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3/7/2025 2:20:38 PM
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HN <br />i <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER . CO. <br />HOM7 TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius maybe 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 givggn 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 l° curve will <br />be the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE. <br />r i <br />�f <br />#! Wanted a Curve with an Ext. of about 12 ft. Angle <br />O <br />of Intersection or I. P.=23* 20' to the R: ai- Station <br />iY, <br />542+72. , <br />£1 Ext. iri 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' fora 10° Cur. =0.16 <br />_ 11S.31+0.16 =118.47 =corrected Tangent. . <br />r <br />— <br />4 (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 />40 4921' _ ". +50- Tan. = 1 .18.47 <br />r. 7° = " « « <br />192' 543 B. C.=sta. 541-X53.53 <br />9° 492'= « u �: +50 <br />11 ° 40, :: :: L. C.= 2 .33.33 <br />• 7 _ — :: 543+ <br />.a <br />E. C.=Sta. 543+86.86 <br />I'56.86 <br />100-53-53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 192'=def. for sta. 542. <br />1 Def. for 50 ft. =2° 30' for a 10° Curve. <br />• <br />Def, for 36.86 ft. =1° 502' for a 10° Curve. <br />i <br />
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