Laserfiche WebLink
-z CURVE TABLES. <br /> Published by KEU"ZL & ESSER CO. <br /> i <br /> HOW TO USE CURVE TABLES. <br /> nearly . <br /> Ext.o an I. contann�Tangents and Externals to a 1°curve. Tan.and <br /> or Ext.opposite y ttheus maybe found <br /> l An b theme van dividing theTan. <br /> rg y g agree of curve. <br /> To find Deg. of Curve,having the Central Angle and Tangent: <br /> Divide Tan.opposite the given Central <br /> - � To fin Anile by the given Tangent. <br /> 30 90 - rr d Deg. of Curve, having the Central Angle and External: <br /> �j ; -/� vE •-.-'J Divide Ext.opposite the given Central Angle by the given External <br /> G� OGTo find Nat.Tan.and Nat.Ex.Sec.for any angle by Table I.:Tan. <br /> _ or Eat.of twice the given angle divided by the radius of a 1°curve will <br /> be the Nat.Tan.or,Nat.Ex.See. <br /> ¢�, ; ✓ si >, EXAMPLE. <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+72. <br /> _ — Ext.in Tab. I o ° <br /> 120.87=12=1007. Say a 0*Curve. ' <br /> Tan.in Tab. I opp.23°20'=1183.1 <br /> _ 1183.1-=-10=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.16 <br /> _ 118.31-{-0.16=118.47=corrected Tangent. <br /> (If corrected Ext.is required find in same way) <br /> Ang.23°20'=23.33 +10=2.3333=L.C. <br /> 2°191'=def.for sta. 542 I. P.=sta. 542+72 <br /> 40 4912'= " " " +50 Tan.= 1 .18.47 <br /> 345 70 19121_ " " " 543 <br /> 904912'= " " " B.C.=sta. 541-}-53.53 <br /> 110 40'= " " 'a 54� L. C.= 2 .33.33 <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°1912'=def.for sta.542. <br /> - Def.for 50 ft.=20 30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°5012'for a 10°Curve. <br /> __ •a <br /> � x) <br /> a IAAn9.23.201 <br /> I� <br /> D1 <br /> 10'Curve <br /> y <br />