1. Alignment
An alignment is a three dimensional refernce line consisting of a series of straight and curved segments. It serves as a reference for linear projects such as roads, pipelines, transmission lines, etc. Most alignments have horizontal and vertical components, Figures A1 and A2, each related but having their own geometric elements and specific design constraints.
Figure A1 
Figure A2 Vertical Component 
Although most contemporary alignment design is done digitally in three dimensions, it is still important to understand the geometry of the individual components.
The remainder of the chapter will concentrate on geometry for road alignments.
2. Stationing
A station is both a dimension and a position. As a dimension it has a fixed length with associated unit. As a position it is the cumulative horizontal distance along the alignment from its beginning.
In the English system, a station is defined as the length of a standard surveyor's tape making it 100.00 ft long. A position is expressed as the number of full stations (100 ft intervals) plus a partial station (less than 100 ft) and is written in the full+partial format. Station 8+35.67 is 835.67 feet along the alignment from its origin. Alignments do not usually start at station 0+00 because subsequent redesign can result in negative stations: 1+45.67 looks confusing, doesn't it?
In the metric system, a station may be 100 m or 1000 m with a corresponding partial station. A point 12,56.02 m along the alignment would be at station 12+26.02 or 1+256.02. We'll use the English system although the same basic logic applies to metric stationing.
Alignments are traditionally staked at full station (100 ft) intervals and changes in horizontal direction, Figure A3. In some cases, such as trenching, stakes can be placed at half or quarterstation intervals (50 ft and 25 ft intervals, respectively).
Figure A3 Stationing 
The horizontal distance between two stations along the alignment is their stationing difference.
Figure A4 Distance 
In Figure A4, the distance from sta 12+38.28 to sta 16+56.85 is:
The distance can be expressed in feet or stations.
The position of a point off the alignment is expressed by a station and an offset to the right or left (defined looking upstation). The position of the CP in Figure A5 is 11+68.26 82.40R
Figure A5 Station and Offset 
A single point can sometimes have two different stations values. The most common situation is where two different alignments cross, Figure A6.
Figure A6 Station Equation 
Point A is at station 12+83 along one alignment and station 26+45 along a second. A Station Equation can be written for point A: Sta 12+83 = Sta 26+45. This may be brought to the readers' attention on a set of plans to ensure the appropriate station value be used for alignment calulations. Another station equation situation can occur in horizontal curve computations and will be covered in that section.
3. Grade
a. Definition
Grade is the slope of a straight line and is computed from:
Eqn A1  
elev diff: elevation difference horiz dist: horizontal distance 
If the elev diff and horiz dist are both in feet (or meters) Equation A1 results in grade as a dimensionless ratio; multiplying it by 100 converts it to a percentage (%).
If elev diff is feet and hori dist is stations, Equation A1 returns grade as a percentage.
Grade is positive (+) going uphill in the direction of stationing, negative () going downhill, Figure A7.
Figure A7 Grade 
b. Examples
(1) The elevation at sta 10+25.00 is 1214.80 ft; at sta 12+75.00 it is 1193.50 ft.
What is the grade of the line between the two stations?
First, draw a sketch:
Using Equation A1:
(2) At sta 16+50.00 the elevation is 867.50 ft. If the grade through 16+50.00 is +3.00%, what is the elevation at sta 19+00.00?
Sketch:
We need to rearrange Equation A1 to solve for the elevation difference between the two stations:
The horizontal distance is  
and elevation difference is  
and elevation at 19+00.00 
According to the sketch, 19+00 is higher than 16+50 so the computed elevation looks correct
(3) Using the data in example (2), what is the elevation at station 14+00?
Sketch:
The horizontal distance is:
Since we're going backwards along the alignment, the distance is negative.
The elevation is:
From the sketch we see that 14+00 is lower than 16+50, so the computed elevation looks correct.
 Details
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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.
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 B10.
a. Crest Curve 
b. Sag Curve 
Figure B10 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}, is determined from Equation B8:


