1.0.0 Horizontal Curves
2.0.0 Vertical Curves SummaryReview Questions 
The center line of a road consists of a series of straight lines interconnected by curves that are used to change the alignment, direction, or slope of the road. Those curves that change the alignment or direction are known as horizontal curves, and those that change the slope are vertical curves. As an EA you may have to assist in the design of these curves. Generally, however, your main concern is to compute for the missing curve elements and parts as problems occur in the field in the actual curve layout. You will find that a thorough knowledge of the properties and behavior of horizontal and vertical curves used in highway work will eliminate delays and unnecessary labor. Careful study of this lesson will alert you to common problems in horizontal and vertical curve layouts.
When you have completed this lesson, you will be able to:
When a highway changes horizontal direction, making the point where it changes direction a point of intersection between two straight lines is not feasible. The change in direction would be too abrupt for the safety of modern highspeed vehicles. Therefore it is necessary to interpose a curve between the straight lines. The straight lines of a road are called tangents because the lines are tangent to the curves used to change direction. On practically all modern highways, the curves are circular curves, or curves that form circular arcs. The smaller the radius of a circular curve, the sharper the curve. For modern highspeed highways, the curves must be flat, rather than sharp. That means they must be largeradius curves. In highway work, the curves needed for the location or improvement of small secondary roads may be worked out in the field. Usually, however, the horizontal curves are computed after the route has been selected, the field surveys have been done, and the survey base line and necessary topographic features have been plotted. In urban work, the curves of streets are designed as an integral part of the preliminary and final layouts, which are usually done on a topographic map. In highway work, the road itself is the end result and the purpose of the design. But in urban work, the streets and their curves are of secondary importance; the best use of the building sites is of primary importance. The principal consideration in the design of a curve is the selection of the length of the radius or the degree of curvature. This selection is based on such considerations as the design speed of the highway and the sight distance as limited by headlights or obstructions (Figure 31). Some typical radii you may encounter are 12,000 feet or longer on an interstate highway, 1,000 feet on a major thoroughfare in a city, 500 feet on an industrial access road, and 150 feet on a minor residential street.
There are four types of horizontal curves. They are described as follows: 1. Simple The simple curve is an arc of a circle (Figure 32, View A). The radius of the circle determines the sharpness or flatness of the curve. Figure 31 – Lines of sight. 34 2. Compound Frequently, the terrain will require the use of the compound curve. This curve normally consists of two simple curves joined together and curving in the same direction (Figure 32, View B). 3. Reverse A reverse curve consists of two simple curves joined together, but curving in opposite direction. For safety reasons, the use of this curve should be avoided when possible (Figure 32, View C). 4. Spiral The spiral is a curve that has a varying radius. It is used on railroads and most modern highways. It provides a transition from the tangent to a simple curve or between simple curves in a compound curve (Figure 32, View D).
The elements of a circular curve are shown in Figure 33. Each element is designated and explained as follows: POINT OF INTERSECTION (PI) The point of intersection is the point where the back and forward tangents intersect. Sometimes the point of intersection is designated as V (vertex). INTERSECTING ANGLE (I) The intersecting angle is the deflection angle at the PI. Its value is either computed from the preliminary traverse angles or measured in the field. RADIUS (R) The radius is the distance from the center of a circle or curve represented as an arc, or segment. The radius is always perpendicular to back and forward tangents. Figure 32 – Horizontal curves. 35 POINT OF CURVATURE (PC) The point of curvature is the point on the back tangent where the circular curve begins. It is sometimes designated as BC (beginning of curve) or TC (tangent to curve). POINT OF TANGENCY (PT) The point of tangency is the point on the forward tangent where the curve ends. It is sometimes designated as EC (end of curve) or CT (curve to tangent). CENTRAL ANGLE (Δ) The central angle is the angle formed by two radii drawn from the center of the circle (O) to the PC and PT. The value of the central angle is equal to the I angle. Some authorities call both the intersecting angle and central angle either I or A. POINT OF CURVE (POC) The point of curve is any point along the curve. LENGTH OF CURVE (L) The length of curve is the distance from the PC to the PT, measured along the curve. TANGENT DISTANCE (T) The tangent distance is the distance along the tangents from the PI to the PC or the PT. These distances are equal on a simple curve. LONG CHORD (LC) The long chord is the straightline distance from the PC to the PT. Other types of chords are designated as follows: C The fullchord distance between adjacent stations (full, half, quarter, or onetenth stations) along a curve C1 The sub chord distance between the PC and the first station on the curve C2 The subchord distance between the last station on the curve and the PT EXTERNAL DISTANCE (E) The external distance (also called the external secant) is the distance from the PI to the midpoint of the curve. The external distance bisects the interior angle at the PI. MIDDLE ORDINATE (M) The middle ordinate is the distance from the midpoint of the curve to the midpoint of the long chord. The extension of the middle ordinate bisects the central angle. DEGREE OF CURVE (D) The degree of curve defines the sharpness or flatness of the curve.
The last of the elements listed above (degree of curve) deserves special attention. Curvature may be expressed by simply stating the length of the radius of the curve. This was done earlier in this lesson when typical radii for various roads were cited. Stating the radius is a common practice in land surveying and in the design of urban roads. For highway and railway work, however, curvature is expressed by the degree of curve. Two definitions are used for the degree of curve. These definitions are discussed in the following sections.
