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60_ yZ 6 <br />N hn, N <br />A <br />Zia <br />79 <br />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 />.d� <br />y <br />W <br />-1� <br />I <br />Ext. to any other radius maybe found nearly enough, bydividing 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 10 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 />o <br />2 /R0N <br />542+72. . <br />Ext. in Tab. I opposite 23' 20' =120.87 <br />120.87+12=10.07. Say a 10° Curve. <br />F_. <br />Tan. in Tab. I opp. 23'21Y=1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 230 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° 194'= def. for sta. 542 I. P. =sta. 542+72 <br />40 4912'= '° " " ; +50 Tan. = 1 .18.47 <br />o r= « �a a <br />90491'= " a +5o B. C.=sta. 541+53.53 <br />110 40'= <' " " 543+ L. C.= 2 .33.33 <br />11 <br />I <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47XT(def. for 1 ft. of 100 Cur.) =139.41'= <br />2° 192 def. for sta. 542. <br />V <br />Def. for 50 ft. =20 30' for a 100 Curve. <br />'Def'. for 36.86 ft. =10 50;' for a 100 Curve. <br />