1. Nomenclature
A vertical curve is used to provide a smooth transition between two different grade lines, Figure B1.

Figure B1 Adding a Vertical Curve 
The parts of the curve, Figure B2, are:
Figure B2 Curve Parts 
PVI  Point of Vertical Intersection (aka PI) 
BVC  Begin Vertical Curve (aka, BC, PVC, PC) 
EVC  End Vertical Curve (aka EC, PVT, PT) 
L  Curve Length 
Distances, including the curve length, are horizontal, not along the grade lines or curve.
An equal tangent vertical curve is used. This places BVC and EVC equidistant from the PVI, Figure B3.
Figure B3 Equal Tangent Vertical Curve 
The curve is tangent to the grade lines at both ends to provide a smooth transition between the grade lines and curve.
The stations and elevations of the BVC and EVC are determined from Equationa B1 and B2:
Equation B1  Equation B2 
Because grade ratio and percent differ by a factor of 100 as do distances in feet or stations, care must be taken to use the correct form of each in Equations B1 and B2. If g is in %, then L must be in stations, if g is a ratio, then L is in feet.
2. Grade Change Rate
A vehicle enters the curve at g_{1} and after a distance of L it departs the curve at g_{2}. The total grade change is (g_{2}g_{1}). The grade change rate, k, is the total grade change divided by the distance, Equation B3.
Equation B3 
For example, a 2.00% grade into a +1.00% grade connected with a 400.00 ft curve has a k of:
This tells us that the grade changes +0.75% per each 100 ft of curve. If we go 100 ft past the BVC, the curve grade is 1.00%+0.75% = +0.25%
Increasing the curve length to 900.00 ft changes k to +0.33%/sta.
That second smaller k means the longer curve is flatter since is spreads the total grade change over a longer distance, Figure B4.
Figure B4 Different Curve Lengths 
Note that the example was a sag curve with a positive k: we went from a negative grade to a positive grade. A crest curve, on the other hand, would have a negative grade change rate, Figure B5.
Figure B5 Crest and Sag Curves 
What about curves whose grades at both ends are different but have the same mathematical sign? Figure B6 shows four different curve situations where the incoming and outgoing grades have the same mathematical sign.
Figure B6 Grades with Same Mathematical Sign 
The two curves on the left are both crest curves having a negative k since the outgoing grades are less than the incoming grades.
The two curves on the right are sag curves having a positive k since their outgoing grades are mathematically greater (less negative or more positive) than the incoming grades.
3. Curve Equation
A constant grade change rate means the curve does not have a fixed radius  it's continually changing. This provides a smoother transition when vertical travel direction is changed. A circular arc isn't generally used as a vertical curve because it has constant curvature with a fixed radius. While acceptable at lower speeds, at higher speeds there is a tendency for a vehicle to "fly" at the highest point of a crest curve or "bottom out" at the lowest point of a sag curve. While desirable for roller coaster design, upsetting passenger stomachs is generally frowned upon in road design.
Instead, a parabolic arc is used. A parabola tends to flatten at the vertical direction change making for a more comfortable transition.
The general equation for a parabolic curve, Figure B7, is Equation B4.


