Easements

Abstract

An Easement is a right to use the land of another person. Easements are not possessory interests in land.

Appendices

Appendix

A.1 Introduction

The earlier portions of this book intentionally limit the presentation of mathematics in order for the book to be readable by the widest possible audience. Some readers, however, may wish to dig a bit deeper into the underlying theory. If the reader is a student of land surveying, a scientist, engineer, architect or someone with a technical background, then this supplement will provide a more in-depth treatment of the technical principles presented earlier in this book. Because the focus of the book is boundary surveying, this supplement continues in this vein and concentrates on those issues that are relevant to the discipline. Some areas of surveying, such as construction surveying, earthwork, astronomical observations and topographic surveying, for example, are not particularly relevant to boundary surveying so they will not be discussed.

The mathematics presented in this supplement is straightforward and the author has attempted to provide clearly worded explanations of each subject. Although a background in algebra, geometry, trigonometry and statistics would be helpful, none of these is strictly necessary in order for the reader to work through these materials.

Computers and software have made surveying calculations very quick and easy. This does not relieve a land surveyor from understanding the principles underlying a specific task or calculation. On the other hand, most of us lead busy lives so our available time is limited. This supplement therefore takes a pragmatic approach to the subjects. The material presented is what a boundary surveyor should know—what he or she needs to know. We will not spend much time on abstractions and things that are interesting and nice to know, but really not that essential to a working knowledge of the subject. The subjects presented here have been somewhat constrained by limits on the size of the appendix, so, not every subject has been covered. If there are specific areas that, you, the reader, would like to see covered in future editions, the author would love to hear from you.

So, let’s get started.

A.2 Angles and Trigonometry

An angle is a measurement that describes the geometric relationship between two intersecting lines. Land surveyors in the U.S. use the sexigesimal system of angle measurement. Angles are measured in degrees, minutes and seconds. There are 360° (degrees) in a circle, 60′ (minutes) in a degree and 60″ (seconds) in a minute. As can be seen from Fig. A.1, angles are always measured from a reference line and can be measured to the right (clockwise) or to the left (counterclockwise).

Fig. A.1

Measurement of angles

18.3.1 A.2.1 Adding and Subtracting Angles

When adding and subtracting angles, it is easiest to start with seconds, then minutes, then degrees. In other words, start at the right and move left. For minutes and seconds it is customary to show a maximum value of 59 so if the value is 60 or more, we will subtract 60 and add 1 to the value to the left. For example, let’s add the following two angles:

• 95° 44′ 32″ and 10° 22′ 45″

• First we add the seconds: 32″ + 45″ = 77″. Because this is greater than 59 seconds we subtract 60 seconds: 77″ − 60″ = 1′ 17″.

• Nest we add the minutes: 44′ + 22′ + 1′ = 67′. Because this is greater than 59 min we subtract 60 min: 67 − 60 = 1° 7′.

• Next we add the degrees: 95° + 10° + 1° = 106°.

• Our answer is: 106° 07′ 17″.

The above example assumes that the value of seconds and minutes is less than 120. It is entirely possible that the value of minutes or seconds is substantially greater. This often happens when adding many angles together as in the following example:

$$ \begin{array}{*{20}l} {39^\circ\;14^{\prime } \;45^{\prime \prime } } \hfill \\ {12^\circ\; 02^{\prime } \;56^{\prime \prime } } \hfill \\ {19^\circ \;15^{\prime }\;49^{\prime \prime } } \hfill \\ {54^\circ\;07^{\prime } \;39^{\prime \prime } } \hfill \\ {\overline{{124^\circ\; 38^{\prime } \;189^{\prime \prime } }} } \hfill \\ \end{array} $$

In such cases we just need to recognize that the value is some multiple of 60. In the present example, 189 seconds is greater than 180 seconds (3 * 60 = 180), so we would subtract 180 seconds and add 3 min resulting in a sum of 124° 41′ 09″. We would use the same approach for large values of minutes.

Subtracting angles uses a similar approach. Again, we start from the right. Let’s perform the following subtraction:

• 102° 12′ 22″ − 11° 23′ 36″

First we subtract 36″ from 24″. Because we do not want to end up with a negative number we first need to borrow 60 seconds from the minute column. This will decrease the minutes by one.

• 22″ + 60″ = 82″ so, 82″ − 36″ = 46″

Next we subtract the minutes. Instead of 12′ we now have 11′. Because 23′ is greater than 11′ we need to borrow 60′ from the degrees column and reduce the degrees by one.

• 11′ + 60′ = 71′ so, 71′ − 23′ = 48′

• Next we subtract the degrees. 101° − 11° = 90°

• Our answer is: 90° 48′ 46″

18.3.2 A.2.2 Converting DMS to Decimal Degrees

Most Scientific calculators require decimal degrees in order to produce trigonometric functions. These calculators are usually able to convert degrees-minutes-seconds (DMS) to decimal degrees and to reverse the procedure. Sometimes the function is listed as converting hours-minutes-seconds to decimal hours. If you need to do it on a calculator without a conversion function, you can easily do so. Consider the angle: 34° 29′ 45″. Start with the seconds. We want to convert the seconds into decimal minutes. There are 60 seconds in a minute so:

$$ 45^{\prime\prime} = \frac{45}{60} = 0.750\;{ \hbox{min} } $$

Next we add the decimal minutes to the minutes and divide the result by 60 to get decimal degrees:

$$ Decimal\;portion\;of\;degrees = \frac{{29^{\prime } + 0.750^{\prime } }}{60} = 0.4958^\circ $$

Finally, we add the decimal portion of the degrees to the degrees to get decimal degrees.

$$ Decimal \;degrees = 34.4958^\circ $$

18.3.3 A.2.3 Converting Decimal Degrees to DMS

Suppose you have decimal degrees and want to convert it to degrees, minutes and seconds. Using the same angle as above, let’s convert 34.4958° into DMS. First, multiply the decimal part by 60 to get minutes:

$$ Minutes = 0.4958*60 = 29.748^{\prime} $$

We now have 29 min and need the seconds. So, multiply the decimal part of the minutes by 60 to get seconds.

$$ Seconds = 0.748*60 = 44.88^{\prime\prime}\;rounded\;off\;to\;45^{\prime\prime} $$

Our result is 34° 29′ 45″.

18.3.4 A.2.4 Trigonometric Functions

Trigonometric functions enable us to calculate relationships using angles. The best way to understand these functions is by means of the unit circle shown in Fig. A.2. A unit circle has a radius of 1 unit. It doesn’t matter what the units are. It could be 1 foot or 1 m. If we draw a line between the center of the circle and the circumference it will have a length of 1. In the figure, we measure the angle of the line from the horizontal axis in a counterclockwise direction. In this case the angle is 60°. If we draw a vertical line from the end of our 1 unit line down to the horizontal X axis, and accurately measured this line it would have a length of 0.87. We call this value a sine. If we measure from the center of the circle to this line along the horizontal axis it will measure 0.50. We call this value a cosine. Sine is abbreviated sin and cosine is abbreviated cos.

Fig. A.2

Sin and cosine

As we rotate our unit line, the sines and cosines will obviously change. Figure A.3 shows how rotating our line to an angle of 35° changes the sine to 0.57 and the cosine to 0.82.

Fig. A.3

Sin and cosine

If we were to rotate our unit line to the left so that the angle became 90° and the line was vertical; the sine would be 1 and the cosine 0. If we were to rotate it to the right until the angle became 0 and the line was horizontal, the sine would be 0 and the cosine 1. We can therefore see from our unit circle that both the sine and cosine will depend on the angle and will vary between 0 and 1. We can also see from this example that if the angle is 45° the sine and cosine will be the same number (0.7071). An additional function to know is the tangent (tan) which is simply the sin divided by the cosine.

