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 A-1 and A-2, each related but having their own geometric elements and specific design constraints.

 gen 01

 Figure A-1
Horizontal Component

 gen 02
 Figure A-2
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 A-3. In some cases, such as trenching, stakes can be placed at half- or quarter-station intervals (50 ft and 25 ft intervals, respectively).

 gen 03
Figure A-3
Stationing

 

The horizontal distance between two stations along the alignment is their stationing difference.

 gen 04
Figure A-4
Distance

 

In Figure A-4, the distance from sta 12+38.28 to sta 16+56.85 is:

gen 04b

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 A-5 is 11+68.26 82.40R

gen 05
Figure A-5
Station and Offset

 

A single point can sometimes have two different stations values. The most common situation is where two different alignments cross, Figure A-6.

gen 06
Figure A-6
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 01       Eqn A-1
elev diff: elevation difference
horiz dist: horizontal distance

 

If the elev diff and horiz dist are both in feet (or meters) Equation A-1 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 A-1 returns grade as a percentage.

Grade is positive (+) going uphill in the direction of stationing, negative (-) going downhill, Figure A-7.

gen 07
Figure A-7
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:

gr ex1

Using Equation A-1:

gr ex1b

(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:

gr ex2

We need to rearrange Equation A-1 to solve for the elevation difference between the two stations:

 grade07

 The horizontal distance is  gr ex2c 
 and elevation difference is   gr ex2d
 and elevation at 19+00.00   gr ex2e

 

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:

gr ex3

The horizontal distance is:

gr ex3 dist

Since we're going backwards along the alignment, the distance is negative.

The elevation is:

gr ex3 elev

From the sketch we see that 14+00 is lower than 16+50, so the computed elevation looks correct.

 

 


 

1. Nomenclature

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

 vcurv01

vcurv02

Figure B-1
Adding a Vertical Curve

 

The parts of the curve, Figure B-2, are:

vcurv03
Figure B-2
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 B-3.

 vcurv04
Figure B-3
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 B-1 and B-2:

BVC EVC sta   BVC EVC elev
Equation B-1        Equation B-2

 

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 B-1 and B-2. 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 g1 and after a distance of L it departs the curve at g2. The total grade change is (g2-g1). The grade change rate, k, is the total grade change divided by the distance, Equation B-3.

k
Equation B-3

 

For example, a -2.00% grade into a +1.00% grade connected with a 400.00 ft curve has a k of:

k2

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 B-4.

 vcurv06b
Figure B-4
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 B-5.

vcurv05 vcurv06
Figure B-5
Crest and Sag Curves

 

What about curves whose grades at both ends are different but have the same mathematical sign? Figure B-6 shows four different curve situations where the incoming and outgoing grades have the same mathematical sign.

vcurv06 c
Figure B-6
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 B-7, is Equation B-4.

parabola1
parabola eqn 
 Equation B-4
Figure B-7
Parabola
 

 

The equation consists of two parts, Figure B-8:

  • 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, ax2, which is the vertical distance from the tangent to the curve.
parabola3
Figure B-8
Tangent and Offset

 

Figure B-9 shows the parabolic arc in terms of vertical curve nomenclature.

parabola4
Figure B-9
Curve Terms

 

The curve begins at the BVC station and elevation.
The tangent line slope is g1, the incoming grade.

At distance d:

  • the tangent elevation is g1d+ElevBVC
  • the tangent offset is ad2, where a is a function of k

The parabolic equation written in curve terms is the Curve Equation, Equation B-5.

CurveEqn       Equation B-5


(For a complete graphic derivation of the Curve Equation see the Family of Curves section.)

di is the distance from the BVC to any point i on the curve. It is computed from Equation B-6 and ranges from 0 to L.

dist eqn       Equation B-6


Recall the previous warning on grade and distance units. In Equation B-5 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 B-7.

grade eqn       Equation B-7

 

Using Equations B-5 through B-7 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 B-10.

vcurv09

a. Crest Curve

vcurv08

b. Sag Curve

Figure B-10
High and Low Point Conditions

 

On a crest curve, the tangent grade begins at g1 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, dp, is determined from Equation B-8:

vcurv10      
LowHi 
Equation B-8 
Figure B-11
Distance to High/Low Point
 


The high/low point elevation can be computed by substituting dp in Equation B-5.

