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�Sv <br />�. <br />• <br />yrb: ' <br />6 <br />mac <br />c— <br />I <br />O <br />' <br />—d _ <br />3 it ; <br />r� <br />'V6 <br />w 1 <br />' 1 <br />z'r <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 />Ext. to any other radius may be 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. 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 'Intersectiou.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° 2(' =1183.1 S <br />IJ 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: r �' <br />(If�corrected F_xt. is required find in same way) <br />Aug. 23° 20'=23.33'-10 =2.3333 =L. C. <br />2°19'=def. for sta. 542 I.P.=sta. ' 542-1-72 <br />40 4921'— " if '0 +50 Tan. = 1 .18.47 <br />70 1912'= F" " rc 543 Jd <br />° , . « B. C.'= sta. 541 X53.53 — <br />90491, <br />49 1z- _ " [I -1-50 <br />11° 40'= " rc " 543+ L. C.= . 2 .33.33 <br />- 86.86 E. C.=Sta. 543+86.86 . <br />100 —53.53 = 46.47 X3'(def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 19 z'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.56 ft. =1° 50.1.1-.' for a 10° Curve. yq <br />