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3 <br />mcp <br />95-'- J_� q 7. 3 T I <br />.2.% <br />S7,, <br />CURVE :TABLES ._ ,C' "6-.-.-_1, <br />Published by KEUFFEL & ESSER. CO.: <br />o a . -HOW .TO USE- CURVE TABLES <br />Table h. 'contains Tangents and Externals to a.l° curve: Tan;' and <br />Ext. to any other radius may be found 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 Cilrve,: having the :Central: -Angle;aind .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 l° curve will <br />be the Nat. Tan. or Nat. Ex. Sea <br />7` EXAMPLE = <br />✓ .� Wanted a Curve with an Ezt. of about 12 1t{Angle LO <br />'Q � of- Intersection 'or I. P.'=-23° 20' to-tlie. R. at' Station <br />J f'i542+72. <br />Ext. in Tab. I opposite 23'•20' =120.87 C <br />- : 120.87.=.12=10.07: ;'SayI a 10° -Curve. <br />Tan. in Tab. I� opp. 23° 20' =1183.1 <br />f'o1183.1=10=118.31 <br />Co <br />rrectiori for.+A. 23°20' -for a 10° Cur.=0.16.: + ? <br />118.31-{-0.16=118.47=corrected Tangent.,:.; <br />(If correcte'd-Ei:t. is required find in saihi way)-:•- <br />Ang. 23'.20'=23.33o =10 = 2.3333 = L. C. i <br />2° 19^=def. forsta;: 542- I. P.,=sta.' 542 }72 " <br />40 49,'= " '° -b50 .Tan. ;= 1 .18.47 <br />70.19 - 543 • B C. =sta: 541:53,53 <br />. .. 9° 491'= - R t!'Q+50 - <br />110 40' _ " « .543+ L. C. = 2..33.33 <br />86.86 'E. C.=Sta. 543-}-86:86 <br />100-53:53 =46.47j(3'(def. for. l:ft. of 10°'Cur.)�=139.41' <br />2°-19'=def : for sfa: 542:.' -:;•; <br />Def. -for 504t;-29 30' for a 10' Curve. -=,t - - • = <br />Def. for 36.86 ft. =1° 50;' for a 10° Curve. <br />LP,An9.23'201 <br />.. ri <br />Al <br />io• curve <br />