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FIELDS 7 <br /> - CURVE TABLES. <br /> Published by KEUFFEL 8s ESSER CO. <br /> --—_- - -— — - HOW TO USE CURVE TABLES. <br /> - ---- — Table I. contains Tangents and Externals to a 1°curve. Tan.and <br /> • Ext.to any other radius maybe found nearly enough,by dividing the Tan. <br /> a' or ETo find opposite <br /> t e given C Curve, Angle <br /> the Cent althe given degree <br /> nd Tangent: <br /> curve. <br /> -- —l _ - - - -- - - Divide Tan. opposite the given Central Angle by the 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 /> — - - ye -- - - — — -- EXAMPLE. <br /> Wanted a Curve with an Ext.of about 12 ft. Angle <br /> 0 <br /> _z-Y 6 - _- ---- G 5 -- <br /> 542+72. <br /> 7. Tab. I oppos to 23°20 h t Station <br /> 120.8? <br /> �- <br /> y 9,7 7, 120.87=12=10.07. Say a 10°Curve. <br /> _2 S z D Tan.in Tab. I opp.23°20'=1183.1 <br /> i 1183.1=10=118.31. <br /> Correction for A.23°20'for a 10°Cur.=0.16 I <br /> 118.31-x-0.16=118.47-corrected Tangent. <br /> (If corrected Ext.is required find in same way) <br /> _ ._ -__ .__ --- ----- b. -------- Ang.23°20'=23.33 =10=2.3333=L.C. <br /> 2°19F=def.for sta. 542 I. P.=sta. 542x-72 <br /> +50 Tan.= 1 .18.47 <br /> 70 191'= " " <br /> ---- Z 543,- <br /> B. C.=ata. 541+53.53g° = +50 <br /> rc a a L. C.= 2 .33.33 <br /> 7 eI { 11o 40'= 543+ <br /> ---- _- - --— 86.86 E.C.=Sta. 543+86.M <br /> 7 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> ----_-_-____ 2°19Y=def.for sta.542. . <br /> Def.for 50 ft.=2°30'for a 10°Curve. <br /> Def.for 36.86 ft.=1°50j'for a 10°Curve. <br /> ----- ---- ___-- x�8 LP.An9.23•Y0 <br /> 1 qy� <br /> ----- —— 10•Curet <br /> 4 <br /> ` - - <br /> A2'.. - - <br /> • <br /> .' - , . <br />