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0 <br />I <br />CURVE TABLES . <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />contains Tangents and Externals to a 1° curve. Tan. and <br />Cher radius maybe found nearly enough, by dividing theTan. <br />site the given Central Angle by the given degree of curve. <br />Deg. of Curve, having the. Central Angle and Tangent: <br />opposite the given Central Angle by the given Tangent. <br />Deg. of Curve, having the Central Angle and External: <br />opposite the given Central Angle by the given External. <br />Nat_ Tan. and Nat. Ex. Sec. for any angle by Table T.: Tan. <br />rice the given angle divided by the radius of a 1° curve will <br />Tan. or Nat. Ex. Sec. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />ntersection or I. P.=23° 20' to the R. at Station <br />+72. <br />Ext. in Tab. I opposite 23° 20' =120.8 7 <br />t20.87+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.I6=118.47=corrected Tangent. <br />(if corrected Ext. is required find in sante way) <br />Ang.23° 20'=23.33°_10=2.3333=L. C. <br />9'1'=def. for sta.542 1. P.=sta. 542+72 <br />i91, _ +50 an. = i .1$.47 <br />l9#'— 543 B. C.=sta. 541-53.53 <br />+50 L. C. = 2 .33.33 <br />40'— 543+ <br />86.86 E. C. T Sta. 543+86.86 <br />—53.53=46.47x3'(def. for 1 ft. of 10° Cur.)=Y39•41'= <br />2° 19'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10, Curve., <br />Def. for 36.86 ft. =1° 501' for a 10` Curve. <br />a <br />r4� <br />77 <br />Table L <br />A. to any o <br />Ext, oppe <br />To find <br />wide Tan_ <br />To find <br />c _ <br />:- <br />< <br />j <br />2,n a E e <br />rvide Ext. <br />G N� To find <br />Ext. of 0 <br />i the Nat. <br />;' <br />of <br />542 <br />F' <br />2° <br />4° <br />7' <br />11' <br />100 <br />r <br />Ole <br />I <br />CURVE TABLES . <br />Published by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />contains Tangents and Externals to a 1° curve. Tan. and <br />Cher radius maybe found nearly enough, by dividing theTan. <br />site the given Central Angle by the given degree of curve. <br />Deg. of Curve, having the. Central Angle and Tangent: <br />opposite the given Central Angle by the given Tangent. <br />Deg. of Curve, having the Central Angle and External: <br />opposite the given Central Angle by the given External. <br />Nat_ Tan. and Nat. Ex. Sec. for any angle by Table T.: Tan. <br />rice the given angle divided by the radius of a 1° curve will <br />Tan. or Nat. Ex. Sec. <br />EXAMPLE <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />ntersection or I. P.=23° 20' to the R. at Station <br />+72. <br />Ext. in Tab. I opposite 23° 20' =120.8 7 <br />t20.87+12=10:07. Say a 10° Curve. <br />Tan. in Tab. I opp. 23° 20'= 1183.1 <br />1183.1-10 =118.31. <br />Correction for A. 23° 20' for a 10° Cur. =0.16 . <br />118.31+0.I6=118.47=corrected Tangent. <br />(if corrected Ext. is required find in sante way) <br />Ang.23° 20'=23.33°_10=2.3333=L. C. <br />9'1'=def. for sta.542 1. P.=sta. 542+72 <br />i91, _ +50 an. = i .1$.47 <br />l9#'— 543 B. C.=sta. 541-53.53 <br />+50 L. C. = 2 .33.33 <br />40'— 543+ <br />86.86 E. C. T Sta. 543+86.86 <br />—53.53=46.47x3'(def. for 1 ft. of 10° Cur.)=Y39•41'= <br />2° 19'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a 10, Curve., <br />Def. for 36.86 ft. =1° 501' for a 10` Curve. <br />