Chapter 15, Problem 17P

To determine

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

Answer to Problem 17P

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

Station	Deflection angle	Chord length
$\text{671+35}\text{.99}$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

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 $\Delta ={75}^{\circ }$

Angle of first deflection curve ${\Delta }_1={45}^{\circ }$

Radius of first curve ${\text{R}}_{\text{1}}\text{=600ft}$

Radius of second curve ${\text{R}}_2\text{=450ft}$

PCC station $(675+35.25)$

Calculation:

Deflection for second curve is given by,

  $\begin{array}{l}\text{Δ=}{\text{Δ}}_{\text{1}}\text{+}{\text{Δ}}_{\text{2}}\\ {\text{Δ}}_{\text{1}}\text{=4}{\text{5}}^{\text{o}}\text{=Angle of first deflection curve}\\ \text{Δ=7}{\text{5}}^{\text{o}}\text{=Intersection angle}\\ \text{7}{\text{5}}^{\text{o}}\text{=4}{\text{5}}^{\text{o}}\text{+}{\text{Δ}}_{\text{2}}\\ {\text{Δ}}_{\text{2}}\text{=3}{\text{0}}^{\text{o}}\end{array}$

Tangent length of first curve is given by,

  $\begin{array}{l}{\text{t}}_{\text{1}}\text{=}{\text{R}}_{\text{1}}\text{tan}\left(\frac{{\text{Δ}}_{\text{1}}}{\text{2}}\right)\\ {\text{R}}_1=600ft=Radiusoffirstcurve\\ {\text{Δ}}_1\text{=4}{\text{5}}^{\text{o}}=\text{Angle of first deflection curve}\\ {\text{t}}_{\text{1}}\text{=600tan}\left(\frac{\text{4}{\text{5}}^{\text{o}}}{\text{2}}\right)\\ {\text{t}}_{\text{1}}\text{=248}\text{.53ft}\end{array}$

Similarly tangent length for second curve is given by,

  $\begin{array}{l}{\text{t}}_2\text{=}{\text{R}}_2\text{tan}\left(\frac{{\text{Δ}}_2}{\text{2}}\right)\\ {\text{R}}_2=450ft=Radiusofsecondcurve\\ {\text{Δ}}_{\text{2}}\text{=3}{\text{0}}^{\text{o}}=\text{Angle of second deflection curve}\\ {\text{t}}_2\text{=450tan}\left(\frac{{30}^{\text{o}}}{\text{2}}\right)\\ {\text{t}}_2\text{=120}\text{.58ft}\end{array}$

Length of first curve is given by,

  $\begin{array}{l}{\text{L}}_{\text{1}}\text{=}\frac{{\text{R}}_{\text{1}}{\text{Δ}}_{\text{1}}\text{π}}{\text{180}}\\ {\text{L}}_{\text{1}}\text{=}\frac{\text{600×4}{\text{5}}^{\text{o}}\text{×π}}{\text{180}}\\ {\text{L}}_{\text{1}}\text{=471}\text{.24ft}\end{array}$

Length of second curve is given by,

  $\begin{array}{l}{\text{L}}_2\text{=}\frac{{\text{R}}_2{\text{Δ}}_2\text{π}}{\text{180}}\\ {\text{L}}_2\text{=}\frac{\text{450×3}{\text{0}}^{\text{o}}\text{×π}}{\text{180}}\\ {\text{L}}_2\text{=235}\text{.62ft}\end{array}$

Station of first curve at PC is given by,

  $\begin{array}{l}\text{Station of PC=Station of PCC-}{\text{L}}_1\\ \text{Station of PC=}\left(675+35.25\right)\text{-}\left(471.24\right)\end{array}$

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

  $\begin{array}{l}\text{Station of PC=}\left(675+35.25\right)\text{-}\left(4+71.24\right)\\ \text{Station of PC=}\left(\left(675-4\right)+\left(35.25-71.24\right)\right)\\ \text{Station of PC}=\left(671+35.99\right)\end{array}$

Station at point of tangency is given by,

  $\begin{array}{l}\text{Station of PT=Station of PCC+}{\text{L}}_2\\ \text{Station of PT=}\left(675+35.25\right)+235.62\\ \text{Station of PT=}\left(675+35.25\right)+\left(2+35.62\right)\\ \text{Station of PT=}\left(675+2\right)+\left(35.25+35.62\right)\\ \text{Station of PT}=\left(677+70.87\right)\end{array}$

