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© 6� <br />a <br />1)Z' <br />L <br />— — X <br />r <br />CURVE TABLES. <br />Published by.-KEUFFEL & ESSER CO. - <br />HOW;,TO;USE CURVE TABLES. <br />Table'L' contains Tangents and Externals to a•10 curve!'* Tein. and <br />Ext., <br />xt. to any other radius may be found nearly enough, bydividing• fltle Tan. <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 T'aagent: <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 Extemal. <br />To find Nat. Tan. and Nat. Ex. Sec. for any angle by,,TableJ.: Tan. <br />or Est. of twice the given angle divided by the radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. See.-, . <br />-EXAMPLE: <br />Wanted a Curve with an Ezt. 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 10P Curve. <br />Tan. in Tab. I'opp. 23° 20'=1183.1 <br />• �' 1183.1=:10=118.31.- <br />C6rrection for A. 23° 20' for a"'101' Cur. =0.16. <br />118.31-}•0.16='118.47 ' <br />,=correcied Tangent. ' <br />(If corrected Ext: is -required find insame way) <br />Ang.23°20'=_23.33=10=2.3333=L. C.° <br />2°19;'=def. forsta' ' 542• I. P.=sta. 542+72 <br />40491;_' +50 Tan. 1 .18:47 <br />t 7° 19j'= u « ' 543 . <br />9°49;'=+50 <br />11° 40' _' . " u 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° Cuc.) =139.41'= <br />20 19;'=def. .for sta.542. ' <br />Def: for, 50 ft. =2° 30' fora 10°Curve. <br />Def. for 36.86 ft. =1° 50F for a 10° Curm <br />x� <br />d I.P.An9.23'EO% <br />