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Yr 7 <br />CURVE FABLES, <br />t <br />;� Published "by KEUFFEL 8s, ESSER CO._ <br />.'HOW TO. USE CMVE STABLES. <br />1 = Table I. contains Tangents and Externals�to i 1° curve. Tan, and - <br />'7 Ext. to any other radius maybe found nearly enough, bydividing the Tan. <br />i 9 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 the given angle divided by the radius of a 1° curve will <br />-be- the Nat. Tan. or. Nat: Ex. Sec. = <br />- <br />(�7 fA 4 Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I_P.=23' P. =23° 20' , to the R: at Station <br />f / ® 542+72: <br />Ext. in Tab. I opposite 230'201 „120.87 <br />120.87,,--.12 =10.07. -Say, a 10° Curve. <br />.0 y Tan. in Tab. I opp. 23' 20'= 1,183. 1 <br />.. °1183,1=10=:118.31. <br />Correction for A. 23° 20' for X 10° Cur. =0.16 <br />118.31+0.16=118.47=corrected Tangent.• =- <br />�i > / (If corrected Ext: is required find in same way) <br />'Ang.23`20'. =23.33°-10=2.3333=L. C.'•` <br />2° 19- = def: for sta. 542 .1. P. =sta. - 542+72 <br />i 4°492'= " ''a -f -0 Tan. = �l .18.47 <br />/ t1 70 192' = « u +r 543 <br />9049z'— ° 50 B. C.=sta_ 541+53.53 <br />�� 11°40'= « .° +-'=53} L.C.= .2 .'33.33' <br />86.86 E. C.=Sta. 543+86.86 <br />_ 100-53.53= 45.47+X3'(def. for 1 ft. of 10° Cur.) =139.41'=1 <br />2°,192' =def. #or sta. 543, <br />6. Def. -for 50 ft.=2°`30' far a 10° Curve_ <br />co " Def. for 36.86 ft. =1' 502' for a 10' Curve. <br />a <br />tyy <br />�,1 <br />