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CURVE TABLES <br />- ,Published by KEUFFEL & ESSER CO. <br />!! HOW JO USE CURVE TABLES <br />-Table I. contains Tangentsand Externals to a l° curve. Tan: and <br />Ext. to any other radius:may lie found nearly enough, by dividing the Tan. <br />or Ext. opposite the given. Central Angle by the given degree of curve. <br />d 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 />J or Ext.'of twice the given angle divided by the radius of a .1° curve will <br />I: be the Nat. Tan. or Nat. Ex. See. - <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. =0.16 <br />() 118.31-1-0.16 =118.47 =corrected Tangent.' <br />4 (If corrected Ext. is required find in same way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C: <br />2°.192'=def. for sta.. 542 I. P. =sta. 542+72 <br />'404911— 2 44 It +50 Tan. = 1 .18.47 <br />'7° 192'= " " " 543 <br />9° 491' "` +50 B.' C. =sta. 541+53.53 <br />L <br />11°'40, ". " 543+ L. C. 2 .33.33 <br />86.86 E. C. = Sta. 543 86.86 <br />t.' 100-53.53=46.47X3'(def.•for 1 ft.:of 10° Cur.) =139.41'=. <br />- 2° 1921'=def. for sta. 542. <br />Def, for'50 ft. =20 30' for a 10° Curve. <br />i Def. for 36.86 ft. =1° 50" fora 10° Curve. <br />i Sq • <br />? IAAn9.23°20' <br />) <br />c -- <br />Zf.�Gt� 0.7 �i J <br />4//a/ ANN (7 <br />-+ <br />f <br />\ �••� <br />J <br />N <br />� o <br />6/ <br />vt,;, <br />oz' <br />v-zro� uji <br />fi <br />tb <br />i <br />ti k <br />k k k <br />• L <br />N Ut w <br />Cp pa o i <br />o <br />\ <br />u <br />� <br />v <br />e <br />% m IN <br />/ <br />Z <br />Zt <br />V G <br />v <br />�A <br />I otR <br />CURVE TABLES <br />- ,Published by KEUFFEL & ESSER CO. <br />!! HOW JO USE CURVE TABLES <br />-Table I. contains Tangentsand Externals to a l° curve. Tan: and <br />Ext. to any other radius:may lie found nearly enough, by dividing the Tan. <br />or Ext. opposite the given. Central Angle by the given degree of curve. <br />d 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 />J or Ext.'of twice the given angle divided by the radius of a .1° curve will <br />I: be the Nat. Tan. or Nat. Ex. See. - <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. =0.16 <br />() 118.31-1-0.16 =118.47 =corrected Tangent.' <br />4 (If corrected Ext. is required find in same way) <br />Ang. 23° 20'=23.33°=10=2.3333=L. C: <br />2°.192'=def. for sta.. 542 I. P. =sta. 542+72 <br />'404911— 2 44 It +50 Tan. = 1 .18.47 <br />'7° 192'= " " " 543 <br />9° 491' "` +50 B.' C. =sta. 541+53.53 <br />L <br />11°'40, ". " 543+ L. C. 2 .33.33 <br />86.86 E. C. = Sta. 543 86.86 <br />t.' 100-53.53=46.47X3'(def.•for 1 ft.:of 10° Cur.) =139.41'=. <br />- 2° 1921'=def. for sta. 542. <br />Def, for'50 ft. =20 30' for a 10° Curve. <br />i Def. for 36.86 ft. =1° 50" fora 10° Curve. <br />i Sq • <br />? IAAn9.23°20' <br />