So, of what value is this? If we were to use a circle with a radius of any real number, we could use the sins and cosines derived above as factors to calculate the X and Y values. Using the example in Fig. A.3, if our unit line were actually 325 ft. long, the X value would be:

$$ 325* \sin 35^\circ = 186.41\,{\text{ft.}} $$

And our Y value would be:

$$ 325* \cos 35^\circ = 266.22\,{\text{ft.}} $$

We have seen that every angle has a unique sine and cosine. This means that every sine and cosine must have a unique angle. If the sine of 35° is 0.57, then the angle associated with the sine value of 0.57 must be 35°. In order to differentiate the two cases we say that the arcsine of 0.57 is 35° and the arccosine of 0.82 is 35°. These are abbreviated asin and acos. The exponential notation is a more common way to express these values. So, arcsine is written sin⁻¹ and arcos is written cos⁻¹.

18.3.5 A.2.5 Right Triangles

A triangle is a geometric figure having three sides and three interior angles. A right triangle means that one of the interior angles is 90°. In any triangle, the sum of the three interior angles must equal 180°. So, in the case of a right triangle where one of the angles is 90°, the sum of the other two must also equal 90°.

The most basic, and probably the most commonly used trigonometric solution, in surveying, is a right triangle. Right triangles form the basis of coordinate geometry, which is commonly used in boundary surveying calculations. Right triangles are also used to calculate horizontal distances from slope distances. The reader is encouraged to memorize the three solutions and become adept at solving right triangles. We used right triangles in our unit circles above to demonstrate how sins and cosines are derived.

A right triangle is shown in Fig. A.4. The convention is that angles are noted with upper case letters and the length of the sides with lower case letters. Angle C is 90°. The sides are sometimes given names. The hypotenuse is opposite the 90° angle. It is always the longest side. The side next to angle θ is called adjacent. The side opposite angle θ is called, believe it or not, opposite.

Fig. A.4

Right triangle

The following three equations will allow you to solve any right triangle:

$$ \sin A = \frac{a}{c} $$

(A.1)

$$ \cos A = \frac{b}{c} $$

(A.2)

$$ \tan A = \frac{a}{b} $$

(A.3)

The above equations can be rearranged as necessary in order to provide the needed solution. If any two of the variables are given, a right triangle can be solved.

Let us look at a few examples of solutions. One common solution illustrated in Fig. A.5, is when a slope distance is measured with a total station and we need the horizontal distance. In addition to the slope distance, the total station provides us with a vertical angle. Normally vertical angles are measured from the zenith (zero is directly overhead), so we will need to subtract the vertical angle from 90° to get the angle from the horizontal plane.

Fig. A.5

Slope distance measured with an EDM

Given: Slope Distance = 442.58 and Zenith Angle = 78°

First, we calculate the vertical angle from horizontal: 90° − 78° = 12°.

Looking at Figs. A.5 and A.6, the slope distance would be side c and the vertical angle, as measured from the horizontal, would be angle A. Imagine the total station at point A measuring to a prism at point B. Line A to C represents the horizontal plane. So, we have c and angle A, and we need distance b, the horizontal distance.

Fig. A.6

Calculate horizontal distance from slope distance

From the equations above we select the one containing all of the three variables: A, b, c:

$$ \cos A = \frac{b}{c} $$

(A.4)

Because we need to solve for b we rearrange the equation as follows:

$$ b = c*\cos A $$

Substituting our values in the variables:

$$ b = 442.58*\cos 12^\circ \left( {0.9781} \right) = 432.909\, {\text{ft.}} $$

Recall that the horizontal distance is always smaller than the slope distance. We could also solve for the vertical distance, which would be side a. Because we now have sides b and c, we could use either Eq. A.1 or Eq. A.3.

Here is a second example. You need to measure the distance from the instrument to a utility pole but the EDM on the total station stopped working. You measure the angle to the top of the pole (15°) and know the height of the pole (30 ft.).

Looking at Fig. A.4, you know A and a. First select the correct equation.

$$ \tan A = \frac{a}{b} $$

Rearrange to solve for b.

$$ \begin{aligned} b & = \frac{a}{\tan A} \\ b & = \frac{30}{\tan 15} = \frac{30}{0.268} = 111.96 \,{\text{ft.}} \\ \end{aligned} $$

Another useful equation to know when calculating right triangles is:

$$ c^{2} = a^{2} + b^{2} $$

(A.5)

Knowing the length of any two sides will allow the third side to be calculated. Using Fig. A.6 as an example, assume that we have sides a and b, and we want to solve for side c. We would rearrange Eq. A.5 as follows:

$$ c = \sqrt {a^{2} + b^{2} } \quad {\text{so}}\quad c = \sqrt {92.02^{2} + 432.91^{2} } = { 442}. 5 8 $$

Let’s try another example using Fig. A.6. Assume that we know sides c and b and need to calculate side a. Rearrange Eq. A.5 as follows:

$$ \begin{aligned} a^{2} & = c^{2} - b^{2} \\ a & = \sqrt {c^{2} - b^{2} } \\ a & = \sqrt {442.58^{2} - 432.91^{2} } = 9 2.0 2\\ \end{aligned} $$

18.3.6 A.2.6 Oblique Triangles

Oblique triangles can be more difficult to solve than right triangles. As with right triangles, it is necessary to have three pieces of information in order to solve an oblique triangle. However, with right triangles we already know that one of the angles is 90° so there are really only two unknowns. One possible solution is to break the triangle into two right triangles. Figure A.7 shows an oblique triangle with a line running from B to the base which divides the oblique triangle into two right triangles. However, we shall see that with oblique triangles, it is not always possible to divide and conquer.

Fig. A.7

Oblique triangle

The equations governing the solution of oblique triangles, known as the law of sines, are:

$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$

(A.6)

The law of sines can be used when two angles and one side are known. Let us consider the example shown in Fig. A.8.

Fig. A.8

Oblique triangle example

Although this triangle could be solved using two right triangles we will use the law of sines. We know angle A, side a, and angle C, so:

$$ \frac{a}{\sin A} = \frac{c}{\sin C} $$

Rearranging:

$$ c = \frac{a}{\sin A}*\sin C $$

(A.7)

$$ c = \frac{103.18}{\sin 35^\circ }*\sin 70^\circ = \frac{103.18}{0.574}*0.940 = 168.97 $$

We could then solve for side b by continuing to use the law of sines. Recall that in a triangle, the sum of the interior angles is always 180° so angle B can be easily calculated.

$$ B = 180 - \left( {35 + 70} \right) = 75^\circ $$

If we wanted to solve the previous example using two right triangles we could do so by drawing a line from B to a point on side b such that the line would form a 90° angle with side b. We would then have two known variables: Angle C and side a. Using this information we could calculate the length of our newly created line. Once this length was known, we could use angle A and this length to calculate the remaining angles and sides.

Oblique triangles have an ambiguous case where two sides and the angle opposite are known. A number of solutions are possible but they depend on whether the triangle is acute or obtuse. If the angle opposite is acute five possible solutions exist. If the angle opposite is obtuse there are three possible solutions. It is rarely necessary to know how to solve these in boundary surveying, so they are beyond the scope of this chapter. As we will see further on, nearly all of the calculations which need to be performed for the purposes of a boundary survey will be done using coordinate geometry. If the reader wishes to pursue these solutions there are many texts and online resources relating to trigonometry which deal with the subject at length.

A.3 Bearings

Let us first review bearing quadrants. Referring to Fig. A.9, we see there are four bearing quadrants. Each quadrant is named after the two primary directions such as North and East or South and East. Bearing directions are always measured from either the north direction or the south direction. So, as shown in Fig. A.9, a bearing of N15°E is simply an angle measured from north, 15° to the right. N15°W would be in the NW quadrant measured to the left from north. A bearing of S68°E is an angle measured to the left from south.

