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CURVE-. TABLES <br />!fir <br />iPublishQd by KEUFFEL & ESSER CO. <br />r <br />-,HOW -TO: USE. -CURVE TABLES <br />Tablb I.'contains Tangents and Externals to a'° curve. Tan. and <br />'.hxt. to any other radius may be found nearly enough, by dividing the Tan. <br />a 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 />pDivide Tan: opposite the given Central Angle by the given "Tangent. <br />To find '.Deg. of Curve; -having the ,Central Angle land External: <br />; ivide 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 />; <br />y <br />; <br />T <br />Ext. of twice the given angle divided by the,radius ofa 1°• curve will <br />0e the Nat Tan. or Nat. Ex. Sec. ? <br />Wanted"a Curve with an Ext. of about 12 ft. Angle <br />of Intersection'or 1. P.=23`20' -to ,the ;R; at Station <br />542.x-72. -= <br />' <br />Ext. in Tab' I opposite 23° 20 =120.87 <br />- 120.87=12; 10.07." Say a 10' Curve. i <br />! <br />a <br />i <br />Tan. in Tab. I'' opp. 23° 20'= 1183.1 <br />1183.1=10=118.311 <br />Correction for A. 223° 20' for a 10° Cur. =0.16 <br />118.31-0.16=118.47.=corrected Tangent. <br />`' <br />"`(If corrected'Ext. isrequired find in same way) <br />' Ang..23° 20' =23.33 -10=2.3333=L. C. <br />2°.`192.' Idef. for sta.. 542 _1. P. =sta. 542+72 <br />-1-50' Tan. = 1 .18.47 <br />' <br />7° 19= �: u u - 543 <br />9° 492; _ ` " S. C. =sta. 541-1-53.53 <br />} a0 L. C. = 2 .33.33 <br />11°; 40' '� �i ..a u.. 543-1- <br />.ti <br />- <br />86.86 E. C. =Sta. 543+86.86 • <br />100_53.53.=46.47X3'(def. for 1 ft. of 10° Cur.)=MAI'= <br />- <br />2° 197'=def. for sta. 542. <br />Dei. for 50 ft. =2° 30' for a 10° Curve. -" <br />w <br />Def. for 36.86 ft. =1° 501' for a 10' Curve. <br />dx <br />LBAng.23"20' <br />- <br />` <br />- �0•Gurre <br />_ xi <br />