# Math Help - Am I getting the right quadratic formula?

1. ## Am I getting the right quadratic formula?

The question:

When a road is being built, it usually has straight sections, all with the same grade, that must be linked to each other by curves. (By this we mean curves up and down rather than side to side, which would be another matter.) It's important that as the road changes from one grade to another, the rate of change of grade between the two be constant. The curve linking one grade to another grade is called a vertical curve.

Surveyors mark distances by means of stations that are 100 feet apart. To link a straight grade of $g_1$ to a straight grade of $g_2$, the elevations of the stations are given by the following equation.

$y = \frac {g_2-g_1}{2L}x^2+g_1x+E- \frac {g_1L}{2}$

Here $y$ is the elevation of the vertical curve in feet, $g_1$ and $g_2$ are percents, $L$ is the length of the vertical curve in hundreds of feet, $x$ is the number of the station, and $E$ is the elevation in feet of the intersection where the two grades would meet. (See the figure shown below.) The station $x = 0$ is the very beginning of the vertical curve, so the station $x = 0$ lies where the straight section with grade $g_1$ meets the vertical curve. The last station of the vertical curve is $x = L$, which lies where the vertical curve meets the straight section with grade $g_2$.

Assume that the vertical curve you want to design goes over a slight rise, joining a straight section of grade 1.34% to a straight section of grade –1.77%. Assume that the length of the curve is to be 500 feet (so L = 5) and that the elevation of the intersection is 1070.64 feet.

(a) What phrase in the first paragraph of this exercise assures you that a quadratic model is appropriate?

(I think it's: The last station of the vertical curve is x = L, which lies where the vertical curve meets the straight section with grade g2.)

(b) What is the equation for the vertical curve described above? Don't round the coefficients.

(For the formula, I'm getting:
$y= \frac {-0.0177-0.0134}{2(5)}x^2+0.0134x+1070.64- \frac {0.0134(5)}{2}$

after cleaning it up I got:
$y = -0.07775x^2+0.0134x+535.2865$

I'm sure this is not right.)

(c) What are the elevations of the stations for the vertical curve? (Round your answers to two decimal places.)

station number | elevation
0 | ____ ft
1 | ____ ft
2 | ____ ft
3 | ____ ft
4 | ____ ft
5 | ____ ft

(d) Where is the highest point of the road on the vertical curve? (Give the distance along the vertical curve and the elevation. Round your answers to two decimal places.)

We find the maximum at [____] feet along the vertical curve with an elevation of E = [____] feet.

(I assume that answers to (c) and (d) should come easy after the right formula is found, unless I need to take some extra steps...)

2. Hi MathBane:

I used to actually do this in real life. I was a surveyor for road and bridge construction contractors and this is how vertical curves are laid out.
We would use 50 foot stations though and 25 foot for fine grade.

The beginning of the vertical curve is called the PVC, Point of Vertical Curvature.

Where it ends is called the PVT, Point of Vertical Tangency.

The intersection point, where a straight line off of the curves intersects is called the PVI, Point of Vertical Intersection.

I wrote a program years back for my calculator that works these grades. But we used to use an HP-48GX with the Construction V software that done the calculations in the field for traversing, horizontal curves, vertical curves, or whatever was needed. They are obsolete now. Technology has advanced since thos days.

Using 0+00 as the PVC, its elevation is 1067.29

1+00 is elev. 1068.32

2+00 is elev. 1068.726

part d, The high point is at station 2+15.43 with elev.1068.733

2+50 is the PVI with elev. 1070.64, but the road elevation at that point is

1068.696. There is a 1.94 foot difference in the PVI elevation and the road elevation at that point.

Note as we round the top of the curve the grades are very similar.

3+00 is elev. 1068.511

4+00 is elev. 1067.674

5+00 at the PVT is elev. 1066.215