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3 <br />/ X00.7. <br />3 <br />17 <br />v <br />.3 f0 0 <br />000, <br />{0 <br />7' <br />s <br />aoilc <br />•� <br />2,7'00 <br />/74af <br />7. <br />ti' <br />C <br />y <br />S <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HO`OV.•-TO USE CURVE TABLES. <br />Table I. _contains Tangents and Externals to a 10 curve. Tan. and <br />Ext. to any other radius may be f ound 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 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.=23* 20' to the R. at Station <br />542-L72. r:r•. <br />Ext. in Tab. I opposite 23° 20' =120.87 <br />120.87 —12 =10.07. Say a loo Curve. <br />Tan. in Tab. I opp. 23' 20' =1183.1 <br />1183.1+10=1f8.31., <br />Correction for A. 230 20' for a 10° Cur. =0.16 <br />118.31+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° 1912'-- def. for sta. 542 J. P.=sta: 542+72 <br />40 4912'•= « « a +50 'Tan. = 1 .18.47 <br />70 19,111 « « a 543 <br />9° 4911= "- " aF80 B. C. =sta. 541 153.53 <br />11* 40' " 543 } L. C.— 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47XX(def. for 1 ft. of 10° Cur.) = 139.41'= <br />2° 19"=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =10 501' for loo Curve. <br />