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CURVE TABLES. <br />published'by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a b d iding the Tan. <br />Ext, <br />to any other radius may be f ound nearly enough, y <br />or Ext. opposite the given Central Angle by the given degree of curve. <br />To find Deg. of Curve, having the Central Angle en Tangent nd t <br />Divide Tan. opposite the given Central Angle by <br />To find Deg. giveof 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 ice t or given Se divided by the radius of a 1° curve will <br />he Nat at. Ex. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 13 ft. Angle <br />of Intersection or I. P.=230 20' to the R. at Station <br />542+72. <br />Ext. in Tab. f opposite 23° 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 />C 18.31+0.16 A. <br />'o.16 <br />47 �cotd rr cTangent. <br />If corrected Ext. is re wired find in same way) <br />33 <br />Ang.23°20'=23.=10=2.3333=L. C. ' <br />542+72 <br />2' 102' =def. for sta. 542 I. P. =sta. 1 .18.47 <br />4° 492' _ « +50 Tan. = <br />l 7° 192'= « « a 543 B. C.=std. 54t� 53.53 <br />9° 491' _ « « 1 50 2 .33.33 <br />L. C. _ <br />110 4;t a cc 543+ <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' l92'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a10* Curve - <br />Curve. <br />Def for 36.86ft— OZ <br />sa <br />ax <br />P'1 V <br />`\9 <br />t0° Curve <br />• N <br />s 4°% <br />\\ )c O j <br />'P ' <br />I <br />-- - <br />---moi <br />-�-- <br />- <br />CURVE TABLES. <br />published'by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES. <br />Table I. contains Tangents and Externals to a b d iding the Tan. <br />Ext, <br />to any other radius may be f ound nearly enough, y <br />or Ext. opposite the given Central Angle by the given degree of curve. <br />To find Deg. of Curve, having the Central Angle en Tangent nd t <br />Divide Tan. opposite the given Central Angle by <br />To find Deg. giveof 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 ice t or given Se divided by the radius of a 1° curve will <br />he Nat at. Ex. <br />EXAMPLE. <br />Wanted a Curve with an Ext. of about 13 ft. Angle <br />of Intersection or I. P.=230 20' to the R. at Station <br />542+72. <br />Ext. in Tab. f opposite 23° 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 />C 18.31+0.16 A. <br />'o.16 <br />47 �cotd rr cTangent. <br />If corrected Ext. is re wired find in same way) <br />33 <br />Ang.23°20'=23.=10=2.3333=L. C. ' <br />542+72 <br />2' 102' =def. for sta. 542 I. P. =sta. 1 .18.47 <br />4° 492' _ « +50 Tan. = <br />l 7° 192'= « « a 543 B. C.=std. 54t� 53.53 <br />9° 491' _ « « 1 50 2 .33.33 <br />L. C. _ <br />110 4;t a cc 543+ <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' l92'=def. for sta. 542. <br />Def. for 50 ft. =2° 30' for a10* Curve - <br />Curve. <br />Def for 36.86ft— OZ <br />sa <br />ax <br />P'1 V <br />`\9 <br />t0° Curve <br />• N <br />s 4°% <br />\\ )c O j <br />'P ' <br />