Traverse Computations
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A. Overview
A traverse is a series of connected lines whose lengths and angular relationships are measured.
1. Traverse configuration
A traverse is either a loop or a link:
Loop Link
begins and ends on the same point begins on one point and ends on another
2. Traverse Type
A traverse is either:
• Open – no mathematical checks
• Closed – includes at least one mathematical check
A loop traverse is closed if all its angles and distances are measured.
In order for a link traverse to be closed it must begin and end on two points whose positions are known in the same system.
B. Computation Steps
Traverse computations is the process of taking field measurement through a series of mathematical calculations to determine final traverse size and configuration. These calculations include error compensation as well as reformation to determine quantities not directly measured.
Traditional traverse computation steps are:
1. Balance (adjust) angles
2. Determine line directions
3. Compute latitudes and departures
4. Traverse Adjustment: Balance latitudes and departures
5. Determine adjusted line lengths and directions
6. Compute coordinates
7. Compute area
8. Closed Link Traverse
9. Traverse Adjustment by Coordinates
10. Forcing Closure
The order of some steps can be changed. For example, steps 1 and 2 would be reversed for closed link traverses with directions at both ends. Balancing angles would normally not be done If a least squares adjustment is used at step 4.
A traditional approach does not need coordinates, even for area determination. Most software, however does. Whether done manually or with software, the process starts with coordinate computations; at each successive step those initial coordinates are updated reflecting the operation.
Our process will follow the traditional computation steps.
Balancing Angles
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A. Angular Condition
1.Interior Angles
For a closed non-crossing loop traverse, the angle condition is:
A loop traverse which crosses itself doesn’t have conventional “interior” angles so the previous equation won’t work. Some other angle condition must be established for such cases.
2. Deflection Angles
If deflection angles are used, the angle condition depends on how many times, if any, the traverse crosses itself.
Crosses even times: Crosses odd times:
Right deflection angles are positive (+), left are negative (-)
Angle condition equations are exact so their 0°, 180°, and 360°, respectively, are also exact.
3. Link Traverse
In order to determine angular misclosure for a closed link traverses, known directions are required at both ends:
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B. Angular Misclosure
Angular misclosure is the difference between the total measured angles on a traverse and the angle condition for the traverse configuration.
The amount of allowable angular misclosure is dependant on the survey purpose. The limits are generally expressed as an error of a series:
C: allowable misclosure, seconds
k: expected error in each angle, seconds
n: number of angles
Traditional formal standards from the FGCS Standards and Specifications for Geodetic Control Networks are:
Order Class k (sec)
First – 1.7
Second I 3.0
II 4.5
Third I 10.0
II 12.0
Example
What is the allowable angular misclosure for a five-sided traverse if it is to meet Second Order Class II standards?
From the previous table, k = 4.5″.
Because this is a five-sided loop traverse, the angles should sum to:
Therefore, the total should be in the range 539°59’50” to 540°00’10”
Informal standards can be created which reflect equipment, experience, conditions, project needs, etc.
Example
A control network is to be established for a construction project. Sight distances will be relatively short (under 300 ft) so measuring angles consistently will be a challenge. Testing under these conditions, our survey crew was able to measure an angle to a consistency of ±0°00’12”. For this project, we will use an allowable misclosure of:
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C. Angular Misclosure Distribution
Normally, the first step in traverse computations is distribution of the angular misclosure in order to meet the angular condition. This is referred to as angle adjustment. If performing a least squares traverse adjustment, angles are not adjusted since the angular misclosure is part of the overall adjustment.
Because angular misclosure is due to random errors, a distribution method which approximates their behavior would be the best to use. There are a few simplified methods which can be applied, each having advantages and disadvantages:
Method Process Advantage Disadvantage
Don’t distribute – – Simple; Can be replicated. Unless using a least squares adjustment, this method does not try to reconcile error accumulation.
Arbitrary Put all error into random angle. None. Doesn’t model errors well; Difficult to replicate.
Equal distribution Apply equal correction to each angle. Simple; Can be replicated. Assumes error is same in each angle; Can imply incorrect angle accuracy.
Judgment Apply different corrections to each angle depending on measurement repetition and conditions. Relatively simple; Better error modeling. Requires direct knowledge of measurement conditions.
One thing we do not base corrections on is angle size. An angle is the difference between a total station’s initial and final readings; size of the angle has no impact on its error.
Example: Given the following traverse and measured angles:
Point Angle
A 125°30’20”
B 88°21’31”
C 112°38’35”
D 104°21’40”
E 109°07’41”
sum: 539°59’47”
Since the angle total is short of the angle condition, corrections must be added.
Using an equal distribution
Apply correction:
Point Angle corr’n Corr’d Angle
A 125°30’20” +0°00’02.6″ 125°30’22.6″
B 88°21’31” +0°00’02.6″ 88°21’33.6″
C 112°38’35” +0°00’02.6″ 112°38’37.6″
D 104°21’40” +0°00’02.6″ 104°21’42.6″
E 109°07’41” +0°00’02.6″ 109°07’43.6″
sums: 539°59’47” +0°00’13.0″ 540°00’00.0″
check check
While the corrected angles total the angle condition, each is shown to 0.1″ which implies a higher accuracy than shown in the original angles. Rounding each to the nearest second raises the total to 540°00’05”.
Using judgment
In this case, we take into account measurement conditions.
Consider point E, which has short sights on both sides. Shorter sights generally have higher pointing errors (given same target sizes).
Also, the field crew reported larger repeated direct and reverse angle spreads at point C.
The equal distribution’s correction was +0°00’02.6″ per angle. We can drop that to +0°00’02” per angle for our three strongest angles (+0°00’06” total) and divide the remaining 0°00’07” among the two weaker ones.
