 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 and your study toward becoming a member of California's professional land surveying community. Hello. My name is Don Hunter. I work with Caltrans in the headquarters office of Geometronics. During this session, we'll be covering areas and volumes. We'll begin our discussion with a definition of area. Area is defined as the surface within a set of lines. The area within a triangle is the surface within three sides. The area of a circle is the surface within the circumference. Now, area is expressed in square units, square inches, square feet, square miles, et cetera. One object of land surveys is to determine the area of a parcel. The area may be determined by any of the following methods. One, by plotting boundaries to scale and by plotting the boundaries to scale, the area of a parcel may be found by use of a planimeter, or it may be calculated by dividing the parcel into triangles, quadrilaterals, and computing their areas mathematically. Two, for parcels with irregular or curved boundaries, the area may be determined by the trapezoid rule or Simpson's one-third rule. Three, by calculating the area from coordinates, or meridian distances and parallel distances of the corners of the parcel. By calculating the area by double meridian distance and the latitudes of the sides of the parcel. Now, accuracy. Basically, the accuracy of any of the results of these four methods are in order in which they were presented to you. The most accurate would be by the coordinate method, any of the three which we will be covering later. The following figures and formulas appear in appendix f1 through f3 and g1 through g2 of the Caltrans survey manual. Here we have the area of a circle, pi r squared, r, of course, being the radius of the circle. Now we see a sector of a circle. The formula there is area equals pi r squared times theta over 360. The central angle, or theta, is in degrees and decimal of a degree. A segment, the area is equal to pi r squared theta over 360, just as it was in the sector of a circle, but we add minus r squared sine theta over 2. Here again, the central angle in degrees and decimal of a degree. A polygon, a closed plane figure bounded by straight lines or sides. Here we have two examples of a polygon, one a triangle and one a square with all sides equal. The area for the solution of this type of a figure is 1 quarter n s squared, where n is the number of sides and s is the length of the side, times the cotangent of 180 over n. Here are two more examples of a regular polygon, all sides equal. The same formula. Here we have a triangle, which is a three-sided polygon. The area of a triangle is 1 half base times the height. Now there's a second formula for the area of a triangle as shown in the graphic. This is when we do not have any angles, but we know all of the sides of the triangle, and that is the square root of s times the quantity s minus a times the quantity s minus b times the quantity xs minus c. This formula is used only when the three sides of a triangle are given. Here are two more examples of triangles. Assuming all sides are known, the right triangle on the left would be solved by the formula a equals 1 half base times height, since the base and the height are known sides of the triangle. The triangle on the right would be solved by the formula the square root of s times s minus a times the quantity s minus b times the quantity s minus c because the height h must be calculated, and the information to do so is not given. A parallelogram is a quadrilateral with opposite sides equal and parallel, and the formula for getting the area is base times the height. A rectangle is a parallelogram with 90-degree angles. A parallelogram, which is also a quadrilateral with opposite sides equal and parallel and has 90-degree angles. The next figure is a square which, like a rectangle, has opposite sides equal and parallel with 90-degree angles. But added to that, all sides are of equal length. Our next figure is a trapezoid. The solution for the area of a trapezoid is h times the quantity b sub 1 plus b sub 2 over 2. The b sub 1 plus b sub 2 over 2 will give you the average length of the bases times the height. The next figure is a trapezoid, which is a quadrilateral without any parallel sides. And the formula for computing the area of a trapezoid is the quantity 1 half of dh plus within brackets 1 quarter b times the quantity h sub 1 plus h sub 2. Now that we have reviewed the methods of determining the area of a geometric figure, let's put the formulas to work for us. Here we have an exercise. Find the acreage of parcel A. There are a number of methods we could use to determine the acreage of the parcel. I would like you to turn off the video, refer to the diagram in your workbook, and determine the method you would