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

  $\begin{array}{l}{\text{R}}_{\text{1}}\text{=}\frac{\text{5729}\text{.6}}{{\text{D}}_{\text{1}}}\\ {\text{D}}_{\text{1}}\text{=degree of curve}\\ \text{600=}\frac{\text{5729}\text{.6}}{{\text{D}}_{\text{1}}}\\ {\text{D}}_{\text{1}}\text{=9}\text{.54}{\text{9}}^{\text{o}}\end{array}$

Deflection angle for first full station is given by,

  $\begin{array}{l}\frac{{\delta }_1}{{\Delta }_1}=\frac{{\delta }_1}{L_1}\\ \frac{{\delta }_1}{{45}^{\circ }}=\frac{\left(671-\left(670+64.01\right)\right)}{471.24}\\ {\delta }_1={3.437}^{\circ }\end{array}$

Deflection angle is given by,

  $\begin{array}{l}\text{Deflection angle=}\frac{{\text{δ}}_{\text{1}}}{\text{2}}\\ \text{Deflection angle=}\frac{{3.437}^{\circ }}{\text{2}}\\ \text{Deflection angle=}{1.7185}^{\circ }\end{array}$

For the first curve chord length is given by,

  $\begin{array}{l}{\text{C}}_{\text{1}}\text{=2}{\text{R}}_1\text{sin}\frac{{\text{δ}}_{\text{1}}}{\text{2}}\\ {\text{C}}_{\text{1}}\text{=2×600sin}\frac{\text{3}\text{.43}{\text{7}}^{\text{o}}}{\text{2}}\\ {\text{C}}_{\text{1}}\text{=35}\text{.987ft}\end{array}$

Deflection angle for last full station is given by,

  $\begin{array}{l}\frac{{\delta }_2}{{\Delta }_1}=\frac{{\delta }_2}{L_1}\\ \frac{{\delta }_2}{{45}^{\circ }}=\frac{35.25}{471.24}\\ {\delta }_2={3.366}^{\circ }\end{array}$

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

Station	Deflection angle	Chord length
$\text{671+35}\text{.99}$	$0$	$0$
$671$	$1.7185$	$35.99$
$672$	$6.4931$	$99.89$
$673$	$11.268$	$99.89$
$674$	$16.042$	$99.89$
$675$	$20.817$	$99.89$
$675+35.25$	$22.50$	$35.25$

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

  $\begin{array}{l}{R}_2=\frac{5729.6}{D_2}\\ 450=\frac{5729.6}{D_1}\\ D_2={12.732}^{\circ }\end{array}$

Deflection angle for first full station is given by,

  $\begin{array}{l}\frac{{\delta }_2}{{\Delta }_2}=\frac{l_1}{L_2}\\ \frac{{\delta }_1}{{30}^{\circ }}=\frac{\left(676-\left(675+35.25\right)\right)}{235.62}\\ {\delta }_1={8.244}^{\circ }\end{array}$

Deflection angle is given by,

  $\begin{array}{l}\text{Deflection angle=}\frac{{\text{δ}}_{\text{1}}}{\text{2}}\\ \text{Deflection angle=}\frac{{8.244}^{\circ }}{\text{2}}\\ \text{Deflection angle=4}\text{.12}{\text{2}}^{\circ }\end{array}$

For the second curve chord length is given by,

  $\begin{array}{l}{\text{C}}_2\text{=2}{\text{R}}_2\text{sin}\frac{{\text{δ}}_{\text{1}}}{\text{2}}\\ {\text{C}}_2\text{=2×450sin}\frac{\text{8}\text{.24}{\text{4}}^{\text{o}}}{\text{2}}\\ {\text{C}}_2\text{=64}\text{.69ft}\end{array}$

Deflection angle for last full station is given by,

  $\begin{array}{l}\frac{{\delta }_2}{{\Delta }_2}=\frac{l_2}{L_2}\\ \frac{{\delta }_2}{{30}^{\circ }}=\frac{70.87}{235.62}\\ {\delta }_2=0.300\times {30}^{\circ }\\ {\delta }_2={9.023}^{\circ }\end{array}$

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

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