Figure B7 Parabola 
The equation consists of two parts, Figure B8:
 a straight line, bx+c, which is tangent to the curve at its beginning (b is the slope, c is the y intercept), and,
 an offset, ax^{2}, which is the vertical distance from the tangent to the curve.
Figure B8 Tangent and Offset 
Figure B9 shows the parabolic arc in terms of vertical curve nomenclature.
Figure B9 Curve Terms 
The curve begins at the BVC station and elevation.
The tangent line slope is g_{1}, the incoming grade.
At distance d:
 the tangent elevation is g_{1}d+Elev_{BVC}
 the tangent offset is ad^{2}, where a is a function of k
The parabolic equation written in curve terms is the Curve Equation, Equation B5.
Equation B5 
(For a complete graphic derivation of the Curve Equation see the Family of Curves section.)
d_{i} is the distance from the BVC to any point i on the curve. It is computed from Equation B6 and ranges from 0 to L.
Equation B6 
Recall the previous warning on grade and distance units. In Equation B5 the combination should be either grades in percent with distances in stations, or grades as a ratio with distances in feet. Remember that curve length, L, is a distance and must play by the same rules.
The grade at any point on the curve is a function of the beginning grade along with k and the distance from the BVC, Equation B7.
Equation B7 
Using Equations B5 through B7 allows us to compute the elevation and grade at any point along the curve. To compute the elevation of the curve's midpoint, L/2 would be used for d_{i} in Equation B5.
A geometric characteristic of a parabolic vertical curve is at its midpoint the vertical offsets from the PVI to the curve and from the curve to the chord, o in Figure B10, are the same.
Figure B10 Midcurve Offsets 
Besides using Equation B5, another way to compute the midpoint elevation is from the following:
 The chord's midpoint elevation is the average elevation of the BVC and EVC.
 The curve's midpoint elevation is the average elevation of the chord midpoint and PVI.
The curve midpoint generally isn't any more or less important than other curve points. Being able to compute its elevation two ways, however, can be used as a partial math check
4. High or Low Point
When asked where a curve's highest, or lowest, point is, most people assume it is at the middle of the curve. This is true only under a specific condition. The high or low point occurs where the curve changes direction vertically. This happens where the curve tangent has a 0% grade, Figure B11.
a. Crest Curve 
b. Sag Curve 
Figure B11 High and Low Point Conditions 
On a crest curve, the tangent grade begins at g_{1} and deceases uniformly (by k). As long as the grade is positive, the curve elevation keeps increasing. At 0%, the curve reaches its maximum elevation and after that the grade decreases, causing curve elevations to drop.
The opposite is true for a sag curve. As the grade decreases, while it is still negative curve elevations decrease. When the grade is 0%, the elevation decrease stops and as the tangent slope increases so do curve elevations.
The only condition under which the high or low point occurs at the center of the curve is when incoming and out going tangents are numerically equal but have opposite mathematical signs (eg, +2% into 2%, 3% into +3%).
The distance from the BVC to the high/low point, d_{p} in Figure B12, is determined from Equation B8:


Figure B12 Distance to High/Low Point 
The high/low point elevation can be computed by substituting d_{p} in Equation B5.
In order for Equation B8 to return a value that makes sense, the grades must have opposite mathematical signs. Consider a +3.00% grade connected to a +1.00% grade with a 500.00 ft long curve. Using Eqation B8 the distance from the BVC to the high/low point is:
The distance exceeds the curve length. What's up with that? Let's sketch it, Figure B13.
Figure B13 High Point After the EVC 
Remember, we are using only part of the parabolic curve. 7.5000 stations is the distance to the curve's highest point, it's just on that part of the curve we're not using. Our curve's actual highest point is at the EVC.
Similarly for a +0.50% grade connected to a +2.00% with a 400.00 ft curve:
This means the high/low point occurs before the BVC, Figure B14.
Figure B14 Low Point Before the BVC 
The lowest point for this curve is at the BVC.
5. Example Problems
a. Example 1
A +3.00% grade intersects a 2.40% at station 46+70.00 and elevation 853.48 ft. A 400.00 ft curve will be used to connect the two grades. Compute:
(1) Station and elevation for the curve's endpoints
(2) Elevations and grades at full stations
(3) Station and elevation of the curve's highest and lowest points
First, draw a sketch:
Figure B15 Example 1 Sketch 
For our computations, we'll use grades in percent and distances in stations.
(1) Use Equations B1 and B2 to compute the BVC and EVC
(2) Set up equations B3, and B5 to B7 to compute curve elevations.
Equation B3 

