Laserfiche WebLink
�- <br />J- 1 <br />c <br />I <br />I <br />I <br />I <br />I <br />ti I <br />vl� <br />y <br />' <br />I <br />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 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 thegiven 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. =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. Say 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. is required find ,in same way) <br />Ang. 23° 20'=23.33* =10=2.3333 = L. C. <br />2° 192' = def. for sta. 542 1. P. = sta. 542+72 <br />4° 492`= " " +50 Tan. = 1 .18.47 <br />0 2r_ u " <br />9 49 't" " +50 B. C. sta. 541+53.53 <br />53.53 <br />-21' <br />110 40,= " it " 543+ L. C.= 2 .33.33 <br />86.86 E. C. = Sta. 543+86-86 <br />100-53.53=46.47X3'(de.f. for 1 ft. of 10° Cur.)=139.41'= <br />2° 10"—def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50:'.-' for a 10° Curve. <br />