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CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a 11 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. 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.=231 20' to the R. at Station <br />542+72. <br />Ext. in Tab. -I opposite 23° 20' =120.87 <br />120M --'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'=23.330-10=2.3333=L. C. <br />2°.191'=def. for sta. 542 I. P. =sta. 542+72 <br />40 491' _ f° +50 Tan. = 1 .18.47 <br />7° 192'= " " " 543 <br />9°492'_ „ " B. C.=sta. 541+5153- <br />11- <br />41+53.5311° 40'= " `/ " 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° 191'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50;' for a 10° Curve. <br />@qg 1 <br />IAAn9.23 °201 <br />N // <br />A4 <br />10° curve <br />pCl <br />G <br />?° <br />3. <br />�'. 5-3 <br />✓ r- <br />S <br />TL,' <br />1-41 <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />Table I. contains Tangents and Externals to a 11 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. 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.=231 20' to the R. at Station <br />542+72. <br />Ext. in Tab. -I opposite 23° 20' =120.87 <br />120M --'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'=23.330-10=2.3333=L. C. <br />2°.191'=def. for sta. 542 I. P. =sta. 542+72 <br />40 491' _ f° +50 Tan. = 1 .18.47 <br />7° 192'= " " " 543 <br />9°492'_ „ " B. C.=sta. 541+5153- <br />11- <br />41+53.5311° 40'= " `/ " 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° 191'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50;' for a 10° Curve. <br />@qg 1 <br />IAAn9.23 °201 <br />N // <br />A4 <br />10° curve <br />pCl <br />