Figure B11 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 B12.
Figure B12 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 B13.
Figure B13 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 B14 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) 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 B15.
Figure B15 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 B16. 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 B16 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 B17.
(a) Increasing Sag Curve Length 
(b) Increasing Crest Curve Length 
Figure B17 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 B18.
Figure B18 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 B19.
Figure 19 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 B13 
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 B20.
Figure 20 
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 B21. 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 B21 Multiple Design Points 
Or one of the points could be treated as the most critical and used to design a curve, Figure B22.
Figure B22 Critical Point Fit 
Another possible approach is to use a compound vertical curve which will be covered next.
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 B23 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 B23 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 B24.
Figure B24 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 B24:
(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 B25. The CRC (Curve to Reverse Curve) is the EVC of the first curve and BVC of the second.
Figure B25 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.
 Details
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1. General 

2. Compound curve 

3. Reverse curve 
1. General
A multicenter curve consists of two or more backtoback curves that are tangential. The simplest consists of two curves either in the same direction (compound) or opposite directions (reverse), Figure D1.
a. Compound curve  b. Reverse curve 
Figure D1 Multicenter curves 
Multicenter curves provide greater design flexibility. For example, Figure D2 shows a typical cloverleaf freeway interchange.
Figure D2 Cloverleaf interchange 
To travel southeast, southwestbound traffic changes direction approximately 280° using the southwestern loop. This is accomplished using a flat curve for initial deceleration, followed by sharper curves for most of the direction change, and finally shallower acceleration curves. Figure D3 shows the southwestern loop with overlaid circular arcs.
Figure D3 Multiple curves 
Using a multicenter curve makes the interchange more space efficient; consider how much area would be needed for a single radius curve in this situation.
A reverse curve also provides design flexibility, particularly in tight areas.
For example, in Public Land Survey (PLS) states, northsouth Township lines were offset along Standard Parallels to account for meridian convergence, Figure D4a. Roads commonly followed Section and Township lines. At an offset Township line, a sharp left turn shortly after a sharp right turn was perfectly acceptable at horse drawn wagon speed, Figure D4b. As vehicle speeds increased, flatter curves were necessary but had to fit between the Township line segments. This could be done using a reverse curve, Figure D4c.
a. Convergence offset  b. Low speed curves  c. High speed curves 
Figure D4 Offset Township lines 
A reverse curve can also be used to work an alignment around an obstruction. Figure D5 shows what were it not for the presence of a lake would be a simple curve situation .
Figure D5 Obstruction 
Neither a long nor short radius single curve will miss the lake. However, a curve to the right could be fit to an extension of the incoming tangent followed by a curve to the left to meet the original outgoing tangent, Figure D6.
Figure D6 Working around obstruction 
If multicenter curves are so flexible, why don't we see more of them?
Primarily because of the curvature change at each curvetocurve transition.
On a simple curve, once the driver sets the steering wheel angle, it is maintained through the entire curve. A multicenter curve requires a steering wheel angle change at the beginning of each curve. Except for a reverse curve, it may not be visually apparent where one curve ends and the next begins. This makes it easy to over or undershoot successive curves.
Traveling around a circular arc introduces centrifugal force which pushes the vehicle (and its occupants) to the outside of the curve. The sharper the curve or the higher the speed, the greater the force. One way to offset this centrifugal effect is to use superelvation  pavement crossslope inclination. As a vehicle traverses differing radii curves, different superelevation rates are needed. Superelevation change should be introduced gradually or passengers may feel like they are being tossed to the side. This is more pronounced with a large radii difference from one curve to the next. On a reverse curve the condition can be worse since the crossslope direction must be changed.
Figure D7 shows a threecenter curve system with 5 section view labels.
Figure D7 Threecenter curve 
Table D1 depicts the steering wheel angle and superelevation at each section view going through the curve system.
Table D1 

Section  Steering Wheel  Superelevation 
AA  
BB  
CC  
DD  
EE 
Note that the greatest wheel angle and superelvetion occur on the third curve and must be removed entirely when the exit tangent is reached. That abrupt change could cause safety and comfort issues.
A reverse curve provides an additional chanllenge, Figure D8 and Table D2.
Figure D8 Reverse Curve 
Table D2 

