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y <br />- <br />CURVE TABLES.. '3 <br />a 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.toany other radius maybe found nearly enough, bydividing 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 />\r 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 />F ! EXAMPLE: <br />4 <br />Wantedd-a Curve with an Ext. of about 12 ft. An <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 10° Curve. <br />Tan. in Tab. I opp. 23° 20'=1183.1 <br />Correction for A. 23° 20' for a 10° Cur. <br />118.31+0.16=118.47 corrected Tangent. <br />(If corrected'Ext. is required find in same way) �Z <br />f Ang.,23°20'=23.33°.=10=2.3333=I... C. . <br />20 19y =def. for sta. - 542. I. P. =sta. 542+-72 <br />40 492' = " +50 . Tan. = 1 .18.47 ,/^ <br />70 19'-,' = u 11 It 543 3 c p d C <br />`— 9° 4921'= It it« +50 B. C: =,sta. 541 } 53:53 <br />11° 40' _ " f0 " 543 d L. C. = 2 .33.33 °a ' <br />s" <br />.86-86 'E. C: =Sta. 543+86.86 _ <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'=j <br />J 2° 1921'=def. for sta. 542. <br />1 Def. for 50 ft. =2° 30'.for a 10° Curve. <br />Def. for 36.86 ft. 1° 50r' for a 10° Curve. �ry <br />41 <br />)4-67 <br />Ig <br />�, o <br />z <br />yi <br />Z- <br />.3 1-07 <br />,, <br />1 � <br />t <br />1 <br />/ <br />1 <br />_ <br />I <br />} <br />F <br />y <br />- <br />CURVE TABLES.. '3 <br />a 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.toany other radius maybe found nearly enough, bydividing 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 />\r 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 />F ! EXAMPLE: <br />4 <br />Wantedd-a Curve with an Ext. of about 12 ft. An <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 10° Curve. <br />Tan. in Tab. I opp. 23° 20'=1183.1 <br />Correction for A. 23° 20' for a 10° Cur. <br />118.31+0.16=118.47 corrected Tangent. <br />(If corrected'Ext. is required find in same way) �Z <br />f Ang.,23°20'=23.33°.=10=2.3333=I... C. . <br />20 19y =def. for sta. - 542. I. P. =sta. 542+-72 <br />40 492' = " +50 . Tan. = 1 .18.47 ,/^ <br />70 19'-,' = u 11 It 543 3 c p d C <br />`— 9° 4921'= It it« +50 B. C: =,sta. 541 } 53:53 <br />11° 40' _ " f0 " 543 d L. C. = 2 .33.33 °a ' <br />s" <br />.86-86 'E. C: =Sta. 543+86.86 _ <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'=j <br />J 2° 1921'=def. for sta. 542. <br />1 Def. for 50 ft. =2° 30'.for a 10° Curve. <br />Def. for 36.86 ft. 1° 50r' for a 10° Curve. �ry <br />