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CURVE. TABLES.' - <br />Published by KEUFFEL 8r. ESSER CO. <br />Z- Z-1_ <br />HOW TO , USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a " curve. Tan. and <br />- Ext. to any other radius maybe 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 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 I. P. =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 />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 />Aug. 23°20'=23.33' 10=2.3333=L. C. <br />2°19'21=def. for sta. 542 I.P.=sta. 542+72 <br />_ <br />V 492"= " " +50 Tan. = 1 .18.47 <br />- <br />--- 70 193'= " " " 543 53.53 <br />B. C. =sta. 541+53.53 <br />90 49'2'- " " a - +50 <br />110 401= " " t/ 543+ L. C. = 2 .33.33 <br />86.86. E. C. =Sta. _ 543+86.86 <br />- -- 100-53.53=46.47X3'(def. for l ft. of 10° Cur.)=139.41'= ; <br />2° 19'2'=def: forsta. 542. <br />- Def, for 50 ft. =2° 30' for a 10° Curve. <br />_ Def. for 36.86 ft. =1° 501' for a 10° Curved <br />