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it <br />�FZ .�-�� <br />i g <br />CURVE TABLES. <br />Published by KEUFFEL 8s 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 may be found nearly enough, bydividing 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 CentralAn le 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 1(1P 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-x-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° 19V= def. for sta. 542 I. P. =sta. 542+72 <br />4° 4912'= " " " +50. Tan. = 1 .18.47 <br />70 1912 <br />--1_ 543 <br />9049.= a a it B. C.=sta. 541+53.53 <br />53.53 <br />11° 40'= ° s a 5433+ 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° 19;'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50j' for a 10° Curved <br />s" f7 <br />3 Z3 gs3 .g3 <br />8S"o oo <br />ce <br />*15. <br />ooF:WeF Gi <br />7— 7;7 <br />9: <br />qgo G� oc <br />' <br />AR <br />44,6 <br />L ow PT'Qi✓ <br />� D <br />�/ <br />cQr.F+ <br />QdT. <br />S <br />1 /NGoe.n/ <br />5. <br />"Leg <br />;od / <br />zso 1 <br />nu <br />r <br />(o/11 <br />47. Z <br />3S`o• - y�r <br />C rru <br />% <br />S �" <br />Z3 . �O� <br />it <br />�FZ .�-�� <br />i g <br />CURVE TABLES. <br />Published by KEUFFEL 8s 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 may be found nearly enough, bydividing 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 CentralAn le 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 1(1P 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-x-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° 19V= def. for sta. 542 I. P. =sta. 542+72 <br />4° 4912'= " " " +50. Tan. = 1 .18.47 <br />70 1912 <br />--1_ 543 <br />9049.= a a it B. C.=sta. 541+53.53 <br />53.53 <br />11° 40'= ° s a 5433+ 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° 19;'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 50j' for a 10° Curved <br />