Point Angle corr’n Corr’d Angle
A 125°30’20” +0°00’02” 125°30’22”
B 88°21’31” +0°00’02” 88°21’33”
C 112°38’35” +0°00’04” 112°38’39”
D 104°21’40” +0°00’02” 104°21’42”
E 109°07’41” +0°00’03” 109°07’44”
sums: 539°59’47” +0°00’13” 540°00’00”
check check
This takes into account individual conditions. The resulting angles do not imply accuracy beyond the original measurements.
Area
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A. General
Once a loop traverse has been adjusted, how can its area be determined?
We could divide the complex polygon into a series of triangles, compute the area of each triangle, then sum them.
There are quite a few possible triangle combinations:
The more points on the traverse, the more triangles and combinations.
It becomes even more interesting on traverses with concave sides:
Regardless, there are quite a few calculations involved using triangles.
As with other surveying calculations, we need a systematic approach to determine traverse area.
B. Area By Coordinates
1. Concepts
There is a general equation for the area of a closed non-crossing polygon using the coordinates of its vertices. This equation is:
In terms of N and E coordinates, the equation is:
The equation looks complex, more so if you expand all the terms. Rather them remember the equation, it’s easier to remember its pattern:
• Select a start point (doesn’t matter which)
• Going in sequence around the exterior (direction doesn’t matter) list each coordinate pair in two columns: one for X (or E), the other for Y (or N).
• Repeat the first coordinate pair at the bottom.
Using X/Y coords Using N/E coords
• Cross multiply the coordinates and sum the products
Using X/Y coords
Using N/E coords
The units of the cross products are square linear units – if coordinates are in ft, cross products are sq ft.
• Area is half the difference of the sums
The absolute value is used since the area could be negative depending on the combination of direction around the traverse, coordinate pairing (eg, X/Y vs Y/X), and cross multiplication order.
Doesn’t look much simpler than memorizing the equation, does it? Actually, it’s pretty easy to remember once you do it.
2. Examples
Although we carried an additional digit in all previous computations, we’ll carry a few more extra here. We’ll discuss the area error at the end of this section but for now we want to over-compute the area accuracy then report it to an appropriate resolution after we analyze it. If we don’t carry additional digits, we could easily increase error due to rounding. To be on the safe side, we’ll carry the computations to 0.1 which should be lower than the expected error.
a. Traverse 1
Continuing with the first traverse from the Coordinates section:
The previously computed coordinates of the traverse points are:
Point North (ft) East (ft)
A 500.000 2000.000
B 323.614 1561.426
C 526.996 1488.321
D 719.336 1686.956
Step (1) Start at point A and going clockwise around the traverse list the coordinates:
Point N (ft) E (ft)
A 500.000 2000.000
B 323.614 1561.426
C 526.996 1488.321
D 719.336 1686.956
A 500.000 2000.000 remember to return to A
Step (2) Cross multiply in one direction:
Point N (ft) E (ft) (/) (sq ft)
A 500.000 2000.000 647,228.0
B 323.614 1561.426 822,865.2
C 526.996 1488.321 1,070,602.9
D 719.336 1686.956 843,478.0
A 500.000 2000.000
Step (3) Cross multiply in the other direction
Point N (ft) E (ft) (/) (sq ft) () (sq ft)
A 500.000 2000.000 647,228.0
B 323.614 1561.426 822,865.2 780,713.0
C 526.996 1488.321 1,070,602.9 481,641.5
D 719.336 1686.956 843,478.0 889,019.1
A 500.000 2000.000 1,438,672.0
Step (4) Add up the columns
Point N (ft) E (ft) (/) (sq ft) () (sq ft)
A 500.000 2000.000 647,228.0
B 323.614 1561.426 822,865.2 780,713.0
C 526.996 1488.321 1,070,602.9 481,641.5
D 719.336 1686.956 843,478.0 889,019.1
A 500.000 2000.000 1,438,672.0
sums: 3,384,174.1 3,590,045.6
Step (5) Compute the area
Until we discuss area accuracy more fully, we’ll state the area as 102,935.8 sq ft.
b. Traverse 2
Continuing with the second traverse from the Coordinates section:
The previously computed coordinates of the traverse points are:
Point North (ft) East (ft)
E 1000.000 200.000
F 689.206 532.694
G 896.890 627.584
H 692.474 257.460
With a crossing traverse, one must be careful when listing the coordinates. In this case, if you list the coordinates along the original traverse path, E-F-G-H-E, you will be able to compute an area but it will be nonsensical. The traverse turns itself inside out.
Recall that this survey was on a four-sided parcel having two obstructed lines:
We desire the area of the parcel, not the traverse.
Step (1) Start at point E and going clockwise around the parcel list the coordinates:
Point N (ft) E (ft)
E 1000.000 200.000
G 896.890 627.584
F 689.206 532.694
H 692.474 257.460
E 1000.000 200.000 remember to return to E
Step (2)-(4) Cross multiply in both directions; sum the columns:
Point N (ft) E (ft) (/) (sq ft) () sq ft
E 1000.000 200.000 179,378.0
G 896.890 627.584 432,534.7 624,584.0
F 689.206 532.694 368,876.7 477,767.9
H 692.474 257.460 257,460.0 177,443.0
E 1000.000 200.000 138,494.8
1,238,252.4 1,418,289.7
Step (5) Compute the area
C. Area By DMDs
1. Concept
A traditional method of computing area of a closed polygonal traverse is by DMDs: Double Meridian Distances.
A meridian distance (MD) is the distance from a reference meridian to the center of a line: measured in the East (X) direction:
The reference meridian can be placed anywhere, but generally, it passes thru the first part of the traverse. Triangular and trapezoidal areas can be computed by multiplying a meridian distance by the latitude of the respective line. The base of each triangle or trapezoid coincides with the meridian. Adding the areas, some of which are negative based on the latitude and meridian distance, results in the area of the polygon.
