Laserfiche WebLink
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 /> Ext.to any other radius may,befound nearly enough,bydividing theTan. <br /> or Ext.opposite the given Central Angle by thegiven degree of curve. <br /> --- - . <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 /> - _ -- <br /> EXAMPLE. <br /> Wanted a Curve with an Ext. of about 12 ft. Angle <br /> -- <br /> - - - - of intersection or I. P.=23° 20' to the R. at Station <br /> 542-1-72. <br /> Ext.in Tab. I opposite 231 20,=120.87 <br /> -- — - 120.87 : 12=10.07. Say a 10°Curve. <br /> Tan.in Tab. I o 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 /> 3� y6 > d y (If corrected Ext.is required find in same way) <br /> Ang.23°20'=23.33 =10=2.3333=L.C. <br /> 2°19j'=def.for sta. 542 I. P.=sta. 542 -72 <br /> - -- - - -- - p D'f — 40 4912'_ " « It -1-50 Tan.= 1 .18.47 <br /> If 2 70 1912'_ " « a 543 <br /> 904912'= a " a +50 B.C.=sta. 541 F53.53 <br /> 11°40'= " " " 543+ L. C.= 2 .33.33 <br /> 104 7 86.86 E.C.=Sta. 543x-•86.86 <br /> 6 3foo _. <br /> 7.49 _ 7_. 100-53.53=46.47X3'(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 2°1912'=def.for sta.542. <br /> -- Def.for 50 ft.=20 30'for a 10°Curve. <br /> /L Z_fvo 01, � Def.for 36.86 ft.=10 5012'for a 10°Curve. <br /> e, <br /> >� <br /> I.P.A"q*23•R0' <br /> �I•g 7i3:� eya 4> <br /> 10•Cur" - <br /> IIIOr•�j G'��� ';a o <br /> 1G_OfQ�_ lod.r3 7.9 S 0- � � <br /> w <br />