Laserfiche WebLink
CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />ROW TO. USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve. Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing theTan. <br />or Ext. opposite the given Central Angle by the given degree of curve. <br />To find Deg. of Curve, having the Central Angie 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 V curve will <br />be the Nat. Tan. or Nat. Ex. Sec. <br />EXAMPLE. <br />1i 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'=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° 192"= def. for sta. 542 I. P. = sta. 542,+72 <br />40 49 z'= " " « +50 Tan. = 1 .18.47 <br />7° 19,'' = " " " 543 <br />,_ « « «. B. C.=sta. 541-}-53.53-. <br />9 491 -+50 <br />11° 40'= it 543 I L. C.= 2 .33.33 <br />86.86' E. C. = Sta. 543 x-86.86 <br />100-53.53=46.47XT(def. for 1 ft. of 10° Cur.) =139.41'= <br />2° 19a'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. . <br />De£. for 36.86 ft. =1° 50i' fora 10° Curv& <br />