The arc definition is most frequently used in highway design. This definition, illustrated in Figure 34, states that the degree of curve is the central angle formed by two radii that extend from the center of a circle to the ends of an arc measuring 100 feet long (or 100 meters long if you are using metric units). Therefore, if you take a sharp curve, mark off a portion so that the distance along the arc is exactly 100 feet, and determine that the central angle is 12°, the degree of curvature is 12°. It is referred to as a 12° curve. Figure 34 illustrates that the ratio between the degree of curvature (D) and 360° is the same as the ratio between 100 feet of arc and the circumference (C) of a circle having the same radius. Figure 33 – Elements of a horizontal curve. 37 That may be expressed as follows: . 100 360 C D = ° Since the circumference of a circle equals 2πR, the above expression can be written as . 2 100 360 R D π = ° Solving this expression for R: D R 5729.58 = and also D: R D 5729.58 = For a 1° curve, D = 1; therefore R = 5,729.58 feet, or meters, depending upon the system of units you are using. In practice, the design engineer usually selects the degree of curvature on the basis of such factors as the design speed and allowable super elevation. Then the radius is calculated. Figure 34 – Degree of curve (arc definition). 38
The chord definition (Figure 35) is used in railway practice and in some highway work. This definition states that the degree of curve is the central angle formed by two radii drawn from the center of the circle to the ends of a chord 100 feet (or 100 meters) long. If you take a flat curve, mark a 100foot chord, and determine the central angle to be 0°30’, then you have a 30minute curve (chord definition). From observation of Figure 35, you can see the following trigonometric relationship: . 50 2 sin R D = Then, solving for R: . sin 2/1 50 D R = For a 10 curve (chord definition), D = 1; therefore R = 5,729.65 feet, or meters, depending upon the system of units you are using. Notice that in both the arc definition and the chord definition, the radius of curvature is inversely proportional to the degree of curvature. In other words, the larger the degree of curve, the shorter the radius; for example, using the arc definition, the radius of a 1° curve is 5,729.58 units, and the radius of a 5° curve is 1,145.92 units. Under the chord Figure 35 – Degree of curve (chord definition). 39 definition, the radius of a 1° curve is 5,729.65 units, and the radius of a 5° curve is 1,146.28 units. <p>14.0 Curve Formulas The relationship between the elements of a curve is expressed in a variety of formulas. The formulas for radius (R) and degree of curve (D), as they apply to both the arc and chord definitions, were given in the preceding discussion of the degree of curvature. Additional formulas used in the computations for a curve are discussed in the following sections.
By studying Figure 36, you can see that the solution for the tangent distance (T) is a simple righttriangle solution. In the figure, both T and R are sides of a right triangle, with T being opposite to angle Δ/2. Therefore, from your knowledge of trigonometry, to solve for T: . 2 tan ∆ = RT
As illustrated in Figure 37, the solution for the length of a chord, either a full chord(C) or the long chord (LC), is also a simple righttriangle solution. As shown in the figure, C/2 is one side of a right triangle and is opposite angle Δ/2. The radius (R) is the hypotenuse of the same triangle. Figure 36 – Tangent distance. 310 Therefore, R C 2/ 2 sin ∆ and solving for C: 2 2 sin ∆ = RC Figure 37 – Chord distance.
In the arc definition of the degree of curvature, length is measured along the arc, as shown in Figure 38, View A. In this figure the relationship between D, L, and a 100foot arc length may be expressed as follows: . 100 D L ∆ = Then, solving for L: L = 100 D ∆ 311 This expression is also applicable to the chord definition. However, L in this case is not the true arc length, because under the chord definition, the length of curve is the sum of the chord lengths (each of which is usually 100 feet or 100 meters). As an example, if, as shown in Figure 38, View B, the central angle (A) is equal to three times the degree of curve (D), then there are three 100foot chords and the length of “curve” is 300 feet.
Two commonly used formulas for the middle ordinate (M) and the external distance (E) are as follows: − ∆ = ∆ = ∆ −= 1 cos 2/ 1 4 tan 2 1 cos RTE RM <p>1.5.0 Deflection Angles and Chords From the preceding discussions, you may think that laying out a curve is simply a matter of locating the center of a circle where two known or computed radii intersect, and then swinging the arc of the circular curve with a tape. For some applications, that can be done. However, what if you are laying out a road with a 1,000, 12,000, or even 40,000 foot radius? Obviously, it would be impracticable to swing such radii with a tape. In usual practice, the stakeout of a longradius curve involves a combination of turning deflection angles and measuring the length of chords (C1, C2, or C3 as appropriate). A transit is set up at the PC, a sight is taken along the tangent, and each point is located by turning deflection angles and measuring the chord distance between stations. This procedure is illustrated in Figure 39. In this figure, a portion of a curve starts at the PC and runs through points (stations) A, B, and C. To establish the location of point A on this curve, you should set up your instrument at the PC, turn the required deflection angle (all/2), and then measure the required chord distance from PC to point A. Then, to establish point B, you turn deflection angle D/2 and measure the required chord distance from A to B. Point C is located similarly. Figure 38 – Length of curve. 