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CURVE TABLES <br />Published by.KEUFFEL & ESSER CO. <br />' HOW TO USE CURVE TABLES j <br />Table L' 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 Ea,.: opposite the given Central Angle by the given degree of curve. <br />1'o find Deg. of Curve, having the Central Angle and Tangent: I <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 1 <br />be the Nat. Tan. or Nat. Ex.' Sec. <br />, <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'= 120.81' <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 saine way) <br />Ang. 23'20`=23.33°=10=2.3333=L. C. <br />2° 192'=def. for sta. 542 1. P. =sta. 542+72 <br />4° 492' = " +50 Tan. = 1 .18.47 <br />7, 19#, - " ." " 543 <br />9°492'= " " +50 B. C.=sta . 541-53:53 <br />II- 40'= v " " 513+ 1— C'= 2 .33.33 <br />/ 86.86 E, C. =Sta. 543 X86.86 <br />- ( -100-53.53=46.47XT(def. for 1 ft. of 10' Cur.) =739.41' <br />2' 192 def. for sta. 542. <br />Def. for. 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.56 ft. 1° 502V for a 10' Curve. <br />� LP.A»9.23.2D <br />i X10 ? , <br />�jq♦6,1 J � <br />f <br />i <br />I <br />1 <br />'J <br />CURVE TABLES <br />Published by.KEUFFEL & ESSER CO. <br />' HOW TO USE CURVE TABLES j <br />Table L' 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 Ea,.: opposite the given Central Angle by the given degree of curve. <br />1'o find Deg. of Curve, having the Central Angle and Tangent: I <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 1 <br />be the Nat. Tan. or Nat. Ex.' Sec. <br />, <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'= 120.81' <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 saine way) <br />Ang. 23'20`=23.33°=10=2.3333=L. C. <br />2° 192'=def. for sta. 542 1. P. =sta. 542+72 <br />4° 492' = " +50 Tan. = 1 .18.47 <br />7, 19#, - " ." " 543 <br />9°492'= " " +50 B. C.=sta . 541-53:53 <br />II- 40'= v " " 513+ 1— C'= 2 .33.33 <br />/ 86.86 E, C. =Sta. 543 X86.86 <br />- ( -100-53.53=46.47XT(def. for 1 ft. of 10' Cur.) =739.41' <br />2' 192 def. for sta. 542. <br />Def. for. 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.56 ft. 1° 502V for a 10' Curve. <br />� LP.A»9.23.2D <br />i X10 ? , <br />�jq♦6,1 J � <br />f <br />