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CURVE TABLES. <br />1.4 <br />Published by KEUFFEL 8s ESSER CO. <br />HOW TO USE • CURVE TABLES. <br />' " Table I. contains Tangents and Externals to a 10 curve. Tan. and <br />Ext. to any other radius may be 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. See. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />r' of Intersection or I. P. =230 20' to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 230 20' =120.87 <br />120.87 =12 =10.07. Say a 100 Curve. <br />Tan. in Tab. I opp. 230 20''=1183.1 <br />10=118.31. <br />Correction for A. 230 20' for a 100 Cur. =0.16 <br />118.31+0.16 =118.47 =corrected Tangent. <br />I, (If corrected Fxt. is required find in same way) <br />Ane. 23020'=23.33°=10=2.3333=L. C. <br />20192'=def. for sta. 542 I. P.=sta. 542+72 <br />40492'= " " " -f-50 Tan.= 1 .18.47 <br />O = U <br />Its 7192r • (1 643 B. C.=sta. 541+53.53 <br />90 _ ![ 492 — " " +50 L. C. =, 2 .33.33 <br />f 11040'= " " " 543+ <br />86.86 1 E. C.=Sta. 543+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10,Cur.)=139.41'= <br />2° 10V=def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 100 Curve. <br />Def. for 36.86 ft. =10 502' for a 100 Curve. <br />S•4 0 <br />f.P.An 9.23'20' <br />Al g4� <br />10• Curve <br />s <br />�3 <br />r VA7. <br />1 <br />3 <br />1 1 - <br />J0 <br />JO <br />J G(� <br />i <br />CURVE TABLES. <br />1.4 <br />Published by KEUFFEL 8s ESSER CO. <br />HOW TO USE • CURVE TABLES. <br />' " Table I. contains Tangents and Externals to a 10 curve. Tan. and <br />Ext. to any other radius may be 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. See. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />r' of Intersection or I. P. =230 20' to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 230 20' =120.87 <br />120.87 =12 =10.07. Say a 100 Curve. <br />Tan. in Tab. I opp. 230 20''=1183.1 <br />10=118.31. <br />Correction for A. 230 20' for a 100 Cur. =0.16 <br />118.31+0.16 =118.47 =corrected Tangent. <br />I, (If corrected Fxt. is required find in same way) <br />Ane. 23020'=23.33°=10=2.3333=L. C. <br />20192'=def. for sta. 542 I. P.=sta. 542+72 <br />40492'= " " " -f-50 Tan.= 1 .18.47 <br />O = U <br />Its 7192r • (1 643 B. C.=sta. 541+53.53 <br />90 _ ![ 492 — " " +50 L. C. =, 2 .33.33 <br />f 11040'= " " " 543+ <br />86.86 1 E. C.=Sta. 543+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10,Cur.)=139.41'= <br />2° 10V=def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 100 Curve. <br />Def. for 36.86 ft. =10 502' for a 100 Curve. <br />S•4 0 <br />f.P.An 9.23'20' <br />Al g4� <br />10• Curve <br />s <br />�3 <br />