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_ CURVE TABLES. <br />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 />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. 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 />_ <br />120.87--.12=10.07. Say a 10° Curve. <br />Tan. in Tab. I opp. 23° 20'=1183.1 <br />— 1183.110=118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 <br />— <br />118.31+0.16=118.47 =corrected Tangent. <br />_ (If corrected Ext. is re uired find in same way) <br />Ang. 239 20'=23.33 <br />=10 =2,3333 = L. C. <br />_ 2° 19F=def. for sta. 542 I. P. =sta. 542+72 <br />4° 492"= " " " +50 Tan. = 1 .18.47 <br />7° 19;' _ " 543 <br />90 4911' _ " a is +60 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=46,47X3'(def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 19z'=def. for sta. 542. <br />— I Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1* 50;' for a 10° Curves <br />x <br />