6. Fitting Vertical Curves
a. Introduction
There are many different combinations of fitting a vertical curve to meet design conditions. In this section we will concentrate on two examples of fitting an equal tangent curve between fixed grade lines. Holding grade lines fixed isolates curve selection effects. When a grade is changed, whole sections of the alignment are affected and must be recomputed, Figure B17. Although software can easily handle this, it's the designer's responsibility to ensure solving a problem in one area doesn't adversely affect another.
(a) Preliminary Vertical Alignment 
(b) Holding Grades Fixed 
(c) Changing a Grade 
Figure B17 Fitting Vertical Curves 
In these examples we will compute simple equal tangent curves to meet some elevation condition between fixed grade lines. The next section deals with unequal tangent vertical curves which give additional design flexibility but at the cost of more complex computations.
When computing a curve passing through an elevation, the computed length may be a maximum or minimum. It depends if the specified elevation is a maximum or minimum and whether a sag or crest curve is involved. Increasing the length of a sag curve raises the entire curve; increasing a crest curve's length lowers it, Figure B18.
(a) Increasing Sag Curve Length 
(b) Increasing Crest Curve Length 
Figure B18 Curve Length and Elevations 
To solve lengths requires rearranging and solving the Curve Equation, Equation B7, for the unknown L.
Equation B7 
It looks simple until you realize that d_{i} is dependent on the location of the BVC which is in turn based on the curve length we're tying to compute. Hmmm....
b. No part of a curve may go below {above} a specific elevation
This criteria could come from the elevation of area groundwater, bedrock, or an existing overhead pass. The critical curve part of for this situation is the curve's highest or lowest point, Figure B19.
Figure B19 Low Point Condition 
The distance to the high/low point was given in Equation B10:
Equation B10 
Substituting Equation B10 into Equation B7 and solving for L, we arrive at Equation B9:
Equation B11 
Remember: grades in % and L in stations.
Example
g_{1} & g_{2} are 5.00% and +2.00% respectively; the PVI is at 10+00.00 with an elev of 800.00 ft.
What is the length of curve which does not go below 805.00 ft elev? Is this a maximum or minimum curve length?
Draw a sketch:
Substitute known data into Equation B11 and solve:
The curve should be at least 700.00 ft long. Using a longer length is fine since it raises the curve up from 805.00 ft.
Are there any math checks?
Using L=700.00 ft, set up the Curve Equation and compute the low point's elevation.
c. Curve must pass thru a specific elevation at a specific station
In this case the curve has to pass through a particular elevation at a particular station. For example, an existing road crossing may dictate the station and elevation for an atgrade intersection.
This situation appears to be easier since, unlike the previous example, we know the station at which the elevation must be met, Figure B20.
Figure B20 Elevation at a Station 
Since we know the station of the fixed elevation, we can determine its distance from the PVI. We can also relate the BVC location to the PVI.
If we substitute known and related quantities into Equation B5, we get:
Equation B12 
OK, maybe it's not as simple as it seemed. But the only unknown in the equation is L; unfortunately, it appears as numerator and denominator in the various terms. If we combine and sort terms we see that the equation takes on the form of a (surprise!) parabolic equation, a second degree polynomial. A second degree polynomial has two solutions and can be solved using the quadratic solution.


Second Degree Polynomial Equation B13(a) 
Quadratic Solution 
To simplify Equation B10 and solve it using the quadratic solution we can create the following two sets of equations:
Quadratic Coefficients Equations B14, B15, and B16 
Curve Length Solutions Equation B17 
Equation B17 will return two values for the curve length, one which makes sense in the context of the problem, the other which doesn't.
Example
g_{1} & g_{2} are 4.00% and +1.00% respectively, the PVI is at 14+00.00 with an elev of 900.00 ft.
The curve must go through an elevation of 902.65 at station 15+60.00.
What curve length meets this condition?
Draw a sketch
Compute the quadratic coefficients using Equation B14, 15, 16:
Compute the curve lengths using Equation B17:
Which curve length is correct?
Compute the BVC and EVC stations for each curve length and see if station 15+60.00 falls between them.
Length (sta)  BVC Sta  EVC Sta  15+60 bewteen? 
1.5739  13+21.30  14+78.70  No 
6.5061  10+74.70  17+25.30  Yes 
The correct curve length is 650.61 ft.
Math check?
Set up the Curve Equation, Equation B7, and solve for the elevation at station 15+60.00. It should be the same as the specified elevation of 902.65 ft.
Why are there two possible curve lengths and why does only one fit our design situation?
Remember that these are mathematical formulae independent of physical constraint. We are using only part of a complex curve for our alignment design which does have physical constraints.
There are two parabolic curves which are tangent to the grade lines and pass through the design point. These are shown plotted in Figure B21.
Figure B21 
Both curves are tangent to the two grade lines and pass through sta 15+60 and elev 902.65. Only one of the curves has the design point between the tangent (BVC and EVC) points on the grade lines. Mathematically both curves are correct, but only one meets our design criteria.
d. Multiple design points
It may not be possible to design an equal tangent curve to pass through multiple design points, Figure B22. A single curve can be fit if each of the points can be missed by a specified amount. Least squares could be applied in this case to create a bestfit curve.
Figure B22 Multiple Design Points 
Or one of the points could be treated as the most critical and used to design a curve, Figure B23.
Figure B23 Critical Point Fit 
Another possible approach is to use a compound vertical curve which will be covered next.
e. Spreadsheet
An Excel spreadsheet to fit a vertical curve based on high/low point or station and elevation can be downloaded here. It uses Visual Basic for Applications script so Excel must have macros enabled when loading the sheet.