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CURVE TABLES <br />Published by KEUFFI L & ESSER CO. <br />J i HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a I° curve. Tan. and <br />kt. to oily other radius may be found nearly enough, by dividing the Tan. <br />r 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 />`--`—livide Tan. opposite the given Central Angle by the given Tangent. <br />To find Deg. of Curve, having the Central Angle and External: <br />(vide 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 />Ext. of twice the given angle divided by the, radius of a 1° curve will <br />a the Nat. Tan. or Nat. Ex. Sec. <br />- EXAMPLE <br />t1 Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or 1. F.-230 20' to the R. at Station <br />I� <br />542+72. <br />1 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'= 1193.1 <br />1183.1-.=10=118.31. <br />Correction for A. 23° 20' for a 1.0° Cur. —0.16 <br />i 118.31-}-0.16=118.47=corrected Tangent. <br />corr <br />Ext. is required find in same <br />' (If nge23* 200'=23.331 10=2.3333 L. CaY) <br />2° 192"= def. for sta, 542 I. P. =sta. 542+72 <br />i 4° 49'= ° u x-50 Tan. = 1 _18.47 <br />`r 543 <br />9° 49J!G x-56 B. C. =sta. 541 {53.53 <br />= 4 Y <br />11 11° 40'= u % 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.)=739.41'= <br />2° 191'=def. for sta. 542. <br />i <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Oef. for 36.86 ft. =1° 501}' for a 10° Curve. <br />