Symmetrical Vertical Curves
A symmetrical vertical curve is one in which thehorizontal distance from the PVI to the PVC is equal to the horizontal distance from the PW to the PVT. In other words, l1 equals l2.
The solution of a typical problem dealing with asymmetrical vertical curve will be presented step by step. Assume that you know the following data:
g2 = –7%
L= 400.00´, or 4 stations
The station of thePVI = 30 + 00
The elevation of thePVI = 239.12 feet
The problem is to compute the grade elevation of the curve to the nearest hundredth of a foot at each 50-foot station. Figure 11-17 shows the vertical curve to be solved.
Figure 11-17.—Symmetrical vertical curve.
STEP 1: Prepare a table as shown in figure 11-18.In this figure, column 1 shows the stations; column 2, the elevations on tangent; column 3, the ratio of x/l; column 4, the ratio of (M)*; column 5, the vertical offsets [(x/l)*(e)]; column 6, the grade elevations on the curve; column 7, the first difference; and column 8, the second difference.
STEP 2: Compute the elevations and set thestations on the PVC and the PVT.
Knowing both the gradients at thePVC and PVT and the elevation and station at the PVI, you can compute the elevations and set the stations on the PVC and the PVT. The gradient (g1) of the tangent at the PVC is given as +9 percent. This means a rise in elevation of 9 feet for every 100 feet of horizontal distance. Since L is 400.00 feet and the curve is symmetrical, l1 equals l2 equals 200.00 feet; therefore, there will be a difference of 9 x 2, or 18, feet between the elevation at the PVI and the elevation at the PVC. The elevation at the PVI in this problem is given as 239.12 feet; therefore, the elevation at the PVC is 239.12 – 18 = 221.12 feet.
Calculate the elevation at thePVT in a similar manner. The gradient (g2) of the tangent at the PVT is given as –7 percent. This means a drop in elevation of 7 feet for every 100 feet of horizontal distance. Since l1 equals l2 equals 200 feet, there will be a difference of 7 x 2, or 14, feet between the elevation at the PVI and the elevation at the PVT. The elevation at the PVI therefore is
239.12 – 14 = 225,12 feet.
In setting stations on a vertical curve, rememberthat the length of the curve (L) is always measured as a horizontal distance. The half-length of the curve is the horizontal distance from the PVI to the PVC. In this problem, l1 equals 200 feet. That is equivalent to two 100-foot stations and may be expressed as 2 + 00. Thus the station at the PVC is
30 + 00minus 2 + 00, or 28 + 00.
The station at thePVT is
30 + 00plus 2 + 00, or 32 + 00.
List the stations under column 1.
STEP 3: Calculate the elevations at each 50-footstation on the tangent.
From Step 2, you know there is a 9-foot rise inelevation for every 100 feet of horizontal distance from the PVC to the PVI. Thus, for every 50 feet of horizontal distance, there will be a rise of 4.50 feet in elevation. The elevation on the tangent at station 28 + 50 is
221.12 + 4.50 = 225.62 feet.
The elevation on the tangent at station 29 + 00 is
225.62 + 4.50 = 230.12 feet.
Figure 11-18.—Table of computations of elevations on a symmetrical vertical curve.
The elevation on the tangent at station 29 + 50 is
230.12 + 4.50 = 234.62 feet.
The elevation on the tangent at station 30 + 00 is
234.62 + 4.50 = 239.12 feet.
In this problem, to find the elevation on the tan-gent at any 50-foot station starting at thePVC, add 4.50 to the elevation at the preceding station until you reach the PVI. At this point use a slightly different method to calculate elevations because the curve slopes downward toward the PVT. Think of the eleva-tions as being divided into two groups—one group running from the PVC to the PVI; the other group running from the PVT to the PVI.
Going downhill on a gradient of –7 percent fromthe PVI to the PVT, there will be a drop of 3.50 feet for every 50 feet of horizontal distance. To find the elevations at stations between the PVI to the PVT in this particular problem, subtract 3.50 from the eleva-tion at the preceding station. The elevation on the tangent at station 30 + 50 is
239.12-3.50, or 235.62 feet.
The elevation on the tangent at station 31 + 50 is
235.62-3.50, or 232.12 feet.
The elevation on the tangent at station 31 + 50 is
232.12-3.50, or 228.62 feet.
The elevation on the tangent at station 32+00 (PVT) is
228.62-3.50, or 225.12 feet,
The last subtraction provides a check on the work youhave finished. List the computed elevations under column 2.
STEP 4: Calculate(e), the middle vertical offset at the PVI. First, find the (G), the algebraic difference of the gradients using the formula
G =g2– g1
The middle vertical offset(e) is calculated as follows:
e= LG/8 = [(4)(–16) ]/8 = -8.00 feet.
The negative sign indicates e is to be subtracted fromthe PVI.
STEP 5: Compute the vertical offsets at each50-foot station, using the formula (x/l)2 e. To find
the vertical offset at any point on a vertical curve,first find the ratio x/l; then square it and multiply
bye; for example,
x/l= 50/200 = 1/4.
at station 28 + 50, the ratio of
Therefore, the vertical offset is
(1/4)2 e = (1/16) e.
The vertical offset at station 28 + 50 equals
(1/16)(–8) = –0.50 foot.
Repeat this procedure to find the vertical offset ateach of the 50-foot stations. List the results under columns 3, 4, and 5.
STEP 6: Compute the grade elevation at each ofthe 50-foot stations.
When the curve is on a crest, the sign of the offsetwill be negative; therefore, subtract the vertical offset (the figure in column 5) from the elevation on the tangent (the figure in column 2); for example, the grade elevation at station 29 + 50 is
234.62 – 4.50 = 230.12 feet.
Obtain the grade elevation at each of the stations in asimilar manner. Enter the results under column 6.
Note:When the curve is in a dip, the sign will be positive; therefore, you will add the vertical offset (the figure in column 5) to the elevation on the tangent (the figure in column 2).
STEP 7: Find the turning point on the verticalcurve.
When the curve is on a crest, the turning point isthe highest point on the curve. When the curve is in a dip, the turning point is the lowest point on the curve. The turning point will be directly above or below the PVI only when both tangents have the same percent of slope (ignoring the algebraic sign); otherwise, the turning point will be on the same side of the curve as the tangent with the least percent of slope.
The horizontal location of the turning point is either measured from thePVC if the tangent with the lesser slope begins there or from the PVT if the tangent with the lesser slope ends there. The horizontal loca-tion is found by the formula:
xt= distance of turning point from PVC or PVT
g= lesser slope (ignoring signs)
L= length of curve in stations
G= algebraic difference of slopes.
For the curve we are calculating, the computations would be (7 x 4)/16 = 1.75 feet; therefore, the turning point is 1.75 stations, or 175 feet, from thePVT (station 30 + 25).
The vertical offset for the turning point is foundby the formula:
For this curve, then, the computation is ( 1.75/2)2 x 8 = 6.12 feet.
The elevation of the POVT at 30 + 25 would be 237.37, calculated as explained earlier. The elevation on the curve would be
237.37-6.12 = 231.25. STEP 8: Check your work.
One of the characteristics of a symmetrical parabolic curve is that the second differences between successive grade elevations at full stations are constant. In computing the first and second differences (columns 7 and 8), you must consider the plus or minus signs. When you round off your grade elevation figures following the degree of precision required, you introduce an error that will cause the second difference to vary slightly from the first difference; however, the slight variation does not detract from the value of the second difference as a check on your computations. You are cautioned that the second difference will not always come out exactly even and equal. It is merely a coincidence that the second difference has come out exactly the same in this particular problem.