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CURVE TABLES. <br /> Published by KEUFFEL ESSER CO. <br /> ----_-___ HOW TO USE CURVE TABLES. <br /> Table I. contains Tangents and Externals to a P curve. Tan,and <br /> --- - Fxt.to any other radius maybe found nearly enough,bydividing theTan. <br /> ne 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 Centralle by the given Tangent. <br /> To find Deg. of Curve havingthe entry given <br /> Angle and External: <br /> Divide Ext.opposite the given Central le by te External. <br /> To find Nat.Tan.and Nat.Ex.Sec.for an angle by Table I.:Tan. <br /> or Ext.of twice the given angle divided by the radius of a V 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-i-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 /> i (If corrected Ext.is required find in same way) <br /> 27Ang.23'20'=23.33 =10=2.3333=L.C. <br /> 2°19+}'=def.for sta. 542 1.P.=sta. 542+72 <br /> 40 49}'= " " " +50 Tan.= 1 .18.47 <br /> 3. g U L .9 7 7°19j'= " " " 543 B.C.=sta. 541 53.53 <br /> tr' _ 9°494'= " " " +50 <br /> 3, l .b 4 e le 1p gOr�,Z- " a 543+ L.C.= 2 .33.33 <br /> S Me ..''09r is -A"S ices 86.86 E.C.=Sta. 543+86.86 <br /> 7.7 C/ ehi5*eA <br /> - — J00 53= .47X3'(def.for 1 ft.of 10'Cur.)=139.41'= <br /> ';04.f.-le ,s tiI a 2'191'=def.for sta.542. <br /> -$A&pia Ijef.'for 50 ft.=2'30'for a 10'Curve. <br /> S-. g 1L_ 7. D L Def.for 36.86 ft.=1*50j'for a 10'Curve. <br /> r , �5 ;7 oF�od S,d` a)(, <br /> j we 41,o 9 B IAAn9.23.201 <br /> N <br /> _ 3 J O 0 S :� 10•curve <br /> — — ls: <br /> �' <br />