use to solve this problem. When you have determined the method you would use, turn the video on, and we will review a method I recommend. Did you come up with a method? Let's run through my recommended solution and see how we compare. Here we have a parcel again. Our first step is to circumscribe the parcel with a rectangle, with at least one side of the parcel coincident with a side of the rectangle and a maximum number of angle points of the parcel on the perimeter of the rectangle. The orientation of the rectangle must be such that the length of the size of the rectangle can easily be determined. Step two is to determine the dimensions of the rectangle. Side AD is the length 220 feet plus 280 feet, equaling 500 feet. The side AB is 70 feet plus 225 feet plus 120 feet. The 120 feet is the radius of the circle, giving a total of 415 feet. The area of the rectangle is 500 feet times 415 feet, equaling 207,500 square feet. The next step in the solution is to determine the area of geometric figures between the perimeter of the parcel and the constructed rectangle. The first area we will deal with is the shaded portion of the diagram designated number one. Since this is a right triangle, the base and the height is given. This diagram shows the solution of the problem. If the triangle was not a right triangle, or if we did not know it was, we would have to use the second formula shown in the diagram. The next area to calculate is the shaded area 2, which is a square minus a sector with a radius of 120 feet. Here is the solution to determine the area of area 2. Area 2 equals the area of the square minus the area of the sector. Solving for the square, we have 120 feet times 120 feet for 14,400 feet. For the sector, we have pi r squared times x over 360. x equals degrees and decimal equivalent of minutes and seconds. So therefore, from the given, we know it's a 90 degree delta. The sector would be 3.1416 or pi times 120 squared, the radius, times 90 over 360, which would give us 11,309.76 square feet. Now taking the area of the square, the 14,400 feet, minus 11,309.76 feet, we come up with 3,090.24 square feet for the area of this sector. Area 3 on our diagram is a rectangle designated A, B, C and D. Here is the solution for area 3. Side AB is 160 feet long, which is, we get that from the radius of the circle adjacent to it on the lower side. Side AD equals 415 feet minus 125 feet plus 160 feet or 130 feet. The area of the rectangle would be then 160 feet times 130 feet or 20,800 square feet. Area 4 is another rectangle shown as A, B, C, D on the diagram. And here is the solution of area 4. The side AB is 500 feet minus the quantity 120 feet plus 150 feet plus 160 feet. We can get this right off of the diagram, which would equal 70 feet. Side AD is 415 feet minus the 125 feet or 290 feet and the area would be 70 feet times 290 feet equaling 20,300 square feet. Now the last area is area 5. It is a segment of a circle with a radius of 160 feet. Here we see the solution for area 5 and the square footage of 20,106.24 square feet. We have calculated the areas for each of the five parcels. Now we add them to get a total excess, which is 71,996.48 square feet. Now we can determine the area of parcel A. We take the area of the large rectangle circumscribing the parcel A, which has 207,500 square feet and subtract the total area of parcels 1,2,3,4, and 5, which is an area of 71,996.48 square feet. This leaves a remainder of 135,503.52 square feet. If we divide the remainder by 43,560 square feet, we will arrive at 3.11 acres for parcel A. Next, we will calculate the area of a parcel with an irregular boundary. We will calculate the same area three times by the same basic method, but the last two by a more refined formula. At the end, we'll compare the results. The area formed between the traverse line and the lake will be determined by calculating the areas of the triangles, the trapezoids formed by the traverse line, the offset lines, and the edge of the lake. The area on the other side of the traverse line may be determined by a number of different methods. If the traverse and the offsets are accomplished by a field survey, the areas calculated will be satisfactory for most requirements. If the traverse and the offsets are plotted and scaled without the benefit of a field survey, the accuracy of the areas determined will be limited in their usefulness. This figure shows the stations and offsets. What we have to calculate using this data is, from left to right, a triangle, five trapezoids, and another triangle. Turn the video off and work this problem. The information on this screen is also in your workbook. After you've completed the problem, turn the video back on and we'll compare notes. Okay, let's compare our results. On the diagram shown, we have the triangle with a square footage of 