In order for Equation B-8 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 B-8 the distance from the BVC to the high/low point is:

LowHi2

The distance exceeds the curve length. What's up with that? Let's sketch it, Figure B-12.

 vcurv12
Figure B-12
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:

LowHi3

This means the high/low point occurs before the BVC, Figure B-13.

vcurv13 
Figure B-13
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:

Ex1 
Figure B-14
Example 1 Sketch

 

For our computations, we'll use grades in percent and distances in stations.

(1) Use Equations B-1 and B-2 to compute the BVC and EVC

 Ex1 BVC Sta
 Ex1 EVC Sta
 Ex1 BVC Elev
 Ex1 EVC Elev

 

(2) Set up equations B-3, and B-5 to B-7 to compute curve elevations.

Equation B-3         

 Ex1 k
 Equation B-5  Ex1 Elev
 Equation B-6  Ex1 d
 Equation B-7  Ex1 g

 

 Set up the Curve Table:

Curve Table
      Equation B-6         Equation B-5         Equation B-7    
Station, i Dist, di (sta) Elevi (ft) Grade, gi (%)
 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:

Ex1 4470

sta 45+00:

Ex1 4500

and so on to the EVC.At the EVC each computed value has a math check indicated in red.

sta 48+70.00:

Ex1 4570

The completed Curve Table:

Curve Table
      Equation B-6         Equation B-5         Equation B-7    
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

Elevi (ft)

First Difference

Second Difference

48+70.00 EVC

848.680

 

 

48+00

850.029

850.029-850.809=-0.780

-0.780-0.570=-1.350

47+00

850.809

850.809-850.239=+0.570

+0.570-1.920=-1.350

46+00

850.239

850.239-848.319=+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 B-8 to compute the distance from the BVC:

Ex1 Hdist

Substitute into the Curve Equation to determine its elevation:

Ex1 HEl

Add the distance to the BVC station to get the point's station:

Ex1 HS

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 B-15.

upsidedown
Figure B-15
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.

Ex2 table

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 B-16. 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.

FitCurve1

(a) Preliminary Vertical Alignment

FitCurve2

(b) Holding Grades Fixed

FitCurve3

(c) Changing a Grade

Figure B-16
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 B-17.

FitCurve4

(a) Increasing Sag Curve Length

FitCurve5

(b) Increasing Crest Curve Length

Figure B-17
Curve Length and Elevations

 

To solve lengths requires rearranging and solving the Curve Equation, Equation B-5, for the unknown L.

 CurveEqn
Equation B-5


It looks simple until you realize that di 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 B-18.

FitCurve6
Figure B-18
Low Point Condition

 

The distance to the high/low point was given in Equation B-8:

LowHi
Equation B-8

 

Substituting Equation B-8 into Equation B-5 and solving for L, we arrive at Equation B-9:

FitCurveEqn1
Equation B-9

 

Remember: grades in % and L in stations.

Example

g1 & g2 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:

FitCurve7

Substitute known data into Equation B-9 and solve:

FitCurve8

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 at-grade intersection.

This situation appears to be easier since, unlike the previous example, we know the station at which the elevation must be met, Figure B-19.

FitCurve9
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 B-5, we get:

FitCurve10
Equation B-10

 

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.

FitCurve11

          FitCurve12           

Second Degree Polynomial
Equation B-13 

Quadratic Solution
Equation B-12 

 

 To simplify Equation B-10 and solve it using the quadratic solution we can create the following two sets of equations:

FitCurve13   FitCurve14
Quadratic Coefficients
Equation B-13
     Curve Length Solutions     
Equation B-14

 

Equation B-14 will return two values for the curve length, one which makes sense in the context of the problem, the other which doesn't.

Example

g1 & g2 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

FitCurve15

Compute the quadratic coefficients using Equation B-13.

FitCurve16

Compute the curve lengths using Equation B-14.

FitCurve17

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 B-5, 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 B-20.

FitCurve18

Figure 20
Two Curve Solutions

 

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 B-21. 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 best-fit curve.

FitCurve19
Figure B-21
Multiple Design Points

 

Or one of the points could be treated as the most critical and used to design a curve, Figure B-22.

