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.CURVE TABLES <br />ublished by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />mins Tangents and Externals to a 1' curve. Tan. and <br />adius may be found nearly enough, 6ydividing theTan. <br />he given Central Angie by the given degree of curve. <br />of Curve, having the Central Angle .and Tangent: <br />iite the given Central Angle by the given Tangent. <br />of Curve, having the Central Angle and External: <br />ite the given Central Angle by the given External. <br />Pan. and Nat. Ex, Sec. for any angle by.Table I.: Tan. <br />ie given angle divided by the radius of a 1' curve will <br />)r Nat. Ex. Sec. <br />EXAMPLE <br />ed a Curve with an Ext. of about 12 ft. Angle <br />ction or I. P. =23' 20' to the R. at Station, <br />:xt. in Tab. I opposite 23' 20' =120,87 <br />120.87-t-12=10.07. Say a 10° Curve. <br />Tan. in Tab. I opp. 23' 20' =1183.1 <br />1193.1 X10=118.31. <br />rection for A. 23' 20' for a 10' Cur. =0.16 <br />;.31+Q.16'= 118.47 =corrected Tangent. <br />orrected Ext. is mired find in same way) <br />Tu <br />ung. 23' 20'=23"33 1.10=2.3333=L. C. <br />ef. for sta. 542 L P. =sta. 542+72 <br />" +50 Tan. = 1 .18.47 <br />543 B. C. sta. 541+53.53 <br />q u u +50 <br />a „ 548+ L. C. = 2 .33.33 <br />86.86 E. C.=Sta. 543+86M <br />—46.47XT(def. for 1 ft. of 10' Cur.) =139,41' = <br />j+)&2' :m.2' 1911' = def. for sta. 542. <br />)cf. for 50 ft. =2' 30' for a 10' Curve. <br />d. for 36.86 ft.= I* 50}' for a i0' Curve. <br />a 8 LP.An9.231Ytl' -� <br />Q <br />/ i•s <br />iG•Gury� O% `, <br />J <br />\ a C3 <br />4% <br />ll <br />P <br />r <br />Table I. cont <br />Ext. to any other t <br />,I�" e <br />E/ - <br />p Ext. opposite t <br />To find. Deg. <br />fl <br />A� <br />Divide Tan; oppo <br />To find Deg. <br />%Divide Ext. oppos <br />}� <br />t t To find Nat. <br />Ext. of twice tl <br />�*j! <br />A�zY <br />°o j <br />- ; 1 ±be the Nat. Tan. <br />Want <br />;_ <br />of Inters( <br />542 { 72. <br />_ <br />i <br />33•b <br />71 <br />LIO Oa. <br />17'? <br />M <br />— <br />_ <br />oye <br />1 i <br />Cor <br />(If c <br />Q <br />2' 19"=d <br />. <br />N 5� iT LEUEY N <br />Q <br />` 4' 49� <br />7° 19,E = <br />11' 40'= <br />100-53.53 <br />G v <br />a <br />- - <br />I ` <br />l <br />ems® ra�rmsma <br />.CURVE TABLES <br />ublished by KEUFFEL & ESSER CO. <br />HOW TO USE CURVE TABLES <br />mins Tangents and Externals to a 1' curve. Tan. and <br />adius may be found nearly enough, 6ydividing theTan. <br />he given Central Angie by the given degree of curve. <br />of Curve, having the Central Angle .and Tangent: <br />iite the given Central Angle by the given Tangent. <br />of Curve, having the Central Angle and External: <br />ite the given Central Angle by the given External. <br />Pan. and Nat. Ex, Sec. for any angle by.Table I.: Tan. <br />ie given angle divided by the radius of a 1' curve will <br />)r Nat. Ex. Sec. <br />EXAMPLE <br />ed a Curve with an Ext. of about 12 ft. Angle <br />ction or I. P. =23' 20' to the R. at Station, <br />:xt. in Tab. I opposite 23' 20' =120,87 <br />120.87-t-12=10.07. Say a 10° Curve. <br />Tan. in Tab. I opp. 23' 20' =1183.1 <br />1193.1 X10=118.31. <br />rection for A. 23' 20' for a 10' Cur. =0.16 <br />;.31+Q.16'= 118.47 =corrected Tangent. <br />orrected Ext. is mired find in same way) <br />Tu <br />ung. 23' 20'=23"33 1.10=2.3333=L. C. <br />ef. for sta. 542 L P. =sta. 542+72 <br />" +50 Tan. = 1 .18.47 <br />543 B. C. sta. 541+53.53 <br />q u u +50 <br />a „ 548+ L. C. = 2 .33.33 <br />86.86 E. C.=Sta. 543+86M <br />—46.47XT(def. for 1 ft. of 10' Cur.) =139,41' = <br />j+)&2' :m.2' 1911' = def. for sta. 542. <br />)cf. for 50 ft. =2' 30' for a 10' Curve. <br />d. for 36.86 ft.= I* 50}' for a i0' Curve. <br />a 8 LP.An9.231Ytl' -� <br />Q <br />/ i•s <br />iG•Gury� O% `, <br />J <br />\ a C3 <br />4% <br />