Fig. A.9

Bearing quadrants

Bearings are generally expressed as Degrees, Minutes and Seconds as in N34° 22′ 12″E. In order to reverse the direction of a bearing we need only specify the opposite quadrant. For example a bearing having a direction of N22°E can be reversed by relabeling it S22°W.

The basis of the bearing system in use for any particular project was discussed at length in the main portion of this book. However, as a reminder, bearings can be based on true north, grid north, magnetic north or an assumed north.

18.4.1 A.3.1 Calculating a New Bearing from an Existing Bearing and an Angle

Now, let us take a look at some calculations using bearings. When calculating a new bearing from a bearing and an angle, we obviously always need a bearing to start with. When working with bearings and angles, there are a number of possible solutions, depending on the magnitudes and of the bearings and angles. We will start with the easiest example first.

In Fig. A.10 we have a starting bearing of N09° 26′ 32″E. We want to add the angle 65° 41′ 58″. The angle is to the right of the bearing. The calculation is straightforward.

$$ N09^\circ \;26^{\prime }\; 32{\text{E}} + 65^\circ\; 41^{\prime } \;58 = N75^\circ\; 08^{\prime }\; 30^{\prime \prime }E $$

Fig. A.10

Calculate a new bearing from a bearing and angle

Notice that the resulting bearing is less than 90°. If it were greater than 90° it would place the bearing in the SE quadrant.

In the next example shown in Fig. A.11, we start with a bearing of N26° 22′ 34″E and add an angle to the left of 65° 41′ 58″. We notice that the angle is greater than the bearing so the new bearing has to be in the NW quadrant. In this case we will want to subtract the 26° 22′ 34″ from the angle in order to get the new bearing.

$$ 65^\circ\; 41^{\prime } \;58 - 26^\circ\; 22^{\prime }\; 34 = N39^\circ\; 19^{\prime } \;24^{\prime\prime }W $$

Fig. A.11

Calculate a new bearing from a bearing and angle

The third possible solution is illustrated in Fig. A.12. In this case we start with a bearing of N75° 11′ 39″E and add an angle to the right of 65° 41′ 58″. Adding these together we have:

$$ 75^\circ \;11^{\prime } \;39^{\prime\prime } + 65^\circ \;41^{\prime }\; 58 = 140^\circ\; 53^{\prime }\; 37^{\prime\prime } $$

Fig. A.12

Calculate a new bearing from a bearing and angle

We notice that starting from north we have a total angle of 140° 53′ 37″. Because the resulting angle is greater than 90° but less than 180°, we know that the new bearing must be in the SE quadrant. We therefore calculate our bearing by subtracting the angle from 180°:

$$ 180^\circ - 140^\circ\; 53^{\prime } \;38^{\prime\prime } = {\textrm{S}}39^\circ\; 06^{\prime } \;23^{\prime\prime }{\textrm{E}} $$

The next situation is where the new bearing ends up in an opposite quadrant (Fig. A.13). Here we have an initial bearing of S49° 11′ 56″E and an angle to the right of 163° 28′ 28″. We know that the new bearing is in the NW quadrant because the angle is between 139° and 229°. The 139° is the sum of 90° + 49° 11″ 56″ = 139° (more or less). The number of degrees in the SW quadrant is 90°. The 229° is the sum of 180° + 49° 11′ 56″ = 229° (more or less). The sum of the degrees in the SW and NW quadrants is 180°.

Fig. A.13

Calculate a new bearing from a bearing and angle

In this example, an easy way to calculate the bearing is to reverse the quadrant of the SE bearing so that it becomes NW. We are simply extending the bearing line into the NW quadrant so that the bearing becomes N49° 11′ 56″W. Because these two bearings are exactly 180° apart, we can subtract the angle and add the difference to the NW bearing.

$$ \begin{aligned} 180^\circ - 163^\circ\; 28^{\prime } \;28^{\prime{\prime }} & = 16^\circ \;31^{\prime }\; 32^{\prime\prime } \\ {\text{N}}49^\circ\; 11^{\prime }\; 56{\text{W}} + 16^\circ\; 31^{\prime }\; 32 & = {\text{N}}65^\circ\; 43^{\prime }\; 28^{\prime{\prime }}{\text{W}} \\ \end{aligned} $$

If you are confused by this procedure, it is always possible to plot the lines on graph paper or draw them in a CAD program until you are comfortable working with the various solutions.

18.4.2 A.3.2 Calculating an Angle from Two Bearings

Here we discuss calculating the angle between two bearings. We can use the same examples we used above.

In Fig. A.14 we are given two bearings in the NE quadrant: N75° 08′ 30″E and N09° 26′ 32″E. The angle is calculated by subtracting the smaller bearing from the larger bearing.

$$ 75^\circ\; 08^{\prime } \;30^{\prime{\prime }} - 09^\circ \;26^{\prime } \;32^{\prime{\prime }} = 65^\circ\; 41^{\prime }\; 58^{\prime{\prime }} $$

Fig. A.14

Angle to bearing

Next, we consider two bearings in adjacent N or S quadrants as in Fig. A.15. Here we simply add both bearings together to calculate the angle.

Fig. A.15

Angle to bearing

$$ 26^\circ\; 22^{\prime }\; 34 + 39^\circ \;19^{\prime }\; 24 = 65^\circ\; 41^{\prime } \;58^{\prime{\prime }} $$

The third example is in Fig. A.16. Here we need to add both bearings together and subtract the sum from 180°.

$$ 75^\circ\; 11^{\prime } \;39 + 39^\circ \;06^{\prime } 23 = 114^\circ\; 18^{\prime }\; 02^{\prime{\prime }} $$

$$ 180^\circ - 114^\circ \;18^{\prime }\; 02^{\prime{\prime }} = 65^\circ\; 41^{\prime } \;58^{\prime{\prime }} $$

Fig. A.16

Angle to bearing

Our last example uses two bearings from opposite quadrants in Fig. A.17. We reverse the bearing S49° 11′ 56″E so that it becomes N49° 11′ 56″W. Now we can simply subtract it from the second bearing.

$$ 65^\circ\; 43^{\prime } \;28^{\prime{\prime }} - 49^\circ \;11^{\prime } \;56^{\prime{\prime }} = 16^\circ \;31^{\prime } \;32^{\prime{\prime }} $$

Fig. A.17

Angle to bearing

We then subtract the angle from 180° and we have our angle.

$$ 180^\circ - 16^\circ \;31^{\prime } \;32^{\prime{\prime }} = 163^\circ \;28^{\prime } \;28^{\prime{\prime }} $$

18.4.3 A.3.3 Azimuths

Azimuths are angles measured from either north or south. In most cases they are measured clockwise from north, but in some instances they are measured clockwise from south.

As can be seen from Fig. A.18, azimuths are measured from 0° to 360°. There are no quadrants as with bearings. With a bearing, if you need to calculate the reverse direction it is only necessary to change the quadrant designation. The numbers do not change. When using azimuths, it is necessary to add or subtract 180°. The reverse azimuth direction is sometimes called the back azimuth. If the azimuth is greater than 180°, subtract 180°. If the azimuth is less than 180°, add 180°. For example, you need to calculate the back azimuth of 89°. It is less than 90°, so you would add 180°, giving 269°.

Fig. A.18

Azimuths with 0° at north

18.4.4 A.3.4 Deflection Angles

Deflection angles are angles measured from an extension of the previous line.

Deflection angles may be measured to the right or to the left. Figure A.19 shows a deflection angle to the right.

Fig. A.19

Deflection angle to the right

Figure A.20 shows a deflection angle to the left. In both of the preceding images the deflection angles deflect from an extension of line 1–2 and the deflection angle is 15°.

Fig. A.20

Deflection angle to the left

18.4.5 A.3.5 Angular Closure of a Parcel or Traverse

Sometimes we need to check the mathematical correctness of the deed description of a parcel of land described using interior angles. In other cases we need to adjust a closed traverse. In both cases we have a geometric figure having some number of sides and the same number of interior angles. Sometimes, particularly when working with traverses, some of the angles will be exterior and some interior. In this case, we will need to convert all angles to interior angles before proceeding with the angular closure calculation.