Section  Steering Wheel  Superelevation 
AA  
BB  
CC  
DD 
At the point of reverse curvature, the steering angle must be changed to the other side. The superelevation also changes direction so the vehicle is "rolled" from one side to the other. Coupled with superelevation, the centrifugal force changes direction. Allinall quite a bit of stuff going on, especially when the vehicle is cruising along at a good rate.
So while multicenter curves provide greater design flexibility, they do have drawbacks. particularly in high speed situations. Except for special cases such as interchanges, their use should be limited to lower speeds with reasonable radii differences.
2. Compound curve
a. Nomenclature
Starting with two tangent lines intersecting at a PI, a compound curve is fit between them, Figure D9.
Figure D9 Compound curve 
Figure D10 shows major parts of the compound curve.
Figure D10 Compound curve parts 
The PCC (Point of Curve to Curve) is the EC of the first curve and BC of the second. A tangent line at the PCC intersects the incoming tangent at the PI_{1} and the outgoing tangent at PI_{2}.
T_{i} is the distance from the BC_{1} to the PI; T_{o} the distance from the PI to the EC_{2}.
There are seven parts to a compound curve: T_{i}, T_{o}, R_{1}, R_{2}, Δ_{1}, Δ_{2}, and Δ. Because the tangents deflect Δ at the PI and the compound curve transitions from the incoming to the outgoing tangent:
Δ = Δ_{1}+ Δ_{2}  Equation D1 
That means there are six independent parts. For a unique solution four parts must be fixed, including at least one angle and at least two distances.
Because they share a point (PCC) and tangent line, once all six parts are determined, the compound curve can be computed as two simple curves, Figure D11. Individual curve attributes can be computed using the single curve equations in Chapter C.
Figure D11 Two simple curves 
Depending on which parts are initially fixed, there are different ways to compute the remaining parts.
One way is solving the vertex triangle, Figure D12, created by the three PIs.
Figure D12 Vertex triangle 
Another method is to use a closed loop traverse through points O_{1}BC_{1}PIEC_{2}O_{2}, Figure D13.
Figure D13 Traverse method 
The direction of one line is assumed to be in a cardinal direction and the others written in terms of 90° angles and Δs. Distances are Rs and Ts. Latitudes and departures are computed, summed, are set eqaul to zero to force closure. This results in two equations in two unknowns which can then be simultaneously solved.
b. Examples
(1) Example 1
A PI is located at station 38+00.00 with a left deflection of 72°00'00"L. The compound curve begins at sta 33+50.00. The first curve has a 700.00 ft radius and 30°00'00" central angle.
Determine the radius and central angle of the second curve and the length of both curves.
Sketch:
We'll try a vertex triangle solution. Isolate the triangle and label the tangents:
Since we have the radius and central angle of the first curve, we can compute its tangent:
Compute Δ_{2} using Equation D1
Determine the distance from the PI_{1} to the PI which is a side of the triangle.
The distance between PI_{1} and PI_{2} is the sum of the curve tangents. Using the Law of Sines and the known T_{1}, we can compute T_{2}.
Using T_{2} and Δ_{2}, R_{2} can be determined.
Finally, compute each curve's length.
(2) Example 2
A PI is located at 55+69.23 with a deflection angle between tangents of 85°00'00"R. The compound curve must begin 463.64 ft before the PI and end 405.60 ft after. Central angles of the two curves are 30°00'00" and 55°00'00" respectivley.
Determine the radius of each curve,
Sketch:
Becasue we only have angles of the vertex triangle with no distances (and insufficient given data to compute any), the curve system can't be solved that way. Instead, we will use the traverse method.
To start, the curve system is rotated to make the initial radial line run North. Then using right angles and Δs, bearings of the other lines can be determined.
Updated sketch with bearings:
Compute Latitudes and Departures:
Line  Bearing  Length  Latitude  Departure 
O_{1}BC_{1}  North  R_{1}  R_{1}  0.00 
BC_{1}PI  East  463.64  0.00  463.64 
PIEC_{2}  S 5°00'00"E  405.60  405.60 x cos(5°00'00")  +405.60 x sin(5°00'00") 
EC_{2}O_{2}  S 85°00'00"W  R_{2}  R_{2} x cos(85°00'00")  R_{2} x sin(85°00'00") 