312 As you are aware, the actual distance along an arc is greater than the length of a corresponding chord; therefore, when using the arc definition, either apply a correction for the difference between arc length and chord length, or use shorter chords to make the error resulting from the difference negligible. In the latter case, the following chord lengths are commonly used for the degrees of curve shown: 100 feet—0 to 3 degrees of curve 50 feet—3 to 8 degrees of curve 25 feet—8 to 16 degrees of curve 10 feet—over 16 degrees of curve The above chord lengths are the maximum distances in which the discrepancy between the arc length and chord length will fall within the allowable error for taping. The allowable error is 0.02 foot per 100 feet on most construction surveys; however, based on terrain conditions or other factors, the design or project engineer may determine that chord lengths other than those recommended above should be used for curve stakeout. The following formulas relate to deflection angles. (To simplify the formulas and further discussions of deflection angles, the deflection angle is designated simply as d rather than d/2.) = 1002 CD d Where: d = Deflection angle (expressed in degrees) C = Chord length D = Degree of curve d = 0.3 CD Where: d = Deflection angle (expressed in minutes) C = Chord length D = Degree of curve R C d 2 sin = Figure 39 – Deflection angles and chords. 313 Where: d = Deflection angle (expressed in degrees) C = Chord length R = Radius. <p>1.6.0 Solving and Laying Out a Simple Curve Now let’s solve and lay out a simple curve using the arc definition, which is the definition you will more often use as an EA. In Figure 310, let’s assume that the directions of the back and forward tangents and the location of the PI have previously been staked, but the tangent distances have not been measured. Let’s also assume that stations have been set as far as Station 18 + 00. The specified degree of curve (D) is 15°, arc definition. Our job is to stake halfstations on the curve. 1.6.1 Solving a Simple Curve We will begin by first determining the distance from Station 18 + 00 to the location of the PI. Since these points have been staked, we can determine the distance by field measurement. Let’s assume we have measured this distance and found it to be 300.89 feet. Next, we set up a transit at the PI and determine that deflection angle I is 75°. Since I always equals ∆, then ∆ is also 75°. Now we can compute the radius of the curve, the tangent distance, and the length of curve as follows: R = 5,729.58/D = 381.97 feet. T = R tan Δ/2 = 293.09 feet. L = 100 Δ/D = 500 feet. From these computed values, we can determine the stations of the PI, PC, and PT as follows: Station at Pl = (Sta. 18 + 00) + 300.89 = 21 + 00.89 Tangent distance = Station at PC 18 + 07.80 () 2 + 93.09 Length of curve = Station at PT 23 + 07.80 (+) 5 + 00.00 Figure 310 – Laying out a simple curve. 314 By studying Figure 310 and remembering that our task is to stake halfstation intervals, you can see that the first halfstation after the PC is Station 18 + 50 and the last halfstation before the PT is 23+ 00; therefore, the distance from the PC to Station 18 + 00 is 42.2 feet [(18 + 50)  (18 + 07.80)]. Similarly, the distance from Station 23+ 00 to the PT is 7.8 feet. These distances are used to compute the deflection angles for the subchords using the formula for deflection angles (d= .3CD) as follows: Deflection angle d1 = .3 x 7.8 x 15 = 189.9' = 3°09.9' Deflection angle d2 = .3 x 7.8 x 15 = 35.1' = 0°35.1' A convenient method of determining the deflection angle (d) for each full chord is to remember that d equals 1/2D for 100foot chords, 1/4D for 50foot chords, 1/8D for 25 foot chords, and 1/20D for 10foot chords. In this case, since we are staking 50foot stations, d = 15/4, or 3°45'. Previously, we discussed the difference in length between arcs and chords. In that discussion, you learned that to be within allowable error, the recommended chord length for an 8 to 16degree curve is 25 feet. Since in this example we are using 50foot chords, the length of the chords must be adjusted. The adjusted lengths are computed using a rearrangement of the formula for the sine of deflection angles as follows: C1 = 2R sin d1 = 2 x 381.97 x sin 3°09.9' = 42.18 feet. C2 = 2R sin d2 = 2 x 381.97 x sin 0°35.1' = 7.79 feet. C = 2R sin d2 = 2 x 381.97 x sin 3°45' = 49.96 feet. As you can see, in this case there is little difference between the original and adjusted chord lengths; however, if we were using 100foot stations rather than 50foot stations, the adjusted difference for each full chord would be substantial (over 3 inches). Now, remembering our previous discussion of deflection angles and chords, you know that all of the deflection angles are usually turned using a transit that is set up at the PC. The deflection angles that we turn are found by cumulating the individual deflection angles from the PC to the PT as shown below: 315 Station Chord Deflection angle PC 18 + 07.80  0°00.0' 18 +50 C1 42.18 3°09.9' 19 + 00 49.96 6°54.9' 19 + 50 49.96 10°39.9' 20 + 00 49.96 14°24.9' 20 + 50 49.96 18°09.9' 21 + 00 49.96 21°54.9' 21 + 50 49.96 25°39.9' 22 + 00 49.96 29°24.9' 22 + 50 49.96 33°09.9' 23 + 00 49.96 36°54.9' PT 23 + 07.80 C2 07.79 37°30' Notice that the deflection angle at the PT is equal to one half of the I angle. That serves as a check of your computations. Had the deflection angle been anything different than one half of the I angle, then you would have made a mistake. Since the total of the deflection angles should be onehalf of the I angle, a problem arises when the I angle contains an odd number of minutes and the instrument used is a 1minute transit. Since the PT is normally staked before the curve is run, the total deflection will be a check on the PC; therefore, it should be computed to the nearest 0.5 degree. If the total deflection checks to the nearest minute in the field, it can be considered correct. The curve that was just solved had an I angle of 75° and a degree of curve of 15°. When the I angle and degree of curve consist of both degrees and minutes, the procedure in solving the curve does not change, but you must be careful in substituting these values into the formulas for length and deflection angles. For example, if I = 42°15’and D = 5°37’, the minutes in each angle must be changed to a decimal part of a degree. To obtain the required accuracy, you should convert them to five decimal places, but an alternate method for computing the length is to convert the I angle and degree of curve to minutes; thus, 42°15’ = 2,535 minutes and 5°37’ = 337 minutes. Substituting this information into the length formula gives the following: .23.752 337 535,2 = 100 xL = feet This method yields an exact result. By converting the minutes to a decimal part of a degree to the nearest five places, you obtain the same result. 