 Welcome to Caltrans LSIT LS exam preparation course. One aid in your preparation for California licensure examinations. A word of caution, don't use this course as your only preparation. Devise and follow a regular schedule of study which begins months before the test. Work many problems in each area, not just those in this course's workbook, but problems from other sources as well. This course is funded by Caltrans, but you and I owe a profound thanks to others, the courses instructors from the academic community, the private sector, other public agencies, and from Caltrans as well. We wish you well in your study toward becoming a member of California's professional land surveying community. Hello, my name is Luke Wilson. I'm a land surveyor with Caltrans Geometronics Branch in Sacramento. My associates Ed Zimmerman, CJ Vandergrift, and I will be presenting this unit of the Caltrans land surveyor examination preparation course on route surveying. I'll begin our presentation with an introduction to the topic and the route surveying process. And then Ed will discuss horizontal curves followed by CJ's discussion of vertical curves. After that, I'll be back to discuss a recent land surveyor exam problem. Route surveying includes all surveying necessary for designing and constructing engineering works. Here at Caltrans, we normally think of route surveying in relation to road construction projects. In fact, highway construction probably accounts for the bulk of route surveying work. Other examples of route surveying are surveys for railroads, pipelines, transmission lines, canals, and airport runways. Other civil engineering projects like subdivision design, grading plans, and site plans, though not route surveys per se, utilize the basic techniques of the route surveyor. The route surveying system includes four steps. Reconnaissance and planning, design, right of way acquisition, and construction. During the planning stage, alternate routes are identified from small scale mapping. Then preliminary studies are undertaken to determine potential engineering problems, environmental concerns, and to estimate costs of various alternatives. The planning process culminates with the selection of a preferred alternative. The design phase of the project is undertaken with large scale mapping, generally produced by photogrammetric methods. Traditionally, designers have used paper maps, and designs have been done on hard copies. But this method is being rapidly replaced by the use of three-dimensional digital train models and computer-aided design and drafting systems. The first step in the design phase of the project is to determine an alignment. This requires the survey of a preliminary alignment, either on the ground or using large scale mapping, or in the CAD system. The second step is to design a typical finished cross-section for the work. The typical section is compared to each actual ground cross-section, and earthwork quantities are computed. The alignment is then adjusted until an optimal design for safety, economy, and utility is obtained. The next phase of the project is the right-of-way phase. Surveys are undertaken to develop survey information needed to acquire property rights for the project. Right-of-way surveying problems are covered in our units on boundary surveying and legal description writing. The construction phase is the final phase of route surveying. In this phase, land surveyors are called upon to set points on the ground to control the work of building the project. In the course of computing the location of improvements and the setting of construction stakes, conditions are often found that require design revisions. This is a favored source of exam questions. Routes are described in relation to a control line. For a highway, the control line is typically the center line. Distances are measured along the center line from the beginning of the project in units called stations. Each full station is 100 feet. Distances are given along the center line in stations plus feet between stations. For instance, if the measurement along the control line is 1,050 feet from the beginning of the project, the station would be 10 plus 50, which is 1,050 divided by 100 plus the remainder. Distances measured perpendicularly or radially from the control line are called offsets. Offsets are generally designated as left or right of the center line. Left and right are determined by your view of the project if you are standing on the center line looking in the direction of travel or increasing stationing. Normally offsets to one side of the control line are given negative values, while offsets to those of the other side are given positive values. Together, stationing and offsets define a plain coordinate system for the project. Any point within the project can be easily located by station and offset. Elevations are added to the system to give it three dimensions. Elevations are either referred to a standard datum like the North American Vertical Datum of 1929 or to the control line. When referred to the control line, elevations are given as cuts and fills. Cuts and fills are differences in elevation between an identified point on the control line and any other point in the project. A cut from the control line to another point is a decrease in elevation, while a fill is an increase in elevation. During