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i i CURVE "TABLES. Published by KEUFFEL &, ESSER CO. HOW TO USE CURVE TABLES. Tattle I:' contains Tangents and Externals to a'1° curve. Tan: and Ext. to any other radius may be found nearly enough, by dividing theTan. or Ext. opposite the given Central An le by the given degree of curve. To find Deg. of Curve, having the Central Angle and Tangent: Divide Tan. opposite the given Central Angle by the given Tangent. To find 'Deg. of Curve,: having'the Central Angle and External: Divide Ext: opposite the given Central Angle by the given External. To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. or Ext. of twice the given angle divided' by the radius of a V curve will be the Nat. -Tan. or Nat. Ex. Sec. EXAMPLE. Wanted a Curve with an Ext. of about 12 ft. Angle of Intersection or I. P. =23° 20' to the R. at Station Ext. in Tab. I opposite 23° 20' =120.87 120.87 =12 =10.07. Say a IOP Curve. Tan. in Tab. I opp. 23' 20'=1183.1 1183.1-10=118.31. Correction for A. 23a� 20' for a 10° Cur. =0.16 118.31-{-0:16 =118.47 =corrected Tangent. '(If corrected Ext: is.repired find in same way) Ang.23°20'=23.33 a -10=2.3333=L. C. 2°191'=def. for sta. 542 :I. P.=sta. , 542+72 4049i1 _ +50 Tan. = 1 .18.47 7-1012' _ 543 9°'491'= "' " " " +50 B. C.=sta. 541+53.53 11° 40'= " " 543-1. L. C.= , 2 .33.33 86.86 E. C.=Sta. 543+86.86 100=53.53=46.47Xq'(def."for 1 ft. of'.10° Cur.)=139.41'= 2°.191'=def. for std: 542. , Def: for 50 ft. =2° 30' for -a 10° Curve: Def. for 36.86 ft. =1° 501' for a 10° Curve. . . I.P.An9.23W90" N 10' Gut W 4.0 %/ / N V 3 �L9 V Cv O — U i ID i CURVE "TABLES. Published by KEUFFEL &, ESSER CO. HOW TO USE CURVE TABLES. Tattle I:' contains Tangents and Externals to a'1° curve. Tan: and Ext. to any other radius may be found nearly enough, by dividing theTan. or Ext. opposite the given Central An le by the given degree of curve. To find Deg. of Curve, having the Central Angle and Tangent: Divide Tan. opposite the given Central Angle by the given Tangent. To find 'Deg. of Curve,: having'the Central Angle and External: Divide Ext: opposite the given Central Angle by the given External. To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. or Ext. of twice the given angle divided' by the radius of a V curve will be the Nat. -Tan. or Nat. Ex. Sec. EXAMPLE. Wanted a Curve with an Ext. of about 12 ft. Angle of Intersection or I. P. =23° 20' to the R. at Station Ext. in Tab. I opposite 23° 20' =120.87 120.87 =12 =10.07. Say a IOP Curve. Tan. in Tab. I opp. 23' 20'=1183.1 1183.1-10=118.31. Correction for A. 23a� 20' for a 10° Cur. =0.16 118.31-{-0:16 =118.47 =corrected Tangent. '(If corrected Ext: is.repired find in same way) Ang.23°20'=23.33 a -10=2.3333=L. C. 2°191'=def. for sta. 542 :I. P.=sta. , 542+72 4049i1 _ +50 Tan. = 1 .18.47 7-1012' _ 543 9°'491'= "' " " " +50 B. C.=sta. 541+53.53 11° 40'= " " 543-1. L. C.= , 2 .33.33 86.86 E. C.=Sta. 543+86.86 100=53.53=46.47Xq'(def."for 1 ft. of'.10° Cur.)=139.41'= 2°.191'=def. for std: 542. , Def: for 50 ft. =2° 30' for -a 10° Curve: Def. for 36.86 ft. =1° 501' for a 10° Curve. . . I.P.An9.23W90" N 10' Gut W 4.0 %/ / N