FitCurve20
Figure B-22
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 B-23 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.

compound1
Figure B-23
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 back-to-back equal tangent curves. This is referred to as a compound curve, Figure B-24.

compound2 
Figure B-24
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, g3, 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 B-4 and B-5 through B-7. 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 B-3     compound7  compound11
Equation B-5        compound8       compound12   
Equation B-6     compound9  compound13
Equation B-7     compound10  compound14

 

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 B-8 is:

LowHi
Equation B-8

 

Setting up Equation B-8 and hi/low point station for each curve:

  Distance Station
Curve 1        compound3     compound5  
Curve 2        compound4     compound6  

 

Remember: the high/low point occurs only on one curve or the other. Should g3 be 0%, then the low/high point would be at the CVC. If all three grades are positive or negative, then Equation B-8 does not apply as neither curve tops/bottoms out.

d. Example

Using the data given in Figure B-24:

(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

compound2

(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:

 CompEx1

Since the PVIs are halfway along the grade lines, their positions are the averages of the grade line ends.

CompEx2

CompEx3

g3 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.

CompEx4

Compute the CVC elevation from PVI1.

CompEx5

The CVC elevation can also be computed from PVI2. We'll do it as a math check.

CompEx6

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

CompEx7

Curve 2

CompEx8

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.

CompEx9

CompEx10

 

e. Reverse Curve

A reverse curve is a compound curve except that the two curves have opposite curvature. It consists of either a sag-crest or crest-sag curves sequence, Figure B-25. The CRC (Curve to Reverse Curve) is the EVC of the first curve and BVC of the second.

reverse1 
Figure B-25
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 B-25, 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.

 

 

       mc02      

 1. General

 2. Compound curve

 3. Reverse curve


1. General

A multi-center curve consists of two or more back-to-back curves that are tangential. The simplest consists of two curves either in the same direction (compound) or opposite directions (reverse), Figure D-1.

mc01  mc02 
a. Compound curve b. Reverse curve
Figure D-1
Multi-center curves

 

Multi-center curves provide greater design flexibility. For example, Figure D-2 shows a typical cloverleaf freeway interchange.

mc03a
Figure D-2
Cloverleaf interchange

 

To travel southeast, southwest-bound 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 D-3 shows the southwestern loop with overlaid circular arcs.

mc03b
Figure D-3
Multiple curves

 

Using a multi-center 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, north-south Township lines were offset along Standard Parallels to account for meridian convergence, Figure D-4a. 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 D-4b. 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 D-4c.

mc04   mc04b mc05 
    a. Convergence offset       b. Low speed curves       c. High speed curves   
Figure D-4
Offset Township lines


A reverse curve can also be used to work an alignment around an obstruction. Figure D-5 shows what were it not for the presence of a lake would be a simple curve situation .

mc06
Figure D-5
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 D-6.

mc07 
Figure D-6
Working around obstruction

 

If multi-center curves are so flexible, why don't we see more of them?

Primarily because of the curvature change at each curve-to-curve transition.

On a simple curve, once the driver sets the steering wheel angle, it is maintained through the entire curve. A multi-center 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 under-shoot 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 cross-slope 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 cross-slope direction must be changed.

Figure D-7 shows a three-center curve system with 5 section view labels.

mc08
Figure D-7
Three-center curve

 

Table D-1 depicts the steering wheel angle and superelevation at each section view going through the curve system.

Table D-1
 Section  Steering Wheel  Superelevation
 A-A  mc09a  mc10a
 B-B  mc09b  mc10b
 C-C  mc09c  mc10c
 D-D  mc09d  mc10d
 E-E  mc09e  mc10e

 

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 D-8 and Table D-2.

RC02
Figure D-8
Reverse Curve

 

Table D-2
 Section  Steering Wheel  Superelevation
 A-A RC03a  RC03b 
 B-B RC04a  RC04b 
 C-C  RC05a RC05b 
 D-D  RC06a RC06b 

 

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. All-in-all quite a bit of stuff going on, especially when the vehicle is cruising along at a good rate.

So while multi-center 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 D-9.

 mc22
Figure D-9
Compound curve


Figure D-10 shows major parts of the compound curve.

 mc23
Figure D-10
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 PI1 and the outgoing tangent at PI2.

Ti is the distance from the BC1 to the PI; To the distance from the PI to the EC2.

