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V C� i /.i.rc].c.a Ot. <br />D /0 s 13 <br />-,4 S <br />1 1 I <br />4 01-79 0, C79 97/ ..91 � �r,�is� <br />9..3.0 <br />11 <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 may be 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 />or Ext. of twice the given angle divided by the radius of`a 19 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 />542-72. <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'= 1193.1 <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 saine way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C. <br />2° 1921'= def. for sta. 542 I. P. =sta. 542-x-72 <br />4049 <br />° 492'= It +50 Tan. = 1 :18.47 <br />.7° 192'= " it " 543 <br />9 49a"= " " +50 B. C. =sta. 541+53.53 <br />110 40t= " " ", 543+ L. C.= 2 .33.33 <br />86.86 E. C.=Sta. 543-86.86 <br />100-53.53=46.47X31(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° 502' for a 10° Curve. <br />@4 <br />dy <br />I.P.An9.23 °20 _ <br />V - <br />i0 j <br />Ali! <br />