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_ t ` N�0 - _ _ __ CURVE TABLES. <br /> L s 2 2 s= Published by KEUFFEL 86 ESSER CO. <br /> HOW TO USE CURVE TABLES. <br /> 2 _ 3 AX' ? Table I. contains Tangents and Externals to a'1°curve. Tan.and <br /> - Ext.to any other radius maybe found nearly enough,bydividing the Tan. <br /> or Ext.opposite the given Central Angle by the given degree of curve. <br /> - - - - To find Deg. of Curve, having the CentrEl Angle and Tangent: <br /> Divide Tan.opposite the given Central Angle by the given Tangent. <br /> To find Deg. of Curve, having the Central Angle and Exterhal: <br /> Divide Ext.opposite the given Central Angle by the given External. <br /> Q P 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 1°curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> - - 3,11 ' - EXAMPLE. <br /> St Wanted a Curve with an Ext.of about 12 ft. Angle <br /> of Intersection or I. P.=230 20' to the R. at Station <br /> j - 542+72. <br /> Ext.in Tab. I 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 /> C. _ 1183.1=10=118.31. <br /> Correction <br /> 31-{-0.16 118.47� 16 <br /> =corrected Tangent. <br /> ! �> (If corrected Ext.is required find in same way) <br /> �a W Ang.23°20'=23.33 =10=2.3333=L.C. <br /> 2°19j'=def.for sta. 542 .L P.=sta. 542+72 <br /> 4°49,'= ,� u a <br /> i x-50 Tan.= 1 .18.47 <br /> ,j 70 19x1 cc 543 <br /> 9049211= « u « +50B. C.=sta. 541 -53.53 <br /> t , CIA r1O 11°40'= k u 543+ L. C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543+86.88 <br /> - j - - - ---- 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'=) <br /> } 2°19;'=def.for sta.542. <br /> 0C,ern 1 Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°50}'for a 10°Curve. <br /> TYa vJ u s <br /> � <br /> S. -- <br /> q' � ✓� � ! I.P.An9.23•20•' <br /> q <br /> 3� i `�Zd . <br />