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4 <br />�•5 7 <br />�I e-, <br />o <br />rt.w,40V <br />e, <br />-54.- <br />D <br />L— <br />N.E, <br /># <br />1 2 — <br />U. <br />t- <br />_ <br />41 R6 <br />3 <br />S <br />1 <br />103.43.1, <br />-- <br />0 SD+.C.o✓, <br />4.7 1 <br />I <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />'\/,,I OW. To USE . CURVE TABL, ES. <br />Table I. contains Tangents and Externa]s to a 1° an. and <br />airy` dine. ' theTan. <br />I.xt no g ve~ degree of curve. <br />r Ext. opposite the given Central Angle by Tangent: <br />To find Deg. of Curve, having the Central Angle and n Tangent. P <br />)nide Tan. opposite the given Central Angle by the g <br />To find Deg. of Curve, having the Central Ar iveand xt. External. <br />Divide Ext. opposite -the given Central Angle by the g <br />To find Nat. Tan. and Nat. Ex. See. for any angle by Table I.: Tan. <br />Ir Ext. of twice the given angle divided by the radius of a 11 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 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. 231 20'= 1183.1 <br />1183.1=10 =118.31. <br />Correction for A. 23' 118 31-1-0.16 118.47 20' <br />=corrected16 <br />Tangent.. <br />(If corrected Ext. is required find in same way) - <br />Ang.23'201=23.33°=]0=2.3333 <br />21 191,1=def. for sta. 542 1. P. =sta. .542+72 <br />_{ 40 491' _ u 11 +50 Tan. = 1 •18.47 <br />7° 191'= " 543 B. C.=sta. 541-1-53.53 <br />9° 491' = " " f1 +50 2 .33.33 <br />11° 40'= « « « 543+. L. C. _.. <br />86.86 E. C. =Sta. 543{8 8... <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) <br />2° 19x' =clef. for sta. 542. <br />«(� Def. dor 50 ft. =2° 30' for 10° Curve. <br />Def. for 36.86 ft.=1° 501' for a 10° Curve. <br />sq <br />- `.P.An9.23°20 <br />to, Curve <br />x <br />A <br />