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..CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />1...: <br />HOW TO USE CURVE TABLES.:. <br />-,Tablb I. contains Tangents and Externals to a V curve. Taii. and <br />Ext. to any other radius may be found nearly enough, by dividing theT 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 Ankle 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 0. Tan. and Nat. Ex. See. 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 1. P. =230 201 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° Curve. <br />Tan. in Tab. Lopp. 236 20' = 1183.1 <br />1383.1=10=118.31. <br />Correction f6k A. 23* 20' for a 10P Cur. =0.16 <br />118.91+0.,16 = 118.47 = corrected Tangent. <br />(If corrected Ext. isrNuired find in same way) <br />'Ang. 23* 20'=23.33=10=2.3333=L.' C. <br />2* 19i'—def., for sia". 542 • I. P. = sia. 542+72 <br />40'49' = +50 Tan. = 1-18.47 <br />7o 19j'=If' 543 <br />9- 491'_� . +50 B.C. sta.. 541+53.53 <br />110 40f 543+ L. C. = 2..313-33 <br />86-86 E. C. =Sta.- 543+86.86 <br />100 —53-53 46.47,X3'(def. for I ft. of 10* Cur.) 139.41'= <br />2*49-21' = dd:,f6r sta. 542. <br />Def. for'50 ft. =2* 30'fora 10* -Curve. <br />Def. for 36.86 ft.,= lo 50j' for a 10° Curve. <br />