Equation B5  
Equation B6  
Equation B7 
Set up the Curve Table:
Equation B6  Equation B5  Equation B7  
Station, i  Dist, d_{i} (sta)  Elev_{i} (ft)  Grade, g_{i} (%) 
48+70.00 EVC  
48+00  
47+00  
46+00  
45+00  
44+70.00 BVC 
Why is the table upside down? We'll get to that in a second.
For each station, starting at the BVC, compute each column using the equation identified at the column top. Elevations will be computed to an additional decimal place to minimize rounding error.
sta 44+70.00:
sta 45+00:
and so on to the EVC. At the EVC each computed value has a math check indicated in red.
sta 48+70.00:
The completed Curve Table:
Equation B6  Equation B5  Equation B7  
Station, i  Dist, di (sta)  Elevi (ft)  Grade, gi (%) 
48+70.00 EVC  4.0000  848.680  2.40 
48+00  3.3000  850.029  1.46 
47+00  2.3000  850.809  0.10 
46+00  1.3000  850.239  +1.24 
45+00  0.3000  848.319  +2.60 
44+70.00 BVC  0.0000  847.480  +3.00 
There are a few other math checks in addition to those at the EVC.
In the Dist column, the difference between successive d's at full stations should be 1.0000
In the Grade column, the difference between successive g's at full stations should be k=1.35
Some texts make reference to Second Differences or Double Differences as another check. When computations were all done manually, as many checks as possible were used. We'll include it here just to show the reader how it worked.
A Second Difference is the difference between the elevation differences at full stations. Difference of a difference, got it? For this problem, we compute second differences this way:
Station 
Elev_{i} (ft) 
First Difference 
Second Difference 
48+70.00 EVC 
848.680 


48+00 
850.029 
850.029850.809=0.780 
0.7800.570=1.350 
47+00 
850.809 
850.809850.239=+0.570 
+0.5701.920=1.350 
46+00 
850.239 
850.239848.319=+1.92 

