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Y 7'.3 <br />- . CURVE TABLES <br />-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 Arid Tangent: <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />!3_ pL C 0, fTo 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 the radius of a;l° curve will <br />G Q $ 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.=231 20' to the R. at Station <br />3. In O f7 3 542 = 72. ' <br />Ext. in;Tab. I opposite 23° 20' =120.87 <br />i 120.87=12=10.07. Say alot. Curve. <br />Tan. in Tab. I opp. 23° 20'=1183.1 <br />.3. aZ e jJ^ j� 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 />C? (If corrected. Ext. is required find in same way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C: • <br />:2° 192' = def. for sta. . 542 I. P. = sta. 542 x-•72 <br />4° 492• = " " ; It +50 Tan. = 1 .18.47 <br />�� �C// � 3 • (. 3 8�-� > 3 D . 7° 192' - it " ` « 543 B. C. =stat. 541+53.53 <br />9'49"= " " . +50 _ <br />11° 40' = " " `f 543 } L. C. 2 .33.33 <br />�86-86 <br />�7 g7 3� 3 86.86 E.. of 1060 543-.41'= <br />�C 100-53.53=46.47X3'(def. for 1 ft. of 10' Cur.)=139.41'= <br />/_3 1W3( 3. <br />2°'192'=def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 502' fora 10° Curve. <br />a4A <br />�x �� <br />tl, I! <br />IIS � <br />I' <br />f I <br />Y 7'.3 <br />- . CURVE TABLES <br />-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 Arid Tangent: <br />Divide Tan. opposite the given Central Angle by the given Tangent. <br />!3_ pL C 0, fTo 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 the radius of a;l° curve will <br />G Q $ 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.=231 20' to the R. at Station <br />3. In O f7 3 542 = 72. ' <br />Ext. in;Tab. I opposite 23° 20' =120.87 <br />i 120.87=12=10.07. Say alot. Curve. <br />Tan. in Tab. I opp. 23° 20'=1183.1 <br />.3. aZ e jJ^ j� 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 />C? (If corrected. Ext. is required find in same way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C: • <br />:2° 192' = def. for sta. . 542 I. P. = sta. 542 x-•72 <br />4° 492• = " " ; It +50 Tan. = 1 .18.47 <br />�� �C// � 3 • (. 3 8�-� > 3 D . 7° 192' - it " ` « 543 B. C. =stat. 541+53.53 <br />9'49"= " " . +50 _ <br />11° 40' = " " `f 543 } L. C. 2 .33.33 <br />�86-86 <br />�7 g7 3� 3 86.86 E.. of 1060 543-.41'= <br />�C 100-53.53=46.47X3'(def. for 1 ft. of 10' Cur.)=139.41'= <br />/_3 1W3( 3. <br />2°'192'=def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 10° Curve. <br />Def. for 36.86 ft. =1° 502' fora 10° Curve. <br />a4A <br />�x �� <br />