The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Question

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Answer 1

Question

Chapter 15, Problem 17P

To determine

The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Expert Solution & Answer

Answer to Problem 17P

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Explanation of Solution

Given information:

Intersection angle $Δ=75∘$

Angle of first deflection curve $Δ1=45∘$

Radius of first curve $R1=600ft$

Radius of second curve $R2=450ft$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

$Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o$

Tangent length of first curve is given by,

$t1=R1tan( Δ 1 2)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan( 4 5 o 2)t1=248.53ft$

Similarly tangent length for second curve is given by,

$t2=R2tan( Δ 2 2)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan( 30 o 2)t2=120.58ft$

Length of first curve is given by,

$L1=R1Δ1π180L1=600×45o×π180L1=471.24ft$

Length of second curve is given by,

$L2=R2Δ2π180L2=450×30o×π180L2=235.62ft$

Station of first curve at PC is given by,

$Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)$

The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above $100ft$ intervals into single decimal.

$Station of PC=(675+35.25)-(4+71.24)Station of PC=(( 675−4)+( 35.25−71.24))Station of PC=(671+35.99)$

Station at point of tangency is given by,

$Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)$

For the first curve calculate degree of first curve is given by,

$R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o$

Deflection angle for first full station is given by,

$δ1Δ1=δ1L1δ1 45∘=( 671−( 670+64.01 ))471.24δ1=3.437∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 3.437∘2Deflection angle=1.7185∘$

For the first curve chord length is given by,

$C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft$

Deflection angle for last full station is given by,

$δ2Δ1=δ2L1δ2 45∘=35.25471.24δ2=3.366∘$

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve calculate degree of second curve is given by,

$R2=5729.6D2450=5729.6D1D2=12.732∘$

Deflection angle for first full station is given by,

$δ2Δ2=l1L2δ1 30∘=( 676−( 675+35.25 ))235.62δ1=8.244∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 8.244∘2Deflection angle=4.122∘$

For the second curve chord length is given by,

$C2=2R2sinδ12C2=2×450sin8.244o2C2=64.69ft$

Deflection angle for last full station is given by,

$δ2Δ2=l2L2δ2 30∘=70.87235.62δ2=0.300×30∘δ2=9.023∘$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

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Answer 2

Question

Chapter 15, Problem 17P

To determine

The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Expert Solution & Answer

Answer to Problem 17P

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Explanation of Solution

Given information:

Intersection angle $Δ=75∘$

Angle of first deflection curve $Δ1=45∘$

Radius of first curve $R1=600ft$

Radius of second curve $R2=450ft$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

$Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o$

Tangent length of first curve is given by,

$t1=R1tan( Δ 1 2)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan( 4 5 o 2)t1=248.53ft$

Similarly tangent length for second curve is given by,

$t2=R2tan( Δ 2 2)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan( 30 o 2)t2=120.58ft$

Length of first curve is given by,

$L1=R1Δ1π180L1=600×45o×π180L1=471.24ft$

Length of second curve is given by,

$L2=R2Δ2π180L2=450×30o×π180L2=235.62ft$

Station of first curve at PC is given by,

$Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)$

The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above $100ft$ intervals into single decimal.

$Station of PC=(675+35.25)-(4+71.24)Station of PC=(( 675−4)+( 35.25−71.24))Station of PC=(671+35.99)$

Station at point of tangency is given by,

$Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)$

For the first curve calculate degree of first curve is given by,

$R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o$

Deflection angle for first full station is given by,

$δ1Δ1=δ1L1δ1 45∘=( 671−( 670+64.01 ))471.24δ1=3.437∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 3.437∘2Deflection angle=1.7185∘$

For the first curve chord length is given by,

$C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft$

Deflection angle for last full station is given by,

$δ2Δ1=δ2L1δ2 45∘=35.25471.24δ2=3.366∘$

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve calculate degree of second curve is given by,

$R2=5729.6D2450=5729.6D1D2=12.732∘$

Deflection angle for first full station is given by,

$δ2Δ2=l1L2δ1 30∘=( 676−( 675+35.25 ))235.62δ1=8.244∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 8.244∘2Deflection angle=4.122∘$

