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CURVE TABLE -S. <br />Published by-KEUFFEL 8s ESSER CO. — <br />HOW TO USE CURVE. TABLES. <br />Table 1. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius may be found nearly enou gh, by dividing the Tan. <br />it Ext. opposite the given Central Angle by tbe;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 L: Tan. <br />)r 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.abcut 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. Sziy 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. i- required find in sane way) <br />Ana. 23° 20'=23.33° : 10=2.3333=1... C: <br />2° 19 def. for sta. 542 1: P. =sta. 542-72 <br />4° 491' _ . It" it +50 Tan. = f .18.47 <br />70 191'= « 6 it 543 <br />9° 49,'= " i 50 B. C. =sta. 541-53.53 <br />110 401= " fA 40 643+ - L. C. = 2 .33.33 <br />86.86 F. C. =Sia. 5443+86.86 <br />100--33.53=46.47X31(def. for 1 ft. of 10° Cur.)=139.41'= <br />o°' 19,' =def. for sta. 542. <br />Def. for 50 ft: =2° 30' for a 10° Curve. <br />Def, for 36.86 ft. =1° 5M' for a 10° Curve: <br />