Why then the Double Meridian Distance? Well if we start at point A in the example traverse and begin computing meridian distances, we begin to see a pattern develop:
See all those 1/2’s? By removing them, we would be computing Double Meridian Distances (DMDs).
Notice that the DMD of the last line is the same as the departure of that line except with an opposite math sign. That’s the math check.
Multiplying each line’s DMD by its latitude and summing the results, we wind up with double the area.
Just like area by coordinates it looks pretty confusing, but there is a pattern that helps guide the calculations.
You’ll sometimes see reference to Area by DPDs (Double Parallel Distance). This is identical to the DMD method except everything is rotated 90°. Parallel distances are determined from an E/W reference line (a parallel) and multiplied by the latitudes.
2. Examples
a. Traverse 1
Using the same loop traverse as the first area by coordinates example
Adjusted
Line Lat (ft) Dep (ft)
AB -176.386 -438.574
BC +203.382 -73.105
CD +192.340 +198.635
DE -219.336 +313.044
Step (1) Compute the DMDs
Starting with line AB
If you look at the Deps and DMDs in the table, you notice a pattern to the computations. Follow the colored arrows starting with the dep of the first line, adding as you go:
Adjusted
Line Lat (ft) Dep (ft) DMD (ft)
AB -176.386 -438.574 -438.574
BC +203.382 -73.105 -950.253
CD +192.340 +198.635 -824.723
DE -219.336 +313.044 -313.044
Step (2) Multiply DMDs by Lats; sum the products
Adjusted
Line Lat (ft) Dep (ft) DMD (ft) DMD x Lat (sq ft)
AB -176.386 -438.574 -438.574 +77,358.3
BC +203.382 -73.105 -950.253 -193,264.4
CD +192.340 +198.635 -824.723 -158,627.2
DE -219.336 +313.044 -313.044 +68,661.8
sum: -205,871.5
Step (3) Compute the area
Note that this is the same as computed by coordinates.
b. Traverse 2
What about the crossing traverse?
Well, the parcel area can’t be determined by DMDs without additional computations. By definition, Area by DMDs is limited to travel along the traverse path so you would be computing the area of E-F-G-H-E. To compute the area of the parcel by DMDs, you would need to determine the lat and dep of lines EF and FH. By the time that was done, the area could have been computed by coordinates.
D. Closing
1. Computation Method Comparison
Method Advantages Disadvantages
DMDs o Systematic approach
o No additional calculations besides the area computations
o Simple math check: last DMD o Limited to traverse path
Coordinates o Systematic approach
o Flexible: not limited to traverse path o More calculations: need coordinates before area computations
o Numbers can be huge if using large coordinates
o No math checks
2. Accuracy Revisited
a. How Accurate?
Which method is most accurate? Because both area calculation methods use the same adjusted latitudes and departures, they should have equal accuracy. The best way to determine area accuracy is to propagate all measurement and setup errors into it. Even with a computer that can be a daunting task. If surveying software is available, a sensitivity analysis can be done: vary multiple measurements by introducing reasonable errors into them and examine how the area changes.
A simplified approach can be used which uses basic error propagation along with the traverse precision. Envision a rectangular parcel which includes the same area as the traverse. Because the area of a rectangle is length times width, the Error of a Product can be used to approximate an accuracy.
The equation for Error of a Product is:
in terms of the rectangle
EW and EL are the expected errors in the width and length. What do we use for EW and EL? We can approximate the errors using the traverse precision. The precision summarizes up the overall random error effects for that particular traverse. If we multiply a distance by the precision, p, it gives us an expected error in that distance.
Substituting that back into the equation and reducing:
This simplified method takes into account the overall quality of the traverse in the form of its precision.
b. Examples
Traverse 1
Prior to adjustment we determined its precision to be 1/11,900; area computed by coordinates is 102,935.8 sq ft.
Recall that throughout our calculations, up to that of the area, we carried an additional digit to minimize rounding errors. That means our area uncertainty would be expressed to two sig fig: ±12 sq ft.
The area should be expressed to a level compatible with the uncertainty: 102,936 sq ft.
Traverse 2
Prior to adjustment we determined its precision to be 1/10,100; area computed by coordinates is 90,018.76 sq ft.
Uncertainty is ±12 sq ft; show area as 190,018 sq ft.
Although both examples have an area error of ±12 sq ft, that’s only because size and precision of both traverses are close to each other.
c. A final word
We generally have a tendency to overstate area accuracy: it’s not uncommon to see a computer generated map with areas expressed to 0.01 sq ft. Surveyors sometimes get caught up in equipment specifications and forget about the propagated errors when multiple measurements are combined. After all, our total station reports distances to 0.01 ft and angles to 01″. That should result in an uncertainty smaller than 12 sq ft, shouldn’t it?
Consider this: How would the area of Traverse 1 be affected if we offset line AB by 0.01 ft?
We could approximate the area change with a 0.01′ wide by 472.72 ft long rectangle – a difference of about 4.7 sq ft. That 0.01 ft offset would represent relatively small random errors in both lines CB and DA. Instrument and reflector centering errors and measurement errors could easily exceed 0.01 ft error per line. Include pointing errors measuring angles and the potential error can be substantial. Using similar equipment and procedures measuring larger traverses or traverses with more points will also increase the error in the area. So an error of 12 sq ft doesn’t seem so bad now, does it?
Using the simplified error propagation discussed here allows the surveyor to express area to a reasonable level of uncertainty.
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Line Directions
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A. Meridian
A meridian is a north-south reference line. It is used as a basis for a direction, which describes a line’s orientation.