the route surveying process, elevations are generally used to begin with, while cuts and fills are the reference system used when the process moves from design into the construction stage. Most questions on the land surveyor exams focus on the design and construction portions of the route surveying process. So let's walk through the design and construction phases of a typical project. Reconnaissance and initial studies for our highway project have identified the saddle shown here on our large-scale aerial mapping as a portion of the route. The first step in the design process is to construct a preliminary horizontal alignment. This is done by laying out tangents as shown here. Changes in direction of the route are designated as PIs or points of intersection. Positioning of these preliminary tangents is determined by topography and the change in elevation between the various PIs. In our example, the PI at the top of the hill in the saddle is placed along with the other PIs along the route so that changes in elevation between them meet design criteria. Elevations for the PIs are measured in the field or are obtained from a map or digital train model and grades are calculated between PIs. Grades express the change in elevation along a line and are quantified as a percentage of slope. The relationship is change in elevation divided by length of the line. A change in elevation of one foot between two PIs that are 100 feet apart would yield a grade of 1%, one foot divided by 100 feet. If elevation increases with increasing stationing, grade is expressed as a positive value. If elevation decreases with increasing stationing, grade is expressed as a negative value. Next, horizontal curves are designed to provide a transition for each change in direction. Design considerations such as site distance and speed of travel determine the selection of radius length for the curve. Using the mathematics of circular curves which Ed is going to discuss, the length between the PI and the beginning and end of the curve that is the BC and EC of the curve and the distance along the curve can be determined. With this information, stationing can be calculated along the curve. Next, elevations on the ground are plotted along the alignment. This can be done by staking the alignment and running levels or by plotting the intersection of the alignment with contours from a map or a digital terrain model. What you get is a ground profile with elevation plotted in relation to station, like the one shown. Next, a vertical alignment is plotted to fit the ground profile. In the case shown here, the positive grade line coming up the hill into the saddle intersects the negative grade line going on down the slope out of the saddle. The point where these design grades intersect is called the point of vertical intersection or PVI. For design considerations, such as site distances, smooth parabolic transition curves called vertical curves are designed to connect the grades. The beginning station, BVC and ending station, EVC of the curve and elevations along the curve are calculated and plotted on the profile using the mathematical relationships that CJ will be explaining. Rarely is an alignment designed in a direct linear fashion. The method is more of a series of loops in which information obtained in one step is fed back into the process requiring changes to other steps. Adjusting PI locations, changing curve radii and changing grades and links of vertical curves all demand an intimate knowledge of route surveying calculations. The alignment when complete is a mathematical relationship based on tangents, curves, grades and vertical curves. When the alignment is complete and stationed, the second part of the design process begins. A typical cross section for the work is designed. The typical cross section represents a view of the finished work at right angles or radially to the control line. Shown here is the typical section for our highway project. Dimensions or offsets are given from the center line to hinge points or changes in grade on the section. Typical hinge points shown here, moving from right to left, are located at the top of cut, flow line of ditch, center line, shoulder and toe of fill. Elevations of the hinge points can be calculated by using the elevation of the center line from the vertical alignment and changes in elevation shown on or calculated from the typical section. The change in elevation from the center line to the left shoulder in our example could be shown as a grade from the center line. In this case, grade times offset distance will equal the change in elevation. Slopes like the side slope shown on the right side of the section are typically noted as ratios. For instance, a one to one ratio indicates that for each one foot increase in offset, there is a one foot change in elevation. Ratios and grades shown on the cross section can be used in conjunction with offset distances to calculate cuts and fills in relation to the control line. After the typical cross section is developed, ground cross sections are plotted from direct measurements in the field or are generated from paper maps or digital terrain models. At given stations, the ground cross section and the typical design cross section are plotted together like the one shown here. At this particular station, there's an area that represents material that