For the second curve chord length is given by,

$C2=2R2sinδ12C2=2×450sin8.244o2C2=64.69ft$

Deflection angle for last full station is given by,

$δ2Δ2=l2L2δ2 30∘=70.87235.62δ2=0.300×30∘δ2=9.023∘$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

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Answer 3

Question

Chapter 15, Problem 17P

To determine

The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Expert Solution & Answer

Answer to Problem 17P

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Explanation of Solution

Given information:

Intersection angle $Δ=75∘$

Angle of first deflection curve $Δ1=45∘$

Radius of first curve $R1=600ft$

Radius of second curve $R2=450ft$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

$Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o$

Tangent length of first curve is given by,

$t1=R1tan( Δ 1 2)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan( 4 5 o 2)t1=248.53ft$

Similarly tangent length for second curve is given by,

$t2=R2tan( Δ 2 2)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan( 30 o 2)t2=120.58ft$

Length of first curve is given by,

$L1=R1Δ1π180L1=600×45o×π180L1=471.24ft$

Length of second curve is given by,

$L2=R2Δ2π180L2=450×30o×π180L2=235.62ft$

Station of first curve at PC is given by,

$Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)$

The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above $100ft$ intervals into single decimal.

$Station of PC=(675+35.25)-(4+71.24)Station of PC=(( 675−4)+( 35.25−71.24))Station of PC=(671+35.99)$

Station at point of tangency is given by,

$Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)$

For the first curve calculate degree of first curve is given by,

$R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o$

Deflection angle for first full station is given by,

$δ1Δ1=δ1L1δ1 45∘=( 671−( 670+64.01 ))471.24δ1=3.437∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 3.437∘2Deflection angle=1.7185∘$

For the first curve chord length is given by,

$C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft$

Deflection angle for last full station is given by,

$δ2Δ1=δ2L1δ2 45∘=35.25471.24δ2=3.366∘$

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve calculate degree of second curve is given by,

$R2=5729.6D2450=5729.6D1D2=12.732∘$

Deflection angle for first full station is given by,

$δ2Δ2=l1L2δ1 30∘=( 676−( 675+35.25 ))235.62δ1=8.244∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 8.244∘2Deflection angle=4.122∘$

For the second curve chord length is given by,

$C2=2R2sinδ12C2=2×450sin8.244o2C2=64.69ft$

Deflection angle for last full station is given by,

$δ2Δ2=l2L2δ2 30∘=70.87235.62δ2=0.300×30∘δ2=9.023∘$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

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Answer 4

Question

Chapter 15, Problem 17P

To determine

The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Expert Solution & Answer

Answer to Problem 17P

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Explanation of Solution

Given information:

Intersection angle $Δ=75∘$

Angle of first deflection curve $Δ1=45∘$

Radius of first curve $R1=600ft$

Radius of second curve $R2=450ft$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

$Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o$

Tangent length of first curve is given by,

$t1=R1tan( Δ 1 2)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan( 4 5 o 2)t1=248.53ft$

Similarly tangent length for second curve is given by,

$t2=R2tan( Δ 2 2)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan( 30 o 2)t2=120.58ft$

Length of first curve is given by,

$L1=R1Δ1π180L1=600×45o×π180L1=471.24ft$

Length of second curve is given by,

$L2=R2Δ2π180L2=450×30o×π180L2=235.62ft$

Station of first curve at PC is given by,

$Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)$

The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above $100ft$ intervals into single decimal.