316 1.6.2 Simple Curve Layout To lay out the simple curve (arc definition) just computed above, you should usually use the procedure that follows. 1. With the instrument placed at the PI, the instrumentman sights on the preceding PI or at a distant station and keeps the chainman on the line while the tangent distance is measured to locate the PC. After the PC has been staked out, the instrumentman then trains the instrument on the forward PI to locate the PT. 2. The instrumentman then sets up at the PC and measures the angle from the PI to the PT. This angle should be equal to one half of the I angle; if it is not, either the PC or the PT has been located in the wrong position. 3. With the first deflection angle (3°10’) set on the plates, the instrumentman keeps the chainman on line as the first subchord distance (42.18 feet) is measured from the PC. 4. Without touching the lower motion screw, the instrumentman sets the second deflection angle (6°55’) on the plates. The chainman measures the chord from the previous station while the instrumentman keeps the head chainman on line. 5. The crew stakes out the succeeding stations in the same manner. If the work is done correctly, the last deflection angle will point on the PT. That distance will be the subchord length (7.79 feet) from the last station before the PT. When it is impossible to stake out the entire curve from the PC, a modified method of the procedure described above is used. Stake out the curve as far as possible from the PC. If a station cannot be seen from the PC for some reason, move the transit forward and set up over a station along the curve. Pick a station for a backsight and set the deflection angle for that station on the plates. Sight on this station with the telescope plates, the instrumentman keeps the chainman on line in the reverse position. Plunge the telescope and set the remainder of the stations in the same way as you would if the transit were set over the PC. If the setup in the curve has been made but the next stake cannot be set because of obstructions, the curve can be backed in. To back in a curve, occupy the PT. Sight on the PI and set one half of the I angle of the plates. The transit is now oriented so that, if the PC is observed, the plates will read zero, which is the deflection angle shown in the notes for that station. The curve stakes can then be set in the same order shown in the notes or in the reverse order. Remember to use the deflection angles and chords from the top of the column or from the bottom of the column. Although the backin method has been set up as a way to avoid obstructions, it is also very widely used as a method for laying out curves. The method is to proceed to the approximate midpoint of the curve by laying out the deflection angles and chords from the PC and then laying out the remainder of the curve from the PT. If this method is used, any error in the curve is in the center where it is less noticeable. So far in our discussions, we have begun staking out curves by setting up the transit at the PI. But what do you do if the PI is inaccessible? This condition is illustrated in Figure 311. In this situation, you locate the curve elements using the following steps: 1. As shown in Figure 311, mark two intervisible points A and B on the tangents so that line AB clears the obstacle. 317 2. Measure angles a and b by setting up at both A and B. 3. Measure the distance AB. 4. Compute inaccessible distance AV and BV using the formulas given in Figure 311. 5. Determine the tangent distance from the PI to the PC on the basis of the degree of curve or other given limiting factor. 6. Locate the PC at a distance T minus AV from the point A and the PT at a distance T minus BV from point B. Figure 311 – Inaccessible PI. 1.6.3 Field Notes Figure 312 shows field notes for the curve we solved and staked out above. By now you should be familiar enough with field notes to preclude the necessity for a complete discussion of everything shown in these notes. You should notice, however, that the stations are entered in reverse order (bottom to top). In this manner the data is presented as it appears in the field when you are sighting ahead on the line. This same practice applies to the sketch shown on the righthand page of the field notes. For information about other situations involving inaccessible points or the uses of external and middle ordinate distance, spiral transitions, and other types of horizontal curves, study books such as those mentioned at the beginning of this lesson. 318 Figure 3 12 – Field notes for laying out a simple curve.
Test Your Knowledge 1. A highway is composed of a series of curves and straight lines called _______.
2. What type of curve consists of two simple curves joined together and curving in the same direction?
3. The first step in staking out a simple curve is to set the instrument up at what point?

In addition to horizontal curves that go to the right or left, roads also have vertical curves that go up or down. Vertical curves at a crest or the top of a hill are called summit curves, or oververticals. Vertical curves at the bottom of a hill or dip are called sag curves, or underverticals.