There are seven parts to a compound curve: Ti, To, R1, R2, Δ1, Δ2, and Δ. Because the tangents deflect Δ at the PI and the compound curve transitions from the incoming to the outgoing tangent:

Δ = Δ1+ Δ2         Equation D-1 


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 D-11. Individual curve attributes can be computed using the single curve equations in Chapter C.

mc24
Figure D-11
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 D-12, created by the three PIs.

mc26a
Figure D-12
Vertex triangle


Another method is to use a closed loop traverse through points O1-BC1-PI-EC2-O2, Figure D-13.

 mc27a
Figure D-13
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:

mc30 1

We'll try a vertex triangle solution. Isolate the triangle and label the tangents:

mc30 2

Since we have the radius and central angle of the first curve, we can compute its tangent:

mc30 4

Compute Δ2 using Equation D-1

mc30 3

Determine the distance from the PI1 to the PI which is a side of the triangle.

mc30 5

The distance between PI1 and PI2 is the sum of the curve tangents. Using the Law of Sines and the known T1, we can compute T2.

mc30 6

Using T2 and Δ2, R2 can be determined.

mc30 7

Finally, compute each curve's length.

mc30 8

(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:

mc35 1

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:

mc35 2

Compute Latitudes and Departures:

Line Bearing Length Latitude Departure
O1-BC1 North R1 R1 0.00
BC1-PI East 463.64 0.00 463.64
PI-EC2 S 5°00'00"E 405.60 -405.60 x cos(5°00'00") +405.60 x sin(5°00'00") 
EC2-O2 S 85°00'00"W R2 -R2 x cos(85°00'00") -R2 x sin(85°00'00")
O2-O1 S 30°00'00"W R1-R2 -(R1-R2) x cos(30°00'00") -(R1-R2) x sin(30°00'00") 

 

 Sum and reduce the Latitudes:

mc35 3

Sum and reduce the Departures:

mc35 5

We have two equatiosn with unkowns R1 and R2. Solve them simultaneously. We'll use substitution.

Solve the Latitide summation for R1

mc35 4

Substiute the equation for R1 in the Departures summation and solve R2:

mc35 6

Solve R1:

mc35 7

With R1 and R2 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:

MC40            Equation D-2
MC41   Equation D-3
MC42   Equation D-4
MC43   Equation D-5

 

 The stationing for Example (2)  is:

MC49

 

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 Ti and To from the PI along the tangents to set the BC1 and EC2.

Stake Curve 1

Setup instrument on BC1.

Sight PI with 0º00'00" on the horizontal circle.

Stake curve points using the curve table.

Stake Curve 2

Setup on PCC

Sight BC1 with -(Δ1/2) = -15º00'00"=345°00'00" on the horizontal circle, Figure D-14a.

Inverse the scope to sight away from the BC1, Figure D-14b.

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 EC2.

MC50 

a. Sight BC1 with  -(Δ1/2) reading

MC51
b. Inverse telescope
MC52
c. Rotate to 00°00'00"
Figure D-14
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.

RC01
Figure D-15
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 PI1 to PI2 is T1 + T2. 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:

RC10a            Equation D-6
RC10b   Equation D-7
RC10c   Equation D-8
RC10d   Equation D-9

 

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:

RC15Ex1

RC15Ex2

Ta-daa!

 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 C-1.

horiz02b

Figure C-1
Line-Curve Tangency Condition

 

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 C-2 (a) and (b).

horiz01
(a) Deflection Angles
horiz01c
(b) Curves
Figure C-2
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 C-3.

horiz02a
Figure C-3
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 C-4. 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.

horiz03b
Figure C-4
Spiral Arc

 

Attaching a second sprial with reversed radius change creates an entrance-exit spiral condition where the driver gradually increases then gradually decreases steering wheel angle, Figure C-5.

horiz04
Figure C-5
Combined Spirals

 

A circular arc can be combined with two spirals, Figure C-6.

 horiz04b

Figure C-6
Spiralled Horizontal Curve

 

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 tire-pavement 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 C-7:

horiz11
Figure C-7
Different Arcs

 

Both curves must accommodate a total direction change of Δ but since C1 is shorter it is sharper than C2.

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, Da, is the subtended angle for a 100.00 ft arc; Chord definition, Dc, is the subtended angle for a 100.00 ft chord, Figure C-8.

 horiz08
   (a) Arc Definition   
horiz09
   (b) Chord Definition   
Figure C-8
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 C-9. The larger D is, the sharper the curve.

horiz10
Figure C-9
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 C-1.

horiz21      Equation C-1

 

b. Curve Components; Equations

Figure C-10 shows a tangent circular arc with some basic components labeled. 

 horiz06
Figure C-10
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 C-11 includes additional curve components.

 horiz07
Figure C-11
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:

horiz22       Equation C-2
 horiz23       Equation C-3
 horiz24       Equation C-4
 horiz25       Equation C-5
 horiz26       Equation C-6

 

Although it may look like it in Figure C-11, 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 C-12. Curve fitting comes later.

horiz32
 Figure C-12
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 C-13.

