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- .,CURVE: yTABLES . . <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 maybe found nearly enough, by dividing the Tan. <br />'or Ext. opposite the given Central.Angle by the given degree of curve. <br />l ' 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'l' curve will <br />jbe 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;A. - P."=23° 20' to ""the- R. <br />at. Station <br />542+72.. <br />Ezt. in�Tab:"I opposite 23°-20'=120.87 <br />120.87 .=12 =10.07:.- Say a 10° Curve. <br />Tan. in Tab. Il opp. 23° 20'= 1-183.1 j <br />'1183.1 +'1 t �•'% ;- .f. <br />Correction for A. 23° 20' fora 10° Cur.. =0.16 t- <br />118.31 -F0.16 =118.47 =corrected Tangent. <br />•(If corrected Ext. is required find in sante way) <br />Ang. 23° 20'=23.33°=10=2.3333.„L. C. >- <br />2° 192'=def.-forstL 542” 1. P.=sta. 542 X72 <br />49'' _ +50 Tan. = 1 '.18.47 <br />7°:192,= " " '" = 54.3 <br />B. C: =sta. 2 i 541-}:53.53 <br />9-4912.1= It a -1-50 11 1 <br />11? 40' _ " '.543+,L. C. = 2 .33.33 <br />86.86 . E. ,C. =Sta. <br />543-+86.86 <br />-100-53.53 =46.47X3'(def. for l ft. of 10° Cur:)"=139.41'=. <br />2',19 F = def.'for sta-. 542: _ <br />r- Def: for -50 ft. =2° 30 -for a 10° Curve. <br />Def. for 36.86 ft. =10 502' for a 10° Curve. <br />2340, <br />I <br />I <br />i� <br />