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-CURVE TABL S - 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 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 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 all' curve will be the Nat. Tan. or `Nat: Ex. Seca 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 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°;192'=def-for sta. 542 I. P. =sta. 542+72 40 492' _ " " " x-50 Tan: = 1 .18.47 70 191' _ " 4'' " 543 B . C. =sta. 541+53.53 53.53 9° 491' - +50 _ 11° 40'= 543+ L. C. 2 .33.33 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 10° Curve. Def. for 36:86 ft. 501' for a 10° Curve. 2 96 a. 3,3 i j 9S6 . V t • l ,�G , � 8'SG . b � Ss . 9 0 � gs'S, o Ss' eti[( gs Ft51�. :o n, a.s 14 (, , 15 3 ..s" 9 76 o- 61 �- _ c- 857•a Y ,o0 0. AA. S C o. 36 5 as6 9 �v• / 8'.s � . a est, z� o- s�� �� i X36 y l •ted e S15. -1s' e o - ,SZ (o. � �s..�. 9 8.� 4 • 3 3 c o • 7• r i -CURVE TABL S - 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 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 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 all' curve will be the Nat. Tan. or `Nat: Ex. Seca 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 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°;192'=def-for sta. 542 I. P. =sta. 542+72 40 492' _ " " " x-50 Tan: = 1 .18.47 70 191' _ " 4'' " 543 B . C. =sta. 541+53.53 53.53 9° 491' - +50 _ 11° 40'= 543+ L. C. 2 .33.33 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 10° Curve. Def. for 36:86 ft. 501' for a 10° Curve. 2