In any closed figure, the sum of the interior angles is given by the equation:

$$ Sum\; of\;Angles = \left( {n - 2} \right)*180^\circ $$

(A.8)

where n is equal to the number of angles. We have already learned that the sum of the interior angles in a triangle is 180°. We can prove this using the above equation:

$$ Sum \;of\;Angles = \left( {3 - 2} \right)*180^\circ = 180^\circ $$

Let us now check a deed description by summing the angles. As an example, let’s say that you plan to perform the boundary survey pictured in Fig. A.21.

Fig. A.21

Parcel defined by angles

The image shows the interior angles that were used to define the boundary lines in a deed description. We can check to see if the angles close using our equation. There are a total of four angles so:

$$ Sum \;of\;Angles = \left( {4 - 2} \right)*180^\circ = 360^\circ $$

The sum of the interior angles in our parcel should total 360°. Starting with angle A and adding the angles we have:

$$ \begin{array}{*{20}l} {98^\circ\; 40^{\prime } \;35^{\prime\prime } } \hfill \\ {104^\circ \;08^{\prime } \;14^{\prime\prime } } \hfill \\ {84^\circ \;57^{\prime } \;48^{\prime\prime } } \hfill \\ {72^\circ\; 13^{\prime }\;23^{\prime\prime } } \hfill \\ {\overline{{360^\circ\; 00^{\prime } \; 00^{\prime\prime } }} } \hfill \\ \end{array} $$

So, the sum of the angles is correct and our angular closure is perfect.

When a surveyor runs a traverse, the angles are not likely to close perfectly because, even with the most careful work, there will always be small inconsistencies. Consider the traverse illustrated in Fig. A.22.

Fig. A.22

Traverse angular closure

There are a total of 7 angles so the sum of the angles should be:

$$ Sum \;of\;Angles = \left( {7 - 2} \right)*180^\circ = 900^\circ $$

We add all the interior angles (Table A.1):

Table A.1 Angles as measured in the field

Instead of 900° the sum is 42 seconds off (60″ − 18″ = 42″). If the angles closed perfectly, the sum should read 899° 59′ 60″ which is the same as 900°.

Before we can proceed with adjusting our traverse, we should adjust the angles so that they close perfectly. One way to do this would be to distribute the 42 seconds error into each angle equally. Another possibility would be to place the error more heavily into one or more of the angles. This is where the experience of the surveyor really comes into play. If experience in the field indicated that one of the setups was on soft ground and there was a high likelihood that the angle was off, the surveyor could go back to the field and measure the angle again. When traversing, surveyors often “double” the angles—turn the angle twice (or even more times) and average the results. This gives greater precision. If the error in a particular set is greater than the other sets, this may be an indication of a lack of precision in the measurement. If so, this angle may be a good candidate for adjustment. Angular errors often occur when the angle is at the intersection of a short line and a long line. For best accuracy and precision, traverses should have the longest lines possible. If the surveyor is relatively certain that the problem is with one particular angle or a couple of angles, the error could be distributed only in these angles, rather than distributing the error equally among all of the angles.

In our example, we will assume that there were no suspect angles and will distribute the error equally into each of the seven angles. So, 42 seconds divided by 7 angles means that the value each angle will need to be increased by 6 seconds. The results are shown in Table A.2.

Table A.2 Adjusted angles

A.4 Coordinate Geometry

Understanding coordinate geometry is fundamental to performing boundary surveying calculations. During the deed research portion of a survey project, we may want to plot the metes and bounds of deed descriptions. We can use coordinate geometry to do this and to make a determination if the parcel geometry closes mathematically. When we complete our field traverse and have locations of physical evidence, we will use coordinate geometry to plot the points and determine the relationships between record evidence and physical evidence. When we calculate the boundaries of our parcel, we will use coordinate geometry to plot them, perform a lot closure and calculate the area of our parcel.

When using a total station with a data collector, we have the capability of collecting data in three dimensions. To do so however, requires more time during the field portion of the survey because we must constantly keep track of the instrument height and prism height. If using GPS, three dimensional data will automatically be collected. For most boundary surveys, only two dimensional data will be required, because slope distances will be converted to horizontal distances. We will therefore confine our discussion to two dimensional coordinate geometry. Should the need for elevation arise, it can be easily accommodated, but this decision will need to be made prior to starting field work, as instrument and prism heights for every setup will need to be recorded.

18.5.1 A.4.1 The Survey Coordinate System

The first step in beginning our calculations will be to decide on the bearing system to be used. This may have already been decided during the field work phase of our project. For example, if we used GPS to tie into the state plane coordinate system, this will likely be our preferred choice. If we simply used a compass to measure the magnetic direction of one or more of our traverse lines, we will use a magnetic bearing system. If our survey tied into a previously performed survey which made use of a true north bearing system, we may want to adopt that system.

The second decision will be which coordinate system to use. If we are tied into state plane, the decision is easy; we will use the state plane coordinate system. For most small boundary surveys, such as a retracement survey of a house lot it, will usually be more economical just to use an assumed coordinate system.

The coordinate systems we are interested in exploring are based on the Cartesian coordinate system. It is a two dimensional coordinate system having two axes 90° apart. These axes are traditionally labeled as X for the horizontal axis and Y for the vertical axis. Values increase up and to the right.

In surveying, the Y axis is labeled the Latitude (Northing) and the X axis the Departure (Easting). Latitudes increase in a northerly direction. Departures increase in an easterly direction.

Let us consider a simple line as shown in Fig. A.23. The coordinate system is an assumed system beginning with N5,000, E5,000. The actual values of the starting coordinates do not matter. We could have chosen N100, E100. However, in order to avoid negative coordinate values it is best to pick a large enough starting value, so that all coordinates will remain positive. In a real survey, we may find that, although we thought our field work was nearly complete, we need to extend a traverse down a road 3,000 ft. to the west in order to tie into a monument which has just been discovered. If we had chosen our starting coordinates of N1,000, E1,000 we would end up with negative departure values in the −2,000 range. There is nothing inherently wrong with negative coordinate values, but it is easy to make a mistake by forgetting to include the minus sign. Insuring that all values are positive makes things a bit easier. For small lot surveys, 5,000 is a good choice. For larger tracts you could start with a larger value such as 10,000.

Fig. A.23

Coordinates of a line

In our example, we have chosen to assign Point 1 the starting value of N5,000, E5,000. Our first (and only) line begins at Point 1 and runs N26° 33′ 54″E, 134.16 ft. to point 2. The purpose of this example is to calculate the coordinates of Point 2.

We have already discussed the importance of right triangles in surveying. Right triangles are an essential part of coordinate geometry. We will use a right triangle to calculate the latitude and departure of our line. This will enable us to calculate coordinates for point 2.

Figure A.23 shows the latitude and departure of our line. The latitude and departure and the survey line form a right triangle. In Fig. A.24 we have taken the right triangle and transposed it into the familiar orientation for clarity. In this case, we are given side c (134.16 ft.) and angle A (26° 33′ 54″). Notice that angle A was taken from our bearing N26° 33′ 54″E. If we reverse the quadrant of the bearing to the SW quadrant we have S26° 33′ 54″W. Because our 120.00 foot line runs exactly north and south, the bearing is merely the angle between the north–south grid line and our survey line.

