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FIELD g <br /> OK 1457. <br /> CURVE TABLES. <br /> Published by KEUFFEL & ESSER CO. <br /> HOW TO USE CURVE TABLES. <br /> Table I. contains Tangtnto and Externals to a 1°curve. Tan.and <br /> Ext.to any other radius may befound nearly enough,bydividing the Tan. <br /> i <br /> or Ext.opposite the given Central Anile by the gi'ven degree of curve. <br /> To find Deg, of Curve,having the Central 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 External: <br /> — Divide Ext. o asite the <br /> p ven Central Angle by the given External. <br /> I <br /> To find Nat.T and Nat.Ex.Sec.for any angle by Table L:Tan. <br /> -- or Ext.of twice the even angle divided by the radius of a V curve will <br /> be the Nat.Tan.or Nat.Ex.Sec. <br /> - <br /> EXAMPLE. <br /> --- - - Wanted a Curve with an Ext. of about 12 ft. Angle <br /> of Intersection or 1. P.=230 20' to the R. at Station <br /> 542+72. <br /> Ext.in Tab. I opposite 23°20'=120.87 <br /> -_.---- - __--- <br /> 120.87+12=10.07. Say a 10°Curve. <br /> r 6 J-3 Tan.in Tab. I opp.23°20'=1183.1 <br /> (o.ter': l '!. —' 1183.1+10=118.31. <br /> Correction <br /> 16 <br /> 0.16 118.4 corrected Tangent. <br /> 4"/ !r.5 / (If corrected Ext,is!/ required find in same way)Ang.23°20'=23.33 =10=2.3333=L.C. <br /> 2°19Y-def.for its. 542 I. P.=sta. 542+72 <br /> 40 491'= a a a +50 Tan.= 1 .18.47 <br /> 70 19j'= a a a 543 <br /> — 90491'= a a a B.C.=sta. 541+53.53 <br /> �j'_Z 11°40'= a a a 5a+ L.C.= 2 .33.33 <br /> 86.86 E.C.=Sta. 543+86.86 <br /> _-. <br /> 100-53.53=46.47 X 3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 2°191'=def.for sta.542. <br /> �:4✓rw,7�w i __ <br /> Def.for 60 ft.-2°30'for a 10°Curve. <br /> •^�-' = . ` ._d. _1 I C?, <br /> IT- Def.for 36.86 ft.=1°501'for a 10°Curve. <br /> s� <br /> ?x <br /> .P.An9.23°E0' <br /> 9c 1 ^ .K i �'. Fr Lr v a <br /> X67_ -- ro*o� <br />