Vertical curves are used to connect stretches of road that go up or down at a constant slope. These lines of constant slope are called grade tangents (Figure 313). The rate of slope is called the gradient, or simply the grade. (Do not confuse this use of the term grade with other meanings, such as the design elevation of a finished surface at a given point or the actual elevation of the existing ground at a given point.) Grades that ascend in the direction of the stationing are designated as plus; those that descend in the direction of the stationing are designated as minus. Grades are measured in terms of percent, that is, the number of feet of rise or fall in a 100foot horizontal stretch of the road. After the location of a road has been determined and the necessary fieldwork has been obtained, the engineer designs or fixes (sets) the grades. A number of factors are considered, including the intended use and importance of the road and the existing topography. If a road is too steep, the comfort and safety of the users and fuel consumption of the vehicles will be adversely affected; therefore, the design criteria will specify maximum grades. Typical maximum grades are a 4percent desired maximum and a 6percent absolute maximum for a primary road. (The 6 percent means, as indicated before, a 6foot rise for each 100 feet ahead on the road.) For a secondary road or a major street, the maximum grades might be a 5percent desired and an 8percent absolute maximum, and for a tertiary road or a secondary street, an 8percent desired and a 10percent (or perhaps a 12percent) absolute maximum. Conditions may sometimes demand that grades or ramps, driveways, or short access streets go as high as 20 percent. The engineer must also consider minimum grades. A street with curb and gutter must have enough fall so that the storm water will drain to the inlets; 0.5 percent is a typical minimum grade for curb and gutter, that is, 1/2 foot minimum fall for each 100 feet ahead. For roads with side ditches, the desired minimum grade might be 1 percent, but since ditches may slope at a grade different from the pavement, a road may be designed with a zeropercent grade. Zeropercent grades are not unusual, particularly through plains or tidewater areas. Another factor considered in designing the finished profile of a road is the Figure 313 – A vertical curve. 320 earthwork balance. The grades should be set so that all the soil cut off of the hills may be economically hauled to fill in the low areas. In the design of urban streets, the best use of the building sites next to the street will generally be more important than seeking an earthwork balance. 2.2.0 Computing Vertical Curves As you have learned earlier, the horizontal curves used in highway work are generally the arcs of circles. But vertical curves are usually parabolic. The parabola is used primarily because its shape provides a transition and also lends itself to the computational methods described in the next section of this lesson. Designing a vertical curve consists principally of deciding on the proper length of the curve. As indicated in Figure 313, the length of a vertical curve is the horizontal distance from the beginning to the end of the curve; the length of the curve is NOT the distance along the parabola itself. The longer a curve is, the more gradual the transition will be from one grade to the next; the shorter the curve, the more abrupt the change will be. The change must be gradual enough to provide the required sight distance (Figure 314). The sight distance requirement will depend on the speed for which the road is designed, the passing or nonpassing distance requirements, and other assumptions such as a driver’s reaction time, braking time, stopping distance, eye level, and the height of objects. A typical eye level used for designs is 4.5 feet or, more recently, 3.75 feet; typical object heights are 4 inches to 1.5 feet. For a sag curve, the sight distance will usually not be significant during daylight, but the nighttime sight distance must be considered when the reach of headlights may be limited by the abruptness of the curve. Figure 314 – Sight distance. 321 2.3.0 Elements of Vertical Curves Figure 315 shows the elements of a vertical curve. The meaning of the symbols and the units of measurement usually assigned to them follow: PVC  Point of vertical curvature; the place where the curve begins. PVI  Point of vertical intersection; where the grade tangents intersect. PVT  Point of vertical tangency; where the curve ends. POVC  Point on vertical curve; applies to any point on the parabola. POVT  Point on vertical tangent; applies to any point on either tangent. g1  Grade of the tangent on which the PVC is located; measured in percent of slope. g2  Grade of the tangent on which the PVT is located; measured in percent of slope. G The algebraic difference of the grades: G = g2 –g1 Plus values are assigned to uphill grades and minus values to downhill grades; examples of various algebraic differences are shown later in this section. L  Length of t he curve; the horizontal length measured in 100foot stations from t he PVC to the PVT. This length may be computed using the formula L = G/r, where r is the rate o f change ( usually given i n t he design cr iteria). When t he r ate o f ch ange i s not given, L (in stations) can be computed as follows: for a summit curve, L = 125 x G/4; for a sag curve, L = 100 x G/4. If L does not come out to a whole number of stations using these formulas, then it is usually extended to t he nearest whole number. You should note that these formulas for length are for road design only, NOT railway. l1  Horizontal length of the portion of the PVC to the PVI; measured in feet. l2  Horizontal length of the portion of the curve from the PVI to the PVT; measured in feet. e  Vertical (external) distance from the PVI to the curve; measured in feet. This distance is computed using the formula e = LG/8, where L is the total length in stations and G is the algebraic difference of the grades in percent. X  Horizontal distance from the PVC to any POVC or POVT back of the PVI, or the distance from the PVT to any POVC or POVT ahead of the PW; measured in feet. Figure 315 – Elements of a vertical curve. 322 y  Vertical distance (offset) from any POVT to the corresponding POVC; measured in feet: y = (x/l)2 (e), which is the fundamental relationship of the parabola that permits convenient calculation of the vertical offsets. The vertical curve computation takes place after the grades have been set and the curve designed. Therefore, at the beginning of the detailed computations, the following are known: g1, g2, l1, l2, L, and the elevation of the PVI. The general procedure is to compute the elevations of certain POVTs and then to use the foregoing formulas to compute G, then e, and then the Ys that correspond to the selected POVTs. When the y is added or subtracted from the elevation of the POVT, the result is the elevation of the POVC. The POVC is the finished elevation on the road, which is the end result being sought. In Figure 315, the y is subtracted from the elevation of the POVT to get the elevation of the curve; however, in the case of a sag curve, the y is added to the POVT elevation to obtain the POVC elevation. The computation of G requires careful attention to the signs of g1 and g2. Vertical curves are used at changes of grade other than at the top or bottom of a hill; for example, an uphill grade that intersects an even steeper uphill grade will be eased by a vertical curve. The six possible combinations of plus and minus grades, together with sample computations of G, are shown in Figure 316. Note that the algebraic sign for G indicates whether to add or subtract y from a POVT. The selection of the points at which to compute the y and the elevations of the POVT and POVC is generally based on the stationing. The horizontal alignment of a road is often staked out on 50foot or 100foot stations. Customarily, the elevations are computed at these same points so that both horizontal and vertical information for construction will be provided at the same point. The PVC, PVI, and PVT are usually set at full stations or half stations. In urban work, elevations are sometimes computed and staked every 25 feet on vertical curves. The same or even closer intervals may be used on complex ramps and interchanges. The application of the foregoing fundamentals will be presented in the next two sections under symmetrical and unsymmetrical curves. 323 Figure 316 – Algebraic differences of grades.