Fig. A.24

Using a right triangle to calculate latitude and departure

We want to solve the triangle for side b and side a. Let’s solve for side “a” first. We are given angle A and side c so we will want to use Eq. A.1:

$$ \sin A = \frac{a}{c} $$

Rearranging:

$$ a = c*\sin A $$

Substituting values:

$$ a = 134.16*\sin(26^\circ \;33^{\prime } \;54^{\prime{\prime }} ) = 60.00 $$

Now, let’s solve for side “b”. We will use angle “A” and side “c” so the following equation will be used:

$$ \cos A = \frac{b}{c} $$

Rearranging:

$$ b = c*\cos A $$

Substituting values:

$$ b = 134.16*\cos(26^\circ\; 33^{\prime } \;54^{\prime{\prime }}) = 120.00 $$

If we have a large survey with many courses and distances, using right triangles, as we did above, to calculate latitudes and departures, would be very cumbersome and time consuming. It is much more convenient to perform our coordinate calculations using a tabular format.

Looking at Fig. A.23, we see that side “b” of our right triangle is the latitude and side “a” is the departure. We can use this information to come to the following conclusions:

$$ \begin{aligned} Latitude & = Distance*\cos(Bearing) \\ Departure & = Distance*\sin (Bearing) \\ \end{aligned} $$

Using the line shown on Fig. A.23, we can create a table as in Table A.3. In this table, you can see how the start coordinates, the latitude and departure and the end coordinates are calculated and arranged. It should be apparent that using such a table is much simpler and more convenient than figuring out right triangles for each course.

Table A.3 Latitudes and departures used to calculate coordinates

Notice that the latitudes and departures are labeled + and − so that we will know whether to add or subtract them from the coordinate of the previous point. If, as in our example, the line runs North–East, both values will be positive and both the North and East coordinates of the point being calculated will increase. If our line ran North–West, the north coordinate would increase but the East coordinate would decrease.

Now that we have an understanding of basic coordinate geometry, let us take a look at a slightly more complex example. The example shown in Fig. A.25 is a parcel of land having 4 boundaries.

Fig. A.25

A parcel of land with four boundaries

The coordinate table is shown in Table A.4. Notice that north latitudes increase the values of the north coordinates and south latitudes decrease their values. Notice that east departures increase the values of east coordinates and west values decrease their values. Also notice that the sums of the north and south latitudes are the same, but with opposite signs to they add up to zero. This is because we started at Point 1 and returned exactly to Point 1. The same is true for the sums of the east and west departures. Also notice that the coordinates for Point 1 are exactly the same at the beginning and end. This tells us that the boundaries of this parcel close perfectly. Said in another way, the closure error is zero.

Table A.4 Latitudes and departures

18.5.2 A.4.2 Inversing—Calculating the Bearing and Distance Between Two Points

Now that we are comfortable calculating coordinates from bearings and distances, consider the parcel shown in Fig. A.26. Let’s assume that we do not know the bearing and distance of the line from 1 to 2, shown as a dashed line. However, we do know the coordinates for points 1 and 2. We need to calculate the bearing and distance between these points. This procedure is called an inverse. In our example, we want the direction of the line to start at 2 and run to 1.

Fig. A.26

An inverse

We first need to calculate the latitude and departure of the line from the given coordinates. Table A.5 shows a table containing the coordinates of points 1 and 2. Because we want the direction of the line to run from 2 to 1 we need to be aware of whether the coordinates increase or decrease from 2 to 1. The north coordinate for Point 1 is larger than the coordinate for point 2, so the line must run in a northerly direction. The east coordinate for Point 2 is larger than the east coordinate for Point 1 so the line must run in an easterly direction. The plot of the line in Fig. A.26 visually confirms this.

Table A.5 An inverse

We calculate the values for the latitude and departure by taking the difference between the coordinates. These are recorded in the proper columns in the table, remembering that if the latitude increases it is recorded in the north column. If it decreased, we would record it in the South column. Similarly, if the Departure increases we record it in the East column and if it decreases we put it in the West column. Figure A.27 illustrates the calculation in the form of our familiar right triangle.

Fig. A.27

Right triangle used to visualize an inverse

Next, we need to calculate the bearing and distance of the unknown line. We know “a” and “b” so we can calculate angle A. Referring to our right triangle equations we choose the following:

$$ \tan A = \frac{a}{b}\,\,or, $$

$$ \tan\;Bearing = \frac{Departure}{Latitude} $$

(A.9)

Substituting:

$$ \tan \;Bearing = \frac{18.701}{137.953 } = 0.1356 $$

Now that we have the tangent of the bearing, we need the bearing itself, so we use the inverse tangent (tan⁻¹) to find it. Calculators and spreadsheets will return an angle in decimal degrees. In this case the decimal angle is: 7.7200°, which converts to 7° 43′ 12″. We have already determined that the bearing runs northeast from point 2 to point 1 so our bearing is N7° 43′ 12″E.

Now that we know the angle, it is a simple matter to calculate the distance.

From Eq. A.5 recall that:

$$ c^{2} = a^{2} + b^{2} $$

So:

$$ c = \sqrt {a^{2} + b^{2} } $$

Applying the previous equation to latitude and departures:

$$ distance = \sqrt {latitude^{2} + departure^{2} } $$

(A.10)

Substituting:

$$ distance = \sqrt {137.953^{2} + 18.701^{2} } = 139.21 $$

Using the methods described above we can calculate coordinates for all of our survey points. We can also calculate the bearing and distance of a line if we know the beginning and end coordinates. If we needed to work with elevation information, this would require adding a third dimension to our coordinate system. The principles used in the calculations would be identical to those already covered. We essentially would take our coordinate grid and rotate it 90° vertically, so that, in addition to an X and Y axis we would have a Z axis.

A.5 Traverse Adjustment

After a traverse has been run in the field it will be necessary to adjust that traverse. Some data collector software will allow this to be done in the field. Many surveyors, however, prefer to perform this phase of the survey in the office where they have access to a desktop computer with a large screen. Offices usually have air conditioning and heat so this is sometimes an added incentive to leave the field in favor of the office.

All traverses will contain some error, no matter how carefully the work was performed. Errors can fall into a number of categories. One category consists of blunders or mistakes. These can be caused by writing wrong numbers into a field book. Numbers are sometimes transposed or misread. For example 199.98 might mistakenly be written as 198.99. Sometimes measurements are called out over a walkie-talkie and misunderstood. Although relatively uncommon, the memory of a data collector could become corrupted, or a bad cable connection between a data collector and the instrument might cause a corrupted data transfer. Data collectors have substantially reduced blunders compared to the days when every measurement had to be written in a field book. Normal traverse adjustment procedures will not correct blunders.

Other types of errors are systematic errors. This type of error is not random but occurs as a result of some uniform process such as an instrument that is not adjusted properly, a measuring tape that is defective or an instrument person who consistently sights a target incorrectly. Systematic errors are generally not corrected by traverse adjustment.

Random errors can result from a lack of precision in making measurements or in setting up tripods or tribrachs. These errors are usually small and can result in positive or negative errors so that they might cancel each other out. For example, a tripod is set up 0.01′ to the right of point 2. On the next setup the tripod is set up 0.02′ to the left of point 2. Because surveyors work outdoors, sometimes in inclement weather and on difficult terrain, random errors are very common in land surveying. We adjust our traverses and measurements primarily to reduce the effects of random errors.

There are several methods which can be used to adjust a traverse. One is the Compass Rule. This method assumes that both the angles and distances were measured with similar precision. The compass rule works well when modern instruments, such as total stations, are used, so the method is widely favored for traverse adjustment.

The Transit Rule is another method. This method assumes that angles were measured with greater precision than distances. The Crandall Method is another option where the angles are adjusted first and held fixed while the distances are adjusted using a least squares method. It is a relatively time consuming method.

Another method of adjustment is the Least Squares method of adjustment. This is a more difficult and time consuming method, and before computers it was used primarily for large and complex control traverses where highly precise results were needed. The advent of computers makes the least squares method more commonly used than it was in years past. Most land surveyors now have software that will perform least squares adjustments, and because it is likely to provide a more refined adjustment it is the preferred method for surveys in which a high order of precision is desired. The least squares adjustment uses the theory of probability to determine the statistically most probable coordinate location for each point in a network. The adjustment provides a statistic best-fit for each point. A least squares adjustment has the additional advantage of providing statistics which tells the surveyor something about the degree of confidence of the calculated position of each point. The surveyor can then accept or reject the adjustment.

