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-? - <br />-'.'.,CURVE: TALES. <br />Published by KEUFFEL & ESSER <br />HOW TO USE CURVE TABLESDxtern' <br />. to An IL contains y other radius may be fou d nearly enough, by dividing the Tan- <br />-ents and I <br />'Ext. opposite the given Central Angle by the given degree of curve. <br />To find,Deg. of Curve; . having the Central ' nghe given Tangent- <br />ii <br />angle and t Tangent: <br />iivide Tan. opposite the given Central Angle by <br />id find Deg—of Curve,. -having the Centralh nglee-i EzExternal: <br />)ivide Ext. opposite the given Central Angle by Table 1.: Tan. <br />To find.Nat. Tan. and -N - Ex. Sec: fortune anglradius of a' 1° :curve will <br />r.Ext. of twice the given Angle divided by <br />e the Nat. Tan. or Nat. Ex: Sec, j <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 230 20'C Orbe <br />120.87-12=10:07: `_8ay,a10 , <br />Tan. in Tab. 1, 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°.192'=def. forsta:: 542' I. P.:=sta. 542 }72 <br />« <br />4 492' = «; . <br />-x-50' `Tan. = 1..18.47 <br />.7°`192, _ « u ' u 543.: B. C: =sta: 541+53.53 <br />9° 49, = ,« +50 2 _.33,33 <br />it ,11 40',--:- <br />u ;« <br />86.86 E. C. Sta. 543 { 86:86 <br />100-53.53 =.46.47X3'(aef. for lift. of 10° Cur. 139641'. <br />20 192' =.def. for -sta, 542. ' <br />Def. for- 50 ft. 29 30' for a-10° Curve. <br />Def. for 36.86 ft—V 502' for a 10° Curve. <br />IO° curve <br />� <br />F <br />�E <br />�I <br />I <br />I <br />II <br />,I <br />e <br />�IY <br />\ <br />fff,'II <br />III <br />tt�' <br />i <br />-? - <br />-'.'.,CURVE: TALES. <br />Published by KEUFFEL & ESSER <br />HOW TO USE CURVE TABLESDxtern' <br />. to An IL contains y other radius may be fou d nearly enough, by dividing the Tan- <br />-ents and I <br />'Ext. opposite the given Central Angle by the given degree of curve. <br />To find,Deg. of Curve; . having the Central ' nghe given Tangent- <br />ii <br />angle and t Tangent: <br />iivide Tan. opposite the given Central Angle by <br />id find Deg—of Curve,. -having the Centralh nglee-i EzExternal: <br />)ivide Ext. opposite the given Central Angle by Table 1.: Tan. <br />To find.Nat. Tan. and -N - Ex. Sec: fortune anglradius of a' 1° :curve will <br />r.Ext. of twice the given Angle divided by <br />e the Nat. Tan. or Nat. Ex: Sec, j <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 230 20'C Orbe <br />120.87-12=10:07: `_8ay,a10 , <br />Tan. in Tab. 1, 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°.192'=def. forsta:: 542' I. P.:=sta. 542 }72 <br />« <br />4 492' = «; . <br />-x-50' `Tan. = 1..18.47 <br />.7°`192, _ « u ' u 543.: B. C: =sta: 541+53.53 <br />9° 49, = ,« +50 2 _.33,33 <br />it ,11 40',--:- <br />u ;« <br />86.86 E. C. Sta. 543 { 86:86 <br />100-53.53 =.46.47X3'(aef. for lift. of 10° Cur. 139641'. <br />20 192' =.def. for -sta, 542. ' <br />Def. for- 50 ft. 29 30' for a-10° Curve. <br />Def. for 36.86 ft—V 502' for a 10° Curve. <br />IO° curve <br />
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