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ME <br />MEW <br />�M <br />0 <br />— _/ <br />CURVE TABLES.. Js <br />Published by KEUFFEL & 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, bydividing 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 />q-fye Divide Tan. opposite the given Central Angle by the glven Tangent. <br />Ze-�- 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 />642+72. <br />Ext. in Tab. I opposite 23° 20Y=120.87 <br />120.87 •. 12 =10.07. Say a 10P 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 sway) <br />Ang. 23°20'= 23.33 :-10=2.3333=L. C. <br />2° 197'=def. for stall .542 I. P.=sta. 542-x-72, <br />4° 4921'= " +56. Tan. _. 1 .18.47 <br />70 1911= if " 543 <br />B. C.=sta. 541+53.53 <br />-6�C Z"?l 9° 497'= " " " -x-50 <br />110 40'= " " " 543 } L. C. _ ' 2 .33.33 <br />86.861 E. C.-Sta. 543+86.86 <br />100-53:53=46.47X3'(def. for 1.ft, of 10° Cur.)=139.41'= <br />_ 2° 19'=def. for sta. 542. <br />Def, for 50 ft. =2° 30' for a 10° Curve. <br />— i Def. for 36.86 ft.= 10 504" for a 10° Curve <br />>a <br />I:P.An9:23.20• _ <br />P4 ,> <br />10• Curve <br />�--. d <br />s" L <br />ME <br />MEW <br />�M <br />0 <br />— _/ <br />CURVE TABLES.. Js <br />Published by KEUFFEL & 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, bydividing 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 />q-fye Divide Tan. opposite the given Central Angle by the glven Tangent. <br />Ze-�- 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 />642+72. <br />Ext. in Tab. I opposite 23° 20Y=120.87 <br />120.87 •. 12 =10.07. Say a 10P 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 sway) <br />Ang. 23°20'= 23.33 :-10=2.3333=L. C. <br />2° 197'=def. for stall .542 I. P.=sta. 542-x-72, <br />4° 4921'= " +56. Tan. _. 1 .18.47 <br />70 1911= if " 543 <br />B. C.=sta. 541+53.53 <br />-6�C Z"?l 9° 497'= " " " -x-50 <br />110 40'= " " " 543 } L. C. _ ' 2 .33.33 <br />86.861 E. C.-Sta. 543+86.86 <br />100-53:53=46.47X3'(def. for 1.ft, of 10° Cur.)=139.41'= <br />_ 2° 19'=def. for sta. 542. <br />Def, for 50 ft. =2° 30' for a 10° Curve. <br />— i Def. for 36.86 ft.= 10 504" for a 10° Curve <br />>a <br />I:P.An9:23.20• _ <br />P4 ,> <br />10• Curve <br />