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WA <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />ow TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a,1'curve. ~Tari. And <br />Ext. to any other radius may befound nearly enough,, bydividing the Tan. <br />or Ext. opposite the given Central "Je`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 />i To find Nat. Tan. and Nat. Ex. See. for any angle by Table L: 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. See. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of ahout 12 ft. Angle <br />of Intersection or 1. P.=23* 20' to the . R. at Station <br />642+72. <br />Est, 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 />115.31+0.16 = 118.47 =corrected Tangent. <br />(If corrected Ext. is required %nd in same way) <br />Ang. 23° 20'=23.33'-- 10 =2.3333 = L. C. <br />2' NY=dcf. for sta. 542 L P.=sta. 542-{ 72 <br />4' 49 V = K CP +50 "Tan. = 1 .18.47 <br />7' 191' = a or <i 543 <br />90 491,= « « « +50 B. C. =sta. 541+53.53 <br />11' 40'= 543+ L. C. = 2 .33.33 <br />86.86 E. C.=Sta, 543+86,,6 <br />100--53.53=46.47XT(def. for 1 ft. of 10' Cur.)=139.41'= <br />2°.191'=def. for sta. 542. <br />Def. for 50 ft. =2' 30' for a 10' Curve. <br />Def. for 36.86 ft.=1' 50',' for a 10' Curve. <br />