will have to be removed during construction. Here labeled the cut area. Some of this material can be used to place the embankment necessary in the area shown as fill area. Volumes of earthwork can be computed using these cut and fill areas. On this sketch, the red lines represent the ground cross section while the yellow lines represent the typical design section. This case is an example of a through cut which is a section without embankment, only excavation. Quantities are calculated by first computing the areas of sections A sub one and A sub two. This can be done by using the coordinate method for computing the area of a polygon. Corners of the polygon are defined by hinge points, grade breaks in the ground cross section and catch points, which are at the intersections of the typical section and the ground cross section. The volume is calculated by multiplying the average end areas by the difference in stationing. The formula shown will yield an area in cubic yards. After volumes are calculated for the route, adjustments are made to the alignment and the volumes are computed again. This process is repeated until the design is optimized. When the design is complete, the construction phase of the project begins. Construction slope stakes are set at catch points at the intersection of the design section side slopes and the ground section. Position of the slope stakes can be approximated graphically from the sections or computed by solving an intersecting line problem. Because ground lines are usually measured to the nearest few tenths of a foot, normally adjustments to the calculated positions must be made in the field when the stakes are set. You should be thoroughly familiar with alignment, earthwork and slope staking problems. The LSIT exam typically has a question on each of these computations. The LS exam you take will probably have at least one problem that will require use of several of these computations. I'll return to discuss a typical LS exam problem, but first, Ed and CJ will review the problems posed by alignment design. Thank you for the introduction Luke and welcome viewers to this segment of the Caltrans route surveying video. I'm a licensed land surveyor and work in the survey section of the Geometronics branch at Caltrans headquarters in Sacramento. In the past I have worked both for a private surveying firm and for the city of Sacramento. I also teach surveying at Sacramento City College where highway curves are among my favorite subjects. And now to the business at hand, understanding and most importantly, simplifying and working with horizontal curves. I want to emphasize that I will only address simple constant radius curves. However, an understanding of the simple curve is basic geometry, calculation methods and stakeout procedure is necessary to understand the more complex forms of curves that you will eventually encounter in your surveying career. Horizontal or circular curves lie in a horizontal plane and are used to connect tangents. Tangents are used for control and alignment of straight sections of various public works such as streets and highways, railroads, pipelines and canals. It is important to remember that all points lying on a circular curve are a constant distance or radius from a central locating point known as the radius point. Consider any horizontal curve as a portion of a complete circle. The math computations required for solutions of circular curves is not difficult. Almost all curve problems and solutions can be done using basic principles of trig, algebra and or geometry. The trigonometry involved in those curve computations and solutions is based largely on the sides and angles of a right triangle. All the curve formulas and equations to follow are presented in a format handy for use than electronic handheld calculator. Handhelds should be capable of trigonometric functions including radian conversion. The simplest form of a circular curve and the one most often encountered in surveying is tangent two and connects two tangent sections of a route alignment. The basic components will be shown in the next few graphics. There is no universally accepted terminology for the parts of a curve in surveying practice. Every surveying company or public agency seems to have different curve terms and abbreviations. Therefore, I will be using terminology and notation that I am comfortable with having lived with them for the last 30 or 40 years. Let's first look at the intersection of back and forward tangents of a soon to be formed circular curve. Keeping in mind that in our example the direction of travel and or stationing is from left to right, the tangent to the left of the point of intersection or PI is called the back tangent. The tangent to the right is the forward tangent. Back and forward tangents intersect at the PI forming a deflection angle I. Deflection angle I between the tangents is generally measured in the field or computed. Usually the PI will have either an assigned or measured station designation as part of the preliminary survey. The next element of our curve to think about is the radius or dimension of sharpness of the curve. For any pair of tangents, an infinite number of curves may be designed. Field conditions generally dictate the choice of size or radius of the curve. For instance, a larger radius curve would be required for a high