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��0_1131 +ts 4 �?a I _ CURVE TABLES + Published by KEUFFEL & ESSER CO. HOW TO USE CURVE TABLES Table I. contains Tangents and Externals to a •1° curve. Tan: and - Ext. to any other radius maybe 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 Angle by the given Tangent. -To find Deg. of Curve, -having the Central. Angle .and External: Divide Ext. opposite the given Central Angle by the given External. - To find Nat. Tan. and -Nat. Ex. Sec. for any angle byy Table I.: Tan. or Ext. of twice the given - angle divided by the radius -of a•1° curve will be the Nat. Tan. 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+0.1-6 =118.47 =corrected Tangent. (If corrected Ext. is required find in same way) Ang. 236 20'=23.33°=10=2.3333=L. C. 2° 19 z'=def. for std. 542 I. P. =sta. 542+72 4049-'= a+50 Tan._ 1 .18.47 7° 192'= " " 543 1,— " B. C.=sta. 541+53.53 9 '491 - +50 L. C. = 2 .33.33 11° 40'= " :" 543+ 86.86 E. C.=Sta. 543-+86.86 100=53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= 2° 192'=def. for sta. 542. Def: for 50 ft. =2° 30' for a 1W Curve. Def. for 36.86 ft. =10 502' for a 100 Curve. 4� x' • A IAAn9.23°20' _ I Al 1 to°curve 4 . i. -�� i, a ,lig+i i. ��0_1131 +ts 4 �?a I _ CURVE TABLES + Published by KEUFFEL & ESSER CO. HOW TO USE CURVE TABLES Table I. contains Tangents and Externals to a •1° curve. Tan: and - Ext. to any other radius maybe 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 Angle by the given Tangent. -To find Deg. of Curve, -having the Central. Angle .and External: Divide Ext. opposite the given Central Angle by the given External. - To find Nat. Tan. and -Nat. Ex. Sec. for any angle byy Table I.: Tan. or Ext. of twice the given - angle divided by the radius -of a•1° curve will be the Nat. Tan. 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+0.1-6 =118.47 =corrected Tangent. (If corrected Ext. is required find in same way) Ang. 236 20'=23.33°=10=2.3333=L. C. 2° 19 z'=def. for std. 542 I. P. =sta. 542+72 4049-'= a+50 Tan._ 1 .18.47 7° 192'= " " 543 1,— " B. C.=sta. 541+53.53 9 '491 - +50 L. C. = 2 .33.33 11° 40'= " :" 543+ 86.86 E. C.=Sta. 543-+86.86 100=53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= 2° 192'=def. for sta. 542. Def: for 50 ft. =2° 30' for a 1W Curve. Def. for 36.86 ft. =10 502' for a 100 Curve. 4� x' • A IAAn9.23°20' _ I Al 1 to°curve 4 . i.