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:..CURVE TABLES. <br />I <br />Published by K(�EUFFEL & 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 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. 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 I., P. =23' 20' to ' the R. at Station <br />542-1-72. - <br />Ext. in Tab. I opposite 23' 20Y =120.87 <br />120.87--12=10.07. Say a 1(Y' 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-x-0.16 =118.47 -4-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 />4' 491 a " +50 Tan. = 1 .18.47 <br />70 1911= i, a a 543' . <br />' B. C.=sta. 541+53.53 <br />9° 49,'= f° d -x-50 <br />11' 40' = a a 543 - L. C.= : 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46;47jC3'(def. for 1 ft. of 10' Cur.) = 139.41'= <br />_ 2''191'=def. forsta. 542. <br />Def, for 50 ft. =2' 30' for a 10' Curve. <br />Def. -for 36.86 ft.=1' 50j' for a 10' Curve., _ <br />_ <br />10• cuj" <br />it 01 <br />LSA <br />� z ':'r <br />-7 9 <br />2 2-i. <br />/ 7-3 <br />S'32 <br />/ 3 <br />N <br />= 3z <br />S <br />O <br />0 <br />3 <br />S 2,0 <br />_32 <br />3 <br />32,00°. <br />31 9 2- <br />,T <br />taO <br />S <br />,6 <br />o <br />G b <br />D <br />of <br />00/ <br />:..CURVE TABLES. <br />I <br />Published by K(�EUFFEL & 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 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. 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 I., P. =23' 20' to ' the R. at Station <br />542-1-72. - <br />Ext. in Tab. I opposite 23' 20Y =120.87 <br />120.87--12=10.07. Say a 1(Y' 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-x-0.16 =118.47 -4-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 />4' 491 a " +50 Tan. = 1 .18.47 <br />70 1911= i, a a 543' . <br />' B. C.=sta. 541+53.53 <br />9° 49,'= f° d -x-50 <br />11' 40' = a a 543 - L. C.= : 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46;47jC3'(def. for 1 ft. of 10' Cur.) = 139.41'= <br />_ 2''191'=def. forsta. 542. <br />Def, for 50 ft. =2' 30' for a 10' Curve. <br />Def. -for 36.86 ft.=1' 50j' for a 10' Curve., _ <br />_ <br />10• cuj" <br />it 01 <br />