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I <br />I <br />4 -'-CURVE TABLES. <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, by dividing 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 />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 1. P. =23° 20' to the R. at Station <br />542-72. <br />Ext. in Tab. I opposite 23° 20' =120,87 <br />120.87=12=10.07. Say'a 1(1" 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 way) <br />Ang. 23°20'=23.33 =10=2.3333=L. C. <br />2° 191'=def. for sta. 542 1. P. =sta. 542-x-72 <br />40 491' = it if " +50 Tan. = 1 .18.47 <br />70191,— '° '° « 543 <br />90 491' u « +50 B• C. =sta. 541 { 53.53 <br />11° 40'= 543+ 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 />20 191'=def. for sta. 542. <br />Def, for 50 ft. =20 30' for a•10° Curve. <br />Def, for 36:86 ft. =1° 501' for a 10° Cusv& <br />J <br />30 <br />-bo <br />- 32 <br />.19 <br />L <br />yS <br />sq9 0.1 6 <br />s- 990 y <br />I <br />4 -'-CURVE TABLES. <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, by dividing 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 />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 1. P. =23° 20' to the R. at Station <br />542-72. <br />Ext. in Tab. I opposite 23° 20' =120,87 <br />120.87=12=10.07. Say'a 1(1" 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 way) <br />Ang. 23°20'=23.33 =10=2.3333=L. C. <br />2° 191'=def. for sta. 542 1. P. =sta. 542-x-72 <br />40 491' = it if " +50 Tan. = 1 .18.47 <br />70191,— '° '° « 543 <br />90 491' u « +50 B• C. =sta. 541 { 53.53 <br />11° 40'= 543+ 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 />20 191'=def. for sta. 542. <br />Def, for 50 ft. =20 30' for a•10° Curve. <br />Def, for 36:86 ft. =1° 501' for a 10° Cusv& <br />