Laserfiche WebLink
CURVE TABLES. <br />Published by KEZJFFEL 85 ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a V curve. Tan. and <br />Ext. to any other radius may be f ound 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 1° curve will <br />be the Nat. Tan._ or Nat. Ex. See. <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-x-72. <br />Ext. in Tab. I opposite 23° 20Y =120.87 <br />120.87=12=10.07. Say a 10P Curve. <br />Tan. in Tab. I opp. 23° 20'= 1183.1 <br />1188.1+10=118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 <br />118.31-x0.16 =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° 1912' = def. for sta. 542 I. P. — sta. 542-x-72 <br />4° 49'2'= it a " -x-50. Tan. = 1 .18.47 <br />70 19'2' = a a " 543 <br />9° 49'2r a a a cc 54 B. C. = sta. 541 +53.53 <br />110 40' = cc a c` 543 I- L. C. = 2 .33.33 <br />86.86 E. C. — Sta. 543x-86.86 <br />100 —53.53 =46.47 X3'(def. for 1 ft. of 10° Cur.) = 139.41'= <br />2°'19'2'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50'2' for a 10° Curved . <br />10. aurvo <br />