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IG <br />C <br />Co <br />a 4A, -I) <br />17— c�1­ <br />'O <br />CZ <br />J51-0 <br />LJ111)9 <br />4- <br />f95 5 a.Al <br />40 0 <br />zr <br />t r_A <br />r <br />- 3G <br />o <br />-1 71 <br />90 - -Ei_ <br />1/- 71( 9r.76 <br />-2 <br />------------- <br />CURVE TABLES.. -- <br />Published by KEUFFEL & ESSER CO. <br />HOW TO ;USE .CURVE TABLES. <br />Table I. contains Tangents and Externals to a 10 curve. Tan. and <br />Ext. to any other radius maybe found nearly enough, by dividing theTan. <br />or Ekt. 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 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. See. for any angle by Table I.: Tan. <br />or Ext. of• twice the given angle divided by the any <br />of a P 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' 201 = 120.87 <br />120.87--12=10.07. Say alOPCurve. <br />Tan. in Tab. I opp. 236 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 />(If corrected Ext. is required find in -same way) <br />. Ang. 23* 20'= 23.33* -- 10 = 2.3333 L. C. <br />2' 19Y= def. for sta. 542 I. P. = sta. -542+72 <br />V 491'= Is a <br />2 +50, Tan. = 1 .18.47 <br />1911= <br />2 543 <br />9°491'= +50 <br />B. C. = sta. 541 + 53.53 <br />. <br />11°40'= 543+ L. C.= 2 .33.33 <br />86.86 E. C.=Sta. 543+86.86 <br />100-53.53=46.47X3'(def. for 1 ft. of 10° Cur.) 139.41'= <br />2* 191',=def. for sta. 542. <br />Def. for 50 ft. 2' 30' for a 10° Curve. <br />Def. for 36.86 ft.=1° 503' fora 10*,Curvp— <br />