Common meridians include, but are not limited to:
• True
• Magnetic
• Astronomic
• Grid
• Geodetic
• Assumed
Depending on the system, meridians may converge or may stay parallel.
Typical of Grid and Assumed systems, meridians are parallel. In a larger earth-based system, such as True, Astronomic, or Geodetic, meridians converge to a point (e.g., North Pole).
For small project areas such as a single lot survey, meridians can assumed to be parallel, regardless of system, without introducing significant error.
Most meridian systems (even Assumed) are constant over time. However, magnetic meridians change location and their change is not consistent. Magnetic meridians are important as the first instrument extensively used for property surveys was the compass. Although True directions were reported in the early Public Land Surveys, they were first measured by compass then converted using solar observation.
A surveyor will select an appropriate meridian for the project at hand and orient all survey lines to it. Sometimes directions based on one meridian must be converted to a different meridian in order to maintain consistency across surveys. We’ll look at that, along with magnetic direction conversions, in Meridian Conversions later.
We will use parallel meridians for the remainder of this discussion – convergency issues will be covered in a later Chapter.
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B. Direction
A direction is angle from a meridian to a line. It is similar to a horizontal angle in a traverse except the backsight is always along the meridian.
There are two different ways to express line direction: bearing and azimuth.
1. Bearing
A bearing is an angle from the North or South end of the meridian turned to the East or West. A bearing has three parts:
• Prefix – N or S indicating which end of the meridian is turned from.
• Angle
• Suffix – E or W indicating turning direction from the meridian to the line.
N 66°40′ E – from the North end of the meridian, turn 66°40′ to the East.
Example
Bearing AB = N 66°40′ E
Bearing AC = S 55°32′ E
Bearing AD = S 44°21′ W
A bearing falls in one of four quadrants so the angle does not exceed 90°. The angle is to the right (clockwise) in the NE and SW quadrants, to the left (counter-clock NW quadrants. A due North direction can be expressed as either N 00°00′ E or N 00°00′ W; due East as N 90°00′ E or S 90°00′ E; similarly for dues South and West.; similarly for dues South and West.
A back-bearing is reverse of a bearing, that is, Bearing BA is the back-bearing of Bearing AB. Because the meridians are parallel at both ends of the line, the bearing angle is the same but quadrant is reverse. This is true only when meridians are parallel. Where meridians converge, the forward and back bearing angles will differ by the total convergence. More on this later.
Bearing AB = N 66°40′ E
Bearing BA = S 66°40′ W
2. Azimuth
An azimuth is an angle to the right (clockwise) from the meridian to the line. In most cases the azimuth is turned from the north meridian end; earlier control surveys used the south end. An azimuth varies from 0° to 360°.
Example
Azimuth AB = 66°40′
Azimuth AC = 124°28′
Azimuth AD = 224°21′
Azimuth AE = 322°26′
A back-azimuth is reverse of a azimuth: Azimuth CA is the back-azimuth of Azimuth AC. Because the meridians are parallel at both ends of the line, the back-azimuth and forward azimuth differ by 180°. As with bearings, this is true only when meridians are parallel. Where meridians converge, the forward and back azimuths will differ by (180° ± total convergence). More on this later.
Example
Azimuth AC = 124°28′
Azimuth CA = 124°28′ + 180°00′ = 304°28′
3. Converting Between Bearings and Azimuths
Since Bearings and Azimuths are both referenced to a meridian it is simple to convert one to the other.
To convert from bearings to azimuths:
Quadrant From Bearing To Azimuth
NE N β E β
SE S β E 180° – β
SW S β W 180° + β
NW N β W 360° – β
Example
Azimuth AB = 66°40′
Azimuth AC = 180°00′ – 55°32′ = 124°28′
Azimuth AD = 180°00′ + 44°21′ = 224°41′
Azimuth AE = 360°00′ – 37°34′ = 322°26′
To convert from an azimuth, α, to a bearing:
Quadrant To Bearing
NE N α E
SE S (180° – α) E
SW S (α – 180°) W
NW N (360° – α) W
Example
Bearing AB = N 64°40′ E
Bearing AC = S (180°00′-124°28′) E = S 55°32′ E
Bearing AD = S (224°21′ – 180°00′) W = S 44°21′ W
Bearing AE = N (360°00′ – 322°26′) W = N 37°34′ W
Rather then memorize tables, drawing a sketch will help determine correct conversion logic to use.
C. Angles and Directions
1. Angles to Directions
Starting with a direction for one traverse line, directions of the others can be computed from the horizontal angles linking them. The process of addition or subtraction is dependent on the type of horizontal angle (interior, deflection, etc), turn direction (clockwise or counterclockwise), and direction type (bearing or azimuth).
Example
The bearing of line GQ is S 42°35′ E. The angle right at Q from G to S is 112°40′. What is the bearing of the line QS?
Sketch:
Add meridian at Q:
At Q, the bearing to G is N 42°35′ W.
Subtracting 42°35′ from 112°40′ gives the angle from North to the East for line QS.
Bearing QS = N 70°05″ E
Example
The azimuth of line WX is 258°13′. At X the deflection angle from W to L is 102°45′ L. What is the azimuth of line XL?
Sketch:
A deflection angle is measured from the extension of a line. The azimuth of the extension is the same as that of the line. To compute the next azimuth , the deflection angle is added directly to the previous azimuth.
Because this is a left deflection angle, you would add a negative angle.
Add meridian at X:
Azimuth XL = 258°13′ + (-102°45′) = 155°28′
Notice how a sketch makes it easier to see the angle and direction relationships.
2. Directions to Angles
Given directions of two adjacent lines, it is a simple matter to determine the angle between the lines.
Example
The bearing of line HT is N 35°16′ W , the bearing of line TB is N 72°54′ E. What is the angle right at T from B to H?
