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I <br />- CURVE TABLES. : <br />t Published by KEUFFEL & 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 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 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 tlie 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 10` Curve. <br />Tan. in Tab. I opp. 236 20'=118.3.1 <br />1183.1=10 =118.31. <br />Correction for A. 23° 20' for a 10" Cur. =0.16. <br />118.31+0.16=118.47=corrected Tangent. <br />(If corrected Ext. is reuired find in same way) <br />�Ang. 23' 20'=23.331 =10 = 2.3333 = L. C. <br />2° 19;'='def. for sta. ' 542 1. P. =sta. 542-1••72 <br />40 49,' _ " " q +50 Tan. <br />7° 19;' _ " " 543 <br />90 49;'= " " " -1-50 B. C.-sta. 541 } 53.53 <br />11° 40' " " 543 { L. C.= , 2 • .33.33 <br />86.86 E. C.=Sta. 543--86.86 <br />,100-53:53=46.47X.4 (def. ford ft' 6'(10° Cur.) 139.41'= <br />2° 19j'=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° Curve. <br />J, <br />z$/ y� / <br />4T <br />S3 <br />. SO �' i2e S 7-&�l 716.k <br />f / <br />� <br />f <br />I1 <br />�" `'s <br />0 <br />ass a g <br />r <br />I <br />�c <br />or <br />Z LP Li <br />� h' � <br />J_ <br />f• �,r l- P �� <br />� <br />c <br />� o <br />I <br />- CURVE TABLES. : <br />t Published by KEUFFEL & 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 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 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 tlie 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 10` Curve. <br />Tan. in Tab. I opp. 236 20'=118.3.1 <br />1183.1=10 =118.31. <br />Correction for A. 23° 20' for a 10" Cur. =0.16. <br />118.31+0.16=118.47=corrected Tangent. <br />(If corrected Ext. is reuired find in same way) <br />�Ang. 23' 20'=23.331 =10 = 2.3333 = L. C. <br />2° 19;'='def. for sta. ' 542 1. P. =sta. 542-1••72 <br />40 49,' _ " " q +50 Tan. <br />7° 19;' _ " " 543 <br />90 49;'= " " " -1-50 B. C.-sta. 541 } 53.53 <br />11° 40' " " 543 { L. C.= , 2 • .33.33 <br />86.86 E. C.=Sta. 543--86.86 <br />,100-53:53=46.47X.4 (def. ford ft' 6'(10° Cur.) 139.41'= <br />2° 19j'=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° Curve. <br />J, <br />