O_{2}O_{1}  S 30°00'00"W  R_{1}R_{2}  (R_{1}R_{2}) x cos(30°00'00")  (R_{1}R_{2}) x sin(30°00'00") 
Sum and reduce the Latitudes:
Sum and reduce the Departures:
We have two equatiosn with unkowns R_{1} and R_{2}. Solve them simultaneously. We'll use substitution.
Solve the Latitide summation for R_{1}
Substiute the equation for R_{1} in the Departures summation and solve R_{2}:
Solve R_{1}:
With R_{1} and R_{2} computed, we have two geometric attributes for each curve. Using Chapter C equations, their remaining attributes can be determined.
c. Stationing
As with a single curve, there are two paths to the EC from the BC. One goes up and down the original tangents, the other along the curves. Whether a station equation is used at the EC depends on how the entire alignment is stationed. Refer to the discussion in Chapter C Section 2.
Equations to compute the curves' endpoints are:
Equation D2  
Equation D3  
Equation D4  
Equation D5 
The stationing for Example (2) is:
e. Radial stakeout
Each curve's table is computed as described in Chapter C.
Following are radial chord tables for each curve in Example (2) (computations aren't shown).
Curve 1  
Sta  Defl ang  Chord  
EC  54+10.64  15º00'00"  301.58 
54+00  14º28'36"  291.29  
53+00  9º33'34"  193.51  
52+00  4º38'32"  94.31  
BC  51+05.59  0º00'00"  0.00 
Curve 2  
Sta  Defl ang  Chord  
EC  58+12.43  27º30'00"  386.54 
58+00  26º38'57"  375.47  
57+00  19º48'17"  283.63  
56+00  12º57'37"  187.77  
55+00  6º06'57"  89.109  
BC  54+10.64  0º00'00"  0.00 
Stakeout procedure
Measure T_{i} and T_{o} from the PI along the tangents to set the BC_{1} and EC_{2}.
Stake Curve 1
Setup instrument on BC_{1}.
Sight PI with 0º00'00" on the horizontal circle.
Stake curve points using the curve table.
Stake Curve 2
Setup on PCC
Sight BC_{1} with (Δ_{1}/2) = 15º00'00"=345°00'00" on the horizontal circle, Figure D14a.
Inverse the scope to sight away from the BC_{1}, Figure D14b.
Rotating right to a horizontal angle of 0º00'00" orients intrument tangent to the curves at the PCC, Figure D14c.
Stake curve points using the curve table.
Last point should match the previously set EC_{2}.
a. Sight BC_{1} with (Δ_{1}/2) reading 
b. Inverse telescope 
c. Rotate to 00°00'00" 
Figure D14 Orienting at PCC 
3. Reverse curve
a. General
A reverse curve consists of two consecutive tangent corves with radius points on opposite sides of the center line. Figure 15 shows basic nomenclature and parts.
Figure D15 Nomenclature 
The PRC is the Point of Reverse Curvature, and is the EC of the first curve and BC of the second.
The distance from PI_{1} to PI_{2} is T_{1} + T_{2}. Because of this relationship, as soon as one curve's geometry is fixed, so is the other's. This makes a reverse curve generally easier to compute than a compound curve.
Once the geometry of both curves are fixed, they can be computed as individual simple curves.
b. Stationing
Stationing can be a little confusing since there can be a station equation at both the PRC and EC of the second curve. Usually, if a stationing is maintained along the tangents, then a station equation only appears at the end of the reverse curve.
To compute stationing:
Equation D6  
Equation D7  
Equation D8  
Equation D9 
b. Example
For a reverse curve, the first PI is at 33+50.15 with a 75°20'00"L deflection angle and second PI is at 40+16.70 with a 83°50'00"R deflection angle.
The first curve has a 375.00' radius. Determine the radius of the second curve.
Sketch:
Tadaa!
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Section 1. Basic Concepts
a. General
While some alignments like an electrical transmission line can be designed with angle points at changes in horizontal direction, alignments for moving commodities with mass must have less abrupt transitions. This is accomplished by linking straight line segments with curves, similar to that in vertical alignments. However in this case the lines and curves linking them are in a horizontal plane.
A simple tangent geometric curve is used to link the two lines. A curve is tangent to a line when its radius is perpendicular to the line, Figure C1.
Figure C1 
The two lines bounding the curve are generically referred to as tangents or tangent lines. The direction of a curve is either right or left based on the deflection direction between the tangents, Figures C2 (a) and (b).
