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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 may befound nearly enough, bydividing theTan. 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 by 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 1. 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 1153.1=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° 1912'- def. for sta. 542 1. P. = sta. 542-72 4° 491"-- " " " -1-50 Tan. = 1 .18.47 7° 191,= 543 B. C. sta. 541•-X53.53 9°491'= it ,,F50 110 40'= " " " 543 } L. C' = 2 .33.33 86.86 E. C.=Sta. 543-1--86.86 •100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) =139.41'= 2° 191'=def. for sta: 542. Def. for 50 ft. =2° 30' for a 10° Curve. Def'. for 36.86 ft.=1° 501' for a 10° Curve. c.., .., A, • a 3 t S /�� TZ � 1�.• h4 •� � - �Bkl Perc. N. S/" A In the G..aSScS/09 Sia 6' H� frt7t.o2•'—' 36�!' .- ,� . P s- P lzPz ., t� y Tyrs /s RGo,v� • Fi % 886 f 9 9. s -----------7 8 �0f 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 may befound nearly enough, bydividing theTan. 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 by 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 1. 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 1153.1=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° 1912'- def. for sta. 542 1. P. = sta. 542-72 4° 491"-- " " " -1-50 Tan. = 1 .18.47 7° 191,= 543 B. C. sta. 541•-X53.53 9°491'= it ,,F50 110 40'= " " " 543 } L. C' = 2 .33.33 86.86 E. C.=Sta. 543-1--86.86 •100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) =139.41'= 2° 191'=def. for sta: 542. Def. for 50 ft. =2° 30' for a 10° Curve. Def'. for 36.86 ft.=1° 501' for a 10° Curve.