Laserfiche WebLink
.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 />x % to any other radius may be found nearly enough, by dividing theTan. <br />Ext: opposite the given Central Angle by fhe given degree of curve. <br />To find peg. of Curve, having the CenQral Ar_gle and Tangent: <br />wide Tan. opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />wide Ext. opposite the given Central Angle.'by the given External. <br />To find Nat. Tan. and at <br />Ex. Sec. for any angle by Table I.: Tan. <br />Ext. of twice the given angle divided by the raduts of a 1° curve will <br />e the Nat. Tan. or Nat. Ex. Se, <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12. ft. Angle <br />of Intersection or I. P.=230 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.1.0° 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 Est. is required find in same way) <br />Ang.23°20'=23.33°=10=2.3333=L. C. <br />2°.191"= def. for sta. 542 1. P.=sta. 542+72 <br />4° 49"= " " " +50 Tan. = 1 .18.47 <br />70 191'= " ". a 543 r <br />x°492 " ° 543 _B. C.=sta. 541+53.53 <br />11° 40' _ +50 <br />" a " 543+ -L. C. = 2 .33.33 <br />86.86 E. C. =5ta. 543+86.86 <br />100_53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= , <br />2° 192'=def. for sta. 542. <br />Def. tor. 50 ft.= 2" 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 502' for a 10° Curve. <br />Jj <br />