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r <br />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 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. =230 20' to the R: at Station' <br />642- 72. <br />Ext. in Tab. I opposite 23',20f -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 />118.31-4-0.16 =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°.191'=def:forsta. 542 I. P.=sta. 542+72 <br />4° 491'= " " " +50, Tan. = 1 .18.47 <br />7°191' cc it 543 <br />9° 491' _ " 's 4' -}-50 B. C. = sta. 541-4-53.53 <br />I Il° 40'= a ca a 554,3+ L. C. = 2 .33.33 <br />86.86 E. C.=Sta. 543-4-86.86 <br />t 100-53.53=46.47X3'(def. for .1 ft. of 10° Cur.). <br />=139.41'= <br />120191'=def. fo_•r sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curved <br />x <br />_ ro• <br />JC <br />N4 <br />d <br />7 <br />U% <br />2� , <br />v <br />,o <br />� <br />d.S <br />O <br />•� <br />i <br />V <br />j <br />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 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. =230 20' to the R: at Station' <br />642- 72. <br />Ext. in Tab. I opposite 23',20f -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 />118.31-4-0.16 =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°.191'=def:forsta. 542 I. P.=sta. 542+72 <br />4° 491'= " " " +50, Tan. = 1 .18.47 <br />7°191' cc it 543 <br />9° 491' _ " 's 4' -}-50 B. C. = sta. 541-4-53.53 <br />I Il° 40'= a ca a 554,3+ L. C. = 2 .33.33 <br />86.86 E. C.=Sta. 543-4-86.86 <br />t 100-53.53=46.47X3'(def. for .1 ft. of 10° Cur.). <br />=139.41'= <br />120191'=def. fo_•r sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curved <br />x <br />_ ro• <br />JC <br />N4 <br />d <br />7 <br />