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CURVE TABLES. <br />Published by KEUFFEL, & ESSER CO. <br />-.HOW. TO, USE:, CURVE TABLES. <br />'Table' I. contains Tangents and Externals to a I' curve. Tan. and <br />Ext. to anyoth6r radius may be found nearly enough, by dividing the Tan. <br />or Ext. opposite th6'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 Ekt. opposite the given Ceintral Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. See. 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. See. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of. Intersection or L- P. =230 201 to - the R. at Station <br />6>42+72. <br />Ext. in Tab. I opposite 23" 20' = 120.87 <br />120.87=12=10.07. Say.alOPCurve. <br />Tan. in Tab. I opp. 23* 20' = 1183.1 <br />1183.1—.-10'=1i8.31. <br />-Correctioii for A. 23" W for a 10° Cur. =0.16 <br />118.31+0.16 = 118.47 —corrected Tangent. <br />7 <br />(If corrected Ext. is required * find in same way) <br />°Ang. 23°20'=23.33 =10=2.3333=L. C. <br />2" 19j'= def. for sta. 542 I. P. = sta. 542+72 <br />I. <br />40491/= aif a +50 Tan. 1 .18.47 <br />70 1911= " It- If 543 B. C. = sta. 541+53.53 <br />90 49V " ft a +50 <br />11. + L. C. = 2 .33.33 <br />86.86 E. C. = Sta. 543+86.86 <br />100-53.53 =46.47X3'(def. for 1 ft. of 10° Cur.) = 139.41'= <br />def. for sta. 542. <br />Def. for 50 ft. =2*30'fora 10° Curve. <br />Def. for 36.86 ft.=10 50'j' for a 10° Curve. <br />III <br />el <br />e <br />—3 2 e <br />35 0 <br />17' <br />3(- <br />38. <br />75- <br />21, <br />7M. H. <br />k <br />7 <br />CURVE TABLES. <br />Published by KEUFFEL, & ESSER CO. <br />-.HOW. TO, USE:, CURVE TABLES. <br />'Table' I. contains Tangents and Externals to a I' curve. Tan. and <br />Ext. to anyoth6r radius may be found nearly enough, by dividing the Tan. <br />or Ext. opposite th6'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 Ekt. opposite the given Ceintral Angle by the given External. <br />To find Nat. Tan. and Nat. Ex. See. 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. See. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of. Intersection or L- P. =230 201 to - the R. at Station <br />6>42+72. <br />Ext. in Tab. I opposite 23" 20' = 120.87 <br />120.87=12=10.07. Say.alOPCurve. <br />Tan. in Tab. I opp. 23* 20' = 1183.1 <br />1183.1—.-10'=1i8.31. <br />-Correctioii for A. 23" W for a 10° Cur. =0.16 <br />118.31+0.16 = 118.47 —corrected Tangent. <br />7 <br />(If corrected Ext. is required * find in same way) <br />°Ang. 23°20'=23.33 =10=2.3333=L. C. <br />2" 19j'= def. for sta. 542 I. P. = sta. 542+72 <br />I. <br />40491/= aif a +50 Tan. 1 .18.47 <br />70 1911= " It- If 543 B. C. = sta. 541+53.53 <br />90 49V " ft a +50 <br />11. + L. C. = 2 .33.33 <br />86.86 E. C. = Sta. 543+86.86 <br />100-53.53 =46.47X3'(def. for 1 ft. of 10° Cur.) = 139.41'= <br />def. for sta. 542. <br />Def. for 50 ft. =2*30'fora 10° Curve. <br />Def. for 36.86 ft.=10 50'j' for a 10° Curve. <br />III <br />