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_ 1 CURVE- TABLES. Published by' KEUFFEL & ESSER CO. : ' HOW TO USE "CURVE • TABLES. -Table I.' contains Tangents and Externals to a'11 curve. Tan. and Ext. to any other radius may be found nearly enough, by dividing the Tan. or Ext. opposite the given Central -Angle by the given degree of curve. .To find Deg. of Curve, having the Central Angle and Tangent: Divide Tan. opposite the given Central An le by the given Tangent. 'To find Deg. of Curve, having the Central Angle and .External: Divide Ext. site the given Central Angle by the given External. To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. or Ext, of twice the given angle divided by the radius of a 1° curve will be the Nat. Tari. or Nat. Ex. Sec. EXAMPLE. Wanted a Curve with an Ext. of about 12 ft. Angle. of Intersection or. I. P. =23° 20' to'.the R.. at' Station 542+72. 'Ext. in Tab. I opposite 23° 20' =120.87 120.87 =12 10.07. Say a 10° Curve. Tan. in Tab. I opp. 23° 20'= 1183.1 1183.1- 10 =118.31. Correction for A. 23° 20' for a 10° Cur. =0.16 118.31+0.16 =118.47 =corrected Tangent. (If corrected Ext. is required find in same way) Ang.23'20'=23.33 =10=2.3333=L. C. - 2° 191'= def. for sta. 542. 1 P.=sta. 542-72 40 493' = " +50 Tan. = 1 .18.47 70 193' = " " u 543 9°4931= " " "+50 B. C.=sta., 541-x-53.53 _ 11° 40'=, a" " L. C.— 2_33.33 86.86 E. C:=Sta. 543 -86.86 100-53.53=46.47X3'(def. ford ft. of 10° Cur.)=139.41'= 2° 191' =def. for sta.-542. - Def.'for 50 ft. =2° 30' fora 10° Curve. Def. for 36.86 ft'. P 503' for a 10° Curve. !x I_ _ 1 CURVE- TABLES. Published by' KEUFFEL & ESSER CO. : ' HOW TO USE "CURVE • TABLES. -Table I.' contains Tangents and Externals to a'11 curve. Tan. and Ext. to any other radius may be found nearly enough, by dividing the Tan. or Ext. opposite the given Central -Angle by the given degree of curve. .To find Deg. of Curve, having the Central Angle and Tangent: Divide Tan. opposite the given Central An le by the given Tangent. 'To find Deg. of Curve, having the Central Angle and .External: Divide Ext. site the given Central Angle by the given External. To find Nat. Tan. and Nat. Ex. Sec. for any angle by Table I.: Tan. or Ext, of twice the given angle divided by the radius of a 1° curve will be the Nat. Tari. or Nat. Ex. Sec. EXAMPLE. Wanted a Curve with an Ext. of about 12 ft. Angle. of Intersection or. I. P. =23° 20' to'.the R.. at' Station 542+72. 'Ext. in Tab. I opposite 23° 20' =120.87 120.87 =12 10.07. Say a 10° Curve. Tan. in Tab. I opp. 23° 20'= 1183.1 1183.1- 10 =118.31. Correction for A. 23° 20' for a 10° Cur. =0.16 118.31+0.16 =118.47 =corrected Tangent. (If corrected Ext. is required find in same way) Ang.23'20'=23.33 =10=2.3333=L. C. - 2° 191'= def. for sta. 542. 1 P.=sta. 542-72 40 493' = " +50 Tan. = 1 .18.47 70 193' = " " u 543 9°4931= " " "+50 B. C.=sta., 541-x-53.53 _ 11° 40'=, a" " L. C.— 2_33.33 86.86 E. C:=Sta. 543 -86.86 100-53.53=46.47X3'(def. ford ft. of 10° Cur.)=139.41'= 2° 191' =def. for sta.-542. - Def.'for 50 ft. =2° 30' fora 10° Curve. Def. for 36.86 ft'. P 503' for a 10° Curve. !x