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~ J <br />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 the Tan, <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 I.- 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 10° 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° 192"= def. for sta. 542 I. P. =sta. 542-1-72 <br />40 492' = " " " +50 Tan. = 1 .18.47 <br />70 192' _ " 543 <br />90 4912'= " it 44 +50 B. C. =stn. 541+53.53 <br />53.53 <br />11°40'= ", rr ra 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 />2° 192'=def, for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curve <br />J2s1N�UA7 <br />J/r <br />i <br />Na2T/J <br />7 <br />r <br />t� <br />z <br />std <br />e <br />/C. <br />fzt,, <br />' ymi— <br />7 i qZ) <br />ILo <br />r <br />~ J <br />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 the Tan, <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 I.- 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 10° 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° 192"= def. for sta. 542 I. P. =sta. 542-1-72 <br />40 492' = " " " +50 Tan. = 1 .18.47 <br />70 192' _ " 543 <br />90 4912'= " it 44 +50 B. C. =stn. 541+53.53 <br />53.53 <br />11°40'= ", rr ra 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 />2° 192'=def, for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 501' for a 10° Curve <br />