$Station of PC=(675+35.25)-(4+71.24)Station of PC=(( 675−4)+( 35.25−71.24))Station of PC=(671+35.99)$

Station at point of tangency is given by,

$Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)$

For the first curve calculate degree of first curve is given by,

$R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o$

Deflection angle for first full station is given by,

$δ1Δ1=δ1L1δ1 45∘=( 671−( 670+64.01 ))471.24δ1=3.437∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 3.437∘2Deflection angle=1.7185∘$

For the first curve chord length is given by,

$C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft$

Deflection angle for last full station is given by,

$δ2Δ1=δ2L1δ2 45∘=35.25471.24δ2=3.366∘$

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve calculate degree of second curve is given by,

$R2=5729.6D2450=5729.6D1D2=12.732∘$

Deflection angle for first full station is given by,

$δ2Δ2=l1L2δ1 30∘=( 676−( 675+35.25 ))235.62δ1=8.244∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 8.244∘2Deflection angle=4.122∘$

For the second curve chord length is given by,

$C2=2R2sinδ12C2=2×450sin8.244o2C2=64.69ft$

Deflection angle for last full station is given by,

$δ2Δ2=l2L2δ2 30∘=70.87235.62δ2=0.300×30∘δ2=9.023∘$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

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Answer 5

Chapter 15, Problem 17P

To determine

The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Expert Solution & Answer

Answer to Problem 17P

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Explanation of Solution

Given information:

Intersection angle $Δ=75∘$

Angle of first deflection curve $Δ1=45∘$

Radius of first curve $R1=600ft$

Radius of second curve $R2=450ft$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

$Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o$

Tangent length of first curve is given by,

$t1=R1tan( Δ 1 2)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan( 4 5 o 2)t1=248.53ft$

Similarly tangent length for second curve is given by,

$t2=R2tan( Δ 2 2)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan( 30 o 2)t2=120.58ft$

Length of first curve is given by,

$L1=R1Δ1π180L1=600×45o×π180L1=471.24ft$

Length of second curve is given by,

$L2=R2Δ2π180L2=450×30o×π180L2=235.62ft$

Station of first curve at PC is given by,

$Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)$

The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above $100ft$ intervals into single decimal.

$Station of PC=(675+35.25)-(4+71.24)Station of PC=(( 675−4)+( 35.25−71.24))Station of PC=(671+35.99)$

Station at point of tangency is given by,

$Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)$

For the first curve calculate degree of first curve is given by,

$R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o$

Deflection angle for first full station is given by,

$δ1Δ1=δ1L1δ1 45∘=( 671−( 670+64.01 ))471.24δ1=3.437∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 3.437∘2Deflection angle=1.7185∘$

For the first curve chord length is given by,

$C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft$

Deflection angle for last full station is given by,

$δ2Δ1=δ2L1δ2 45∘=35.25471.24δ2=3.366∘$

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve calculate degree of second curve is given by,

$R2=5729.6D2450=5729.6D1D2=12.732∘$

Deflection angle for first full station is given by,

$δ2Δ2=l1L2δ1 30∘=( 676−( 675+35.25 ))235.62δ1=8.244∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 8.244∘2Deflection angle=4.122∘$

For the second curve chord length is given by,

$C2=2R2sinδ12C2=2×450sin8.244o2C2=64.69ft$

Deflection angle for last full station is given by,

$δ2Δ2=l2L2δ2 30∘=70.87235.62δ2=0.300×30∘δ2=9.023∘$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Answer 6

Chapter 15, Problem 17P

To determine

The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Expert Solution & Answer

Answer to Problem 17P

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Explanation of Solution

Given information:

Intersection angle $Δ=75∘$

Angle of first deflection curve $Δ1=45∘$

Radius of first curve $R1=600ft$

Radius of second curve $R2=450ft$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

$Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o$

Tangent length of first curve is given by,

$t1=R1tan( Δ 1 2)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan( 4 5 o 2)t1=248.53ft$

Similarly tangent length for second curve is given by,

$t2=R2tan( Δ 2 2)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan( 30 o 2)t2=120.58ft$

Length of first curve is given by,

$L1=R1Δ1π180L1=600×45o×π180L1=471.24ft$

Length of second curve is given by,

$L2=R2Δ2π180L2=450×30o×π180L2=235.62ft$

Station of first curve at PC is given by,

$Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)$

The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above $100ft$ intervals into single decimal.

