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F I f 1 C BOOK 540 .. <br /> - - CURVE TABLES. <br /> Published by KEUFFEL & ESSER CO. <br /> - HOW TO USE CURVE TABLES. <br /> Table I. contains Tangents and Externals to a 1°curve. Tan.and <br /> - -- - l�xt to any other radius maybe found nearly enough,by dividing theTan. <br /> ar 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 Angle by the given Tangent. <br /> To find Deg. of Curve, having the Central Angle and External: <br /> ----__-- -- Divide 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 /> - -- --- 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 /> -- - --- EXAMPLE. <br /> Wanted a Curve with an Ext. of about 12 ft. Angle <br /> of Intersection or I. P.=23* 20' to the R. at Station <br /> -- -- -- 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 /> 1183.1+10=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.18 <br /> --- __---- - -- --- -_--- 118.31+0.16=118.47=corrected Tangent. <br /> (If corrected Ext.is rNuired find in same way) <br /> - - - — Ang.23°20'=23.33 +10=2.3333=L. C. <br /> 2°194'=def.for sta. 542 I. P.=sta. 542+72 <br /> 4°491'= ° u " +50 Tan.= 1 .18.47 <br /> - 7°19}'= a K k 543 <br /> -- - _ - B. C.=sta. 541 -53.53 <br /> 9o4911— u a u ++50 <br /> 1 — - 11°40'_ ° . « 543+ 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°191'=def.for sta.542. <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> 1 Def.for 36.86 ft.=1°50j'for a 10°Curve. <br /> �t p / q d IP.An9.Y3•YO• <br /> _ • vSCjGs <br /> _ i3raSScif 10*Cum* <br />