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.. ,.,�. t <br />I} CURATE TALES. <br />Published by KEUFFEL & ESSER CO. <br />' \ HOW TO USE CURVE ' TABLES. <br />rj. Table I. contains Tangents and Externals to a 1° curve: Tan. and <br />\ <br />4:1g`=!�"7 �- - Ext. to any other radius may befound nearly enough; bydividing the Tan. <br />o'E xt. 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 />' n J 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 />1 or Ext. of twice the given angle divided by the radius of a 1° curve will <br />. T �be the Nat. Tan. or Nat. Ex. Sec. <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 />-L p - * 3�T` ' 542-72. <br />G <br />�� •� ��_% < < �� � ; f ij - Ext. in Tab. I opposite 23° 20' =120.87 <br />1`20.87 -12 =10.07. Sa`y a 10° Curve. <br />Tan. in Tab. Kopp. 23° 20' =1183.1 <br />f -- - -- - \ 1153.1-10 =118.31. <br />{ �— "; i' Correction for A. 23° 20' for a 10° Cur. =0.16 e <br />118.31-}=0.16=115.47=corrected Tangent. <br />_ (If corrected Ext. is required find in same way) <br />Ang. 23°20'=23.33°=10=2.3333=L. C. <br />fig " z <br />. s 2° 192'='dcf. forsta. 543 I. P.=sta. 542-72 <br />-9 4 491 F50 Tan. = 1 .18.47 <br />*' d --- - -- 7° 192'.= « u rs 543 <br />71g--- (.`'1- +— .4 `" « a ° i,__ a B. C.=sta- 541 -53.53 <br />+50 <br />{ J ` °~ r - - --- -- -- ; 4 11° 40'= u 543 1 L. C. = 2 .33.33 <br />0$ '� i t 86.56 E. C.=Sta. 543-86.86 <br />�- -`'y ---- -- - -- 100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= <br />_ - - 2° 19 k'=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 />71 <br />I 1 <br />-1 <br />-A <br />