speed highway allowing traffic to comfortably negotiate the turn. Mountainous terrain considerations may limit curves to smaller radii with a corresponding decrease in speed. As shown on the screen, two lines are generated from the center of the curve or radius point and connect to the beginning of curve BC and to the end of curve EC. The radius or R is the distance from the center of the curve or radius point to any point on the arc of the curve. Since a curve is always symmetrical around the PI, distances from the PI to the BC and EC are equal and are called sub tangent distances or T for a particular curve. In this next graphic, we have added additional basic components of our curve. Note that the central angle or delta for the curve is an angle included between the two radial lines. The delta angle is always equal to angle I at the PI and radial lines from the BC and EC are always perpendicular to the tangents. Some highway agencies and railroads use a degree of curve definition rather than radius to define the sharpness of circular curves. Since degree of curve frequently appears in many land surveyor examinations, I will give you a brief look at this definition and its relationship to radius. The two definitions of degree of curve are arc definition and chord definition. First, let's look at the arc definition. Highway surveying practice defines degree of curve by arc definition as shown in this graphic. This is the definition most commonly used. As shown, this definition of degree of curve is the central angle of an arc of curve subtended by an arc 100 feet in length. It is identified as D or degree of curve by arc definition. Degree of curve expressed by arc definition can be converted to a radius distance by the formula shown on the graphic, 5729.58 divided by degree of curve. D is always expressed in degrees and decimals of degrees when used in this operation. The second definition of degree of curve is by chord definition as shown in the next graphic. The chord definition is most often used by railroads and is defined as the central angle of an arc of curve subtended by a curve with a chord 100 feet long. It is defined as D or degree of curve by chord definition. Degree of curve by chord definition is converted to radius distance by using the formula shown in the graphic now on screen. Radius equals 50 divided by the sign of one half the degree of curve. D is always expressed in degrees and decimals of degrees for this formula. Although the degree of curve and radius are mathematically related and both define sharpness of the curve, the radius definition of circular curves is more prevalent in modern survey practice and will be used in the example calculations for the rest of this presentation. At this point, let's drop back and review two basic facts of horizontal curves. As shown in this graphic, angle I always equals delta or central angle and radio lines from the BC and EC are always perpendicular to the tangents. Let's move on to the next sketch. Note that the line from the radius point to the PI bisects all related lines and angles. Now let's start some calculation procedures by determining the sub tangent distances for our curve. As shown on the screen, the formula for this particular computation is sub tangent distance equals tangent of one half delta times radius. In the sketch, note the right triangle formed by a line from the radius point to the PI or midpoint of curve, hypotenuse, a radial line to the BC adjacent side and the back's sub tangent opposite side. The next thing to do is determine the station of the BC, easily done. Simply subtract the sub tangent distance from the preliminary station of the PI. The length of arc or actual distance traveled around the curve from BC to EC is computed next. To find this distance, multiply delta expressed in radians by the radius distance. To determine the station of the EC, add the length of arc L to the station of the previously calculated BC. The use of radian measure rather than degrees, minutes and seconds, greatly simplifies calculations for arc lanes. Most hand held electronic calculators will perform this conversion. The next element of a circular curve is the length of the long cord or LC. The LC length of long cord is shown in this screen and is a straight line connecting the BC and the EC. Note the right triangle formed by the midline of the curve, the long cord and the radial line to the BC. As shown in the sketch now on screen, the formula for computing the LC is twice the radius times sine of half delta. The graphic now on screen is used to show the angle identified in the sketch as the total deflection angle and its angular relationship to both the angle at I and the delta angle. A survey instrument set up at the BC, zeroed on the PI and then turned to the EC, will read one half of the angular value for angle I. The total deflection angle will always equal half delta. Since sub tangent distances are identical, triangle BC, PI, EC, is considered an isosceles triangle. Note that in the preceding graphics in discussion, each curve, though complete within itself, is a portion or partial arc of a complete circle. Therefore, all elements of the particular curve or arc have a strict geometrical relationship to each other and are unique to a curve of certain radius and central angle. If any two of the curve's elements are