Unsolved Examples - 2.3 | 1. Examples and Solutions | Surveying and Geomatics
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Area Calculation Methods

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0:00
Teacher
Teacher

Today, we are going to explore two important methods for calculating areas: the Trapezoidal rule and Simpson’s rule. Can anyone tell me what these methods are generally used for?

Student 1
Student 1

They are used for finding the area of shapes, especially when dealing with curves.

Teacher
Teacher

Correct! The Trapezoidal rule approximates the area under curves by dividing it into trapezoids, while Simpson’s rule uses parabolic arcs. Let's break down the calculations using our examples.

Student 2
Student 2

How do you decide which method to use?

Teacher
Teacher

Great question, Student_2! Simpson’s rule is generally more accurate than the Trapezoidal rule but requires an even number of intervals. Let's go through the calculations step by step, ensuring clarity in our work.

Teacher
Teacher

For instance, if the offsets are 0, 2.50, 3.50, and so on, using Trapezoidal rule gives us an area of 188 m², while Simpson’s rule yields 196.66 m². Who can summarize why these results might differ?

Student 3
Student 3

The difference may arise due to the shapes of the sections being modeled. Simpson’s rule can model curves better.

Teacher
Teacher

Exactly! Now, let’s summarize the key points: the Trapezoidal rule approximates areas using straight lines, while Simpson’s rule uses curves for a better fit.

Volume of Earthwork Calculations

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Teacher
Teacher

Continuing our conversation, let's discuss how to calculate the volume of earthwork. We have a problem that involves calculating the volume for an embankment. What formulas do you think we should use?

Student 4
Student 4

We should use the Trapezoidal formula and possibly the Prismoidal formula for better accuracy.

Teacher
Teacher

Absolutely correct! The volume of earthwork can be estimated using both. The Trapezoidal formula works well for relatively straight cross-sections, while the Prismoidal formula accounts for curves, making it more suitable for irregular shapes. Let's practice using the provided heights and widths.

Student 1
Student 1

How do we find the cross-sectional areas needed for calculations?

Teacher
Teacher

We find cross-sectional areas through the area formula for trapezoids, which includes base width, height, and side slopes. Use the formula ∆ = (b + sh) * h, where b is the base and sh is the slope height.

Student 2
Student 2

What are typical results when we finish these calculations?

Teacher
Teacher

For example, we calculated volumes of 5096.4 m³ for the Trapezoidal formula and 5142.9 m³ from the Prismoidal formula in our earlier discussion. Each method offers unique insights and levels of precision for our projects.

Teacher
Teacher

Let’s recap: we calculate volume using either formula depending on the shape we're working with.

Understanding Tacheometry

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Teacher
Teacher

Switching gears, let's dive into tacheometry. Does anyone know what tacheometry is used for in surveying?

Student 3
Student 3

It's used to determine distances and elevations without having to directly measure them.

Teacher
Teacher

Exactly! It's a time-saving technique. To calculate the constants used in tacheometry, we equate distances based on staff readings and horizontal distances. Can someone provide a summary of the steps?

Student 4
Student 4

You set up equations to derive K and C constants using readings at different horizontal distances and staff readings.

Teacher
Teacher

Correct! Once we have K and C, we can easily calculate distances to points based on staff readings. For instance, if we have a reading of 2.750 m and a known height, we can find horizontal distances.

Student 1
Student 1

What happens if the staff is held vertically?

Teacher
Teacher

When the staff is vertical, the calculations simplify significantly, as we only need to factor in the height of the instrument. Let’s finish with a recap of these key points.

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section explores various unsolved examples focusing on key surveying concepts, including area computations and tacheometry.

Standard

This section contains a series of unsolved examples in surveying that demonstrate the application of different methods for calculating areas and volumes, as well as addressing practical tacheometric problems.

Detailed

Detailed Summary

This section presents a collection of unsolved examples that encapsulate key concepts in surveying. Each example illustrates the practical application of essential surveying techniques such as the Trapezoidal and Simpson’s rules for area calculations, as well as tacheometry principles.

