# Chapter 3 Horizontal and Vertical Curves

 Contents 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:

1. Describe the different types and methods of calculating horizontal curves.
2. Describe the different types and methods of calculating vertical curves.

## 1.0.0 HORIZONTAL CURVES

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 high-speed 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 high-speed highways, the curves must be flat, rather than sharp. That means they must be large-radius 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 3-1). 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.

### 1.1.0 Types of Horizontal Curves

There are four types of horizontal curves. They are described as follows: 1. Simple- The simple curve is an arc of a circle (Figure 3-2, View A). The radius of the circle determines the sharpness or flatness of the curve. Figure 3-1 – Lines of sight.  3-4 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 3-2, 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 3-2, 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 3-2, View D).

### 1.2.0 Elements of a Horizontal Curve

The elements of a circular curve are shown in Figure 3-3. 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 3-2 – Horizontal curves.  3-5 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 straight-line distance from the PC to the PT. Other types of chords are designated as follows: C The full-chord 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.

### 1.3.0 Degree of Curvature

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.

#### 1.3.1 Degree of Curve (Arc Definition)

The arc definition is most frequently used in highway design. This definition, illustrated in Figure 3-4, 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 3-4 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 3-3 – Elements of a horizontal curve.  3-7 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 3-4 – Degree of curve (arc definition).  3-8

#### 1.3.2 Degree of Curve (Chord Definition)

The chord definition (Figure 3-5) 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 100-foot chord, and determine the central angle to be 0°30’, then you have a 30-minute curve (chord definition). From observation of Figure 3-5, 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 3-5 – Degree of curve (chord definition).  3-9 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.

#### 1.4.1 Tangent Distance

By studying Figure 3-6, you can see that the solution for the tangent distance (T) is a simple right-triangle 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

#### 1.4.2 Chord Distance

As illustrated in Figure 3-7, the solution for the length of a chord, either a full chord(C) or the long chord (LC), is also a simple right-triangle 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 3-6 – Tangent distance.   3-10 Therefore, R C 2/ 2 sin ∆ and solving for C: 2 2 sin ∆ = RC Figure 3-7 – Chord distance.

#### 1.4.3 Length of Curve

In the arc definition of the degree of curvature, length is measured along the arc, as shown in Figure 3-8, View A. In this figure the relationship between D, L, and a 100-foot arc length may be expressed as follows: . 100 D L ∆ = Then, solving for L: L = 100 D ∆  3-11 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 3-8, View B, the central angle (A) is equal to three times the degree of curve (D), then there are three 100-foot chords and the length of “curve” is 300 feet.

#### 1.4.4 Middle Ordinate and External Distance

 Test Your Knowledge1. A highway is composed of a series of curves and straight lines called _______. A. traverses B. radii C. tangents D. center lines 2. What type of curve consists of two simple curves joined together and curving in the same direction? A. Simple B. Compound C. Spiral D. Reverse 3. The first step in staking out a simple curve is to set the instrument up at what point? A. PC B. PI C. PT D. Midpoint

## 2.0.0 VERTICAL CURVES

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.

#### 2.3.1 Symmetrical Vertical Curves

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 3-17 – 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   3-24 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 50-foot station. Figure 3-17 shows the vertical curve to be solved. STEP 1: Prepare a table as shown in Table 3-1. 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 3-1 – 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 half-length 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 100-foot 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.   3-25 List the stations under Column 1. STEP 3: Calculate the elevations at each 50-foot station on the tangent. From Step 2, you know there is a 9-foot 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 50-foot 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 239.12-3.50, or 235.62 feet. The elevation on the tangent at station 31 + 00 is 235.62-3.50, or 232.12 feet. The elevation on the tangent at station 31 + 50 is 232.12-3.50, or 228.62 feet. The elevation on the tangent at station 32+00 (PVT) is 228.62-3.50, 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.   3-26 STEP 5: Compute the vertical offsets at each 50-foot 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 50-foot stations. List the results under Columns 3, 4, and 5. STEP 6: Compute the grade elevation at each of the 50-foot 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).   3-27 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 237.37-6.12 = 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.

#### 2.3.2 Unsymmetrical Vertical Curves

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 3-19 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 3-19 – Unsymmetrical vertical curve.   3-28 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 3-2 as an example. Table 3-2 shows the computations. Set four 100-foot stations on the left side of the PVI (between the PVI and the PVC). Set four 50-foot 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 3-2 – 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.   3-29 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 3-19. 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.

### 2.5.0 Using a Profile Work Sheet

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,   3-30 finished elevations, length of curve, location of PVC, PVI, PVT, and the e. Figure 3-21 shows a portion of a typical work sheet completed up to the point of drawing the curve.

### 2.6.0 Field Stakeout of Vertical Curves

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 3-21 – Profile of worksheet.

 Test Your Knowledge4. What term is used for a vertical curve at the bottom of a hill? A. Summit B. Over vertical C. Sag D. Compound 5.  Vertical curves are used to connect stretches of road that go up or down at a constant slope. A. True B. False 6. When computing the elevations of symmetrical vertical curves, you can check the accuracy of your computation through a derived constant value for the _____. A. second differences in elevations of successive stations B. vertical offsets of successive stations C. second differences in elevations of adjacent stations D. e value at successive stations

## Summary

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.

## Review Questions

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 100-foot 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 backing-in 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.   3-33 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   3-34 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)