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CURVE TABLES. <br />_ Published by KEUFFEL & ESSER CO. <br />�. :.HOW ;T®: USE CURVE TABLES. <br />-Table I. contains Tangents and Externals to a•1° 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 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 />Divide 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 />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. =230 20' to. the R: at Station <br />542-1-72. <br />- Ext. in Tab. I opposite'239 20r =120.87 t <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. foc sta. 542 I. P. =sta. 542+72 <br />40 4911= it" ft +50 Tan. = 1 .18.47 <br />— 70 1911= " " " 43 <br />5 <br />9°491'= " `° -5 R. C. =sta. 541 -53.53 <br />-- 4 11° 40'= " " " 543-1- 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. =20 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curve, <br />. 1 4 <br />