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Jo3.Z�' 97,�ry <br />0.4'? <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. ,4 S 6 ,46 <br />HOW TO USE CURVE TABLES. <br />an. and <br />Table I. contains Tangents and Externals to i 1° dividin �heTan. <br />Ext. to any other r adius maybe f ound nearly enough, by g <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 bythe radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. Seca <br />EXAMPLE. <br />Wanted a Curve .with an Ext. of about 12 ft. Angle <br />of Intersection or 1. P. =23° 20' to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 231 20' =120.87 <br />120.87=12=10.07. Say a 10° 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 10° Cur. =0.16 <br />118.31+0.16= 118.47 =corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang.28°20'=23.33°=10=2.3333=L. C. <br />2° 1912'=def. for sta. 542 1. P. =sta. 542+72- <br />4°491x= `< `: -1x50 Tan.= 1 .18.47 <br />7° IN _ it , C. 543 <br />B. C,=sta- 541-{--53.53 <br />go 49;2'= « a a +50 2 . 33.33 <br />110 40'= « ({ , 543-x- L. C. _ <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 1912'=dcf. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.36 ft. =1° 50' for a 10° Curve. <br />s <br />S -7, 7r' G,70 <br />78 <br />3-9, I8 9E C <br />a <br />ZrL e <br />— <br />J <br />1 S <br />_ <br />1 <br />I <br />.i <br />Jo3.Z�' 97,�ry <br />0.4'? <br />CURVE TABLES. <br />Published by KEUFFEL & ESSER CO. ,4 S 6 ,46 <br />HOW TO USE CURVE TABLES. <br />an. and <br />Table I. contains Tangents and Externals to i 1° dividin �heTan. <br />Ext. to any other r adius maybe f ound nearly enough, by g <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 bythe radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. Seca <br />EXAMPLE. <br />Wanted a Curve .with an Ext. of about 12 ft. Angle <br />of Intersection or 1. P. =23° 20' to the R. at Station <br />542+72. <br />Ext. in Tab. I opposite 231 20' =120.87 <br />120.87=12=10.07. Say a 10° 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 10° Cur. =0.16 <br />118.31+0.16= 118.47 =corrected Tangent. <br />(If corrected Ext. is required find in same way) <br />Ang.28°20'=23.33°=10=2.3333=L. C. <br />2° 1912'=def. for sta. 542 1. P. =sta. 542+72- <br />4°491x= `< `: -1x50 Tan.= 1 .18.47 <br />7° IN _ it , C. 543 <br />B. C,=sta- 541-{--53.53 <br />go 49;2'= « a a +50 2 . 33.33 <br />110 40'= « ({ , 543-x- L. C. _ <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.)=139.41'= <br />2° 1912'=dcf. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.36 ft. =1° 50' for a 10° Curve. <br />