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CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />3 Table I. contains Tangents and Externals to a'1' curve. Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing theTan. <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. See. 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 L ! P.72.r 20' -to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 23° 20P =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 />r 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 />20191'=def. for sta. 542 L P. =sta., 542+72 <br />40 491<= " '1 +50 Tana= 1 .18.47 <br />7° 191' = r. a to 543 <br />9'49' " " 50 13.C:=sta: 541 153.53 <br />V + L. C. = 2 .33.33 <br />110 40'=' " " " .543+ <br />86.86 E. C. =Sta. 543-}-86.86 <br />100-53.53 =46.47X3'(def. for I ft. of 10° Cur.)=139.41`= <br />2° 191'=def. for sta. 542. <br />- Def, for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft.=1° 50-21' for a 10° Curve.. <br />`x - <br />- �2 <br />I.P.An 9.23°PD! _ _• . <br />N ' <br />P� le4y - <br />I D• Curve <br />QS <br />U. R <br />4,3 <br />~�L <br />z t* <br />,� <br />6 <br />Q- <br />• - __-_ <br />T <br />1� f _ <br />_ <br />_ <br />— <br />Y"I+� <br />-r <br />4 - - -- - <br />-----T- <br />I <br />r <br />. <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />3 Table I. contains Tangents and Externals to a'1' curve. Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing theTan. <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. See. 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 L ! P.72.r 20' -to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 23° 20P =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 />r 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 />20191'=def. for sta. 542 L P. =sta., 542+72 <br />40 491<= " '1 +50 Tana= 1 .18.47 <br />7° 191' = r. a to 543 <br />9'49' " " 50 13.C:=sta: 541 153.53 <br />V + L. C. = 2 .33.33 <br />110 40'=' " " " .543+ <br />86.86 E. C. =Sta. 543-}-86.86 <br />100-53.53 =46.47X3'(def. for I ft. of 10° Cur.)=139.41`= <br />2° 191'=def. for sta. 542. <br />- Def, for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft.=1° 50-21' for a 10° Curve.. <br />`x - <br />- �2 <br />I.P.An 9.23°PD! _ _• . <br />N ' <br />P� le4y - <br />I D• Curve <br />