Example: Find the location on curves if $L=157\mathrm{.}07m$ and $=30$ and $PC$ chainage is $792\mathrm{.}1m$ and $($C$)$ chord length is $20m$ and $R=300m?$

Solution:

PT chainage $=PC$ chainage $+L=792\mathrm{.}1+157\mathrm{.}07=949\mathrm{.}17m$

Initial sub chord $=800-792\mathrm{.}1=7\mathrm{.}9m,157\mathrm{.}07-7\mathrm{.}9=149\mathrm{.}17m$

Final sub chord $=149\mathrm{.}17-140=9\mathrm{.}17m$

$\frac{140}{20}=7$ locations $+$initial $+$ final $=9$ locations

Deflection angle for initial sub chord $Si{n}_1=\frac{C}{2}R=\frac{7\mathrm{.}9}{2^{**}}300=0\mathrm{.}0131,a=0\mathrm{.}7545$

$sin{}_2=\frac{20}{2}**300=0\mathrm{.}0333,{}_2=1\mathrm{.}91=a_3=a_1={}_5=a_6={}_7=a_3$

Deflection angle for final sub chord $Si{n}_9=\frac{C}{2}R=\frac{9\mathrm{.}17}{2^{**}}300=0\mathrm{.}0152,=0\mathrm{.}8755$

For checking ; $\frac{}{2}=0\mathrm{.}7545+7(1\mathrm{.}91)+0\mathrm{.}8755=15$ therefore $={15}^{**}2=30$ is given.....OK

Where; $D$ is degree of curvature $D{D}^{}=\frac{573}{R}$

$L$ is curve length $L=\frac{2R}{360}=R{}_{\text{R}\text{A}\text{D}}=10\frac{}{D}$

$T$ is the tangent of curve $T=$Rtan$(\frac{}{2})$

$$ is the angle of curvature$($deflection angle$).$

$M$ is the middle ordinate $M=R-$Rcos$(\frac{}{2})$

$E$ is external distance $E=\frac{R}{cos(\frac{}{2})}-R$

$C$ is the chord length $C=2$Rsin$(\frac{}{2})$

$C_x$ length of chord $x,$

$C_x=2Rsin{d}_x$

Can you please explain how to solve question like this? Please solve in

detail for me

$\backslash$table$[[$Chainage $m,$Chord length $m,$Chord angle $(),$Cumulative angle $)],[PC=792\mathrm{.}1,0,0\mathrm{.}0,0\mathrm{.}0],[800,7\mathrm{.}9,0\mathrm{.}7545,0\mathrm{.}7545],[820,20,1\mathrm{.}91,2\mathrm{.}6645],[840,20,1\mathrm{.}91,4\mathrm{.}5745],[860,1\mathrm{.}91,6\mathrm{.}4845,],[880,20,1\mathrm{.}91,8\mathrm{.}3945],[900,20,1\mathrm{.}91,10\mathrm{.}3045],[920,20,1\mathrm{.}91,12\mathrm{.}2145],[940,20,1\mathrm{.}91,14\mathrm{.}1245],[$PT $=949\mathrm{.}17,20,0\mathrm{.}8755,15\mathrm{.}0=\frac{}{2}$