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Compute departures and latitudes, linear misclosure, and relative precision for the traverse of problem 10.6 if the lengths of the sides (in feet) are as follows: ab = 202.74; bc = 283.87; cd = 498.37; de = 320.33; and ea = 380.78.

User Yokogeri
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Final answer:

The departures and latitudes are as mentioned above, the linear misclosure is 1685.09 feet, and the relative precision is 1.

Step-by-step explanation:

To compute the departures and latitudes, linear misclosure, and relative precision for the traverse problem 10.6, we need to calculate the individual departure and latitude values for each side of the traverse.

Given the lengths of the sides of the traverse:

ab = 202.74 feet

bc = 283.87 feet

cd = 498.37 feet

de = 320.33 feet

ea = 380.78 feet

1. Departures:

The departure of a side is the horizontal distance between the starting and ending points.

- Departure of ab = ab = 202.74 feet

- Departure of bc = bc = 283.87 feet

- Departure of cd = cd = 498.37 feet

- Departure of de = de = 320.33 feet

- Departure of ea = ea = 380.78 feet

2. Latitudes:

The latitude of a side is the vertical distance between the starting and ending points.

- Latitude of ab = 0 feet (since it is a horizontal side)

- Latitude of bc = 0 feet (since it is a horizontal side)

- Latitude of cd = 0 feet (since it is a horizontal side)

- Latitude of de = 0 feet (since it is a horizontal side)

- Latitude of ea = 0 feet (since it is a horizontal side)

3. Linear Misclosure:

The linear misclosure is the algebraic sum of the departures, which should ideally be zero for a closed traverse.

Linear Misclosure = departure of ab + departure of bc + departure of cd + departure of de + departure of ea

Linear Misclosure = 202.74 + 283.87 + 498.37 + 320.33 + 380.78

Linear Misclosure = 1685.09 feet

4. Relative Precision:

The relative precision of the traverse can be calculated by dividing the linear misclosure by the perimeter of the traverse.

Perimeter of the traverse = ab + bc + cd + de + ea

Perimeter of the traverse = 202.74 + 283.87 + 498.37 + 320.33 + 380.78

Perimeter of the traverse = 1685.09 feet

Relative Precision = Linear Misclosure / Perimeter of the traverse

Relative Precision = 1685.09 / 1685.09

Relative Precision = 1

Therefore, for the given traverse problem 10.6, the departures and latitudes are as mentioned above, the linear misclosure is 1685.09 feet, and the relative precision is 1.

User Bigtlb
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