ROUTE SURVEYING

$\alpha\ =\ \frac{\theta}{2}$

$\ tan\ \alpha\ =\ \frac{x}{b}$

$20\ m\ =\ \frac{\pi\ R\ D}{180°}$
$100\ ft\ =\ \frac{\pi\ R\ D}{180°}$

$L_c=\frac{\pi\ R\ I}{180}\ \ \ =\frac{20\ I}{D}$

$Distance\ of\ PI\ \Longleftrightarrow PC$
$Distance\ of\ PI\ \Longleftrightarrow PT$

$R_{min\ }=\ \frac{v^2}{g\ \left(e+\mu\right)}$

$kph\ \ \ \ \ \Rightarrow\ \ \ \ \frac{kph}{3.6}\ \ \ \ \ \Rightarrow\ m/s$

$I.F.\ =\ e+\ \mu\ =\ \frac{v^2}{g\ R}$

$CT=\ T_1+T_2$

$CT=\ T_1+T_2$

$Y\ =\ L-\ \frac{{\ L}^5}{40\ R^2\ {L_s}^2}$

$x\ =\ \frac{L^3}{6\ R\ L_s\ }$

$\theta\ =\ \frac{L^2}{2\ R\ L_s\ }$

$i\ =\ \frac{\theta}{3\ }$

$\frac{i}{{L\ }^2}=\frac{i_s}{{L_s}^2\ }$

$T_s=\ \frac{L_s}{2}+\ \left(R+P\right)\ tan\left(\frac{I}{2}\right)$

$P=\frac{X_c}{4\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P=\ \frac{{L_s}^2}{24\ R}$

$E_s=\frac{R+P}{cos\ \left(\frac{I}{2}\right)}-P $

$\frac{D}{D_C}=\ \frac{L}{L_S}$

$L_S=\ \frac{0.036\ v_{\left(kph\right)}^3}{\ R}$

$e\ =\ tan\ (\theta)$
$tan\ (\theta)\ =\ \frac{v_{\left(m/s\right)}^2}{g\ R_{(m)}}$

$e+\ \mu\ =\ \frac{{v_{max}}^2}{g\ R}$

$e-\ \mu\ =\ \frac{{v_{min}}^2}{g\ R}$

$q_{(m/s^2)}=\ \frac{v_{\left(m/s\right)}^3}{\left(R\ L_S\right)_{(m)}} $

$H=\ \frac{L}{8}\left(\ \left|g_1\right|+\left|g_2\right|\ \right)$

$H=\ \frac{\left(\left|g_1\right|+\left|g_2\right|\right)\ L_1\ L_2}{{2\ L}_T}$
$H=\ \frac{\ L_1\ }{2}\ \left(\left|g_1\right|+\left|g_3\right|\right)$

$h_1=h_2=\ \frac{\ H\ }{4}$

$g_3=g_1-\ \frac{\ 2H\ }{L_1}$
$g_3=\ \frac{\ g_1\ L_1\ +\ g_2\ L_2\ }{L_T}$

$S_2=\ \frac{\ g_2\ \left(L_2\right)^2\ }{2H}$
$S_1=L_T-{\ S}_2$

$S_1=\ \frac{\ g_1\ \left(L_1\right)^2\ }{2H}$
$S_2=L_T-{\ S}_1$

${\ V}_e\ =\ \frac{L}{2}\left({\ A}_1+{\ A}_2\right)\ $

${\ V}_p\ =\ \frac{L}{6}\left({\ A}_1+{\ 4A}_m+{\ A}_2\right)\ $

${\ {{\ C}_p=\ V}_e\ -V}_p$
$C_p=\ \frac{L}{12}\ \left(c_1-c_2\right)\left(d_1-d_2\right)$

$V\ =\ A\left[\frac{1\ \left(\sum{\ h}_1\right)+\ 2\ \left(\sum{\ h}_2\right)+3\ \left(\sum{\ h}_3\right)\ +4\ \left(\sum{\ h}_4\right)}{4}\ \right]\ $