Sketch
Label the back-direction at T and angle to be computed, δ.
Based on the sketch, the desired angle is what’s left over after both bearing angles are subtracted from 180°00′.
δ = 180°00′ – (72°54′ + 35°16′) = 71°50″
Example
The azimuth of line MY is 106°12′, the azimuth of line YF is 234°06′. What is the angle right at Y from F to M?
Sketch
Label the back-azimuth at Y and angle to be computed, ρ.
ρ = 286°12′ – 234°06′ = 52°06′
3. Traverse Direction Computations
Example 1
Given the following traverse and horizontal angles:
Using a bearing of N 36°55′ E for line AB, determine the bearings of the remaining lines clockwise around the traverse.
Many survey texts use a tabular approach to compute traverse line directions from angles. For the beginner this can be confusing and lead to erroneous directions. It helps to instead draw sketches in order to visualize the line relationships.
At point B:
β = 117°19′ – 36°55′ = 80°24′
Brg BC = S 80°24′ E
At point C:
γ = 180°00′ – (80°24′ + 87°42′) = 11°54′
Brg CD = S 11°54′ W
At point D:
η = 93°38′ – 11°54′ = 81°44′
Brg DA = N 81°44′ W
The directions for all four traverse lines have been computed. Angles at B, C, and D have been used, but that at A has not. For a math check, use the Bearing of DA and the angle at A to compute the bearing we started with.
At point A:
α = 180°00′ – (61°21′ + 81°44′) = 36°55′
Brg AB = N 36°55′ E check
If our computed and initial bearings for AB don’t match it means one of two things:
• there is a math error in our computations, or,
• the interior angles weren’t balanced.
For this traverse the angles sum to 360°00′ so there is no angular misclosure. If our math check had failed it would have been due to a math error in our computations.
Had the angles not been balanced and if there were no math errors, the math check would be off by the angular misclosure.
Summary:
Line Bearing
AB N 36°55′ E
BC S 80°24′ E
CD S 11°54′ W
DA N 81°44′ W
Example 2
Given the following traverse and horizontal angles:
Using a azimuth of 68°00′ E for line OP, determine the azimuths of the remaining lines counter-clockwise around the traverse.
At point P:
Line PQ is 92°48′ right of Azimuth PO.
Az PO is the back azimuth of Az OP
Az PQ = (Az OP + 180°00′) + Angle P
Az PQ = (68°00′ + 180°00′) + 92°48′ = 340°48′
At point Q:
Line QR is 112°26′ to the right from Az QP
The Az QP is the back azimuth of Az PQ
Az QR = (Az PQ + 180°00′) + Angle Q
Az QR = (304°48′-180°00′) + 112°26′ = 273°14′
At point R:
Line RO is 67°14′ right from Az RQ
The Az RQ is the back azimuth of Az QR
Az RO = (Az QR + 180°00′) + Angle R
Az RO = (273°14′ – 180°00′) + 67°14′ = 160°28′
The directions for all four traverse lines have been computed. Angles at P, R, and R have been used, but not the angle at O. For a math check, use Azimuth RO and the angle at O to compute the original Az OP.
At point O:
Line OP is 87°32′ right of Az OR.
Az OR is the back azimuth of Az RO.
Az OP = (Az RO + 180°00′) + Angle O
Az OP = (160°28′ + 180°00′) + 87°32′ = 428°00′
Because the azimuth crosses North, subtract 360°00′ to normalize it.
Az OP = 428°00′ – 360°00′ = 68°00′ check!
Summary:
Line Bearing
OP 68°00′
PQ 340°48′
QR 273°14′
RO 160°28′
You’ll note there is a computing pattern:
New Az = (Previous Az + 180°00′) + Angle.
This is true for a loop traverse meeting these conditions:
• Directions are counter-clockwise around the traverse, and,
• Angles are interior to the right.
What if the directions are clockwise around the traverse and interior angles counter-clockwise?
New Az = (Previous Az – 180°00′) – Angle
Other patterns exist for clockwise travel with clockwise interior angles, clockwise exterior angles, counterclockwise exterior angles, etc. Rather than memorize the possible patterns, draw a sketch, and begin computing; the pattern will present itself after a few lines.
Latitudes and Departures
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A. Definition; Equations
Latitude is the north-south component of a line; departure the east-west. North latitudes are positive, South are negative; similarly East departures are positive, West are negative.
The latitude of line AB is North (+), its departure is East (+). The latitude of line CD is South (-), its departure is West (-).
Latitude (Lat) and Departure (Dep) are computed from:
Dir can be either a bearing angle or azimuth angle.
Because a bearing angle never exceeds 90°, the Lat and Dep equations will always return positive values.
Sin(0°) to Sin(90°) ranges from 0 to +1.0
Cos(0°) to Cos(90°) ranges from +1.0 to 0
The correct mathematical sign for the Lat and Dep come from the bearing quadrant.
A bearing of S 47°35′ E has a negative Lat (South) and a positive Dep (East).
An azimuth angle ranges from 0° to 360°, so the sine and cosine return the correct signs on the Lat and Dep.
Examples
Reversing a line direction results in the same magnitude Lat and Dep but reversed signs: Lat and Dep but reversed signs:
Line A to B Line B to A
LatAB is North
DepAB is East LatBA is South
DepBA is West
Top
B. Closure
On a closed loop traverse, the sum of the Lats should equal zero as should the sum of the Deps.
