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-=CURVE TABLES <br />— i _ - Published by KEUFFEL & ESSER CO. <br />HOW TO USE NVE TABLES <br />')?able I: 'contains and Externals to a 1° curve. Tan. and <br />Ext. to any other radius may be found nearly enougivn degree of curve. <br />or Ext. opposite the e given Central Angle by the ge <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 L. 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 />— IEXAMPLE <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 />Ang. 23° 20'=23.33°=10=2.3333=L. C. . <br />2°'19;'=def. for sta. 542 'I. P.=sta. 542+72 <br />4° 49;it it' _ " +50 Tan. = 1 .18.47 <br />_ ?° 19Z, - " " 543. .B: C. sta. ..541+53,53 , <br />9149 <br />`-, /= u ° .... +.50 2 .33.33 <br />_ 11° 40' = 543+. u « u . <br />L. C. _ - <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° 192'=def. for sta. 542.' <br />I -lief. for 50 ft.=2' 30' for a 10' Curve.— <br />Def. for 36.86 ft.=1° 50;' for a 10° Curve. <br />S4A . <br />--`� LP.An9.23°20 <br />