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u <br />Q <br />7s•AkL <br />zo� ?.r I / i <br />CURVE TABLES <br />Published by KEUFFEL & ESSER CO. <br />`HOW TO USE CURVE TABLES b <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 the Tan. <br />I of Ext. opposite the given Central Angle by the given degree. of curve. <br />, • To find Deg. Curve, <br />of 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 th'e 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, l° 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 />120.87-1,12'—_10.07. ' Say a 10° Curve. <br />Tan. in Tab. I opp. 23.° 20'= 1183.1 <br />1183.1'=10=118.31. j <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° 4912' = , +50 Tan.'= 1 .18.47 <br />2- 543 <br />9°'4912'= " ' +50 B.'C.=sta: 541+53:53 .. <br />11° 40' _ " 543+ L. C. = ®' 2 .33.33 <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°• 1912' =def. for sta. 542. <br />- - - Def. for 50 ft. =2° 30' for -a'30° Curve. <br />Def. for 36.86 ft. 1* 5012' for a 10° Curve. <br />ax <br />>e <br />IAAn9.23* 20' <br />P� <br />�. to° Curve . <br />:RA <br />