Closure condition:
Most traverses won’t close perfectly due to the presence of errors. If the field crew followed careful procedures and compensated distance measurements for atmospheric conditions, only random errors prevent achieving perfect closure: event achieving perfect closure:
Actual closure:
The Linear Closure, LC, is the total misclosure distance. It is computed from the Lat and Dep errors:
Traverse precision is a function of the linear closure and the total survey distance. The precision is stated as a relative value, usually as 1 unit of error per D units measured:
What is allowable minimum precision? It depends on the survey purpose. Traditional first order surveys have a required precision of 1/100,000; in Wisconsin a property survey must be at least 1/3000. In order to achieve a formal minimum precision, specific equipment and field procedures must be used which match that precision requirement. Achieving a minimum precision should be by design not by accident.
C. Examples
In the following examples shown, all calculations are shown with an additional significant figure. Because these are generally intermediate computations, carrying an additional digit minimizes roundoff error in subsequent calculations.
When reporting results of an intermediate calculation, those should be stated to the correct number of significant figures so as not to imply an accuracy beyond that of the measurements.
1. Traverse with bearings
Lat and Dep will always compute as positive; must assign correct mathematical sign based on the bearing quadrant.
Line AB
Because the bearing is South and West, the Lat and Dep are -176.357′ and -438.548′ respectively.
Line BC
Because the bearing is North and West, the Lat and Dep are +203.395′ and -73.093′ respectively.
Line CD
Because the bearing is North and East, the Lat and Dep are +192.357′ and +198.651′ respectively.
Line DA
Because the bearing is South and East , the Lat and Dep are -219.312′ and +313.065′ respectively
In tabular form:
Line Bearing Length (ft) Lat (ft) Dep (ft)
AB S 68°05’35″W 472.68 -176.357 -438.548
BC N 19°46’00″W 216.13 +203.395 -73.093
CD N 45°55’20″E 276.52 +192.357 +198.651
DA S 54°59’15″E 382.24 -219.312 +313.065
sums: 1347.57 +0.083 +0.075
Distance Lat err Dep err
too far N too far E
2. Traverse with azimuths
Lat and Dep will always compute directly with the correct sign when using azimuths.
Line ST
Line TU
Line UV
Line VS
Line Azimuth Length (ft) Lat (ft) Dep (ft) Dep (ft) Dep (ft)
ST 309°05’38” 347.00 +218.816 -269.311
TU 258°34’22” 364.55 -72.226 -357.324
UV 128°04’44” 472.74 -291.560 +372.123
VS 60°21’26” 292.94 +144.885 +254.602
sums: 1477.23 -0.085 +0.090
Distance Lat err Dep err
too far S too far E
3. Crossing Traverse
A four-sided parcel has two obstructed lines.
In order to create a closed traverse, the survey crew measures a crossing traverse which connects all four points.
As long as a traverse closes back on its beginning point, the closer condition is still:
regardless of how many times it may cross itself.
Given this traverse data, determine its closure and precision.
Rather than write out each Lat and Dep computation separately, we can simply set up the table and record the computations in it.
Line Azimuth Length (ft) Lat (ft) Dep (ft)
EF 133°02’45” 455.03 -310.780 +332.737
FG 24°33’35” 228.35 +207.691 +94.912
GH 241°05’15” 422.78 -204.403 -370.084
HE 349°25’20” 213.85 +307.534 -57.430
sums: 1419.28 +0.042 +0.135
Distance Lat err Dep err
too far N Too far E
Traverse Adjustment
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A. Adjusting a Traverse
Adjusting a traverse (also known as balancing a traverse) is used to distributed the closure error back into the angle and distance measurements.
Raw field traverse
Adjusted (Balanced) Traverse
The condition for an adjusted traverse is that the adjusted Lats and Deps sum to 0.00. As with other survey adjustments, the method used to balance a traverse should reflect the expected error behavior and be repeatable.
Method Premise Advantage Disadvantage
Ignore Don’t adjust anything. Simple; repeatable Ignores error
Arbitrary Place error in one or more measurements Simple Not repeatable; ignores error behavior
Compass Rule Assumes angles and distances are measured with equal accuracy so error is applied to each. Simple; repeatable; compatible with contemporary measurement methods. Treats random errors systematically
Transit Rule Assumes angles are measured more accurately than distances; distances receive greater adjustment. Simple; repeatable; compatible with older transit-tape surveys. Treats random errors systematically; not compatible with contemporary measurement methods.
Crandall Method Quasi-statistical approach. Angles are held and errors are statistically distributed into the distances. Allows some random error modeling; repeatable. Models only distance errors, not angle errors.
Least squares Full statistical approach. Allows full random error modeling; repeatable; can mix different accuracy and precision measurements; provides measurement uncertainties. Most complicated method
The Compass Rule works sufficiently well for simple surveying projects and is the one we will apply.
B. Compass Rule
The Compass Rule (also known as the Bowditch Rule) applies a proportion of the closure error to each line.
For any line IJ
The Compass Rule distributes closure error based on the proportion of a line’s length to the entire distance surveyed.
C. Adjusted Length and Direction
Regardless of the adjustment method applied, changing a line’s Lat and Dep will in turn change the length and direction of the line.
The adjusted length can be computed from the Pythagorean theorem:
Computing direction is a two-step process: (1) Determine b, the angle from the meridian to the line (2) Convert b into a direction based on the line’s quadrant
To determine b:
b falls in the range of -90° to +90°.
The sign on b indicates the direction of turning from the meridian: (+) is clockwise, (-) is counter-clockwise. The combined signs on the adjusted Lat and Dep will identify the line’s quadrant.
The following diagram shows the relationship for the various mathematic combinations of the adjusted Lat and Dep:
D. Examples
These examples are a continuation of those from the Latitudes and Departures section.
