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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 may be f ound 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. 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 />he 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.=231 20' to the R. at •Station <br />542+72. <br />Ext. in Tab. I opposite 23° 20'= 120.8", <br />120.87-1-12=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 I. P. =sta. 542+72 <br />4° 495"= +50 Tan. = 1 .18.47 <br />7° 191'= 543 E C.=sta. 541+53.53 <br />o ,__ u « <br />+50 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° 5012' for a 10° Curve. <br />