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' <br /> FIELD 1300K 51 : <br /> 47 CURVE TABLES. <br /> Published by KEUFFEL & ESSER CO. <br /> HOW TO USE CURVE TABLES. <br /> y ��r Table I. contains Tangents and Externals to a V 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 /> ,2d iG�iTo find Deg. of Curve, having the Central Angle and External: <br /> — ----- - -- --- - - <br /> Divide Ext. opposite the given Central Angle by the given External. <br /> N y To 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 V curve will <br /> be the Nat.11 Tan..or Nat.Ex.Sec. <br /> E%AMPLE• . <br /> _ - I <br /> Wanted a Curve with an Ext. of about 12 ft. Aii�le <br /> — — - f of Intersection or I. P.=23' 20' to the R at Statio;i <br /> p it 542+72. <br /> Ext.in Tab. I opposite 23°20'=120.87 <br /> _2.v1 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.-0.16 <br /> 118.31- 0.16=118.47-corrected Tangent. <br /> - <br /> (If corrected Ext.is required find in same way) <br /> Ang.23°20'=23.33 T10=2.3333=L.C. <br /> 2°19j'=def.for sta. 542 1. P.=sta. 542+72 <br /> -- - -- 4'491'= a a a +50 Tan.= 1 .18.47 <br /> 7019j'= a a a 543 <br /> - 9-49}'= a a a B. C.=sta. 541 53.53 <br /> 110 40'= a a a 5"} 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 /> 1' <br /> Def.for 250 f.=2 o 030 for a 10a Curve. <br /> Def.for 36.86 ft.=1°509'for a 10°Curve. <br /> e*a <br /> x� <br /> IAAn9.23•201 <br /> e o� <br /> l <br />