3,454 square feet and the five trapezoids with 8,287.50 square feet, 5,135.00 square feet, et cetera, et cetera, et cetera, down to the triangle. The last triangle having an area of 1,797.75 square feet for a total of 32,072.75 square feet. Here is a second approach to this same problem using the same area. Offsets at regular intervals. This is called the trapezoidal rule. Offsets may be established at any desired interval, 25, 50, 100 feet, et cetera. The denser the offsets, the more accurate the results. This type of solution would be based on field survey data. The formula as shown on the diagram is the area equals s times the quantity o sub 1 over 2 plus o sub 2 plus o sub 3 plus o sub 4 on until we get to the end of the trapezoids. In this formula, s equals the regular offset interval, o, the offset distance. The formula shown here appears in your workbook along with the stations and offsets for this problem. Turn the video off and work the problem. After you have completed working the problem, turn me back on and we'll compare results. Thanks for turning me back on. It was getting lonely in here. Well, let's compare results. I have 33,101.50 square feet. And the breakdown is as shown on the diagram. We have one triangle with 1443.75 square feet and the regular trapezoids, using the formula, we have 25 times the quantity 52.5 over 2 plus 88 plus 87 and on and on and ending with 68 feet over 2 and the parentheses. Then we have another trapezoid. Oh, going back to the regular trapezoid, I neglected to say that the total square footage for that formula is 27,206.25 square feet. The second trapezoid and the reason we must calculate it independently is because it has a different base. The difference between the stationing there, you will notice 38 feet. So we have to calculate it separately since it isn't part of the regular trapezoids. And it has an area of 2,641 square feet. The last triangle is 71 feet times 51 feet over 2 or an area of 1,810.50 square feet. Totalling 33,101.50 square feet. The same stations and the same offsets as the previous problem, all of which you found in your workbook. This formula will also be found there with this information solved for the area using Simpson's one-third rule. So turn me off so I can get a little rest and work the problem too. When you get done, come back and we'll compare results. Well, how did it go? I got 33,206.66 square feet. Let's review the formula and plug in the values in their proper niches. Refer the formula in your workbook. S3, where S equals the regular interval, or 25C, feet. Inside the bracket, O1 equals 52.5 feet, the offset distance for the first interval at station O plus 55. O2 equals 68 feet, the offset distance for the last interval at station 5 plus O5. Inside the first parentheses, summation of the even offset distances, station O plus 80, the second offset, equals 88 feet, plus station 1 plus 30, the fourth offset, 78.5 feet, plus the offset of the 6th, 8th, 10th, and on through to station 4 plus 80, the 18th offset, which is 46 feet. Inside the second parentheses, summation of the odd offset distances, we start with station 1 plus O5, the third offset that is 87 feet. We don't incorporate the first odd offset because we use that at the beginning of the formula. And we have plus the next station is the fifth station, 1 plus 55 with an offset of 69.5 plus the seventh station at 2 plus O5, which has an offset of 50.5, and on through the 17th station, 4 plus 55, which has an offset of 27 feet. We don't incorporate the 19th also odd offset because this also was used in the second term of the formula. Remember we have two triangles and the odd ball trapezoid to calculate, and add the results of the formula. Let's compare the results of the three methods. Method one was 32,072.75 square feet. Method two, the trapezoidal rule, was 33,101.50 square feet. Method three, Simpson's one-third rule was 33,206.66 square feet. Hypothetically, method three would be the most accurate. Let's move on now to other and more accurate methods of determining area. Determining area by use of coordinates. Here we have a parcel with coordinates at each corner. Using the coordinates, we will determine the length of the sides of various geometric figures around the perimeter of the parcel and by the methods used in the very first example and arrive at the area of the parcel. Step one, circumscribe the parcel with a rectangle. The sides of the rectangle must be due north and south and east and west. The west side of the rectangle must run through the most westerly corner of the parcel in a north-south direction. The north side of the rectangle must run through the most northerly corner of the parcel in an east-west direction. The east side runs through the most easterly corner in a north-south direction and the south side through the