A final option is to adjust the traverse based on the surveyor’s knowledge of likely sources of error. In this method the error would be placed only in those angles and distances known or believed to have errors. The remaining angles and distances would not be adjusted.

Whatever method is used, the goal is to end up with a traverse that closes perfectly—at least on paper. Because of its long history of use by boundary surveyors and its relative simplicity, our examples in this book will use the compass rule.

18.6.1 A.5.1 Calculating the Error of Closure

Consider the traverse shown in Fig. A.28. This figure contains the raw data which was measured in the field. Because the following is just an example to illustrate the adjustment of latitudes and departures, the angles were not adjusted. However, if this were a real traverse, it would be best to adjust the angles prior to adjusting the traverse. In this exercise all of the angles have already been converted to bearings.

Fig. A.28

Raw data

The data has been compiled in Table A.6 to show the latitudes and departures and the coordinates.

Table A.6 Traverse raw data table

Notice that the closing point is numbered Point 8, not point 1. Although the survey began at point 1 and returned to point 1, small errors in the measurements have caused the ending point to be in a slightly different location than the point of beginning. In order to avoid confusion between the beginning and end points it is usual to assign these points different numbers. Notice that the beginning and ending coordinates are slightly different. This difference represents the error of closure.

The error of closure can be seen by comparing the north and south latitudes (−0.038) and the east and west latitudes (−0.086). This same error is obtained by taking the difference between the starting and ending coordinates. This makes perfect sense because the coordinates are calculated using the latitudes and departures. Comparing the two gives a good check of the math.

We can calculate the bearing and distance of the error by inversing between Point 8 and Point 1. We are already familiar with how to perform an inverse using Eq. A.9:

$$ \tan A = \frac{Departure}{Latitude} $$

Substituting:

$$ \tan A = \frac{ - 0.086}{ - 0.038} = 2.2632 = 66^\circ\; 09^{\prime } \;41^{\prime{\prime }} $$

As a check, if we look at the coordinates for Point 8 and Point 1 we see that both the north and east coordinates for Point 1 are larger than for Point 8. This means that Point 1 must be northeast of Point 8, so the bearing from Point 8 to Point 1 must be pointing northeast. We also notice that the error of the departure is greater than the error of the latitude so the bearing must be greater than 45° (Fig. A.29).

Fig. A.29

Error of closure close-up

Next we calculate the distance between the points using Eq. A.10:

$$ distance = \sqrt {latitude^{2} + departure^{2} } $$

Substituting:

$$ distance = \sqrt {0.038^{2} + 0.086^{2} } = 0.094^{\prime } $$

18.6.2 A.5.2 Ratio of the Error of Closure

The error of closure is usually expressed as a ratio of the error to the total distance traversed. The ratio is always expressed with the numerator as 1 so the ratio will be 1 foot in X feet.

$$ Ratio \frac{error\; of\; closure}{total \;distance \;traversed} $$

So, in our example the error is 0.094 ft. and the total length of the traverse is 600.07 ft. so, the ratio of Error is:

$$ Ratio \frac{0.094}{600.07} = 0.000157 $$

Dividing both the numerator and denominator by the error gives the ratio:

$$ \frac{1\,foot}{ 6 , 3 8 3 {\,{\text{ft.}}}} $$

Hint: after dividing the error by the distance traversed, you can use the 1/x (inverse) key on your calculator to do the division. Sometimes the ratio is expressed to the nearest 100 units so in our case the ratio would be 1 foot in 6,400 ft.

18.6.3 A.5.3 Adjusting the Traverse

The next step is to adjust the traverse. Because the compass rule is one of the most widely used rule for adjusting traverses, we will use it here. We have already calculated the errors in latitude and departure. We will now adjust each latitude and departure so that our traverse closes perfectly.

The compass rule distributes the error to latitudes and departures in proportion to the length of the line. The following equation shows how this is done:

$$ Latitude\; Correction\; Line\; X1 - X2 = \frac{Latitude \;Error*Distance}{Total \;Traverse \;Length} $$

(A.11)

Using the line from Point 1 to Point 2 as an example:

$$ Latitude\; Correction \;Line \;1 - 2 = \frac{ - 0.038*161.20}{600.07} = - 0.010 $$

The same procedure is used for departures.

$$ Departure \;Correction \;Line\; X1 - X2 = \frac{Departure \;Error*Distance}{Total\; Traverse\; Length} $$

(A.12)

The traverse total length is the cumulative length of all traverse lines. In our example the total traverse length is 600.07′.

The adjusted traverse for our example is shown in Table A.7. In order to save space the latitudes and departures have been reduced to a single column each. North and east values are positive and south and west values are negative.

Table A.7 Adjusted traverse

In Table A.7, the Lat. Correction and Dep. Correction columns show the corrections to be made to each latitude and departure in accordance with Eqs. A.11 and A.12. The column of corrections should be added to be sure that the sum is equal to the total correction. For example, the total correction for the departures is −0.038. The sum of the latitude correction column is also −0.038.

After all of the corrections have been calculated, they are applied to the latitudes and departures to create balanced latitudes and departures. Using this procedure, the columns for balanced latitude and balanced departure are completed. The balanced latitudes and balanced departure columns must be added together to ensure that the totals are zero. In other words, for the traverse to close properly the sum of the positive latitudes must equal the sum of the negative latitudes. The same applies to the departures.

After all of the balanced latitudes and departures have been calculated, they are applied to the coordinates. The beginning and end coordinates must be the same. In our example coordinates for Points 8 and 9 are the same. We now have adjusted coordinates for our traverse.

The bearings and distances shown in Table A.7 are no longer correct because the coordinates have changed slightly. We need to calculate new bearings and distances for the traverse lines. This is easily accomplished by inversing between the adjusted coordinates. Our adjusted traverse with the adjusted bearings and distances is shown in Fig. A.30.

Fig. A.30

Adjusted traverse bearings and distances

A.6 Road Geometry

Most parcels of land have frontage on a public or private road. In this chapter we will review the geometry of roads and road layouts. Calculating road geometry requires a background in plane geometry and trigonometry. We have already covered much of the material that you will need in this section; however some new material on curves will be introduced.

Many roads have layouts, sometimes referred to as alignments. A layout or alignment describes the curves and tangents of the roadway. Some of these layouts are approved by a government authority such as a local or state government. Some roads have no layouts so determining the location and width of a road or street can sometimes be a challenge for the boundary surveyor.

There may also be applicable standards which govern the design of roads, such as minimum width, minimum curve radii, grade limits, view obstructions and other features. Because our focus is boundary surveying, our primary discussion will concentrate on the two dimensional features of roads. When we use the term road layout in this section we will use it in its broadest sense to mean the physical dimensions and geometry of a road. The reader should be aware that the term may also have legal meaning referring to a layout approved by a public agency.

18.7.1 A.6.1 Highway and Road Curves

Most roads are composed of straight sections, called tangents, and curves. The majority of curves are simple circular arcs which are segments of a circle. Sometimes spiral curves are used, particularly when it is necessary for vehicles to transition from a high speed to a low speed. Spiral curves start from the tangent with gentle curve which gradually increases until a simple arc is reached.

A tangent line to a curve is a straight line that just touches the curve. Figure A.31 shows a circle which has a straight line tangent to it.

Fig. A.31

Line tangent to curve

If we draw a red dashed line from the center of the circle to a point on the straight line so that the two lines form an angle of 90°, the intersection of the straight lines will be the exact point where the tangent line just touches the circle. The point of tangency of a straight line with a circle or arc will always form a 90° angle between the line and center of the circle or arc.

