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CURVE TABLES <br />_ <br />Published by KEUFFEL & ESSER CO. <br />HOW TA USE CURVE TABLES <br />f, <br />G' <br />i <br />'fable 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 />$$ or Ext. opposite the given Central Angle by the given degree of curve. <br />4 To find Deg. of Curve, having the Central Angle and Tangent: <br />Divide Tan. opposite the givenCentral 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 L: Tan. <br />or Ext. of twice the given angle divided by the radius of a 16 curve will <br />be the Nat. Tan. or Nat. Ex. See. <br />" <br />I,' .� <br />EXAMPLE <br />Wanted a Curve with an Ext, of about 12 ft. Angle <br />of Intersection or, I..[ . =231 20' to the R. at Station <br />1542}-72. <br />— <br />Ext. in Tab: I opposite 23° 20'=120.87 <br />[ 120.87 -12 =10.07. Say. a 10° Curve.. <br />t <br />; <br />I <br />I <br />Tan, in Tab. I opp, 23' 20'=1183-1 <br />1183.1=10=118.31. <br />V <br />Correction for A. 23° 24' fora 10° Cur. =0.16 <br />1118.31 <br />+0.16 =118.47 -corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />w A ng. 23' 20'=23..33'-- 10=2.3333=1.. C. <br />4; ,I <br />2°191'=def. for sta. 542 J. P.=sta. 542-72 <br />4v 4911= , a r� +50 Tan. = 1 •.18.47 <br />_ <br />_ <br />b,= u <br />7° 19, 543: B. C. St 2. 541+53.53 <br />9' 49 . = +50 <br />L. C. = 2 .33.33 <br />119 40'= " `° °° .543+ <br />k. <br />86.86 E. C. — Sta. 543 X86.86 <br />` 100-53.53=46.47x31(def. for "I ft. of 10° Cur.) =139.41'= <br />1 I 2° 19," def. for sta..542. <br />DeL' for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. —1° 50;' for a 16° Curve. <br />LP.An9.23 •@0• <br />' <br />- <br />- - <br />p hq eQt <br />a <br />