(a) Deflection Angles 
(b) Curves 
Figure C2 Curve Direction 
In roadway design horizontal geometry differs from vertical because a driver is responsible for guiding a vehicle from the first tangent to the second one. This requires turning a steering wheel changing the car's direction. Ideally this would be a smooth action to cause minimal discomfort: once into a curve, the steering angle would be maintained until the tangent is reached.
Two different mathematical arcs can be used for horizontal curves, either singularly or in combination: a circular arc and a spiral arc.
b. Circular arc
A circular arc has a fixed radius which means a driver doesn't have to keep adjusting the steering wheel angle as the car traverses the curve. It is a simple curve which is relatively easy to compute, Figure C3.
Figure C3 Circular Arc 
Its primary disadvantage is that the constant curvature must be introduced immediately at curve's beginning. That means a driver would have to instantaneously change the steering wheel angle from 0° to full or the car would overshoot the curve. A similar condition exists ate the curves's end. The higher the vehicle speed and the sharper the curve, the more pronounced the effect.
c. Spiral arc
A spiral has a constantly changing radius, Figure C4. At the spiral's beginning, its radius is infinite; as the vehicle progreeses into the curve, the spiral radius decreases. A spiral provides a more natural direction transition  the driver changes the steering wheel angle uniformly as the car traverses the curve.
Figure C4 Spiral Arc 
Attaching a second sprial with reversed radius change creates an entranceexit spiral condition where the driver gradually increases then gradually decreases steering wheel angle, Figure C5.
Figure C5 Combined Spirals 
A circular arc can be combined with two spirals, Figure C6.
Figure C6 
Combined spirals and spiralled horizontal curves, in conjunction with superelveation, help balance centrifugal forces. For a constant velocity, as the radius decreases the centrifugal force increases. A spiral allows superelevation to be introduced at a uniform rate allowing it to offset increasing centrifugal force. In theory, there exists an equilibrium velocity at which a vehicle could travel from one tangent through the spirals and curve to the second tangent safely even if the road were completely ice covered.
Railroad alignments typically use spiral horizontal curves because of their force balancing nature. Because of the wheel flange to rail connection, a train moving around a curve exerts a force directly to the rails (unlike a vehicle's tirepavement connection which can devolve into a skid). A rail line laid out with a circular arc would shift to a spiral configuration after a train has run through it a number of time at transport speed.
The traditional disadvantage of a spiral is that it is complex to compute, although that has been largely negated by software. While still used for railways, in high speed highway design using long flat circular arcs minimizes the tangent to curve transition so spirals aren't as critical. They can be useful in low speed situations where there is a large direction transition. We'll examine spiral geometry and application in a later section.
Section 2. Nomenclature; Components
a. Degree of Curvature
For any given set of tangents, there are an infinite number of circular arcs which can be fit between them. The arcs differ only in their radii which relates to their "sharpness." Consider the two arcs in Figure C7:
Figure C7 Different Arcs 
Both curves must accommodate a total direction change of Δ but since C_{1} is shorter it is sharper than C_{2}.
Degree of curvature is a traditional way of indicating curve shaprness. There are two different definitions of degree of curvature: arc and chord.
Arc definition, D_{a}, is the subtended angle for a 100.00 ft arc; Chord definition, D_{c}, is the subtended angle for a 100.00 ft chord, Figure C8.
(a) Arc Definition 
(b) Chord Definition 
Figure C8 Degree of Curvature 
Street and road alignments use the arc definition; chord definition is used for railroad alignments. For the remainder of this Chapter we will use the arc defintion and refer to it simply as D.
