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:I <br />a F <br />_ CURVE TABLES <br />Published by KEUFFEL & ESSER CO. <br />_ <br />HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />:t. to any other radius maybe found nearly enough, by dividing theTan. <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 />ivide Tan. opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />.ivide 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 />Ext. of twice the given angle divided by the radius of a 1° curve will <br />o <br />f <br />$ <br />7-,F— <br />f <br />S <br />j -74- <br />—�-1 <br />i the Nat. Tana or Nat. Ex. See. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or 1. 1?. =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 />D Q <br />I <br />f 0 0 <br />Tan. in Tab. I opp. 23° 20'= 1183.1 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 />e <br />(If corrected Ext. is required find in same way) <br />Ang.23°20'=23.33 =10=2.3333=L. C. <br />��, ? ate.%✓, <br />2° 192"= def. for sta. 542 L P. = sta. 542+72 <br />4.491'= ° « +50 Tan. = 1 .18.47 <br />7' 191' = u « 543 <br />o /� � u � B. C, sta. 541-1-53.53 <br />11440' = " u « 54+3 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° 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 />:I <br />a F <br />