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:9 <br />.'._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 />Ext. to any other radius maybe found nearly enough, by dividing theTan. <br />or Ext. opposite the given Central Angle by the given degree of curve. <br />I 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 thd 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-. 1. P. =230 201 to the R. at Station <br />642+72. <br />Ext. in Tab*. I opposite 230 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 />Correction for A. 23' 20' for a 10° Cur: =0.16 <br />118.31+0.16 =:418.47 =corrected Tangent. <br />(If corrected Ext. is required find in sanie way) <br />Ang.23*20'=23,33,=10=2.3333=L.C. - <br />2* 19Y= def. for sta. 542 1. P. sta. 542+72. <br />491'= . . . +50 Tan. = 1 .18.47 <br />70 191,1---: a <br />543 E�. C. =,t.. 541+53.53 <br />+50 1 2 .33.33 <br />11°40' 5u+ L. C.= <br />86.86 :E. C. Sta. f 543+86.86 <br />100-53.53 =46.47X3'(def. for I ft. of 10' Cur.') =139.41'= <br />20 191V=def. for sta.542. <br />Def. for 50 ft. =2° 30' fora 10° -Curve. <br />Def. for 36.86 ft. =1° 501' for a 10* Curve— <br />ILI <br />)0_ e'k <br />2 1 <br />\0 Z <br />r <br />-b <br />A <br />Ot <br />0 <br />:9 <br />.'._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 />Ext. to any other radius maybe found nearly enough, by dividing theTan. <br />or Ext. opposite the given Central Angle by the given degree of curve. <br />I 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 thd 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-. 1. P. =230 201 to the R. at Station <br />642+72. <br />Ext. in Tab*. I opposite 230 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 />Correction for A. 23' 20' for a 10° Cur: =0.16 <br />118.31+0.16 =:418.47 =corrected Tangent. <br />(If corrected Ext. is required find in sanie way) <br />Ang.23*20'=23,33,=10=2.3333=L.C. - <br />2* 19Y= def. for sta. 542 1. P. sta. 542+72. <br />491'= . . . +50 Tan. = 1 .18.47 <br />70 191,1---: a <br />543 E�. C. =,t.. 541+53.53 <br />+50 1 2 .33.33 <br />11°40' 5u+ L. C.= <br />86.86 :E. C. Sta. f 543+86.86 <br />100-53.53 =46.47X3'(def. for I ft. of 10' Cur.') =139.41'= <br />20 191V=def. for sta.542. <br />Def. for 50 ft. =2° 30' fora 10° -Curve. <br />Def. for 36.86 ft. =1° 501' for a 10* Curve— <br />ILI <br />