$Station of PC=(675+35.25)-(4+71.24)Station of PC=(( 675−4)+( 35.25−71.24))Station of PC=(671+35.99)$

Station at point of tangency is given by,

$Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)$

For the first curve calculate degree of first curve is given by,

$R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o$

Deflection angle for first full station is given by,

$δ1Δ1=δ1L1δ1 45∘=( 671−( 670+64.01 ))471.24δ1=3.437∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 3.437∘2Deflection angle=1.7185∘$

For the first curve chord length is given by,

$C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft$

Deflection angle for last full station is given by,

$δ2Δ1=δ2L1δ2 45∘=35.25471.24δ2=3.366∘$

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve calculate degree of second curve is given by,

$R2=5729.6D2450=5729.6D1D2=12.732∘$

Deflection angle for first full station is given by,

$δ2Δ2=l1L2δ1 30∘=( 676−( 675+35.25 ))235.62δ1=8.244∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 8.244∘2Deflection angle=4.122∘$

For the second curve chord length is given by,

$C2=2R2sinδ12C2=2×450sin8.244o2C2=64.69ft$

Deflection angle for last full station is given by,

$δ2Δ2=l2L2δ2 30∘=70.87235.62δ2=0.300×30∘δ2=9.023∘$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Answer 7

Chapter 15, Problem 17P

To determine

The central angles and the corresponding chord lengthsfor a compound curve which connects two tangents.

Answer 8

Expert Solution & Answer

Answer to Problem 17P

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

For the second curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$675+35.25$	$0$	$0$
$676$	$4.122$	$64.69$
$677$	$10.488$	$99.79$
$677+70.87$	$15$	$70.80$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

Intersection angle $Δ=75∘$

Angle of first deflection curve $Δ1=45∘$

Radius of first curve $R1=600ft$

Radius of second curve $R2=450ft$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

$Δ=Δ1+Δ2Δ1=45o=Angle of first deflection curveΔ=75o=Intersection angle75o=45o+Δ2Δ2=30o$

Tangent length of first curve is given by,

$t1=R1tan( Δ 1 2)R1=600ft=Radius of first curveΔ1=45o=Angle of first deflection curvet1=600tan( 4 5 o 2)t1=248.53ft$

Similarly tangent length for second curve is given by,

$t2=R2tan( Δ 2 2)R2=450ft=Radius of second curveΔ2=30o=Angle of second deflection curvet2=450tan( 30 o 2)t2=120.58ft$

Length of first curve is given by,

$L1=R1Δ1π180L1=600×45o×π180L1=471.24ft$

Length of second curve is given by,

$L2=R2Δ2π180L2=450×30o×π180L2=235.62ft$

Station of first curve at PC is given by,

$Station of PC=Station of PCC-L1Station of PC=(675+35.25)-(471.24)$

The station of PCis the point of curve according to the standards of AASHTOit is calculated by dividing the station when it reaches above $100ft$ intervals into single decimal.

$Station of PC=(675+35.25)-(4+71.24)Station of PC=(( 675−4)+( 35.25−71.24))Station of PC=(671+35.99)$

Station at point of tangency is given by,

$Station of PT=Station of PCC+L2Station of PT=(675+35.25)+235.62Station of PT=(675+35.25)+(2+35.62)Station of PT=(675+2)+(35.25+35.62)Station of PT=(677+70.87)$

For the first curve calculate degree of first curve is given by,

$R1=5729.6D1D1=degree of curve600=5729.6D1D1=9.549o$

Deflection angle for first full station is given by,

$δ1Δ1=δ1L1δ1 45∘=( 671−( 670+64.01 ))471.24δ1=3.437∘$

Deflection angle is given by,

$Deflection angle=δ12Deflection angle= 3.437∘2Deflection angle=1.7185∘$

For the first curve chord length is given by,

$C1=2R1sinδ12C1=2×600sin3.437o2C1=35.987ft$

Deflection angle for last full station is given by,

$δ2Δ1=δ2L1δ2 45∘=35.25471.24δ2=3.366∘$

For the first curve central angle and chord length is shown in table below,

Station	Deflection angle	Chord length
$671+35.99$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$