45+00 
848.319 


44+70.00 BVC 
847.480 


Note that values in the Second Difference column are equal to k, that's the math check. For you calculus buffs, the second difference is the second derivative of the Curve Equation.
It looks like a lot of computing for only two checks against k (and it is). But for those two checks to be valid requires that 5 of the 7 elevations be correct. The remaining two are the BVC and EVC.
(3) High and low point
Because this is a crest curve it will have a high point somewhere along it. The lowest point of the curve is at the BVC: 847.48 ft at 44+70.00.
Examining elevations in the completed Curve Table, it looks like the curve tops out between stations 46+00 and 48+00.
Use Equation B8 to compute the distance from the BVC:
Substitute into the Curve Equation to determine its elevation:
Add the distance to the BVC station to get the point's station:
In summary
Point 
Station 
Elevation (ft) 
High 
46+92.22 
850.81 
Low 
44+70.00 BVC 
847.48 
(4) Midpoint
Another math check is computing the curve midpoint elevation two ways
First using the curve equation
Then using offsets
(5) Misc
You're not restricted to computing elevations only at full stations along the curve. Using the equations in step (2) developed for this curve, you can compute the elevation and grade for any point between 0.00' and 400.00' feet from the BVC. If you exceed 400.00', the equations will give you values (after all, they're just equations), but for part of the curve beyond.
OK, so why the upside down table? It's traditional. When standing on the alignment and looking up station, the layout of the table visually matches the stations in front of the surveyor, Figure B16.
Figure B16 Upside Down Curve Table 
b. Example 2
A 3.50% grade intersects a +2.00% at station 12+17.53 and elevation 634.25 ft. A 400.00 ft curve will be used to connect the two grades. Compute:
(1) Station and elevation for the curve's endpoints
(2) Elevations and grades at half stations
(3) Station and elevation of the curves highest and lowest points
The computations are left to the reader. The answers are shown for checking your work.
Low point at sta 12+72.07 and elev 636.80 ft.
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 B5, for the unknown L.
Equation B5 
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 B8:
Equation B8 
Substituting Equation B8 into Equation B5 and solving for L, we arrive at Equation B9:
Equation B9 
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 B9 and solve:
The curve should be at least 700.00 ft long. Using longer lengths is fine since they pull 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 B10 
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 B12(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 Equation B13 
Curve Length Solutions Equation B14 
Equation B14 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 B13.
Compute the curve lengths using Equation B14.
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 B5, 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.
7. Compound Curves
a. Geometry
Sometimes a simple equal tangent vertical curve cannot be fit to a particular design condition. For example, the vertical curve in Figure B24 must start at an existing intersection at sta 20+00 elev 845.25 ft and end at a second intersection at sta 28+00 elev 847.75 ft. To minimize earthwork an incoming grade of +2.50% is followed by an outgoing grade of 1.00%. This places the PVI at sta 23+00 elev 852.75 ft.
Figure B24 Fitting a Curve 
This combination results in a PVI that is not midway between the BVC and EVC. If a 600.00 ft equal tangent curve were used, it would begin at 20+00 and end at 26+00, 200.00 ft back along the 1.00% grade from the specified EVC.
This situation requires an unequal tangent vertical curve. While a mathematical curve other than a parabolic arc could be used, the traditional method is to use two backtoback equal tangent curves. This is referred to as a compound curve, Figure B25.
Figure B25 Compound Curve 
A compound curve isn't as complex as it looks: the key is to break it into two successive equal tangent veridical curves.
PVIs are created for each curve at midpoint on the grade lines. A new grade, g_{3}, is created by connecting the new PVIs. This is the outgoing grade for the first curve and the incoming grade for the second.
The CVC is the Curve to Vertical Curve point which is the EVC of the first curve and the BVC of the second; its station is the same as the overall PVI station. Once we have sufficient information for each curve to set up its Curve Equation, we can compute each curve independently.
The closer the individual curve's k values, the smoother the transition between them, particularly when there is a large grade change over a short distance.
b. Curve Equation
For an equal tangent vertical curve, we set up and solved Equations B4 and B5 through B7. When setting these up for the curves in a compound curve, care must be taken to use the correct grades and lengths in each. Using subscripts i and m for the first and second curves respectively, their equations are:
Curve 1  Curve 2  
Equation B3  
Equation B5  
Equation B6  
Equation B7 
c. High/Low Point
A compound curve has the same high/low point condition as an equal tangent curve: it occurs where the curve grade passes through 0%.
For the compound curve shown in Figure B 24: the first curve has two positive grades so it never tops out, the second begins with a positive and ends with a negative grade. That places the highest point of this compound curve on the second curve.
When using the equation to compute distance to the low/high point, remember:
 use the grades for the curve on which the point occurs, and,
 the distance computed is from the BVC of the curve on which the point occurs.
For example, the distance equation Equation B8 is:
Equation B8 
Setting up Equation B8 and hi/low point station for each curve:
Distance  Station  
Curve 1  
Curve 2 
Remember: the high/low point occurs only on one curve or the other. Should g_{3} be 0%, then the low/high point would be at the CVC. If all three grades are positive or negative, then Equation B8 does not apply as neither curve tops/bottoms out.
d. Example
Using the data given in Figure B25:
(1) Complete the following table below
(2) Setup the Curve Equations for both curves
(3) Determine the station and elevation of the highest curve point.
Sketch
(1) Complete the table
Curve 1 
Component 
Curve 2 
incoming grade (%) 

outgoing grade (%) 

Length (ft) 

BVC Sta 

BVC Elev 

PVI Sta 

PVI Elev 

EVC Sta 

EVC Elev 
Entries that can be made from the given criteria:
Curve 1 
Component 
Curve 2 
+2.50 
incoming grade (%) 

outgoing grade (%) 
1.00  
Length (ft) 

