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IV <br />�1. <br />M <br />7 y r ` <br />!1, <br />.,CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />— xt. to any other radius may be found nearly enopgh, by dividing the Tan. <br />— for 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 />e 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 />_ 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. 23° 20' =1183.1 <br />1183.1-: 10=118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 <br />118.31-1-0.16=118.47=corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang.23°20'=23.33°=10'=2.3333=L. C. <br />_ 2° 191'=def. for sta. 542 I. P. =sta. 542+72 <br />4049121= . " " " +50 Tan. = 1 . 18.47 <br />70 19'= « I: a 543 <br />12 r � <br />o " B. C.=sti. 541= 53.53 <br />9 49i a' _ :, _ \ <br />t <br />110 40'= " " " 543+ L. C.= 2 .33.33 <br />86.86 E. C. = Sta. • 543-1-36.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) =139.41'= <br />' <br />2° 19 =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 />i.P.An9.23°20' <br />_\\7 <br />1• <br />