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CURVE TABLES <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE <br />,�. CURVE TABLES <br />1 Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />4 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 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 V 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 I. P.=230 20' to the R. at Station <br />542+72. <br />Ext. in Tab. I o polite 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' 191'= def. for sta. 542 1. P. =sta. 542+72, <br />4- <br />42+72- <br />4- 49j'= +50 Tan. = 1 . 18.47 <br />o r_ u " <br />543 <br />91 491' = " +50 B. C. =sta. 541+53.53 <br />53.53 <br />11 ° 40' = " " " 543+ L. C.= 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47X31(def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 19`=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 />