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/57 <br />.° <br />�A <br />ni 0 <br />N <br />.o1 <br />µ <br />zr- <br />o <br />C+i <br />09 <br />� <br />17 <br />q <br />� <br />� <br />o <br />M <br />- <br />,tyoa <br />A <br />des <br />4° <br />- <br />v' <br />4 <br />7 s <br />- <br />-CURVE TABLES. <br />Published by KEUFFEL & E SSER CO. <br />HOW 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 found nearly enough, by dividing the Tan. <br />I 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 />1 Divide Tan. opposite the given Central Angle by the given Tangent. <br />1.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. See. for any angle by Table I.: an. <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 ;bout 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 110` 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 16' Cur. =0.16 <br />(( <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=2.5333=L. C. <br />2° 191'=def. for sta. 542 I. P. =sr 542 X72 <br />4049121= " " " +50 Tan. = 1 .18.47 <br />70 1921'= �, " « 543 <br />9° 49-'-'= " `° " +50 R. C. = sta- 541E 53:53 <br />2 L. C. = 2 .33.33 <br />110 40'= " " " 543+ <br />86.86 E. C. =Eta.. 543+86.86 <br />100-53.53=46.47X3'(def. fort ft. of '0° Cur.) = 139.41'= <br />2° 1912'=def. for sta. 542. <br />Def, for 50 ft. =2° 30' for a 1.11° Curve. <br />Def. for 36.86 ft. =1° 502' for a 10° Curve. <br />