The nomenclature used to describe various elements of a curve typically used on roads and highways is shown in Fig. A.32. It can be seen from this figure that the road tangents are 90° to the radius lines. Starting on the left side of the figure, the point of tangency is labeled PC which is an abbreviation for point of curvature. If we follow the curve along to the next point of tangency, we arrive at PT, which stands for point of tangency. All symmetrical road curves have a PC and a PT. When there is a road layout which begins at one end of the road and traverses to the other end, using bearings for example, the term PC will be used for the beginning of a curve and PT for the end. Think of driving your vehicle down the road in the direction of the layout. You will come to PCs before PTs.

Fig. A.32

Curve nomenclature

The curve length, abbreviated L, is the length of the arc between the PC and PT. The radius (R) is the radius of the arc. If a road is designed for slow speed traffic, the radius can be shorter than for high speed traffic. For a given curve angle, as we lengthen the radius, the curve length will increase. Notice the difference in radii and curve lengths between the two examples in Fig. A.33. It would be possible to drive at higher speed along the curve with the larger radius than the one with the smaller radius.

Fig. A.33

Different curve radii

The curve Central Angle (Δ.) is equal to the Deflection Angle (Δ). This can be clearly seen in Fig. A.34.

Fig. A.34

Delta angle = central angle

The Angle between the chord and the Tangent, sometimes called the Tangent Deflection Angle or just Tangent Angle is half the deflection angle. See Fig. A.35.

Fig. A.35

Tangent angle

18.7.2 A.6.2 Calculate Tangent

An example of calculating the length of a Tangent is shown in Fig. A.36. We can see from this figure that the geometry is our familiar right triangle. We have labeled the three angles in the right triangle A, B and C. Recall that in a normal curve the radius line is always tangent to both the street line and the tangent line (the tangent line is simply an extension of the street line).

Fig. A.36

Calculate tangent

In the example, we know the curve radius: 275.00′ and the Central Angle: 40° 10′ 00″. We know that the angle between the radius line and the middle ordinate is half the central angle so we calculate angle A of our right triangle to be 20° 05′ 00″. Using Eq. A.3 we rearrange to solve for side a (tangent line).

$$ \begin{aligned} \tan A & = \frac{a}{b}\quad {\text{so}}\quad a = b*\tan A \\ a & = 275.00^{\prime } *0.3656 = 100.545 \\ \end{aligned} $$

Because angle A is simply $\frac{\varDelta }{2}$ we can write an equation for calculating the tangent as follows:

$$ T = R*\tan \frac{\Delta }{2} $$

(A.13)

18.7.3 A.6.3 Calculate the Length of a Curve

We will next calculate the length of a curve. The curve length is the length of the arc between the two tangent street side lines. We know that a circle contains 360°. Our curve is simply a portion of a circle. The Central Angle (Delta or Δ) tells us what portion of the circle our curve represents. In the curve shown in Fig. A.37, the central angle is 40° 10′ 00″. We can use a proportion to calculate what fraction of a circle our curve represents. We first need to convert our central angle to decimal degrees then proceed as follows:

$$ Fraction = \frac{40.1667^\circ }{360^\circ } = 0.111574 $$

Fig. A.37

Calculate curve length

Next we need to know that the circumference of a circle is calculated as:

$$ C = \pi *D $$

where C is the circumference of the circle and D is the diameter. In our example the radius is 275.00 ft. so the diameter is 550.00 ft. The circumference of our circle is therefore:

$$ C = \pi *550.00^{\prime } = 1,727.876^{\prime } $$

To calculate our curve length we simply multiply the circumference by the fraction of a circle that our curve represents:

$$ L = 1,727.876^{\prime } *0.111574 = 192.786^{\prime } $$

We can generalize our equation to calculate the curve length as follows:

$$ L = 2*\pi *R*\frac{\Delta }{360^\circ } $$

(A.14)

Using our equation to check the above result:

$$ L = 2*\pi *275.00^{\prime } *\frac{40.1667^\circ }{360^\circ } = 192.786^{\prime } $$

Suppose we only have the radius and the curve length and want know the central angle? We can rearrange Eq. A.14:

$$ \Delta = \frac{L*360^\circ }{2*\pi *R} $$

(A.15)

Substituting our values:

$$ \Delta = \frac{192.786*360}{2*\pi *275.00} = 40.166 \;decimal\; degrees $$

And converting from decimal degrees to DMS we have: 40° 10′ 00′.

18.7.4 A.6.4 Curves at Intersections

We will next consider curves typically found at the intersections of streets as shown in Fig. A.38. These curves are really no different than the highway curves which we just examined.

Fig. A.38

Curve at street intersection

The curve geometry will have a Radius, Delta, Chord and Tangent. In many, if not most, cases it is convenient for the radius to be the same for both sides of the intersection, although in some cases it is not. In a typical example, as shown in Fig. A.38, the radius is 30.00 ft. on both sides of Roswell Street. However, it is apparent from the layout that the Delta, Chord and Tangent vary from one side to the other.

In our example, assume that Main Street is straight and that the two sidelines of Roswell St. are parallel. In this geometry, the two Deltas for the curves must add up to 180°. If we do the math and add 115° 39′ 06″ and 64° 20′ 54″ we will get 180°. The remaining values can be calculated using the equations and methods used above.

18.7.5 A.6.5 Compound Curves

A compound curve exists when there are two adjacent curves, each having a different radius. Figure A.39 shows a compound curve.

Fig. A.39

Compound curve

Notice that the curve on the left has a radius of 300.00′ and the curve on the right has a radius of 200.00′. Because both curves share a common radius line, the curves are tangent to each other. The methods for calculating the tangents and curve lengths for each of the curves are the same as for the simple curves that have already been discussed.

The intersection of the tangents of the compound curve does not occur opposite the central radius line because of the different radii. Notice that the lengths of the tangents are different. The left side is 140.53′ while the right side is 116.65. One way to calculate the tangent lengths is to use an oblique triangle solution.

In the example shown in Figs. A.39 and A.40, we know the bearings of the tangents. These bearings are simply the bearings of the street lines extended. We also know the bearing of the center radius line and we know that the tangent at PCC (Point of Compound Curve) is at 90° to the radius line bearing. Because we have three bearings we can calculate the interior angles of the oblique triangle. These are shown in Fig. A.40.

Fig. A.40

Compound curve

We also know that the distance along the base of our triangle is the sum of the lengths of the tangents of each of the curves: 76.06′ + 45.17′ = 121.23′. With this information we can use Eq. A.6 to calculate the unknown side of our triangle. First we combine and rearrange the equations:

$$ a = \frac{b}{\sin B}*\sin A $$

Next, we substitute the values and calculate the length of side a.

$$ a = \frac{121.23^{\prime } } {\sin 126^\circ \;05^{\prime } \;45^{\prime{\prime }} }*\sin 28^\circ \;27^{\prime }\;10 = 71.48^{\prime } $$

The total length of the tangent is the sum of the value we just calculated and the value of the tangent of the curve:

$$ T = 71.48^{\prime } + 45.17^{\prime } = 116.65^{\prime } $$

We would calculate the remaining tangent using the same method.

18.7.6 A.6.6 Reverse Curves

When a curve immediately changes direction, it is called a reverse curve. Figure A.41 shows a reverse curve. Reverse curves can have identical radii or each curve can have a different radius. The calculations are similar to that of the simple curves that we have already covered, and so do not bear repeating here. In order for both curves to be tangent at the point where the curve changes direction, the radius line of each curve running through Point of Reverse Curve (PRC) must form a straight line. This can be seen in the image where the bearing of each line, N39° 30′ 56″E, is identical.

