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L/ <br /> Tz�11'141'1^vz <br /> P411 Al CURVE TABLES <br /> s Published by KEUFFEL 8s ES3gRC <br /> _ <br /> HOW TO USE CURVE TABLES. <br /> Table I. contains Tangents and Externals to a 10 curve. Tan.and <br /> Ext.to any other radius maybe found nearly enough,by dividing the Tan. <br /> _-- or 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 /> Divide Tan.opposite the given Central An le by the given Tangent. <br /> ` ZS CoNr; To find Deg. of Curve, having the Central Angle and External: <br /> 1 Divide Ext. opposite the given Central Angle by the given External <br /> k ID %YG To find Nat.Tan.and Nat.Ex.Sec.for an angle by Table I.•Tan. <br /> I or Eat.of twice the given angle divided by the radius of a l0 curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> j - EXAMPLE. <br /> 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 /> 542-x-72.' <br /> !` Ext.in Tab. I opposite 230 20'=120.87 <br /> 120.87+12=10.07., Say a loo Curve. <br /> ' Tan.in Tab. I opp.23°20'=1183.1 /4 4. tt <br /> } q 1183.1=10-118.31.O —/ 8 Correction for A.230 20'for a 100 Cur.—0.16 /f <br /> 118.31-}-0.16=118.47=corrected Tangent. <br /> (If corrected Ext.is required find in same way) <br /> Ang.230 20'=23.33 =10=2.3333=L.C. <br /> 2019i'=def.for sta. 542 I. P.=sta. 542+72 <br /> _ 40 4911= a a a +50 Tan.= 1 .18.47 <br /> M o '_ a a a <br /> 90 49J'= a a a ,+50 B. C.=sta. 541+53.53. <br /> I 110 40'= 543+ L.C.= 2 .33.33 <br /> j 86.86 E. C.=Sta. 543-x-86.86 <br /> 100-53.53=46.47X3'(def.for 1 ft.of 100 Cur.)=139.41'= <br /> i' 28 19"=def.for sta.542. <br /> Def.for 50 ft.,2Y 30'for a 100 Curve. <br /> Def.for 36.86 ft.=10 50"for a 100 Curve. <br /> y >� <br /> LP.An9.23•E01 <br /> 10•Curve _ <br /> I-VI rise vial <br /> _ _ 04- <br /> 1 <br />: . <br /> ren, • - - - .. <br />