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CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW. TO USE CURVE TABLES. <br />?� Table I. contains Tangents and Externals to a 11 curve. Tan. and <br />Ext. to any otherradius may befound nearly enough,bydividing the 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 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 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 1° 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.=230 20' to ,the R. at Station <br />542+72. <br />�* Ext. in Tab. I opposite 23° 20'=120.87 <br />r 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 />118.31+6.16 = 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° MY= def. for sta. 542 I. P. =sta. - 542+72 <br />4° 492' _ cc " +50 Tan. = 1 .18.47 <br />L <br />7° 1941jr = a a <c 543 <br />90 49,' 1 _ tf « B. C.=sta. 541+53.53 <br />+50 <br />11° 40'= " " " 543 { L. C. = 2 .33.33 <br />86.86 E. C. =Sta. 543+86.86 <br />100-53.53=46.47X31(def. for 1 ft. of 10° Cur.) 139.41'= <br />2° 192'=def. for'sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for.36.86 ft. =1' 503' for a 10° Curve. <br />;� ax <br />LP.An9.23 °20'' <br />1 <br />ID• Curve! <br />4Ni ; <br />b i <br />u <br />