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I <br />CURVE - TABLES. <br />Published by, KEUF, FEL & ESSER CO. <br />HOW TO • USE ,.CURVETABLES. <br />Table I. contains Tangents and Externals to -a P curve. - Tan. and <br />Ext. to Any other radius maybe found nearly enough, by dividing theT,an. <br />or Ext, opposite the given Central Angle by the given degree of cur Oe. <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 Ankle by th'6 given External. <br />y angle by Table I.: Tan. <br />To find Nat. Tan. and Nat. Ex. See. for an <br />or Ext. of twice the given angle'divided bythe any, <br />of a -I* curve will <br />be the Nat. Tan. or Nat. Ex, See. <br />EXAMPM - <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of,, <br />ntersection or I. -P.=230 20' to the R. at' Station <br />542+72. <br />Ext. in Tab. I opposite 23°•20f = 120.87 <br />120.87 7142 =10.07.. Say ;i 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+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* 1921'=def. for sta.".: 542 1. P. = sta. 542+72 <br />404911—, a a a <br />2 +50, Tan. = 1 .18.47 <br />70 19;'= u u u 543 <br />90 49j'a +50 B. C. = sta. 54i }53.53 <br />= <br />110 40'�' u 11 543+ L. C.— 2 .33.33 <br />8636 , E. C. Sta. 543+86.86 <br />100-:-53.53 =46.47XX(def. for I ft. of. 10° Cur.) 139.41'= <br />2`19i' = def.' for sta.- 542. <br />Def. for -50 ft. --2* 30'for-a-10* Curve. - <br />Def. for 36.86 ft. P 50j' for a 10° Curve. <br />7, �F <br />;7 <br />®r <br />4- <br />,_3 <br />6, <br />313-d <br />T: <br />3 70 <br />CURVE - TABLES. <br />Published by, KEUF, FEL & ESSER CO. <br />HOW TO • USE ,.CURVETABLES. <br />Table I. contains Tangents and Externals to -a P curve. - Tan. and <br />Ext. to Any other radius maybe found nearly enough, by dividing theT,an. <br />or Ext, opposite the given Central Angle by the given degree of cur Oe. <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 Ankle by th'6 given External. <br />y angle by Table I.: Tan. <br />To find Nat. Tan. and Nat. Ex. See. for an <br />or Ext. of twice the given angle'divided bythe any, <br />of a -I* curve will <br />be the Nat. Tan. or Nat. Ex, See. <br />EXAMPM - <br />Wanted a Curve with an Ext. of about 12 ft. Angle <br />of,, <br />ntersection or I. -P.=230 20' to the R. at' Station <br />542+72. <br />Ext. in Tab. I opposite 23°•20f = 120.87 <br />120.87 7142 =10.07.. Say ;i 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+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* 1921'=def. for sta.".: 542 1. P. = sta. 542+72 <br />404911—, a a a <br />2 +50, Tan. = 1 .18.47 <br />70 19;'= u u u 543 <br />90 49j'a +50 B. C. = sta. 54i }53.53 <br />= <br />110 40'�' u 11 543+ L. C.— 2 .33.33 <br />8636 , E. C. Sta. 543+86.86 <br />100-:-53.53 =46.47XX(def. for I ft. of. 10° Cur.) 139.41'= <br />2`19i' = def.' for sta.- 542. <br />Def. for -50 ft. --2* 30'for-a-10* Curve. - <br />Def. for 36.86 ft. P 50j' for a 10° Curve. <br />
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