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CURVE TABLES. <br />Published by KEUFFEL 8s ESSER 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 maybe found nearly enough, by dividing the Tan. <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 />y N or Est. of twice the given angle divided by the radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. Sec. <br />36 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 />a 542+72. <br />Ext. in Tab. I opposite 230 20' =120.87 <br />120.87--.12=10.107., Say d10' 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.23°20'=23.33°=10=2.3333=L. C. <br />2° 199"= def. for sta. 542 I. P. =sta. 542-x-72 <br />40 49" _ " " " +50 Tan. = 1 .18.47 <br />701911— f/ "it 543 <br />90 49,' _. " { 50 B. C. =sta. 541-x--53.53 <br />11° 40' _ " " « 543 L. C.= 2 .33.33 <br />86.86 E. C. =Sta. 543+86.86 <br />1 100-53.53=46.4' X3'(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.= 1' 50;' for a 10° Curve <br />