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An existing highway-railway at-grade crossing is being redesigned as grade separated to improve traffic operations. The railway must remain at the same elevation. The highway is being reconstructed to travel under the railway. The underpass will be a sag curve that connects 2% tangent sections on both ends, and the PVI will be centered under the railway (a symmetrical alignment.) The sag curve design speed is 45 mph. How many feet below the railway should the PVI of the curve be located

Respuesta :

Answer:

Resultant elevation=17.6 ft

Explanation:

From the table of design controls for vertical curves that are based on braking sight distance, are as follows:

K=79 and SSD=360 ft

the minimum length of the vertical curve is equal to:

[tex]L=KA=K(|G_{1}-G_{2}|)=79(|-2-(+2)|)=316 ft[/tex]

This is the length of the vertical curve. Due SSD>L, the minimum clearance height is:

[tex]L=2SSD-\frac{800(Hc-5)}{A}\\ L-2SSD=-\frac{800(Hc-5)}{A}[/tex]

Clearing Hc:

[tex]Hc=((2SSD-L)(\frac{|G1-G2|}{800}))+5=((2*360-316)(\frac{-2-2}{800}))+5=7 ft[/tex]

Like 7 ft is not enough, we take 16 ft as desirable minimum height:

[tex]y=\frac{AL}{800}=\frac{|-2-(+2)|316}{800}=1.6 ft[/tex]

the resultant elevation is:

Resultant elevation=16+1.6=17.6 ft