known, the remaining elements can be calculated from the two known values. This graphic is used to show how a particular arc of curve can be divided into a sub arc length, L sub one. Sub arc one starts at the BC and runs along the curve to POC sub one. Therefore, sub arc one will have among its other elements, delta sub one, long chord sub one, and deflection angle sub one. These values are always unique to their particular arc. Before proceeding into curve layout computations and layout procedure, let's take a look at some other elements of a circular curve keeping in mind that all elements of a curve are mathematically related. This graphic shows the external distance. This curve element is the distance from the midpoint of the arc outwards to the curve's PI. The computation to determine this dimension is performed as shown on screen. Determine the quantity one over the cosine of half delta subtract one from it and then multiply the difference by the radius of curve. The next curve element is shown on this graphic and is called the midordinate distance. This is the distance from the midpoint of the long chord to the midpoint of the arc. This measurement is determined by multiplying the quantity one minus the cosine of half delta by the radius distance. Now on with the business of computing and laying out individual points on a curve. The deflection angle between the back tangent and a line or sub chord drawn from the BC to a point on the curve POC is the deflection angle for that particular POC or station. If a particular POC or station's deflection angle and sub chord distance are known, layout of the station from the BC is easily accomplished. After the design elements of a curve have been finalized, deflection angles for each station on the curve that are required to be laid out are computed. Let's take a look at the complete curve ready for stakeout and compute some deflection angles and sub chords to lay out the curve. The radius of this curve is 550 feet and the central angle is 28 degrees, 35 minutes, zero seconds. The total deflection angle for the complete curve is half delta or 14 degrees, 17 minutes and 30 seconds. Calculation of the total arc length is shown on this graphic. The computation is total arc equals 550 feet times 28 degrees, 35 minutes and zero seconds expressed in radians or 0.498873. This equals 274.38 feet. Considering the sub arc defined as L sub one, we determined it has an arc length of 40 feet. This is based on the stationing of the BC being 21 plus 80 and the station of POC sub one being 22 plus 20. Three calculations are shown on this sketch. Determination of delta sub one, determination of half delta and or deflection sub one for the same arc sub one and determination of the chord sub one for the same arc sub one. L sub one's central angle or delta is determined by dividing L sub one 40 feet by the radius 550 feet which equals 0.072727 radians or four degrees, 10 minutes and one second. Deflection angle sub one D sub one equals four degrees, 10 minutes, one second over two or two degrees and five minutes. Next LC sub one is determined by twice the radius times sine of half delta or 1100 feet times 0.036358 which equals 39.99 feet. Therefore, using a theatolite set up the BC and zero to the PI, deflection angle D sub one is set off to the right and the distance 39.99 feet is measured, laid out from the BC on the line produced from BC to POC sub one or long chord sub one correctly placing station 2200 in its position on the arc of the curve. This method of curve layout by deflection angles can be used to stake any portion of a curve on the ground. Now for the final portion of our curve discussion let's take a look at computing and laying out a curve by using coordinates and inversing between coordinates of required curve stations and a setup point remote from the curve. Combining the computing powers of today's electronic calculators and state of the art electronic total stations it can be more efficient to lay out a curve using coordinates and inverse methods. In this approach to curve layout a centrally located or convenient setup station is selected. This point is not necessarily located on the curve but is selected to allow all points of the curve to be staked out from the one setup. As seen in this graphic the setup station is not located on the curve but is tied to the BC and EC by bearing and distance or coordinates. With all elements of the curve located on the coordinate system calculations for the curve data and also the actual layout are reduced to routine inverse calculations. This provides staked out data for bearing and distance layout of all required curve points using a single setup location. In practice the setup point is selected and tied to the curves control system either in the field by actual measurement or by office calculation. As shown in this graphic the BC, RP, EC and setup stations all have coordinate values. To determine a coordinate for POC sub one first determine L sub one by station differences as in the deflection angle method just discussed and then calculate delta sub one by the same process used in the deflection method. For example, if the BC station is 12 plus 46 feet and the station at POC sub one is 12 plus 80 feet L sub one 12 plus 80 minus 12 plus 46 equals 34 feet. Assuming a radius of 550 feet delta sub one equals