <br />7° 19j'= " " " 543 <br />- B. C.=sta. 541-{-53.53 <br />V 49j'= " a " +50 _ _ <br />11* 40'= - " " 543 I I C. 2 33.33 <br />86.86 E. C.=Sta. 543-86.86 <br />100 —53.53 = 46.47 XX(deL for 1 ft. of 10° Cur.) =139.41'= <br />2' 191'=def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 10° Curve. <br />Def. for 36.96 it. =1° 50,' for a 10° Curve. <br />
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