most southerly corner in an east-west direction. The second step is to determine the area large rectangle, 1, 2, 3, 4, from the differences in coordinates. The third step is to calculate the area of triangle A, B, A, rectangle small a, b, small b, 1, the triangle large b, small b, large c, et cetera. Around the perimeter of the parcel. Then determine the area of the parcel. Turn off the video and work the problem using the diagram in your handbook and come back when you're finished. I make the area 43,325 square feet. Let's review the process. The area of the rectangle, 1, 2, 3, 4, take the easting of the most westerly corner, point A, which is 1910, from the easting of the most easterly corner, point E, which is 2,260. This equals 350 feet. Next, take the northern of the most southerly corner, point G, which is 1,500, from the northern of the most northerly corner, point C, which is 1,710. This equals 210 feet. Therefore, the area of rectangle, 1, 2, 3, 4, has 350 feet on one side times 210 feet on the other side equaling 73,500 square feet. Next, calculate the area of triangle, large A, large B, small A. Take the northern of A, 1,600, from the northern of B, 1,670. This equals 70 feet, one leg of the right triangle. Then take the easting of A, 1910, from the easting of B, 1945. This equals 35 feet, the second leg of the right triangle. Therefore, the area of the right triangle, large A, large B, small A, is 1,5 times the base times the height, or 1,5 of 70 times 35 for 1,225 square feet. Calculate the balance of the triangles and rectangles using the same method to determine the length of the legs. When you add all of the areas of these triangles and rectangles together, you should get 30,175 square feet. Take 30,175 square feet from the area of the large rectangle, 1,234, which equals 73,500 square feet, and you will have 43,325 square feet, the area of the parcel. The next method we will use to determine the area of the same parcel with the same coordinate values is known as the coordinate method. There's a rule, and it states, multiply the northing of each corner by the difference between the eastings of the following and the preceding corners, always algebraically subtracting the following from the preceding. One half of the algebraic sum of the resultant products is the area. The formula for this method is shown on the screen and also in your workbook. Use this formula to determine the area of the parcel. Turn off the video and work the problem, and then come back when you're done. That formula is kind of nasty until you've worked it a couple of times. Well, I got 43,325 square feet for the parcel. Let's review the problem and see how I got that answer. You start with any corner, but to keep it simple, as possible, let's start with corner A. From the formula n sub 1 times the quantity e sub n minus e sub 2, let's change that to fit our parcel. So that becomes n sub A times the quantity e sub g minus e sub b. Or the northing of A, 1600, times the quantity, the easting of g 2000 minus the easting of b, 1945, is a plus 55. A plus 1600 times a plus 55 equals a plus 88,000 square feet. Put this in the plus column. Next, n sub 2 times the quantity e sub n minus e sub 3 becomes n sub b times the quantity e sub A minus e sub c, or 1670, the northing of b times the quantity 1910, the easting of A minus 2040, the easting of c. This is a minus 130. A plus 1670 n b times a minus 130, which is the product of e sub A minus e sub c equals a minus 271,000 square feet. Put this in the minus column. Continue around the parcel to point g, which would be ng times the quantity e sub f minus e sub a. Watch the algebraic signs of the product and enter in the proper column, either plus or minus. When you have found all the products, you will have a column of pluses and a column of minuses. Add each column, then take the smaller from the larger, divide the remainder by 2, and you have your area. Simple, huh? The last method we will discuss to determine area is double meridian distance, or DMD. This method uses the latitude and departures of a calculated traverse that determines the coordinates of the angle points of a closed traverse. Here again, we have some rules. Number one, start DMD calculations from the most westerly point in the travers. Two, the DMD of the first course is equal to the departure of the first course. Three, the DMD of the second course through the next to last course is equal to the DMD of the preceding course plus the departure of the preceding course plus the departure of the course itself. Sounds good to me. Algebraic signs must be considered. Four, the DMD of the last course is equal to the departure of the last course with the opposite sign. Five, DMDs are run clockwise. Hint, when working with DMDs, it is best to set up a form to keep track and see what you're doing. The