Degree of curvature is inversely proportional to radius: as D increases, R decreases, Figure C9. The larger D is, the sharper the curve.
Figure C9 D and R Relationship 
Note the importance of 100.00 in both Degree of Curvature versions. This goes back to a standard tape length and stationing interval. However, in the metric system there is no convenient base equivalent of 100.00 ft. While design criteria was traditionally expressed in terms of D, it is more common today to instead use R which works for both the English and metric systems.
The relationship between D and R is expressed by Equation C1.
Equation C1 
b. Curve Components; Equations
Figure C10 shows a tangent circular arc with some basic components labeled.
Figure C10 Basic Components of a Circular Arc 
PI 
Point of Instersection 
BC 
Begin Curve (aka: PC  Point of Curve; TC  Tangent to Curve) 
EC 
End Curve (aka: PT  Point of Tangent; CT  Curve to Tangent) 
Δ 
Defelction angle at PI; also the central angle of the arc 
R 
Arc radius 
L 
Arc Length 
LC 
Long Chord length 
Figure C11 includes additional curve components.
Figure C11 Additional Curve Components 
T 
Tangent distance 
E 
External distance  from PI to midpoint of arc 
M 
Middle ordinate  distance between midpoints of arc and Long Chord 
Equations for the curve components are:
Equation C2  
Equation C3  
Equation C4  
Equation C5  
Equation C6 
Although it may look like it in Figure C11, E and M are not equal.
c. Stationing
As mentioned in Chapter A, an alignment is stationed at consistent intervals from its beginning through its end. On a finished desgin, the stationing should be along the tangents and the fitted curves.
Traditionally, an alignment is stationed along the straight lines thru each PI, Figure C12. Curve fitting comes later.
Figure C12 Stationing Along Tangents 
A circular curve is fit and staked. The stationing along the tangents between curve ends would be replaced by the curve stations, Figure C13.
There are two ways to get from BC to EC:
(1) Up and down the tangents, T+T
(2) Along the curve, L
The distance along the curve is shorter than up and down the tangents: L < T+T
Figure C13 Curve Inserted 
That means for a typical curve there are two stations for the EC:
One along the original tangents
One along the curve.
Figure C14 
The EC station with respect to the original tangents is the EC Ahead. If we are standing on the EC and take a step ahead (upstation), we are on the tangent and its original stationing.
The EC station with respect to the curve is the EC Back. If we are standing on the EC and take a step back, we are on the curve and its stationing.
Chapter A mentioned that a station equation is used where one point has two stations. In this case we have a station equation at the EC: EC Sta Ahead = EC Sta Back. Figure C15 is an example of a station equation indicator on a set of WisDOT highway plans.
Figure C15 Station Equation Indicator 
A station equation represents a stationing discontinuity. We know the distance between two alignment points is their stationing difference. However if the points are on each side of an EC, the discontinuity must be taken into account. For example, the distance between stations 13+00 and 16+00 on the alignment shown in Figure C14 is normally 300.00 ft but there are 48.88 ft "missing" at the EC. The 48.88 ft is the difference between the Ahead and Back stations: (14+82.97)  (14+34.09) = 48.88 ft. So the correct distance from 13+00 to 16+00 is 251.12 ft.
With software, it is possible to avoid station equations by waiting until the curves are fitted before stationing the entire alignment. With computer assisted designs, the PIs could be computed positions, Figure C16.
Figure C16 Computed Tangents and PVI 
A curve is then fit to the tangents, Figure C17.
Figure C17 A Curve is Fit 
Then the alignment is stationed from its beginning to its end through the curves, Figure C18. That way there are no station equations.
Figure C18 Stationing Through Curve 
So this latter method is simpiler and the one that should be used, right? Well, it does have some advantages, but it also has disadvatages. What happens if the alignment design must be changed at some point later in the process?
With traditional stationing, each curve has its own EC station equation. If one curve is altered, only its stationing is affected, no other stations on the alingment change, Figure C19.

