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CURVE TA-BLES. t <br />• -Published by KEUFFEL l ESSER CO. <br />HOW TO USE CURVE TABLES. d <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius may befound nearly enough, bydividing the Tan. <br />or Ext. opposite the given Central An le 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 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 10 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 S ogog <br />of Intersection or 1. P. =230 20' to the R. at Station <br />542+72. <br />r <br />Ext. in Tab. I opposite 23° 20' =120.87 t I e <br />120.87-12=10.07. Say a 100 Curve. <br />Tan. in Tab. I opp. 2320'=1183.1 <br />1183.1-10=118.31. n.,._7 <br />Correction for A. 230 20' for a 100 Cur. =0.16 77• c�M . <br />113.31-1-0.16=118.47=corrected Tangent. /irt`?I <br />(If corrected Fxt. is required find in same way) 53s -N <br />• Ang. 23' 20'=23.33* -10 =2.3333 = L. C. <br />20191'=def. for sta. 542 I. P.=sta. .542-} 72 <br />40 491'= " +50 Tan. = 1 .15.47 <br />70 19 I' = " 19 cc <br />B. C.=sti. 541+53.53 <br />90 4921'-- _ " " r.+50 <br />11040, <br />= • i `° " " 543+ L. C.= 2 .33.33 <br />86.86. E. C.=Sta. 543-1-86.86 <br />100-53.53=•46.47X3'(def. for 1 ft. of 100 Cur.) =139.41'= <br />20 IN' =def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 100 Curve. <br />c. Def. for 36.86 ft. =10 501' for a 100 Curve. . <br />4 4 <br />,1 - � LP.Av9"23.20• - <br />D�.�, <br />qjL <br />F+"'� f <br />77- � <br />�D <br />o <br />b <br />IBC- 6 t <br />s <br />54 <br />RIz <br />? - <br />6 <br />CURVE TA-BLES. t <br />• -Published by KEUFFEL l ESSER CO. <br />HOW TO USE CURVE TABLES. d <br />Table I. contains Tangents and Externals to a 1° curve. Tan. and <br />Ext. to any other radius may befound nearly enough, bydividing the Tan. <br />or Ext. opposite the given Central An le 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 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 10 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 S ogog <br />of Intersection or 1. P. =230 20' to the R. at Station <br />542+72. <br />r <br />Ext. in Tab. I opposite 23° 20' =120.87 t I e <br />120.87-12=10.07. Say a 100 Curve. <br />Tan. in Tab. I opp. 2320'=1183.1 <br />1183.1-10=118.31. n.,._7 <br />Correction for A. 230 20' for a 100 Cur. =0.16 77• c�M . <br />113.31-1-0.16=118.47=corrected Tangent. /irt`?I <br />(If corrected Fxt. is required find in same way) 53s -N <br />• Ang. 23' 20'=23.33* -10 =2.3333 = L. C. <br />20191'=def. for sta. 542 I. P.=sta. .542-} 72 <br />40 491'= " +50 Tan. = 1 .15.47 <br />70 19 I' = " 19 cc <br />B. C.=sti. 541+53.53 <br />90 4921'-- _ " " r.+50 <br />11040, <br />= • i `° " " 543+ L. C.= 2 .33.33 <br />86.86. E. C.=Sta. 543-1-86.86 <br />100-53.53=•46.47X3'(def. for 1 ft. of 100 Cur.) =139.41'= <br />20 IN' =def. for sta. 542. <br />Def. for 50 ft. =20 30' for a 100 Curve. <br />c. Def. for 36.86 ft. =10 501' for a 100 Curve. . <br />4 4 <br />,1 - � LP.Av9"23.20• - <br />