A symmetrical vertical curve is one in which the horizontal distance from the PVI to the PVC is equal to the horizontal distance from the PVI to the PVT. In other words, l1 equals l2. Figure 317 – Symmetrical vertical curves. The solution of a typical problem dealing with a symmetrical vertical curve will be presented step by step. Assume that you know the following data: g1 = +9% g2= –7% L = 400.00', or 4 stations 324 The station of the PVI = 30 + 00 The elevation of the PVI = 239.12 feet. The problem is to compute the grade elevation of the curve to the nearest hundredth of a foot at each 50foot station. Figure 317 shows the vertical curve to be solved. STEP 1: Prepare a table as shown in Table 31. In this figure, Column 1 shows the stations; Column 2, the elevations on tangent; Column 3, the ratio of x/l; Column 4, the ratio of (x/l)2 ; Column 5, the vertical offsets [(x/l)2 (e)]; Column 6, the grade elevations on the curve; Column 7, the first difference; and Column 8, the second difference. Table 31 – Table of computations of elevations on a symmetrical vertical curve. STEP 2: Compute the elevations and set the stations on the PVC and the PVT. Knowing both the gradients at the PVC and PVT and the elevation and station at the PVI, you can compute the elevations and set the stations on the PVC and the PVT. The gradient (g1) of the tangent at the PVC is given as +9 percent. This means a rise in elevation of 9 feet for every 100 feet of horizontal distance. Since L is 400.00 feet and the curve is symmetrical, l1 equals l2 equals 200.00 feet; therefore, there will be a difference of 9 x 2, or 18 feet between the elevation at the PVI and the elevation at the PVC. The elevation at the PVI in this problem is given as 239.12 feet; therefore, the elevation at the PVC is 239.12 – 18 = 221.12 feet. Calculate the elevation at the PVT in a similar manner. The gradient (g2) of the tangent at the PVT is given as –7 percent. This means a drop in elevation of 7 feet for every 100 feet of horizontal distance. Since l1 equals l2 equals 200 feet, there will be a difference of 7 x 2, or 14 feet between the elevation at the PVI and the elevation at the PVT. The elevation at the PVI therefore is 239.12 – 14 = 225.12 feet. In setting stations on a vertical curve, remember that the length of the curve (L) is always measured as a horizontal distance. The halflength of the curve is the horizontal distance from the PVI to the PVC. In this problem, l1 equals 200 feet. That is equivalent to two 100foot stations and may be expressed as 2 + 00. Thus the station at the PVC is 30 + 00 minus 2 + 00, or 28 + 00. The station at the PVT is 30 + 00 plus 2 + 00, or 32 + 00. 325 List the stations under Column 1. STEP 3: Calculate the elevations at each 50foot station on the tangent. From Step 2, you know there is a 9foot rise in elevation for every 100 feet of horizontal distance from the PVC to the PVI. Thus, for every 50 feet of horizontal distance, there will be a rise of 4.50 feet in elevation. The elevation on the tangent at station 28 + 50 is 221.12 + 4.50 = 225.62 feet. The elevation on the tangent at station 29 + 00 is 225.62 + 4.50 = 230.12 feet. The elevation on the tangent at station 29+ 50 is 230.12 + 4.50 = 234.62 feet. The elevation on the tangent at station 30+ 00 is 234.62 + 4.50 = 239.12 feet. In this problem, to find the elevation on the tangent at any 50foot station starting at the PVC, add 4.50 to the elevation at the preceding station until you reach the PVI. At this point use a slightly different method to calculate elevations because the curve slopes downward toward the PVT. Think of the elevations as being divided into two groups— one group running from the PVC to the PVI, the other group running from the PVT to the PVI. Going downhill on a gradient of –7 percent from the PVI to the PVT, there will be a drop of 3.50 feet for every 50 feet of horizontal distance. To find the elevations at stations between the PVI to the PVT in this particular problem, subtract 3.50 from the elevation at the preceding station. The elevation on the tangent at station 30 + 50 is 239.123.50, or 235.62 feet. The elevation on the tangent at station 31 + 00 is 235.623.50, or 232.12 feet. The elevation on the tangent at station 31 + 50 is 232.123.50, or 228.62 feet. The elevation on the tangent at station 32+00 (PVT) is 228.623.50, or 225.12 ft. The last subtraction provides a check on the work you have finished. List the computed elevations under Column 2. STEP 4: Calculate e, the middle vertical offset at the PVI. First, find the G, the algebraic difference of the gradients using the formula G = g2– g1 G= 7 – (+9) G= –16% The middle vertical offset (e) is calculated as follows: e = LG/8 = [(4)(–16) ]/8 = 8.00 feet. The negative sign indicates e is to be subtracted from the PVI. 326 STEP 5: Compute the vertical offsets at each 50foot station, using the formula (x/l)2 e. To find the vertical offset at any point on a vertical curve, first find the ratio x/l; then square it and multiply by e; for example, at station 28 + 50, the ratio of x/l = 50/200 = 1/4. Therefore, the vertical offset is (1/4)2 e = (1/16) e. The vertical offset at station 28 + 50 equals (1/16)(–8) = –0.50 feet. Repeat this procedure to find the vertical offset at each of the 50foot stations. List the results under Columns 3, 4, and 5. STEP 6: Compute the grade elevation at each of the 50foot stations. When the curve is on a crest, the sign of the offset will be negative; therefore, subtract the vertical offset (the figure in Column 5) from the elevation on the tangent (the figure in Column 2); for example, the grade elevation at station 29 + 50 is 234.62 – 4.50 = 230.12 ft. Obtain the grade elevation at each of the stations in a similar manner. Enter the results under Column 6. NOTE When the curve is in a dip, the sign will be positive; therefore, you will add the vertical offset (the figure in Column 5) to the elevation on the tangent (the figure in Column 2). STEP 7: Find the turning point on the vertical curve. When the curve is on a crest, the turning point is the highest point on the curve. When the curve is in a dip, the turning point is the lowest point on the curve. The turning point will be directly above or below the PVI only when both tangents have the same percent of slope (ignoring the algebraic sign); otherwise, the turning point will be on the same side of the curve as the tangent with the least percent of slope. The horizontal location of the turning point is measured either from the PVC if the tangent with the lesser slope begins there or from the PVT if the tangent with the lesser slope ends there. The horizontal location is found by the formula: G gL xt = Where: xt= distance of turning point from PVC or PVT g = lesser slope (ignoring signs) L = length of curve in stations G = algebraic difference of slopes. For the curve we are calculating, the computations would be (7 x 4)/16 = 1.75 feet; therefore, the turning point is 1.75 stations, or 175 feet, from the PVT (station 30 + 25). 