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1'7( ,. CURVE :TABLES. Published by KEUFFEL &'-ESSER CO.= H_ OW: TO USE °:CURVE TABLES: Table 1: "contains Tangents and Externals to 'a V curve.. Tan.'and Ext. to any other radius may be found nearly enough, bydividing the Tan. or: Ext. opposite the given -Central Angle by the, given degree of curve. To find -Deg. 'of Curve; having the Central Angle and Tangknt: Divide Tan. _opposite the given Central Angle by the given Tangent. To, find Deg. :of Curve, -having the Central Angle nd-External: Divide Ext: opposite the given.Cential Ankle by 66 -given` External. To find Nat. Tan. and Nat. Ex. Sec. for any'angle by _Table T.: Tan. or Ext. of twice the given angle divided -by the radius of a 1° curve will be the Nat. Tan. or Nat. Ex. Sec. Wanted a Curve with an Ext. of about 12 ft. Angle: of Intersection or I. P. =32 ' 20' to the R. at ,Station 542+72. Ext. in Tab. I opposite 23° 20' =120.87 12.0.87-- 12 =10.07.' ' Say a 10P Curve. Tan. in Tab. I opp. 23° 20'= 1183.1 1183.100=11831.'-- Correction 183:1=10=118.31.':Correction for A.-23° 20' for a IOP Cur. =0.16 118.31 +0.16 -118.47 =corrected Tangent. (If corrected Ext. is required find in same way) ' Ang.23°20',=23.33 =10=2.3333=L.C. 20 191'=def. for sta. " 542 I. P. =sta. .542+72 40 491' +50 .Tan. 1 .18.47 70 191,= « « " .. 543 9" 49 J, - B. C. = sta..:. 541+53.63 110 40' " u " -} + L. C. = 2 .33.33 = —' .543 + - . 86.86 E. C. =Sta. . ; ' -543+86.86 100-53.53=46.47XX(def. for l ft. of 10° Cur.) =139.41'= 2° 191 =def. for sta. 542. -Def for 50 ft. -=2' 30' for -a 100 -Curve: - - - - - Def. for 36.86 ft. =1° 501' for a 10' Curve. 67_:3 V_ 43-21-1— 7, s /D 4- 11-5- 77- 1-0. z. Ce /41 1 of mei— Zo5'.34 o -ted -�S" is:z 17 7,r 4' -5 a 175.7' 1 8 4- - -3 5- Aa ✓ 176 —2 -7L3-5-7 /z. z ✓ l iv -/57 43-/1 9 171-3 77 J 11.1 �Z-3a i'- S_ 6.7 1,.3 g o. 4 Gd - p 6 24 .2, ✓ ;� r� :a p E�� Irl-'�� z.�. 0 85.. 30 ,. CURVE :TABLES. Published by KEUFFEL &'-ESSER CO.= H_ OW: TO USE °:CURVE TABLES: Table 1: "contains Tangents and Externals to 'a V curve.. Tan.'and Ext. to any other radius may be found nearly enough, bydividing the Tan. or: Ext. opposite the given -Central Angle by the, given degree of curve. To find -Deg. 'of Curve; having the Central Angle and Tangknt: Divide Tan. _opposite the given Central Angle by the given Tangent. To, find Deg. :of Curve, -having the Central Angle nd-External: Divide Ext: opposite the given.Cential Ankle by 66 -given` External. To find Nat. Tan. and Nat. Ex. Sec. for any'angle by _Table T.: Tan. or Ext. of twice the given angle divided -by the radius of a 1° curve will be the Nat. Tan. or Nat. Ex. Sec. Wanted a Curve with an Ext. of about 12 ft. Angle: of Intersection or I. P. =32 ' 20' to the R. at ,Station 542+72. Ext. in Tab. I opposite 23° 20' =120.87 12.0.87-- 12 =10.07.' ' Say a 10P Curve. Tan. in Tab. I opp. 23° 20'= 1183.1 1183.100=11831.'-- Correction 183:1=10=118.31.':Correction for A.-23° 20' for a IOP Cur. =0.16 118.31 +0.16 -118.47 =corrected Tangent. (If corrected Ext. is required find in same way) ' Ang.23°20',=23.33 =10=2.3333=L.C. 20 191'=def. for sta. " 542 I. P. =sta. .542+72 40 491' +50 .Tan. 1 .18.47 70 191,= « « " .. 543 9" 49 J, - B. C. = sta..:. 541+53.63 110 40' " u " -} + L. C. = 2 .33.33 = —' .543 + - . 86.86 E. C. =Sta. . ; ' -543+86.86 100-53.53=46.47XX(def. for l ft. of 10° Cur.) =139.41'= 2° 191 =def. for sta. 542. -Def for 50 ft. -=2' 30' for -a 100 -Curve: - - - - - Def. for 36.86 ft. =1° 501' for a 10' Curve.