L sub one 34 feet divided by radius 550 feet equals 0.061818 radians which converts to three degrees, 32 minutes and 31 seconds. Applying delta sub one to the bearing RP BC the bearing RP POC sub one is determined. Using the coordinates of the RP the coordinate of POC sub one is computed. Inversing between the setup point and POC sub one determines the bearing in distance to be laid out to correctly position POC sub one on the arc. Coordinates are calculated for each POC that is to be set to define the curve by using the bearing radius from the radius point. The bearing of each sub POC is determined by applying its sub delta to the bearing of line RP BC The resulting coordinates of individual POC stations are used to obtain bearings and distances to the POCs from the setup point by the inverse generation process. All calculations should be done in a logical sequence and be neatly tabulated. There is an increase in computation time for the coordinate method when compared to the set point. to the deflection angle method. But this increase is more than made up for by the decrease in time required for field layout. Well, I've pretty well used up my allotted time discussing the basics of circular curves. I realized that only the bare necessities have been addressed in my presentation. Hopefully with study aids such as the workbook and reference books listed in the bibliography you will be able to cope with the mysteries of circular curve computation and layout. Thanks for your attention and I sincerely hope that I was able to make the horizontal curve process a little more user friendly. Good luck with your surveying career in general and with the land surveyors examination processes in particular. I now want to introduce CJ Van de Gryff, a land surveyor for Caltrans, who will discuss the aspects of vertical or parabolic curve computation. CJ, it's all yours. Thanks Ed and hello viewers. My name is CJ Van de Gryff. I'm a surveyor at the Hilltop Field Office in district four. I've been surveying for approximately 12 years and have been licensed since 1987. I will be presenting vertical curves as a portion of the route surveying unit. Simply put, vertical curves are parabolic curves that lie in the vertical plane. Vertical curves are used in design to transition from one grade line to another in order to maintain safe side distances for passing and stopping and to provide comfortable riding conditions. The geometry of parabolic curves allows easy computation of elevations at any point on the vertical curve. Field applications of vertical alignments usually requires that elevations be calculated for specific stations or at regular intervals. These values may be derived by either the tangent offset method or the parabolic curve equation. Both methods are based on the following properties. The offsets from a tangent to a parabola are proportional to the squares of the distances from the point of tangency. This presentation will only discuss the parabolic curve equation. It is less complex and well suited to computer and calculator use. Other curve computation methods are no less correct and some familiarity with them is encouraged. Most of the sources listed in the workbook cover both methods. This graphic shows the two classifications of vertical curves. Curve A points upwards from its point of beginning or BVC and is called a sag curve. Curve B moves downwards from its point of beginning or BVC and is called a summit or crest curve. The elements of a symmetrical or equal tangent curve are as follows. BVC is the beginning of vertical curve sometimes shown as PVC or point of vertical curvature. EVC is the end of vertical curve sometimes shown as PVT or point of vertical tangency. PVI is the point of vertical intersection of the incoming and outgoing tangents. Capital L is the length of the curve in stations measured on the horizontal plane. One full station equals 100 feet. The distance from the BVC to the PVI is always one half the length of the curve. Small L is not shown here but it is often used to define the distance from the BVC to a station of interest on the curve. G sub one and G sub two are the grade rates in percentages with the tension to algebraic sign. G one refers to the tangent grade entering the curve and G two refers to the tangent grade leaving the curve. Little R is the rate of change for a vertical curve and is found by subtracting G one from G two, preserving algebraic signs and dividing by L, the length in full stations. For example, a vertical curve where the incoming G one equals plus 2.75% and the outgoing G two equals minus 4.25% with a length of 800 feet. Little R would equal minus 0.875% per full station. The rate of change is always negative for a crest curve and positive for a sag curve. The value of little R is the rate of change of the slope in percent per station. Since the rate of change for a symmetrical vertical curve is a constant, the following equation can be derived. The equation states the elevation of any given station is equal to one half the rate of change times the distance to the given station from the BBC squared plus G one times the distance to the station plus the elevation at the BBC. This next graphic shows where the elements of the equation come from. Begin with the elevation at the BBC. Then add algebraically the beginning grade multiplied by the