327 The vertical offset for the turning point is found by the formula . 2 e l x y t t = For this curve then, the computation is (1.75/2)2 x 8 = 6.12 feet. The elevation of the POVT at 30 + 25 would be 237.37, calculated as explained earlier. The elevation on the curve would be 237.376.12 = 231.25. STEP 8: Check your work. One of the characteristics of a symmetrical parabolic curve is that the second differences between successive grade elevations at full stations are constant. In computing the first and second differences (Columns 7 and 8), you must consider the plus or minus signs. When you round off your grade elevation figures following the degree of precision required, you introduce an error that will cause the second difference to vary slightly from the first difference; however, the slight variation does not detract from the value of the second difference as a check on your computations. You are cautioned that the second difference will not always come out exactly even and equal. It is merely a coincidence that the second difference has come out exactly the same in this particular problem.
An unsymmetrical vertical curve is a curve in which the horizontal distance from the PVI to the PVC is different from the horizontal distance between the PVI and the PVT. In other words, l1 does NOT equal l2. Unsymmetrical curves are sometimes described as having unequal tangents and are referred to as dog legs. Figure 319 shows an unsymmetrical curve with a horizontal distance of 400 feet on the left and a horizontal distance of 200 feet on the right of the PVI. The gradient of the tangent at the PVC is –4 percent; the gradient of the tangent at the PVT is +6 percent. Note that the curve is in a dip. Figure 319 – Unsymmetrical vertical curve. 328 As an example, let’s assume you are given the following values: Elevation at the PVI is 332.68 Station at the PVI is 42 + 00 l1 is 400 feet. l2 is 200 feet. g1 is –4% g2 is +6% To calculate the grade elevations on the curve to the nearest hundredth foot, use Table 32 as an example. Table 32 shows the computations. Set four 100foot stations on the left side of the PVI (between the PVI and the PVC). Set four 50foot stations on the right side of the PVl (between the PVI and the PVT). The procedure for solving an unsymmetrical curve problem is essentially the same as that used in solving a symmetrical curve. There are, however, important differences you should note. Table 32 – Table of computations of elevations on an unsymmetrical vertical curve. Col. 1 Stations Col. 2 Elevations on tangent Col. 3 x/l Col. 4 4 (x/l)2 Col. 5 Vertical Offsets Col. 6 Grade elevation on curve 38 + 00 (PVC) 39 + 00 40 + 00 4g1 −= 41 + 00 42 + 00 (PVI) 42 + 50 43 + 00 6 2 g += 43 + 50 44 + 00 (PVT) 348.68 344.68 340.68 336.68 332.68 335.68 338.68 341.68 344.68 0 ¼ ½ ¾ 1 ¾ ½ ¼ 0 0 1/16 ¼ 9/16 1 9/16 ¼ 1/16 0 0 +0.42 +1.67 +3.75 +6.67 +3.75 +1.67 +0.42 0 foot50 stations 344.68 342.10 340.35 339.43 339.35 foot100 stations 340.43 345.10 348.68 First, you use a different formula for the calculation of the middle vertical offset at the PVI. For an unsymmetrical curve, the formula is as follows: )( )(2 12 21 21 gg ll ll e − + = In this example then, the middle vertical offset at the PVI is calculated in the following manner: e = [(4 x 2)/2(4 + 2)] x [(+6)  (–4)] = 6.67 feet. Second, you should note that the check on your computations by the use of second difference does NOT work out the same way for unsymmetrical curves as for symmetrical curves. The second difference will not check for the differences that span the PVI. The reason is that an unsymmetrical curve is really two parabolas, one on each side of the PVI, having a common POVC opposite the PVI; however, the second difference will check out back, and ahead of the first station on each side of the PVI. 329 Third, the turning point is not necessarily above or below the tangent with the lesser slope. The horizontal location is found by the use of one of two formulas as follows: from the PVC e gl xt 2 )( 1 2 1 = from the PVT e gl xt 2 )( 2 2 2 = The procedure is to estimate on which side of the PVI the turning point is located and then to use the proper formula to find its location. If the formula indicates that the turning point is on the opposite side of the PVI, you must use the other formula to determine the correct location; for example, you estimate that the turning point is between the PVC and PVI for the curve in Figure 319. Solving the formula: xt= (l1) 2 (g1)/2e xt= [(4)2 (4)]/(2 x 6.67) = 4.80, or station 42 + 80. However, station 42 + 80 is between the PVI and PVT; therefore, use the formula xt= (l2) 2 (g2)//2e. xt= [(2)2 (6)]/(2 x 6.67) = 1.80, or station 42 + 20. Station 42 + 20 is the correct location of the turning point. The elevation of the POVT, the amount of the offset, and the elevation on the curve are determined as previously explained. 2.4.0 Checking the Computation by Plotting Always check your work by plotting the grade tangents and the curve in profile on an exaggerated vertical scale, that is, with the vertical scale perhaps 10 times the horizontal scale. After the POVCs have been plotted, you should be able to draw a smooth parabolic curve through the points with the help of a ship’s curve or some other type of irregular curve; if you can’t, check your computations.
