Laserfiche WebLink
CURVETABLES. " <br />Published by KEUFFEL & ESSER CO. <br />H®W TO USE CURVE TABLES. <br />Table I.- contains Tangents and Externals to. a 19 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 tbe.-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 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 as Ext. of about 12 ft. Angle <br />,)of Intersection or I. P. =230..201 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° 21Y=1183.1 <br />1183.1=10 =118.31. <br />Correction for A: 23° 20' fora 10° Cur. =0.16 <br />118.31+0.16* 118.47 Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang. 23'20'.=23.33 =10=2.3333=L. C. <br />2° 192'=def. for sta. 542 I: P. =sta. 542-72 <br />40,4921'= " `° " • +50 Tan. - 1 .18.47 <br />' 7' 191 <br />9°49 -I-50 191'= [( If R543 <br />$'= " P''C..=sta. 541-1-53.53 <br />11° 40'= °° " « 543+. L' C'= . 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47.X3'(def. for 1 ft. of 10° Cur:) =139.41'= <br />2° 1912'= def. for sta. 542. . <br />Def, for 50 ft. =2`30"for a 10° Curve. - <br />n Def. for 36.86 ft.=1° 501' for a 10° Curve. <br />d'y <br />_ ax <br />�a <br />B <br />. ._. 10°Curve _.. ... <br />