stations along the curve. Then add algebraically the offset from the tangent to the curve. For example, let's say the equation, the station at the BBC of our curve is 10 plus 00 and the elevation is 100 feet. The elevation at station 12 plus 50 would equal minus 0.875% divided by two multiplied by 2.5 squared plus 4.25 times 2.5 plus 100 feet which equals 107.89. In practice, it's not always critical to know all the equations. We often just plug numbers into computers and record answers. But this one relatively simple equation will allow you to compute grades at any station on a vertical curve. Questions involving vertical curve computations are very common on both the LSIT and the LS exam. Since the computations are quite simple, the critical elements to solve the vertical curve are often hidden in with lots of unneeded data. Always identify which of the basic components are given and which are required to find. We've just looked at finding an elevation on a vertical curve given the curve length, the tangent grades, and the stationing. A problem on the 1991 exam was presented as such. A 12-inch water pipe was found at station 18 plus 50. The top of the pipe elevation at centerline was given as 730.92 feet. Required was to design a vertical curve to clear the water pipe. Carefully using all the information given, the length of the curve became the only unknown. It is helpful when setting up an equation to list what is known and therefore identify what must be solved for. In our example question, the elements given are as follows. G1 equals plus 8.75%. G2 equals minus 1.50%. The BVC station is 16 plus 50 and the BVC elevation equals 720.23. The elevation needed to clear the water pipe at station 18 plus 50 is 734.32. And little L equals 2.0 stations from the BVC to the water pipe, left to find the length of the curve. If we plug these values into the basic parabolic equation, it looks like this. Some basic algebra shows the value of L, in this case rounded, to be 6.0 stations or 600 feet. The final aspect I want to cover is locating the station and elevation of the highest and lowest point of the vertical curve. This can be very helpful when checking profiles for conflicts like our water pipe. At the higher low point, the slope of the tangent to the curve equals zero. Therefore, the value of little L, the stations from the BVC to the higher low point is equal to G1 divided by the rate of change. Changing the sign of this value will always yield a positive number. This provides a good check that all the algebraic signs have been carried correctly. Using the values for our water pipe problem, the rate of change equals minus 1.708. Little L then equals 5.120 stations from the BVC. Therefore, the high point station is 21 plus 62.20. The elevation can then be calculated using the parabolic equation. The elevation of the high point in this case does equal 742.64 feet. This concludes my portion, and I want to wish you all success in your pursuits as professional land surveyors. Now back to you, Luke. Thanks, CJ. Now we're going to discuss problem B5 from the 1990 land surveyors exam. The problem is in your workbook. We're provided with information in the form of a street centerline profile, a street section, and the site plan you see, which I've modified for our use. We're required to answer five questions about the site plan. If you want, you can stop the tape now and solve the problem on your own. Then turn the tape back on and compare your method against mine. This site plan problem can be solved using a route surveying methodology because it is based on a control line, in this case, the centerline of the street. The first step in the solution is to read and study the whole problem. The problem statement, you have been provided design criteria as shown on the diagrams below and on the facing page. Problem requirement, answer the following questions using the information provided in the diagrams. One, determine the ground elevations of the back of sidewalk at the following locations. Driveway centerline, southeasterly property corner, southwesternly property corner. Two, provide the grade percentage between point C and the building pad. Three, what is the slope ratio from point A to the toe of slope? Four, calculate the cut from the back of the sidewalk to the sewer lateral invert at the property line. And five, calculate the distance from the north property line to the toe of slope at point B. Notice that questions one and four require back of walk elevations and that the other three questions require use of these back of walk elevations to develop their answers. To solve questions one and four, calculate elevations on the centerline of the street for stations that correspond to the southwest property corner, driveway centerline, southeast property corner, and sewer lateral. To do this, first, stations on the property corners must be calculated. Property corner stations can be calculated using the station on the driveway centerline, the dimension given from the driveway centerline to the west property line, and the width of the lot. The elevations can all be calculated using the street profile that is given. Two of these elevations are in a vertical curve section, so you must be familiar with vertical curves to solve the problem. From the street section, we find that there is a 0.33 foot fill from