1. Example with Bearings
Line Bearing Length (ft) Lat (ft) Dep (ft)
AB S 68°05’35″W 472.68 -176.357 -438.548
BC N 19°46’00″W 216.13 +203.395 -73.093
CD N 45°55’20″E 276.52 +192.357 +198.651
DA S 54°59’15″E 382.24 -219.312 +313.065
sums: 1347.57 +0.083 +0.075
Distance Lat err
too far North Dep err
too far East
a. Adjust the Lats and Deps
Line AB
Line BC
Line CD
Line DA
Check the closure condition
Adjusted
Line Lat (ft) Dep (ft)
AB -176.386 -438.574
BC +203.382 -73.105
CD +192.340 +198.635
DE -219.336 +313.044
sums: 0.000 0.000
check check
A common mistake is to forget to negate the Lat err and Dep err in the correction equations. If that happens, the closure condition will be twice what it originally was as the corrections were applied in the wrong direction.
b. Compute adjusted lengths and directions
Line AB
Adj Lat = -176.386 <- South
Adj Dep = -438.574 <- West
Because it’s the SW quadrant, Brng = S 68°05’27.4″ W.
Line BC
Adj Lat = +203.382 <- North
Adj Dep = -73.105 <- West
Because it’s the NW quadrant, Brng = N 19°46’14.9″ W
Line CD
Adj Lat = +192.340 <- North
Adj Dep = +198.635 <- East
Because it’s the NE quadrant, Brng = N 45°55’20.7″ E
Line DA
Adj Lat = -219.336 <- South
Adj Dep = +313.044 <- East
Because it’s the SE quadrant, Brng = S 54°58’58.0″ E
Adjustment summary
Adjusted Adjusted
Line Lat (ft) Dep (ft) Length Bearing
AB -176.386 -438.574 472.715 S 68°05’27.4″ W
BC +203.382 -73.105 216.122 N 19°46’14.9″ W
CD +192.340 +198.635 276.479 N 45°55’20.7″ E
DE -219.336 +313.044 382.237 S 54°58’58.0″ E
2. Example with Azimuths
Line Azimuth Length (ft) Lat (ft) Dep (ft)
ST 309°05’38” 347.00 +218.816 -269.311
TU 258°34’22” 364.55 -72.226 -357.324
UV 128°04’44” 472.74 -291.560 +372.123
VS 60°21’26” 292.94 +144.885 +254.602
sums: 1477.23 -0.085 +0.090
Distance Lat err
too far South Dep err
too far East
a. Adjust the Lats and Deps
Line ST
Line TU
Line UV
Line VS
Check the closure condition
Adjusted
Line Lat (ft) Dep (ft)
ST +218.836 -269.332
TU -72.205 -357.346
UV -291.533 +372.094
VS +144.902 +254.584
sums: 0.000 0.000
check check
b. Compute adjusted lengths and directions
Line ST
Adj Lat = +218.836 <- North
Adj Dep = -269.332 <- West
Because it’s in the NW quadrant: Az = 360°00’00″+(-50°54’20.4″) =309°05’39.6″
Line TU
Adj Lat = -72.205 <- South
Adj Dep = -357.346 <- West
Because it’s in the SW quadrant: Az = 180°00’00″+(78°34’36.0″) = 258°34’36.0″
Line UV
Adj Lat = -291.533 <- South
Adj Dep = +372.094 <- East
Because it’s in the SE quadrant: Az = 180°00’00″+(-51°55’17.6″) = 128°04’42.4″
Line VT
Adj Lat = +144.902 <- North
Adj Dep = +254.584 <- East
Because it’s in the NE quadrant: Az = 60°21’09.7″
Adjustment summary:
Adjusted Adjusted
Line Lat (ft) Dep (ft) Length (ft) Azimuth
ST +218.836 -269.332 347.029 309°05’39.6″
TU -72.205 -357.346 364.568 258°34’36.0″
UV -291.533 +372.094 472.700 128°04’42.4″
VS +144.902 +254.584 292.933 60°21’09.7″
3. Crossing Loop Traverse
As long as a traverse closes back on its beginning point, it can be adjusted the same as any other loop traverse.
Line Azimuth Length (ft) Lat (ft) Dep (ft)
EF 133°02’45” 455.30 -310.780 +332.737
FG 24°33’35” 228.35 +207.691 +94.912
GH 241°05’15” 422.78 -204.403 -370.084
HE 349°25’20” 312.85 +307.534 -57.430
sums: 1419.28 +0.042 +0.135
Dist Lat err
too far N Dep err
too far E
Adjust and recompute each line.
Equations:
Line EF
Because it’s in the SE quadrant: Az = 180°00’00″+(-46°56’57.1″) = 133°03’02.9″
Line FG
Because it’s in the NE quadrant: Az = 24°33’19.7″
Line GH
Because it’s in the SW quadrant: Az = 180°00’00″+(61°05’18.8″) =241°05’18.8″
Line HE
Because it’s in the NW quadrant: Az = 360°00’00″+(-10°35’00.5″) = 349°24’59.5″
Adjustment summary:
Adjusted Adjusted
Line Lat (ft) Dep (ft) Length (ft) Azimuth
EF -310.794 +332.694 455.278 133°03’02.9″
FG +207.684 +94.890 228.335 24°33’19.7″
GH -204.416 -370.124 422.821 241°05’18.8″
HE +307.525 -57.460 312.847 349°24’59.5′
sums: -0.001 0.000
check (rounding) check
Coordinates
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A. General
Point position can be expressed in relative or absolute terms
1. Relative position
A relative position is the either the (a) length and direction, or, (b) latitude and departure between two points.
Give the following adjusted traverse:
(a) Lengths and directions around a traverse defines the relative location successive traverse points.
Point B is 472.72 feet from A at a bearing of S 68°05’27″W.
Point C is 216.12 feet from B at a bearing of N 19°46’15″W.
(b) In terms of latitudes and departures
Point B is 176.39 ft South and 438.57 ft West of A.