(b) Limited Stationing Changes with Curve Modification 
Figure C19 Alignment with Station Equations 
With continuous stationing, when a curve is altered it affects stations on it as well as all stations after it, Figure C20.
(a) Original Continuous Stationing 
(b) Stationing Changes on and after Modified Curve 
Figure C20 Alignment with Continuous Stationing 
For example, if the redesigned curve is shorter, then all full (+00) station points after the curve increase. For example, 12+00 becomes 12+7.03, 13+00 becomes 13+07.03, etc. If the alignment is already staked then each stake could be renumbered or each could be each be moved back 7.03 feet. Hmm, odd stations or moving stakes.... Maybe station equations aren't so bad after all.
BC and EC stations are computed from the following formulae:
Equation C7  
Equation C8  
Equation C9 
d. Example: Curve Components, Stationing
A PI is located at station 25+00.00. The deflection angle at the PI is 55°00'00" R. A 500.00 ft radius curve will be fit between the tangents.
Compute curve components and endpoint stations.
Start with a sketch:
Use Equations C2 through C6 to compute curve components (carry an additional digit to minimize rounding errors):
Compute degree of curvature using Equation C1:
Use Equations C7 through C9 to compute endpoint stations:
Section 3. Radial Chord Method
a. Circlular Geometry
For any circular arc, the angle between the tangent at one end of the arc and the chord is half the arc's central angle, Figure C21.
Figure C21 Deflection angle 
Angle a/2 is the deflection angle from one end of the arc to the other. The chord's length is computed from:
Equation C10 
In terms of the degree of curvature, Figure C22:
Figure C22 Full station deflection angle 
The deflection angle for a full station is half the degree of curvature. Since the deflection angle is D/2 and it occurs over a 100.00 ft, the deflection rate can be computed from:
Equation C11 
Extending this geometry to the entire curve, Figure C23, the total deflection angle at the BC from the PI to the EC is Δ/2.
Figure C23 Deflection angle for entire curve 
Since the deflection angle occurs across the curve's length, the deflection rate can also be written as Equation C12.
Equation C12 
b. Radial chords
One way to stake a horizontal curve is by the radial chord method, Figure C24.
Figure C24 Radial chord method 
An instrument is set up on the BC and the PI sighted as a backsight. Then to stake each curve point, a defelction angle is turned and chord distance measured. For each point to be staked, we need to compute its deflection angle and chord distance.
The deflection angle to any point i on the curve, Figure C25, can be computed from Equation C13.
Figure C25 Deflection angle and radial chord 
Equation C13 
l_{i} is the arc distance to the point from the BC and is computed using Equation C14.
Equation C14 
Equation C10 can be rewritten using the deflection angle:
Equation C15 
Using Equations C13 to C15, the defelction angle and distance any curve point from the BC can be computed.
c. Example
Determine the radial chord stakeout data at full stations for the example from Section 2.d.
Summary of given and computed curve data:
Δ = 55°00'00"  R = 500.00 ft  
D = 11°27'33.0"  L = 479.965 ft  T = 260.284 ft 
LC = 461.749 ft  E = 63.691 ft  M = 56.494 ft 
Point  Station 
PI  25+00.00 
BC  22+39.716 
EC  27+19.681 Bk = 27+60.284 Ah 
Use Equation C11 to compute the curve's deflection rate:
Set up Equations C13, C14, and C15 for the curve:
Set up the Curve Table:
Curve Point 
Arc dist, l_{i}, (ft) 
Defl angle,δ_{i} 
Radial chord, c_{i} 

