Laserfiche WebLink
CURVE TkBLES <br />Published- by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table"L contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius may befound ttearlyenough, bydividing the Tan. <br />.9r'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 />Pe the Nat. Tan. br Nat. Ex. See. <br />EXAMPLE. <br />Whnted a Curve with an Ext. of about 12 ft. Angle - <br />of Intersection or I. P.=230 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' fora IOP ,Cur. =0.16 <br />1.18.31+0.16'-118.47=corrected Tangent. <br />(If'corrected Ext. is required find in sarne way) <br />Ang. 23'20'=23.33° : 10=2.3333=L. C. <br />2° 1.02'=def. for sta. 542 T. P. =sta. 542+72 <br />40 49'77' — " ` +50 Tan. = 1 .18.47 <br />70102,= « 543 <br />o a - « « . BC. = sta 541 { 53.53 <br />901' <br />, = +.50 <br />- <br />11° 40' _ « . rs « 543 { L. C. = 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47x3'(def. for 1" ft. of 10° Cur.) <br />2° 19"=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 />