Concept:
Length of L vertical curve (L) is given by:
Condition | Summit curve | Valley curve |
S < L | \(\frac{{{\rm{N}}{{\rm{S}}^2}}}{({\sqrt {2{\rm{H}}} + \sqrt {2{\rm{h}}})^2}}\) | \(\frac{{{\rm{N}}{{\rm{S}}^2}}}{{2{{\rm{h}}_1} + 2{\rm{S}}\tan {\rm{\alpha }}}}\) |
S > L | \(\frac{{2{\rm{S}} - {{\left( {\sqrt {2{\rm{H}}} - \sqrt {2{\rm{h}}} } \right)}^2}}}{{\rm{N}}}\) | \(\frac{{2{\rm{S}} - 2{{\rm{h}}_1} + 2{\rm{S}}\tan {\rm{\alpha }}}}{{\rm{N}}}\) |
Where,
N = deflection angle, H = height of driver cycle, h = height of object on road surface,
S = stopping sight distance, and α = beam light angle
Calculation:
SSD = 180 m,
\({n_1} = \frac{1}{{200}}\),
\({n_2} = \left( {\frac{{ - 1}}{{200}}} \right)\)
\(N = \left( {{n_1} - {n_2}} \right) = \left( {\frac{1}{{200}} - \left( {\frac{{ - 1}}{{200}}} \right)} \right) = \frac{2}{{200}} = \frac{1}{{100}}\)
H = 1.2 m, h = 0.15 m (As per IRC)
Assume the length of curve > stopping sight distance
\( \Rightarrow L = \frac{{N{S^2}}}{{{{\left( {\sqrt {2H} + \sqrt 2 \;h} \right)}^2}}}\)
\(\Rightarrow L = \frac{{\frac{1}{{100}} \times {{180}^2}}}{{{{\left( {\sqrt {2 + 1.2} + \sqrt {2 \times 0.15} } \right)}^2}}} = 73.68\;m\)
∵ L < SSD ⇒ Our assumption is wrong.
Applying the another formula,
\(L = 2S - \frac{{4.4}}{N} = 2 \times 180 - \frac{{4.4}}{{\left( {\frac{1}{{100}}} \right)}}\) = -80 m
⇒ The gradient change does not need any summit curve, hence the minimum curve length is provided for aesthetic purpose.
Minimum curve length = 60 m