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11 <br />CURVE TABLES. <br />Published by KEuFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />'- Table -L"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 />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' anjleAivid6dby the radius of a P curve will <br />be the Nat. Taxi. of Nat. Ex. Sec,,,' <br />KkAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />I.%'P.=23* 20' to'the of Intersection or the R., at Station, <br />542+72. <br />Ejit.- in Tab;'I- opposite 23* 2(Y=120.87 <br />;'120.87:A�,121=10.07:! Sdy'alOP 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 10P Cur. =0.16 <br />118-31+0.16 = 118.47 =corrected Tangent. <br />(If corrected Ext. is,req irid find in same way) <br />Ang. 23* 20'=23 .'33' =10=2.3333=L. ' C. <br />f f'-_� 542 x-72 <br />5 <br />2 19Y=de . orsta.-.�', 542 • t P. sta. <br />1 .18.47 <br />40 49211" a' a �'+50 Tan. <br />f .543 B. 541+53.53 <br />70 192'= P , " <br />9°49;'= " " +50 L. C. — 2 .33.33 <br />11? 40'--' - - ,543+ <br />",86.86 _E C. = Sta. 543+86-86 <br />100-53.53 =46.47 X3'(aef. for I ft. of 10° Cur.) 139.41'= <br />2-'-19ii=&f:1oc sta. 542. <br />-Def. for-50tft.,--20 30' for a 10° Curve; <br />Def. for 36.86 ft. V 50j' for a 10° Curve. <br />IAAn9.23-20V <br />C' W✓,' <br />zl-:i��g -0'✓ (r/.Y, C Ir-57a"d <br />4!_ <br />/z/ 00 <br />z/V 'r a of/5 /,9,,"- <br />6 <br />/moo e <br />A/1 r j o AIS 96 <br />-1--eyw 1,eq11'1rS'-eV <br />S77 ✓ "r/ _S <br />r-. 01P <br />CURVE TABLES. <br />Published by KEuFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />'- Table -L"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 />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' anjleAivid6dby the radius of a P curve will <br />be the Nat. Taxi. of Nat. Ex. Sec,,,' <br />KkAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />I.%'P.=23* 20' to'the of Intersection or the R., at Station, <br />542+72. <br />Ejit.- in Tab;'I- opposite 23* 2(Y=120.87 <br />;'120.87:A�,121=10.07:! Sdy'alOP 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 10P Cur. =0.16 <br />118-31+0.16 = 118.47 =corrected Tangent. <br />(If corrected Ext. is,req irid find in same way) <br />Ang. 23* 20'=23 .'33' =10=2.3333=L. ' C. <br />f f'-_� 542 x-72 <br />5 <br />2 19Y=de . orsta.-.�', 542 • t P. sta. <br />1 .18.47 <br />40 49211" a' a �'+50 Tan. <br />f .543 B. 541+53.53 <br />70 192'= P , " <br />9°49;'= " " +50 L. C. — 2 .33.33 <br />11? 40'--' - - ,543+ <br />",86.86 _E C. = Sta. 543+86-86 <br />100-53.53 =46.47 X3'(aef. for I ft. of 10° Cur.) 139.41'= <br />2-'-19ii=&f:1oc sta. 542. <br />-Def. for-50tft.,--20 30' for a 10° Curve; <br />Def. for 36.86 ft. V 50j' for a 10° Curve. <br />IAAn9.23-20V <br />