the centerline to the back of the walk. Adding 0.33 feet to the centerline elevations for each station will give us the answer to question one. Question four asks for the difference in elevation from the back of the walk to the invert of the sewer. Because we planned ahead and already know the back of walk elevation at station 12 plus 67, we need to only know the invert of the pipe at the back of the walk. We're given the invert of the pipe two feet from the house and the grade on the pipe, so we only need the distance from the position of the given invert elevation to the back of the walk to figure the difference in elevation we need. The site plan shows the swale as being five feet from the house and 25 feet from the back of the walk. With this information, we can determine the distance from our given invert to the back of the walk, which is 28 feet. Next, calculate the invert of the pipe and then the cut to the invert of the pipe. Questions two, three, and five all require elevations along the swale. We'll do question two first, because to do it, we will have to figure elevations at each grade break in the swale needed to solve both questions three and five. We begin with the elevation we obtained in question one at the back of the walk and the center line of the driveway. Dimension and grade from this point to the swale grade break in the driveway is given. So we can calculate an elevation at the swale grade break at the center of the drive. The elevation of the flow line at point D can be computed from the grades and dimensions given on the plan. Finding the distance from the swale grade break in the driveway to point F at the flow line of the swale is a bit tricky. Here's where you can use your years of experience pouring over architects' plans that always seem to admit key dimensions. The dimension from the west property line to the center of the swale at the center of the drive can be calculated and is 35 feet. The swale on the east side of the lot is five feet from the property line. Therefore, we can add these two distances, five feet and 35 feet, and subtract the sum from the width of the lot to obtain the distance necessary to calculate the elevation at grade break F in the swale. From point F, distance and grade shown on the plan give us the elevation at point E. Now for point C. With elevations at D and E and grades given from these two points to the grade break in the swale at C, we can solve for the elevation at C in the following way. We can write equations for the elevation of C from both point D and point E. The equation from point D is the elevation of C equals the elevation of D plus 0.015 times the distance D, C. The equation from point E is the elevation of C equals the elevation of E plus 0.02 times the distance C, E. We know that distance D, C plus distance C, E equals 120 feet. So distance D, C equals 120 feet minus distance C, E. Now we can rewrite our equation for C from point D by substituting 120 feet minus distance C, E for distance D, C. Giving us elevation of C equals the elevation of D plus 0.015 times 120 feet minus the distance C, E. Now we can combine the two equations as shown and solve for distance C, E. With distance C, E and the grade of 2%, we can easily calculate the elevation at grade break C and from that, the grade to the building pad. After question two, question three is a breeze. It asks for the slope from point A to the toe. We know the elevation of point F at the flow line of the swale from our work on question two and the fact that the top of the slope and point A are one foot above the flow line from the plan. We've already computed the back-of-walk elevation at the southeast property corner for question one so we know the change in elevation. From the site plan, we know that the dimension from the back-of-walk to flow line is 25 feet and that the top of the slope is five feet from the flow line and the toe of the slope is one foot from the back of the walk. This leaves 19 feet for slope. The grade is the difference in elevation between point A and the elevation of the southeast property corner divided by the distance allowed for slope on the site plan. The last question, number five, asks for the distance from the north property line to the toe of slope at point B. The elevation of the toe of slope is given on the site plan. The top of the slope is one foot above the flow line of the swale at point E, which we calculated for question number two. The change in elevation is easily obtained as a difference between the elevation of top and toe. The ratio of the slope is given as two to one so the distance from the top to the toe is two times the change in elevation. From the site plan, the distance from the top of slope to the property line is 20 feet. So the dimension from the property line to the toe is 20 minus two times the change in elevation between the top and the toe. As you can see, a lot of time and effort can be saved by proper planning and organization. The method you use is very important. Efficient method can be the difference between success and failure in your day-to-day practice as a land surveyor. On the exam, correct method becomes all important because of time restraints. Inefficiency on the job might cause evening or weekend shifts, but on the test situation, it can cause failure. With that in mind, I wish you good planning.