20+00 
BVC Sta 
23+00 
845.25 
BVC Elev 

PVI Sta 

PVI Elev 

23+00 
EVC Sta 
28+00 
EVC Elev 
847.75 
Compute:
Since the PVIs are halfway along the grade lines, their positions are the averages of the grade line ends.
g_{3} is the grade of the line connecting the two curve PVIs. There can be different ways to compute it depending on the given data. In this example, we can use the position of the PVIs.
Compute the CVC elevation from PVI_{1}.
The CVC elevation can also be computed from PVI_{2}. We'll do it as a math check.
Complete the table.
Curve 1 
Component 
Curve 2 
+2.50 
incoming grade (%) 
+0.312 
+0.312 
outgoing grade (%) 
1.00 
300.00 
Length (ft) 
500.00 
20+00 
BVC Sta 
23+00 
845.25 
BVC Elev 
849.47 
21+50 
PVI Sta 
25+50 
849.00 
PVI Elev 
850.25 
23+00 
EVC Sta 
28+00 
849.47 
EVC Elev 
847.75 
(2) Set up curve equations
Curve 1
Curve 2
Once these are set up, a Curve Table can be computed for each curve.
(3) Station and elevation of highest curve point.
The highest point will occur on Curve 2 since its grades go through 0%.
Use the equations for Curve 2.
e. Reverse Curve
A reverse curve is a compound curve except that the two curves have opposite curvature. It consists of either a sagcrest or crestsag curves sequence, Figure B26. The CRC (Curve to Reverse Curve) is the EVC of the first curve and BVC of the second.
Figure B26 Reverse Curve 
Reverse vertical curves can be used to better approximate terrain to reduce earthwork. They also provide additional design flexibility if multiple design points are constrained. Their disadvantage is the vertical direction reversal which can be abrupt at higher speeds, particularly if the designer gets carried away and adds a third, fourth, etc, curve. Regardless, they are computed as equal tangent vertical curves once sufficient parameters for each is fixed or computed.
High/Low point determination can be interesting since both could exist on a reverse curve. For example, in Figure B25, the first curve has a low point and the second has a high point. However, the lowest point on the entire curve is on the second curve at its EVC and the highest point on the entire curve is at the BVC of the first vertical curve.
8. Self Study Problems
Problem (1)
PVI station and location are 74+26.50 and 772.30 ft. What are the endpoints' stations and elevations of a 750.00 ft long curve with incoming and outgoing grades of 4.00% and +2.00%?
For problems (2) through (4), compute the following items
(a) Curve table at full station interval. Include curve point grades.
(b) Stations and elevations of the curve's highest and lowest points?
Problem (2)
PVI station and location are 56+50.00 and 1325.00 ft. Incoming and outgoing grades are +5.0% and 3.5%. A 500.00 ft long vertical curve will be used.
Problem (3)
A 425.00 ft long curve ends at station and elevation 88+00.00 and 1220.00. Its initial and final grades are 4.00% and +2.75%.
Problem (4)
A curve begins at station 32+00.00 and elevation 912.00 ft; it ends at station 38+00.00 and elevation 915.00 ft. The PVI elevation is 918.00 ft.
Problem (5)
A 400.00 ft vertical curve begins at station 29+00.00 and elevation 954.00 ft. Its initial and final grades are 1.00% and 2.25%.
Part (a)
What are the station and elevation of the curve's midpoint?
Part (b)
What are the stations and elevations of the curve's highest and lowest points?
Problem (6)
PVI station and elevation are 41+00.00 and 1262.00 ft. Incoming and outgoing grades are 2.50% and +3.00%. What is the curve length if it must pass through an elevation of 1268.00 ft at station 42+50.00?
Problem (7)
A +1.75% grade intersects a 2.25% grade at 32+50.00 and elevation 1015.00. An existing overpass is located at station 31+56.50. Its underside elevation is 1022.50 ft. What is the minimum curve length which will provide 12 ft of clearance under the overpass?