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- 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 befound 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 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., Sec. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P. =231 201. 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' Cine. <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 hi same way) <br />Ang.23°20'=23.33'-10=2.3333=L.. C. <br />20191'=dcf.forsta:. 542 1. P.=sta. 542+72 <br />4° 4921'= a +50 Tan..= 1 .18.47 <br />70 191' 543 <br />9' 491, „ B. C. =sta. 541+53.53 <br />+50 <br />110 40' _ 543+ I'' C. = 2 .33.33 <br />86.86 . E. C. =Sta. 543+86.86 <br />100-53.53=4(;.47X3'{def. for 1 ft. of 10' Cur.) =139.411= <br />2° 19" =,def, for sta. 542. <br />Def. for 50 ft. =2° 30' far a 10° Curve. <br />Def, for 36.36 ft. =1° 50,' for a l0° Curve. <br />Zv <br />LP.An9.23 °20' <br />P� V e43 <br />- 16°curve <br />4-0 f <br />X <br />A <br />