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,.� I /Oj/ , <br />0/7 .17& <br />q 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 />l( Ext. to any other radius may be found nearly enough, bydividing 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 />7j Divide Tan. opposite the given Central Angle by the given Tangent. <br />C To find Deg. of Curve, having the Central Angle and External: <br />Divide Ext. opposite the given Central Angle by the given External.�3 <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. <br />0 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 G �� <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 230 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 />118.314--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° 192"= def. for sta. 542 I. P. =sta. 542+72 <br />40 4912'= 11 +50 Tan. _ .1 .18.47 <br />70 19,'= " " " 543 <br />9°4911'= " " it +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 = 46.47 XT(def. for 1 ft.•of 10° Cur.) =139.41'= <br />2° 1911'=def . for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10* Curve. <br />Def. for 36.86 ft. =1° 50,' for a 10° Curve. <br />