Fig. A.41

Reverse curve

18.7.7 A.6.7 Road Stations

When designing and laying out roads it is customary to use Stations. Stations are commonly numbered using integers which represent 100 ft. So, the distance between Station 5 and Station 9 would be 400.00 ft. When it is necessary to identify a point between stations, it is designated using a + sign. This is illustrated in Fig. A.42. The intersection of the centerlines of Main St. and Roswell Street occurs at Main Street station 10 + 90.91. The intersection at the centerline of Roswell Street occurs at Roswell station 5 + 31.51.

Fig. A.42

Road stations along centerline

Calculating the distances between stations is straightforward. For example calculate the distance between station 9 and station 10 + 90.91.

$$ \begin{aligned} Station\,9 & = 9*100^{\prime } = 900^{\prime } \\ Station\,10 + 90.91 & = \left( {10*100^{\prime } } \right) + 90.91^{\prime } = 1,090.91^{\prime } \\ Distance & = 1,090.91^{\prime } - 900^{\prime } = 190.91^{\prime } \\ \end{aligned} $$

18.7.8 A.6.8 Road Baselines and Centerlines

Stations can be on a Baseline or Centerline. Both streets in Fig. A.42 have centerlines. Centerlines are equidistant from the street sidelines. In other words, centerlines run down the middle of the road. If the road has a curve, the centerline radius will be halfway between the inner and outer radii of the road (assuming that the curves are concentric). Notice that in Roswell Street, the offsets from the centerline to the sidelines are the same on both sides, 20.00 ft. The centerline is parallel to the sidelines so both would share the same bearing.

In Fig. A.43, the dashed line running down the middle of June Street is a Centerline, but the line running through Wally Way is a baseline.

Fig. A.43

Baselines and centerlines

A Baseline is generally not in the middle of the road layout, nor does it necessarily run parallel to the sidelines. In Fig. A.43, Wally Way has a baseline. Notice that the baseline has an angle point at 29 + 11.03. When an angle occurs in a baseline or centerline, the station numbering does not reset to zero. For example, the distance between station 30 and 29 + 11.03 is not 100 ft. it is 88.97′.

Notice that along the baseline of Wally Way, the offsets from the baseline to the curves at June Street have different dimensions. June Street has a centerline so the offsets are equal (20.00 ft.).

Both baselines and centerlines are reference lines which are measured with great precision. From a practical point of view, working along the sidelines of a road or highway can be difficult because of the many obstructions such as trees, walls and fences. It is therefore much more common to work in the relatively unobstructed area at or near the center of the roadway where baselines and centerlines are located. In roads that do not have medians, great care must be taken because of the danger from moving vehicles, so in these cases it is not uncommon to work on some convenient offset from one of the sidelines. For example, when working in an urban area where there are sidewalks, it is often convenient and relatively safe to run survey lines over the sidewalks.

Offsets are used to locate angle points in the road and at the beginning and ends of curves. When roads are designed and laid out, baselines or centerlines are used by surveyors and engineers to establish and mark the sidelines of the roads. For example, in Fig. A.43, if we wanted to set a point at the beginning of the curve at Wally Way and June Street, we could set our total station on baseline station 28 + 8.26, turn an angle of 90° from the baseline and measure over 33.66′ to set the point. This is a very important point: unless the offset line has a specific bearing and distance, it is assumed to be at 90° to the baseline or centerline.

A.7 Calculating Area

Knowing how to calculate area is very important for land surveyors. The economic value of a parcel of land is related to its area. Some subdivision and zoning regulations require parcels of land to have minimum or maximum areas. The area of a parcel of land being conveyed is generally included in the deed description.

Like distances, which are always measured in the horizontal plane, area is also calculated in a horizontal plane. If a parcel of land is situated on a hill, the actual ground area will be larger than the area calculated on a horizontal plane. When we calculate area in boundary surveying we are generally concerned with area measured in the horizontal plane, so we will confine our discussion to calculating area in this manner.

In the U.S. area is usually measured in square feet or acres. One acre = 43.560 ft.². Often, graphic methods or planimeters were used which were not very accurate. One way to calculate area is to break up a parcel of land into a series of rectangles and triangles. The area of a rectangle is simply its width times it height.

To calculate the area of the parcel in Fig. A.44, we first notice that the all of the interior angles are 90°. If the angles were not 90°, we would have a parallelogram, and the area would be less than a rectangle. The area of a rectangle is given by:

$$ A = L*W = 200*100 = 20{,}000 \,{\text{ft.}}^{2} $$

Fig. A.44

Area of a rectangular parcel

Many corner lots have curves at the intersections. One way to calculate the area of such a parcel is to break it up into rectangles or triangles. The area inside the circular sector is calculated as a portion of a circle.

Consider Fig. A.45 which shows a rectangular parcel bounded by two streets. In order to calculate the area, the parcel is broken up into 3 sections and the area of each is calculated. The area of the circular sector is calculated as a fraction of a circle. In this case the Delta is 90° so we have 1/4 of a circle. The area of a circle is given by:

$$ A = \pi r^{2} $$

Fig. A.45

Area of corner lot

So, for a full circle with a radius of 25.00′ we calculate the area as:

$$ A = \pi *25.00^{2} = 1{,}963.50 \,{\text{ft.}}^2 $$

The area of ¼ of the circle is:

$$ A = 1{,}963.50*0.25 = 491\,{\text{ft.}}^2 $$

If the Delta were not 90° we would simply use a ratio of the Delta to 360° as we did when we calculated the curve length in the section on road geometry. Another way to approach calculating the area of the parcel shown in Fig. A.45 would be to calculate the area of the entire parcel, as if the curve did not exist then subtract the area outside the curve from the lot area. The lot area without the curve would simply be: 100*200 = 20,000 square feet. We have already calculated the curve area of 491 square feet. The area outside the curve would be the difference between the curve area just calculated and a 25 foot square which has an area of 625 square feet (25*25 = 625). So, the area outside the curve would be: 625–491 = 134 square feet. The lot area would therefore be 20,000–134 = 19,866 square feet.

We can also use triangles to calculate area. If we know the lengths of each side of a triangle, we can use the following equations (Heron’s formula) to calculate the area.

$$ Area = \sqrt {s(s - a)(s - b)(s - c)} $$

(A.16)

$$ s = \frac{1}{2}(a + b + c) $$

(A.17)

where a, b and c are the lengths of the sides of the triangle.

Let us use the parcel of land shown in Fig. A.46 to see how we can use triangles to calculate the area of a parcel.

Fig. A.46

Using triangles to calculate area

This is the same parcel of land that is shown in Fig. A.21 with some of the information removed for clarity. We break the parcel into two triangles by inversing between the opposite lot corners. We now have two triangles so we can use Eqs. A.16 and A.17 to calculate the areas.

Triangle 1:

$$ \begin{aligned} s & = \frac{1}{2}\left( {a + b + c} \right) = \frac{1}{2}\left( {150.72 + 196.34 + 207.83} \right) = 277.445 \\ A & = \sqrt {277.445\left( {277.445 - 150.72} \right)\left( {277.445 - 196.34} \right)\left( {277.445 - 207.83} \right)} = 14{,}089.5\,{\text{feet}}^2 \\ \end{aligned} $$

The same procedure would be used to calculate the area of the second triangle, and the two would be added together.

The parcel shown in Fig. A.47 is an example of how the area of an irregularly shaped parcel can be calculated using triangles. As you can see from this exercise, calculating area this way is a lengthy and time consuming process. Because of the large number of calculations, there is also a greater possibility of making a mistake.

Fig. A.47

Area of irregular parcel using triangles

A very common way to calculate area is based on double meridian distances or coordinates. When latitude and departures and coordinates are calculated in tabular format as in the section on adjusting traverses, these methods are convenient because they use this same information to calculate area. However, neither of these methods are used very much by surveyors these days, except when they are a part of computerized survey calculation software. Thankfully, the laborious methods of performing area calculations are now performed by software in seconds rather than the hours it took using manual calculation methods.