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L <br />STi9/ <br />S f0 <br />t-/O.f�2oi.rAc <br />2 CCoGtte..SF <br />57^AD/9 vr,A7r L <br />Q/S7AN�F BS Y E L <br />FS <br />L <br />t, <br />u. <br />9,1J�� <br />b <br />I <br />6. U, G <br />r JL4, <br />G6f�G 57-e <br />6 <br />3,z a o <br />y 7 <br />r <br />T LC s <br />CURVE TABLES. <br />I Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve. Tan. and <br />Est. to any other radius may be found nearly enough, by dividing the T an. <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 -F72. <br />Ext. in Tab. I opposite 23° 20' =120.87 <br />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 />118.31.0.16 =118.47 =corrected 'Tangent. <br />(If corrected Ext. is required find -in same way) <br />Ang. 2320'=23.33°=10=2.3333=L. C.. <br />2°191'=def. for sta. 542 I: P.=sta. 542-x-72 <br />4' 491' = " (t f;+50 "Can. = 1 .18.47 <br />70 1N'= " " ca 543 <br />949 " « B. C.=sta. 541+53.53 <br />i — " . <br />11° 40' _ " " 5 3-i- L. C. - 2 .33.33 <br />86.86 E. C. = S ta. 543+86.86 <br />100-53.53=46.47X3' (def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 1912-'= def. for sta. 542. <br />-Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curve. <br />M <br />