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

 horiz33
Figure C-13
Curve Inserted

 

That means for a typical curve there are two stations for the EC:

One along the original tangents
One along the curve.

horiz34 

Figure C-14
Stationing Along Curve

 

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 (up-station), 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 C-15 is an example of a station equation indicator on a set of WisDOT highway plans.

horiz35
Figure C-15
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 C-14 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 C-16.

 horiz39
Figure C-16
Computed Tangents and PVI

 

A curve is then fit to the tangents, Figure C-17.

 

 horiz40
Figure C-17
A Curve is Fit

 

Then the alignment is stationed from its beginning to its end through the curves, Figure C-18. That way there are no station equations.

 horiz41
Figure C-18
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 C-19.

horiz37
(a) Original Stationing with Station Equations

horiz38
(b) Limited Stationing Changes with Curve Modification 
Figure C-19
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 C-20.

horiz45
(a) Original Continuous Stationing
horiz46
(b) Stationing Changes on and after Modified Curve
Figure C-20
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:

horiz47            Equation C-7 
 horiz48 Equation C-8 
 horiz49 Equation C-9 

 

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:

horiz51

Use Equations C-2 through C-6 to compute curve components (carry an additional digit to minimize rounding errors):

horiz55

horiz56

horiz57

horiz58

horiz59

Compute degree of curvature using Equation C-1:

horiz60

Use Equations C-7 through C-9 to compute endpoint stations:

horiz61

horiz62

horiz63

 

horiz79


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 C-21.

 horiz70
Figure C-21 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:

horiz72

     Equation C-10

 

In terms of the degree of curvature, Figure C-22: 

horiz71
Figure C-22 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:

 horiz73         Equation C-11

 

Extending this geometry to the entire curve, Figure C-23, the total deflection angle at the BC from the PI to the EC is Δ/2.

horiz71b
Figure C-23 Deflection angle for entire curve

 

Since the deflection angle occurs across the curve's length, the deflection rate can also be written as Equation C-12.

 horiz73b         Equation C-12

 

b. Radial chords

One way to stake a horizontal curve is by the radial chord method, Figure C-24.

 horiz78
Figure C-24 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 C-25, can be computed from Equation C-13.

horiz74 
Figure C-25 Deflection angle and radial chord

  

horiz75        Equation C-13

 

li is the arc distance to the point from the BC and is computed using Equation C-14.

horiz76        Equation C-14

 

Equation C-10 can be re-written using the deflection angle:

 

horiz77         Equation C-15

 

Using Equations C-13 to C-15, 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

horiz79

Use Equation C-11 to compute the curve's deflection rate:

horiz80

Set up Equations C-13, C-14, and C-15 for the curve:

horiz81

Set up the Curve Table: 

     Curve Point

Arc dist, li, (ft)

Defl angle,δi

Radial chord, ci

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:

horiz82

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 C-16 and C-17 are general equations for computing coordinates using direction and distance from a known point, Figure C-26.

 horiz86          Equation C-16 
 horiz87   Equation C-17 

 

horiz85
Figure C-26. 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:

horiz88

Figure C-27

      The original tangent lines have directions; PC has coordinates.

 horiz89

Figure C-28

 

A curve is fit to the tangents.

End points are at distance T from the PI along the tangents.

 horiz94

Figure C-29

 

Compute coordinates of BC using back-direction of the tangent BC-PI and T.

horiz97        Equation C-18
horiz98   Equation C-19 

 horiz95

Figure C-30

 

Compute coordinates of EC using direction of the tangent PI-EC and T.

These will be use for a later math check.

horiz99       Equation C-20
 horiz100   Equation C-21

 horiz96

Figure C-31

 

Use a curve point's deflection angle to compute the direction if its radial chord from the BC.

horiz101        Equation C-22 

 

δ is positive for right deflections, negative for left.
Using the direction and chord length, compute the point's coordinates.

horiz102        Equation C-23 
 horiz103   Equation C-24 

 

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

 

horiz79

 

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:

horiz110

horiz111

Compute the coordinates of the EC:

horiz112

horiz112a 

Set up Equatons C-21 through C-24 for this curve.

horiz113

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, Azi North, Ni East, Ei
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:

horiz113

At 24+00:

horiz114

and so on for the rest of the curve points.

The completed curve table is: 

     Curve Point Azimuth, Azi North, Ni East, Ei
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).