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CURVE TABLES. <br />Published by KEUFFEL 8r, ESSER CO. <br />HOW -TO USE CURVE 'TABLES. <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing theTan. <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. 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 />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 1. P. =230 20' to the R. at Station - <br />542-F72., <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+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 />20 19j'— def, for sta. 542 I. P. = sta. 542+72 <br />40 49;,'= «, :� « +50. Tan. = 1-.18.47 <br />7* 1912'=. " " " ' 543 B. C. = sta. 541 -53.53 <br />90 49" " a +50 <br />_ ' L : C. _ ` 2 .33.33 <br />110 40:= "z"-,�" 543+ <br />86.86 E. C. = Sta. 543 86.86 <br />(def. for 1 ft. of 10° Cur.)=139.41'= <br />100-53.53=46.47X( <br />2° 19"=def. for,sta:•542.' - <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft.= l* 501' for a 10` Curve., <br />It, <br />�a- <br />LP.An9,23'EW . <br />10• Cur✓0 <br />i <br />3o <br />a <br />G <br />—f <br />D.2. Q <br />® <br />— <br />7 <br />CURVE TABLES. <br />Published by KEUFFEL 8r, ESSER CO. <br />HOW -TO USE CURVE 'TABLES. <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing theTan. <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. 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 />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 1. P. =230 20' to the R. at Station - <br />542-F72., <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+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 />20 19j'— def, for sta. 542 I. P. = sta. 542+72 <br />40 49;,'= «, :� « +50. Tan. = 1-.18.47 <br />7* 1912'=. " " " ' 543 B. C. = sta. 541 -53.53 <br />90 49" " a +50 <br />_ ' L : C. _ ` 2 .33.33 <br />110 40:= "z"-,�" 543+ <br />86.86 E. C. = Sta. 543 86.86 <br />(def. for 1 ft. of 10° Cur.)=139.41'= <br />100-53.53=46.47X( <br />2° 19"=def. for,sta:•542.' - <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft.= l* 501' for a 10` Curve., <br />It, <br />�a- <br />LP.An9,23'EW . <br />10• Cur✓0 <br />