diagram above shows this form. The first column is your line. Under this heading are listed in a column, the lines of the travers, A, B, C, D, et cetera. The second column is labeled latitude. Under this heading, listed in the column, the latitude for the corresponding line opposite line AB lists plus 70 for the latitude plus because it is headed northerly for line CD minus 30 because it is heading southerly. The third column labeled departure. Under this heading, listed in column, the departure for the corresponding line opposite line AB lists plus 35 for the departure plus because it is heading easterly for line EF minus because it is heading westerly. The fourth column lists DMD. Under this heading, listed in column, the DMD for the corresponding line. The fifth column is square feet or double area. Under this heading, you will have two columns, plus and a minus. Under the appropriate column, plus or minus, lists the double area for the corresponding line. OK, turn off the video. Using the information in your workbook, determine the area of the parcel, then turn us back on for a review. By the DMD method, the area of the parcel is 43,325 square feet. Let's review the procedure. The DMD of line AB is plus 35 by rule number two. The double area is the latitude of AB plus 70 times the DMD plus 35 or plus 2450. The DMD of line BC is plus 35, DMD first line, plus plus 35, departure first line, plus 95, departure of the line itself. This equals plus 165. The double area is the latitude of BC plus 40 times the DMD plus 165 or plus 6600. Proceed through the traverse in the same manner, watching the signs. Then add the pluses of the double areas and all the minuses of the double areas. Take the algebraic difference and divide by 2 and you have your square feet. You will note that all the answers for all three coordinate methods came out the same, 43,325 square feet. So it really doesn't matter which system we use of coordinates. They're all as equally accurate. So we always want to use the simplest one that we can. Now we're going to get into the area of volumes. First we'll define volume. It is the amount of material occupying a certain space measured in cubic units of inches, feet, or yards. Now the geometric figures and volume formulas we'll see these figures and formulas appear in the appendix H-1 and H-2 of the Caltrans Survey Manual. This diagram shows a parallel opi-pad, a six-sided solid, all sides are parallelograms, and the opposite sides are parallel. A right rectangular prism, a parallel opi-d with all angles 90 degrees. For either of these two figures, the volume is calculated by multiplying the base times B times H, where B equals the area of the base and H the height. This diagram shows a prism, a solid whose two ends are parallel, similar and equal, and whose sides are parallelogram. The next diagram shows a prism, a solid whose two ends are parallel, similar and equal, and whose sides are parallelograms. A parallel opi-pad is a prism. My god, I think I finally pronounced it right. The volume of a prism is calculated by taking the area of the base times the height. Here we have a practical application of computing volume. This drawing represents a section of proposed roadway. The centerline of the roadway is shown, and cross sections at two stations, A1 and A2, are also shown. To compute the volume in cubic yards, we would calculate the area of the cross sections in square feet by a method discussed in the area segment of this presentation. The method typically used is the coordinate method. The coordinate method formula given previously can be adapted by substituting offset distances and elevations or cuts and fills for the northings and eastings for all of the points on the section that define the polygon. Add the two cross section areas and divide by two to get the average end area. Next, multiply the distance along the centerline between the two cross sections to get the volume in cubic feet. We then have to divide the cubic feet volume by 27 to get the cubic yard value. Now this represents the amount of material in cubic yards that has to be excavated. That concludes the presentation on areas and volumes. However, I would like to take a moment to talk to you about the LSIT or LS examination, whichever you're taking. I would urge you to solve problems with areas or volumes in them, because they are a very economical way of accruing points. I say economical, meaning the time expended for the amount of points given. The one thing you must know are the formulas and also recognize the problem for what it is and what formulas you have to use to solve it. Above all else, you must be accurate in your calculations because this is the exact science portion of surveys rather than the art. So in any of your endeavors in the LS or LSIT, I wish you the very best of luck.