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<br />'77: 717*CURVE TABLES
<br />Published by'KEUFFEL & ESSER CO.
<br />HOW TO USE CURVE TABLES
<br />Table I. contains Tahgents,and, Externals to a V curve. Tan. and
<br />Ext. to any other radius may be found nearly enough, by dividing the Tan.
<br />or Ext. 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 .Extermal:
<br />Divide Ext. opposite the gNen Central Angle by th& giveii External.
<br />To find Nat. Tan. a:nd.Nat. Ex., Sec.for any angle byTablel': Tan.
<br />0iXI
<br />'AL. of.tivice'the given: angle 'divided by the radius of a_1?curve Will
<br />be. the Nat.:Tan. or Nat. Ex� Sec..
<br />Wanted a Cdrye,with an Ext. of about 12 ft. Anglef
<br />of= Intersection ora I: J P. =23 20' to the R., at Station,-
<br />542+72.
<br />Ext. in 'Tab. I opposite 239 20' 120.87'
<br />120.87 +12 10.07. ' Say' a 10*'.Curve.
<br />Tan. in Tab. I i opp. 23*, 20'- 1183.1 Y,
<br />1183.1'+10-118.31.
<br />Correctionf6i.A. 23' 20' for a 10° Cur. =0.16 *
<br />118M +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.
<br />C
<br />..
<br />2-�1.gi'='def. for sta- 542' L P. = sta. 5412+72-'
<br />-
<br />4* 4912' It it it +50- Tan. = 1 48.41
<br />543
<br />BC. = sta.' - T54i +53.;5 4:153.53
<br />. . 1
<br />go 49'211 it cc +150' .33.33
<br />, I oe it it it
<br />15"4 54�+.
<br />86.86 E. C'=Sta:
<br />54'3+86.86
<br />100,'53.53 _--�46.47 X3'(def. for l ft. of 100.Cur.) 139AV-,
<br />2*. 19 =def.: for sta. 542.
<br />Def; fof-50 ft:=2oM-for a 10° Curve.
<br />Def. for 36.86 ft. 10 50 for a 10° Curve.
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<br />j -1
<br />'77: 717*CURVE TABLES
<br />Published by'KEUFFEL & ESSER CO.
<br />HOW TO USE CURVE TABLES
<br />Table I. contains Tahgents,and, Externals to a V curve. Tan. and
<br />Ext. to any other radius may be found nearly enough, by dividing the Tan.
<br />or Ext. 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 .Extermal:
<br />Divide Ext. opposite the gNen Central Angle by th& giveii External.
<br />To find Nat. Tan. a:nd.Nat. Ex., Sec.for any angle byTablel': Tan.
<br />0iXI
<br />'AL. of.tivice'the given: angle 'divided by the radius of a_1?curve Will
<br />be. the Nat.:Tan. or Nat. Ex� Sec..
<br />Wanted a Cdrye,with an Ext. of about 12 ft. Anglef
<br />of= Intersection ora I: J P. =23 20' to the R., at Station,-
<br />542+72.
<br />Ext. in 'Tab. I opposite 239 20' 120.87'
<br />120.87 +12 10.07. ' Say' a 10*'.Curve.
<br />Tan. in Tab. I i opp. 23*, 20'- 1183.1 Y,
<br />1183.1'+10-118.31.
<br />Correctionf6i.A. 23' 20' for a 10° Cur. =0.16 *
<br />118M +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.
<br />C
<br />..
<br />2-�1.gi'='def. for sta- 542' L P. = sta. 5412+72-'
<br />-
<br />4* 4912' It it it +50- Tan. = 1 48.41
<br />543
<br />BC. = sta.' - T54i +53.;5 4:153.53
<br />. . 1
<br />go 49'211 it cc +150' .33.33
<br />, I oe it it it
<br />15"4 54�+.
<br />86.86 E. C'=Sta:
<br />54'3+86.86
<br />100,'53.53 _--�46.47 X3'(def. for l ft. of 100.Cur.) 139AV-,
<br />2*. 19 =def.: for sta. 542.
<br />Def; fof-50 ft:=2oM-for a 10° Curve.
<br />Def. for 36.86 ft. 10 50 for a 10° Curve.
<br />2
<br />z
<br />I RA,c , 350
<br />>
<br />curve
<br />--------
<br />COA/
<br />Y
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