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CURVE TABLES. <br />Published by KEUFFEL 8s ESSER CO: <br />1 HOW TO USE CURVE TABLES: <br />Table 1. contains Tangents and Externals to a 1° 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 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. f or 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 />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P. =230 201- to_ the R. at Station <br />542-72. <br />j 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 +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° 19' -,'=def. for sta. 542 I. P. =sta. 542+72 <br />4049-211= " " " +50 Tan. = 1 .18.47 <br />7° 19;' = " " " 543 <br />- B. C.=sta.541-}-53:53 <br />i 90 493, = a a a -1-50 - <br />110 40'--' a a a 543+L. C. = 2 .33.33 <br />86.86 E'. C.= ta <br />543 186.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) =139.41:= <br />2° 191'=def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 10° Curve. • <br />Def. for 36.86 ft. =1° 50j' for a 10° Curv& <br />J <br />- <br />— <br />ca <br />j <br />- <br />�c <br />I <br />Q� <br />6 <br />H' <br />o <br />I <br />1 <br />--_ -__ <br />CURVE TABLES. <br />Published by KEUFFEL 8s ESSER CO: <br />1 HOW TO USE CURVE TABLES: <br />Table 1. contains Tangents and Externals to a 1° 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 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. f or 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 />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of Intersection or I. P. =230 201- to_ the R. at Station <br />542-72. <br />j 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 +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° 19' -,'=def. for sta. 542 I. P. =sta. 542+72 <br />4049-211= " " " +50 Tan. = 1 .18.47 <br />7° 19;' = " " " 543 <br />- B. C.=sta.541-}-53:53 <br />i 90 493, = a a a -1-50 - <br />110 40'--' a a a 543+L. C. = 2 .33.33 <br />86.86 E'. C.= ta <br />543 186.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) =139.41:= <br />2° 191'=def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 10° Curve. • <br />Def. for 36.86 ft. =1° 50j' for a 10° Curv& <br />