Where is point C relative to A? Because the two points are not directly connected on the traverse, it requires a little more computing.
Lat A to C: [-472.72 ft x cos(68°05’27”) + 216.12 ft x cos(19°46’15”)] = +26.99 ft
Dep A to C: [-472.72 ft x sin(68°05’27”) – 216.12 ft x sin(19°46’15”)] = -511.68 ft
Point C is 28.99 ft North and 511.68 ft West of point A. We could compute the lats and deps through point D instead of B – the distances from point A to C would be the same.
2. Absolute Position
An absolute position is a distance from a datum. In the case of a traverse point, two horizontal lines serve as the data. One line corresponds with the meridian, the other is perpendicular to it:
The meridional line is called either the Y or North (N) axis; the other the X or East (E) axis.
A point position is expressed as a coordinate pair are represent perpendicular distances from the two axes.
For example:
The coordinates of point P are X=225.64′ and Y=320.95′
In terms of N & E:
The coordinates of point P are E=225.64′ and N=320.95′
B. Coordinates Computations
1. Forward Computation
A forward computation uses a starting coordinate pair along with a distance and direction to determine another coordinate pair.
Starting with coordinates at P, computer coordinates at Q
To compute coordinates:
If X and Y axes are used:
For a complete traverse:
Start with known coordinates at T: NT, ET
Compute coordinates of Q:
Compute coordinates of R:
Compute coordinates of S:
Compute coordinates of T:
Computing back into T gives a math check: your end coordinates should be the same as the start coordinates.
In order for the math check to be met, adjusted lats and deps must be used.
Where do the start coordinates come from? They can be assumed or they could be from a formal coordinate system. We’ll discuss formal coordinate systems in a later chapter.
2. Inverse Computation
An inverse computation is used to determine the distance and direction between two coordinate pairs. The computations involved are basically the same as those for determining a line’s new length and direction from its adjusted lats and deps.
For the traverse shown below, how would we determine the length and direction of the line from point T to R?
Knowing the coordinates of the two points, we can determine the latitude and departure of the line from the coordinate differences:
Note that the differences are the To point minus the From point.
and
;
The mathematic signs on the coordinate differences determine the direction quadrant.
If X and Y coordinates are used, remember that Y corresponds to N and X corresponds to E:
C. Examples
1. Traverse 1
a. Forward Computation
Adjusted
Line Lat (ft) Dep (ft)
AB -176.386 -438.574
BC +203.382 -73.105
CD +192.340 +198.635
DE -219.336 +313.044
The coordinates of point A are 2000.000′ N, 500.000′ E.
Compute the coordinates of the remaining points.
b. Inverse Computation
What are the length ad bearing of the line A to C
Because ΔN is North and ΔE is West: N86°58’47.6″W
Line AC: 512.39′ at N86°58’48″W
2. Traverse 2
a. Forward Computation
The coordinates of point E are 200.000′ X, 1000.000′ Y.
Compute the coordinates of the remaining points.
b. Inverse Computation
Determine the length and azimuth of the line from F to H.
Remember: Y=>N; X=>E
Because ΔY is North and ΔX is West:
Line FH: 275.25′ and azimuth of 270°40’49”.
D. Closing
Although computing coordinates involves addition computations, they provide ability to indirectly determine other values. We saw examples here where the distance and direction can be determined between points not directly connected on a traverse. This is very useful where lines may be obstructed but we still need to determine them. We will see in the next section how coordinates facilitate area computation.
Later still in the Coordinate Geometry chapter, we’ll see how integrating forward and inverse computations with trigonometric principles allow for greater field measurement and computational flexibility.
J. Mahun 06 Jun 2012
Coordinate Geometry Basics – Area of polygons
Welcome today’s lesson. In this lesson, we’ll establish the formula for finding out the area of a polygon, whose vertices are given. I’ll start with the triangle. You’ve already seen one (tedious) method of finding the area, which involved the distance formula. Here is a better one.
Consider a triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3). Lets start by making some construction.
Fig. 1: Area of a Triangle
The idea here is to express the area of the triangle ABC, in terms of other areas which can be calculatedmuch easily. These are the areas of the trapeziums ABED, ADFC and BEFC. The required area , Ar(ABC) = Ar(ABED) + Ar(ADFC) – Ar(BEFC)
Now, Ar(ABED) = ED.(BE + AD)/2 which equals (x2 – x1)(y2 + y3)/2
Similarly, the other areas can be found out, and after putting in the values we get, Ar(ABC) = 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
This can also be written as 1/2[(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y1 – x1y3)].
The former expression is easier to remember. Note that this expression assumes that A,B and C are in anticlockwise order. Had they been clockwise order, this expression would have given the same answer with a negative sign. So we have two options to get the right answer, either ensure that the points are taken in anticlockwise order, or take the modulus of the expression.
Moving on to polygons. The method of finding the area remains the same. I’ll first illustrate finding out the area of a quadrilateral, and generalize the method it to a n-sided polygon.
Fig. 2: Area of a quadrilateral
The required area can again be expressed in the terms of the trapeziums’ areas. Ar(ABCD) = Ar(ABFE) + Ar(AEHD) – Ar(BFGC) – Ar(CGHD)
Putting in the values, we get Ar(ABCD)=1/2[(x1y2 – x2y1) + (x2y3 – x3y2) + (x3y4 – x4y3) + (x4y1 – x1y4)](similar to the second expression I mentioned for the triangle)
Similarly, the area of an n-sided polygon whose vertices (taken in anticlockwise order) are (x1, y1), (x2, y2) … (xn, yn) is given by 1/2[(x1y2 – x2y1) + (x2y3 – x3y2) + …… + (xny1 – x1yn)]
That’s it for now!