After you have had some experience computing curves using a table as shown in the previous examples, you may wish to eliminate the table and write your computations directly on a working print of the profile. The engineer will set the grades and indicate the length of the vertical curves. You may then scale the PVI elevations and compute the grades if the engineer has not done so. Then, using a calculator, compute the POVT elevations at the selected stations. You can store the computations in some calculators. That allows you access to the grades, the stations, and the elevations stored in the calculator from one end of the profile to the other. You can then check the calculator at each previously set PVI elevation. Write the tangent elevation at each station on the work sheet. Then compute each vertical offset: mentally note the x/ 1 ratio; then square it and multiply by e on your calculator. Write the offset on the work print opposite the tangent elevation. Next, add or subtract the offsets from the tangent elevations (either mentally or on the calculator) to get the curve elevations; then record them on the work sheet. Plot the POVC elevations and draw in the curve. Last, put the necessary information on the original tracing. The information generally shown includes grades, 330 finished elevations, length of curve, location of PVC, PVI, PVT, and the e. Figure 321 shows a portion of a typical work sheet completed up to the point of drawing the curve.
The stakeout of a vertical curve consists basically of marking the finished elevations in the field to guide the construction personnel. The method of setting a grade stake is the same whether it is on a tangent or on a curve, so a vertical curve introduces no special problem. As indicated before, stakes are sometimes set closer together on a curve than on a tangent. But that will usually have been foreseen, and the plans will show the finished grade elevations at the required stations. If, however, the field conditions do require a stake at an odd plus on a curve, you may compute the needed POVC elevation in the field using the data given on the plans and the computational methods explained in this lesson.
Figure 321 – Profile of worksheet.
Test Your Knowledge 4. What term is used for a vertical curve at the bottom of a hill?
5. Vertical curves are used to connect stretches of road that go up or down at a constant slope.
6. When computing the elevations of symmetrical vertical curves, you can check the accuracy of your computation through a derived constant value for the _____.

This lesson discussed the types, elements, and formulas used to calculate horizontal and vertical curves. It also addressed some of the common problems associated with horizontal and vertical curve layout.
1. What is the principal consideration in curve design? A. Speed of the highway B. Degree of curvature C. Length of the radius D. Both B and C 2. What term is used for the angle formed by two radii that subtend an arc of 100 feet? A. Degree of curve B. Point of curve C. External distance D. Central angle 3. If you take a flat curve, mark a 100foot cord, and determine the central angle to be 0° 30’, then you have a _______ minute curve. A. 0° 30’ B. 300 C. 30 D. 3 4. The degree of curve and the intersecting angle are both given in degrees and minutes. Which of the following actions should you take during the computation to maintain the degree of accuracy? A. Round off angles to the nearest tenth of a degree. B. Round off angles to the nearest hundredth of a degree. C. Convert angles to minutes for computations. D. Convert angles to seconds for computations. 5. As a check during the stakeout of a simple curve, the angle from the PI to the PT is measured while the instrument is still at the PC. The angle should equal which of these? A. One half of the central angle B. One half of the intersecting angle C. Total of the deflection angles D. All of the above 6. What is gained by using the backingin method of staking out a horizontal curve? A. Fieldwork is accomplished much faster. B. Curve distortion is minimized by applying the error at the center of curve. C. Fewer instrument setups are needed. D. Deflection angles can be turned more accurately. 333 7. A constant slope between curves is known by what term? A. Grade B. Grade tangents C. Gradient D. All of the above 8. Vertical curves are usually what shape? A. Parabolic B. Circular C. Elliptical D. Hyperbolic 9. Elements of vertical curves include all of the following except which one? A. PVC B. PVI C. l2 D. PVT 10. What factor makes a curve symmetrical? A. g1 equals g2 B. 11 equals 12 C. G equals zero D. Both B and C 11. Vertical curve computation should be checked by plotting the curve on an exaggerated scale in which the vertical scale is larger than theBBBBBBBBB A. vertical offset B. horizontal scale C. ship’s curve D. stationing 12. The original tracing of a road profile will contain which of the following information? A. Tangent elevations B. Vertical offsets C. Length of the curve D. x/1 ratio 13. The procedure used to set grade stakes for a POVC differs greatly from the procedure used to set grade stakes for a point on a grade tangent. A. True B. False 334 14. Which of the following terms is another name for “I” when discussing curve data? A. Degree of curvature B. Deflection angle C. Radius D. Interior angle 15. What is the radius ( R ) of a 30° curve? A. 189.90 ft B. 190.98 ft C. 198.90 ft D. 198.98 ft 16. What is the length of the curve if I (∆) = 62°, and D = 30°? A. 206.67 ft B. 206.76 ft C. 207.67 ft D. 207.76 ft 17. In which, if any, of the following ways does a vertical curve differ from a horizontal curve? A. Vertical curves are usually parabolic B. A horizontal curve is measured in a straight line; a vertical curve is measured along the curve. C. Only the vertical curve stations start at 0 + 00. D. Only the horizontal curve is laid out using a constant radius. tional)