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3/10/2025 1:25:11 PM
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CURVE TABLES <br />j published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />'xt. to any other radius maybe found nearly enough, by dividing tbeTan. <br />t 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 />I <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 />�l ivide Ext. opposite the given Central Angle by the given External. <br />i To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />r;, Ext. of twice the given angle divided by the radius of a 1° curve will <br />i+ the Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />j; of Intersection or I. P.=23* 20' to the R. at Station <br />542-72. <br />Ext. in Tab. 1 opposite 23° 20'= 120.87 <br />120.$7 =12 =10.07. Say a l0° Curve. <br />Tan. in Tab. I opp. 23° 20'= 1 L83-1 <br />1183.1=10 <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=1_. C. <br />2° 19,"=def. for sta. 542 L P. =sta. 542-72 <br />4° 49,' = u <� x-50 Tan. — , 1 . 18.47 <br />7° 191,= . « v 543 B. C.=sta. 541+53.53 <br />j 9° 491'= a u +50 2 , 33.33 <br />11 ° 40'= 513+ L. C. _ <br />86.86 F. C. —Sta: 543+86-86 <br />100_53,53=46.47XT(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 it. =1° 50V for a 10° Curve. <br />iV <br />v� <br />
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