EC  27+19.681 Bk 



27+00 




26+00 




25+00 




24+00 




23+00 




BC  22+39.716 
Solve the three equations for each curve point and record the results in the table.
At 22+39.716, we're still at the BC so all three entries are zero.
At 23+00:
And so on until the table is complete.
Curve Point 
Arc dist, li, (ft) 
Defl angle,δi 
Radial chord, ci 

EC  27+19.681 Bk  479.965  27°30'00.0"  461.748 
27+00  460.284  26°22'20.4"  444.203  
26+00  360.284  20°38'33.9"  352.540  
25+00  260.284  14°54'47.4"  257.355  
24+00  160.284  9°11'01.0"  159.599  
23+00  60.284  3°27'14.5"  60.248  
BC  22+39.716  0.000  0°00'00.0"  0.000 
Math checks:
Arc distances between successive full stations differ by 100.00 ft.
At the EC:
Arc distance should equal the curve length
Defl angle should equal Δ/2
Radial chord shoud equal Long Chord
The difference between deflection angles at successive full staions should be D/2.
These checks have been met.
Remember that we carried additonal decimal places to minimize rounding errors. Rounding can become quite pronounced for long flat curves so computation care must be exercised. Once the table has been computed, the final values can be shown to reasonable accuracy levels. For example, distances can be shown to 0.01 ft and deflection angles to 01".
d. Summary
Computing and staking curve points by the radial chord method is simple and straightforward. Using the curve equations, any point on the curve can be computed and staked, not just those included in the curve table.
However, in the field, it may not be the most efficient way to stake a curve using modern instrumentation, particularly with long curves which may have chords thousands of feet long at very small deflection angles. Most contemporay survey computations and fieldwork use coordinates giving greater stakeout flexibility. While the radial chord method may not be used for stake out, it is well adapted to coordinate computations as we'll see in the next section.
Section 4. Curves and Coordinates
a. Coordinate Equations
Equations C16 and C17 are general equations for computing coordinates using direction and distance from a known point, Figure C26.
Equation C16  
Equation C17 
Figure C26. Coordinate Computation 
Direction (Dir) may be either a bearing or azimuth.
Curve point coordinates can be computed using these equations from a base point. Since the radial chord method uses the BC as one end of all the chords, it can also be used as the base point for coordinate computations.
b. Computation Process
Assuming we start with the tangents and PI, then fit a curve, the general process is as follows:
Figure C27 
The original tangent lines have directions; PC has coordinates.  
Figure C28 
A curve is fit to the tangents. End points are at distance T from the PI along the tangents. 

Figure C29 
Compute coordinates of BC using backdirection of the tangent BCPI and T.


Figure C30 
Compute coordinates of EC using direction of the tangent PIEC and T. These will be use for a later math check.


Figure C31 
Use a curve point's deflection angle to compute the direction if its radial chord from the BC.
δ is positive for right deflections, negative for left.

c. Example
Continuing with the previous example problem.
Summary of given and computed curve data:
Δ = 55°00'00"  R = 500.00 ft  
D = 11°27'33.0"  L = 479.965 ft  T = 260.284 ft 
LC = 461.749 ft  E = 63.691 ft  M = 56.494 ft 
Point  Station 
PI  25+00.00 
BC  22+39.716 
EC  27+19.681 Bk = 27+60.284 Ah 
Additional information: Azimuth of the initial tangent is 75°40'10"; coordinates of the PI are 1000.00 N, 5000.00' E.
Compute coordinates of the BC:
Compute the coordinates of the EC:
Set up Equatons C21 through C24 for this curve.
This is the Radial Chord table computed previously:
Curve Point 
Arc dist, li, (ft) 
Defl angle,δi 
Radial chord, ci 

EC  27+19.681 Bk  479.965  27°30'00.0"  461.748 
27+00  460.284  26°22'20.4"  444.203  
26+00  360.284  20°38'33.9"  352.540  
25+00  260.284  14°54'47.4"  257.355  
24+00  160.284  9°11'01.0"  159.599  
23+00  60.284  3°27'14.5"  60.248  
BC  22+39.716  0.000  0°00'00.0"  0.000 
Add three more columns for direction and coordinates:
Curve Point  Azimuth, Az_{i}  North, N_{i}  East, E_{i}  
EC  27+19.681 Bk  
27+00  
26+00  
25+00  
24+00  
23+00  
BC  22+39.716 
Complete the table using the three equations for this curve
At 22+39.716, we're still at the BC so the coordinates don't change.
At 23+00:
At 24+00:
and so on for the rest of the curve points.
The completed curve table is:
Curve Point  Azimuth, Az_{i}  North, N_{i}  East, E_{i}  
EC  27+19.681 Bk  103°10'10.0  830.375  5197.419 
27+00  102°02'30.4"  842.904  5182.244  
26+00  96°18'43.9"  896.816  5098.218  
25+00  90°34'57.4"  932.959  5005.157  
24+00  84°51'11.0"  949.894  4906.770  
23+00  79°07'24.5"  946.944  4806.981  
BC  22+39.716  75°40'10"  935.576  4747.815 
Math check: the coordinates computed for the EC in the table should be the same as the EC coordinates computed from the PI. Within rounding error, that's the case here.
d. Summary
The radial chord method lends itself nicely to computing curve point coordinates. The computations are not complex, although they are admittedly tedious.
Once coordainates are computed, field stakeout is much more flexible using Coodrinate Geometry (COGO).
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