Laserfiche WebLink
i <br />o 7')r <br />Z <br />8 881497 s _ - <br />CURVE -TABLES <br />Published by KEUFFEL & ESSER CO. - <br />8 7 HOW TO" USE CURVE TABLES <br />Table I. contains Tangents and Externals to a 1° curve. Tan: and r <br />Ext. to any other radius may be f ound nearly enough, by dividing &h an. <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 />07 To find. Deg. of Curve,, -having the Central Angle .and External: <br />Divide Ext.'opposlte the given Central Angle by the given External. <br />8 �' 2 �? To find Nat. Tan. and Nat. Ex: Sec. for any angle by -Table I.: Tan. <br />r <br />or Ext. of twice the given. angle divided by the radius of a, l° curve will <br />6 G ' 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 />06 542 -72. <br />d Ext. in Tab. I opposite 23° 20'= 120.87 <br />8 • <br />120.87--12=10.07. Say a 10° Curve. <br />$'1 Tan. in Tab. I opp. 23° 20'=1183.3 <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 />g (� 2° 19 Z' = def. for sta. 542 I. P. = sta. 542 72 <br />4° 49z' Tan. = 1 :18.47 <br />3 !7019'= « " ." 543 <br />B. C.=sta. 541`-{-53.53 = <br />90 4912'= +50 <br />11° 40' = 543+ 1 L. C' 2 .33..33 <br />86.86 E. C..=Sta. 543+86.86 <br />¢� 100-53.53=46.47X3'(def. for 1' ft. of 10° Cur.) =139641'= <br />.l l' 2° 10V=def. for sta. 542. <br />Def. for -50 ft. =2 "30' for a 10° Curve. <br />G Def. for 36.86 ft. =1° 5012' for a 10° Curve. <br />X, <br />Al e4� <br />t0° Curve - <br />- <br />-- <br />/1 <br />