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CURVE TABLES <br />Published by KEUFFEL & ESSER CO.. <br />HOW TO USE CURVE TABUS <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius maybe -found nearly enough, bydividing 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. See. for any angle by Table L: Tan. <br />or Ext. of tvvice'the given angle divided by the radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE <br />Wanted a Curve with an Ext. of abinut 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 23° 20'— 120.87 <br />1.20.87 =12 =10.07. Say a 10' Curve. <br />Tan. in Tab. I app. 23° 20'= I. 183.1 <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 recjuired find in same way) <br />. Ang. 23° 20'=23.33 =10=2.3333=L. C. <br />2' 191,'=clef. for sta. 542 I. P. =sta. 542+72 <br />4°49'j'= " " " +50 Tan.= 1 .18.47 <br />70 .19jr= " a ." 543 <br />9' 49j,= " " " +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=46.47X3'(def. for 1 ft. of 10° Cur•)='139.41'= <br />2" 19,' =def. for sta. 542. _ <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 it. =1° 501' for a 10` Curve. <br />i <br />�r <br />D�'d <br />E . <br />, <br />a <br />a ;.r 3 <br />ft <br />,. v 0 <br />77e,00 <br />.( r <br />i <br />4-'- <br />fte <br />14 <br />�, 50 {? <br />.I <br />r <br />1� <br />R <br />NY! <br />11i <br />i► <br />(r� <br />a� <br />:i <br />CURVE TABLES <br />Published by KEUFFEL & ESSER CO.. <br />HOW TO USE CURVE TABUS <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius maybe -found nearly enough, bydividing 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. See. for any angle by Table L: Tan. <br />or Ext. of tvvice'the given angle divided by the radius of a 1° curve will <br />be the Nat. Tan. or Nat. Ex. See. <br />EXAMPLE <br />Wanted a Curve with an Ext. of abinut 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 23° 20'— 120.87 <br />1.20.87 =12 =10.07. Say a 10' Curve. <br />Tan. in Tab. I app. 23° 20'= I. 183.1 <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 recjuired find in same way) <br />. Ang. 23° 20'=23.33 =10=2.3333=L. C. <br />2' 191,'=clef. for sta. 542 I. P. =sta. 542+72 <br />4°49'j'= " " " +50 Tan.= 1 .18.47 <br />70 .19jr= " a ." 543 <br />9' 49j,= " " " +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=46.47X3'(def. for 1 ft. of 10° Cur•)='139.41'= <br />2" 19,' =def. for sta. 542. _ <br />Def. for 50 ft. =2° 30' for a 10° Curve. <br />Def. for 36.86 it. =1° 501' for a 10` Curve. <br />i <br />�r <br />