Let me show you how this is done in the real life road construction world.
If I interpret the problem correctly, you have a concave up/under vertical
road way with 2 tangent lines intersecting below the curve.
One is tangent to where the curve begins and one where it ends.
For simplicity sake, we can let the point the two lines intersect be (500,0). And the curve begins at (0,30)
Since the curve going in has elevation -.06, then the elevation of the curve at that point is left end point is 500(.06)=30.
The other side has elevation 500(.09)=45
Their coordinates are (0,30) and (1000,45)
You can find the elevations on the respective tangents easy enough by just pro rating the elevations.
Let's use 100' increments beginning with the curve going in, the 6% curve.
Code:
distance tangent elev.
0 30 PVC elev
100 30-(100*.06)=24
200 30-(200*.06)=18
300 30-(300*.06)=12
400 30-(400*.06)=6
500 30-(500*.06)=0 PVI, point where the lines meet.
Now we use the other 9% grade:
600 0+(100*.09)=9
700 0+(200*.09)=18
800 0+(300*.09)=27
900 0+(400*.09)=36
1000 0+(500*.09)=45 elevation where curve comes out, PVT.
We have an elevation we can call B, which is the midpoint between the PVC and PVT. (elev PVC+elev PVT)/2=(30+45)/2=37.5=B
We can now find the elevations along any point on the parabola by using
y=ax^2. where a= (elev B-elev PVI)/2=(37.5-0)/2=18.75=a in \(\displaystyle y=ax^{2}\).
Now, we can use the formula:
\(\displaystyle \L\\18.75(\frac{x}{L/2})^{2}\)
L is the total curve length (1000).
For the first one: \(\displaystyle \L\\18.75(100/500)^{2}=0.75\)
Add this to the tangent line elevation at x=100, which is 24, and we get 24.75.
That is the elevation of the curve at x=100.
elevation at x=200: \(\displaystyle \L\\18+18.75(\frac{200}{500})^{2}=21\)
elevation at x=300: \(\displaystyle \L\\12+18.75(\frac{300}{500})^{2}=18.75\)
elevation at x=400: \(\displaystyle \L\\6+18.75(\frac{400}{500})^{2}=18\)
elevation at x=500: \(\displaystyle \L\\0+18.75(\frac{500}{500})^{2}=18.75\)
Which is what we got for a.
Now, the other portion with the 9% grade. I will just list them out.
x=600, elev.=21
x=700, elev. =24.75
x=800, elev. =30
x=900, elev.=36.75
x=1000, elev. = 45
Notice the low point is at x=400.
This can be found by using the formula:
\(\displaystyle \L\\x=\frac{LG_{1}}{G_{1}-G_{2}}=\frac{1000(-.06)}{-.06-.09}=400\)...............
[1]
400 feet from where we started our curve at x=0
L= total length of curve, G_1 is the grade in and G_2 is the grade out.
The low point is at x=400, which we can see from our elevation table.
When we get a set of blueprints, certain info is given and we can proceed from there. PVI, PVC, PVT, L
If we need any stations in between, we can calculate them.
The low point of the curve is the vertex:
V(400,18)
This gives us a parabola equation of \(\displaystyle \L\\y=\frac{3x^{2}}{40000}-\frac{3x}{50}+30\). Using this coordinate system.
You could use x=-b/2a or differentiate to find the low point. You get the same as I showed in the above formula
[1]. If you differentiate and set to 0, you'll get x=400.
Hope this helps and isn't too confusing. I had fun showing this. It's probably more than you needed to know, but I wanted to show you how they do it in the real world. Of course, in real life the elevations would not be this small. They's probably be in the 1000's.
Tell your teacher a surveyor/road engineer showed you this. Which is
what I done for years.
Here's an exaggerated graph of your curve:
