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C <br />'-CURVE TABLES. - <br />Published by-,KEUFFEL 8r, ESSER CO. <br />-HOW TO USE. CURVE TABLES. <br />-Table I. contains Tangents and Externals to a1° curve. Tam and <br />Ext. to any other radius may be found nearly enough, by dividing theTan. <br />or Ext. opposite the given Cential 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 Ile by the given Tangent. <br />To find Deg.. of Curve; ha�viiig the entral,Angle and External: <br />Divide Ext. opposite the givefi.-C6tral Angle by the given External. <br />Ext. <br />find Nat. Tam and Nat. ;Ex. Sec. for any angle by Table I.: Tan. <br />&Extoftwice the the radius of a 1' curve will <br />be the Nat. Tan. or Nat. <br />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, Sayja10PCurve. <br />Tan. in T . <br />Tab. I opp. 23' 20'= 1183.1 <br />C6rriction for A. 23*20'for a 10P Cur.=0.16 <br />118.31+0.1.6=118.47 =corrected Tangent. <br />(If corrected Ext: is requited find in same way) <br />Ang.,23*20'=23.330--10=2.3333=L..C. <br />2* 19Y def. for sta. 542 Li P. = sta. 542+72 <br />4- 49j'= +50 Tan. = 1 .18.47 <br />19", 543 <br />604911= B. C. sta. ;,541-X53.53 <br />2 , +50 H3 +�, L. C.= 2 .33.33 <br />110 40.y =� I . I - <br />86.86 E. C. - Sta." 543--86.86 <br />100-53.53=46.47X3'(def. for I ft.. of -10° Cur.) =139.41'- <br />20 1911 def. for sta.' 542. <br />2 <br />Def. for 50 ft. 2* 30' for a 10° Cufve. <br />Def. - for 36.86 ft. =1° 501' for a 10° Curvr- <br />Zo <br />0 <br />NN\ <br />00 A/ro C <br />a. <br />/a, <br />3 <br />7 <br />4' r <br />3.9i <br />4- <br />1l <br />1 <br />1A <br />- <br />Rai <br />S' <br />- A, <br />W. b <br />0.2 <br />+5 <br />(-Z.9 <br />tb <br />33 <br />Q, <br />OL <br />/_0004 <br />"a <br />%a <br />1 6'/ <br />I": mtl <br />-1.7S <br />.v <br />SQL7- <br />f <br />V <br />'-CURVE TABLES. - <br />Published by-,KEUFFEL 8r, ESSER CO. <br />-HOW TO USE. CURVE TABLES. <br />-Table I. contains Tangents and Externals to a1° curve. Tam and <br />Ext. to any other radius may be found nearly enough, by dividing theTan. <br />or Ext. opposite the given Cential 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 Ile by the given Tangent. <br />To find Deg.. of Curve; ha�viiig the entral,Angle and External: <br />Divide Ext. opposite the givefi.-C6tral Angle by the given External. <br />Ext. <br />find Nat. Tam and Nat. ;Ex. Sec. for any angle by Table I.: Tan. <br />&Extoftwice the the radius of a 1' curve will <br />be the Nat. Tan. or Nat. <br />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, Sayja10PCurve. <br />Tan. in T . <br />Tab. I opp. 23' 20'= 1183.1 <br />C6rriction for A. 23*20'for a 10P Cur.=0.16 <br />118.31+0.1.6=118.47 =corrected Tangent. <br />(If corrected Ext: is requited find in same way) <br />Ang.,23*20'=23.330--10=2.3333=L..C. <br />2* 19Y def. for sta. 542 Li P. = sta. 542+72 <br />4- 49j'= +50 Tan. = 1 .18.47 <br />19", 543 <br />604911= B. C. sta. ;,541-X53.53 <br />2 , +50 H3 +�, L. C.= 2 .33.33 <br />110 40.y =� I . I - <br />86.86 E. C. - Sta." 543--86.86 <br />100-53.53=46.47X3'(def. for I ft.. of -10° Cur.) =139.41'- <br />20 1911 def. for sta.' 542. <br />2 <br />Def. for 50 ft. 2* 30' for a 10° Cufve. <br />Def. - for 36.86 ft. =1° 501' for a 10° Curvr- <br />Zo <br />0 <br />NN\ <br />