Key Concepts Covered:

  1. Area Computation: Understanding the use of the Trapezoidal and Simpson’s rules in calculating the area between a chain line and an irregular boundary. The section begins with offsets taken from a chain line to an irregular boundary line, leading to practical computations of areas, such as 188 m² using the Trapezoidal rule and 196.66 m² using Simpson's rule.
  2. Volume of Earthwork: The section continues to demonstrate how to compute the volume of earthwork needed for an embankment using both the Trapezoidal and Prismoidal formulas, highlighting the significance of cross-sectional areas in volume calculations.
  3. Tacheometric Measurements: The examples also cover how to calculate horizontal distances and reduced levels using tacheometric readings, demonstrating the concepts of tacheometric constants effectively.
  4. Trigonometric Applications: Lastly, it includes practical applications involving angles of depression and elevation, showcasing real-world scenarios that students may encounter in surveying practice.

Overall, this section emphasizes practical problem-solving skills by illustrating various unsolved examples that require students to think critically about the principles and methods discussed in preceding chapters.

Audio Book

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Example 1.19: Area Computation Using Trapezoidal and Simpson’s Rule

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The following offsets were taken from a chain line to an irregular boundary line at an interval of 10 m: 0, 2.50, 3.50, 5.00, 4.60, 3.20, and 0 m. Compute the area between the chain line, the irregular boundary line and the end of offsets by: (a) the Trapezoidal rule, and (b) Simpson’s rule.

Solution:
By Trapezoidal rule:
Here d=10 m
Required area=10 / 2{0 + 0 + 2(2.50 + 3.50 + 5.00 + 4.60 + 3.20)}
= 5*37.60 =188 m²

By Simpson’s rule:
d=10 m
Required area=10/ 3{0 + 0 + 4(2.50 + 5.00 + 3.20) + 2(3.50 + 4.60)}
= 10/3{42.80 +16.20}=10/3*59.00
= 196.66 m²

Detailed Explanation

This example involves calculating the area between a chain line and an irregular boundary using two methods: the Trapezoidal rule and Simpson’s rule. The Trapezoidal rule is used first, which involves taking the average of the bases (the chain line segments) and multiplying by the interval. Simpson’s rule is then applied, which takes into account the weights of the areas more accurately, especially when the area is not a perfect rectangle. Both methods yield different results for comparison: 188 m² and 196.66 m² respectively.

Examples & Analogies

Imagine you are trying to find the area of a garden that isn't a simple rectangle. You could lay down a long measuring tape (like the chain line) and measure how far the garden grows at regular intervals (like the offsets). Using different methods to estimate the area—one simpler, which just averages the lengths between points, and another that considers the shape more accurately—helps you ensure you cover all the garden and not underestimate its size.

Example 1.20: Volume Calculation of Soil Embankment

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An embankment of width 10 m and side slopes 1½:1 is required to be made on a ground which is level in a direction transverse to the centre line. The central heights at 40 m intervals are as: 0.90,1.25,2.15,2.50,1.85,1.35, and 0.85 m, calculate the volume of earth work according to (i) Trapezoidal formula, and (ii) Prismoidal formula.

Solution:
The cross-sections areas are calculated by ∆= (b + sh)h
∆1= (10 + 1.5
0.90)*0.90 = 10.22 m²
...
V according to trapezoidal formula:
= 40 / 2{10.22 + 9.58 + 2(14.84 + 28.43 + 34.38 + 23.63 + 16.23)}
= 20{19.80 + 235.02} = 5096.4 m³

V calculated in prismoidal formula:
= 40 / 3 {10.22 + 9.58 + 4(14.84 + 34.38 +16.23) + 2(28.43 + 23.63)}
= 40 / 3 (19.80 + 261.80 +104.12) = 5142.9 m³

Detailed Explanation

This example focuses on computing the volume of soil that needs to be moved to create an embankment. The area of the cross-section of the embankment is calculated using the formula for trapezoids based on the width and heights at various intervals. Both the trapezoidal formula and the prismoidal formula are used to calculate the volume, where the trapezoidal formula provides a simpler estimation while the prismoidal formula offers more accuracy considering the shape of the embankment over lengths.

Examples & Analogies

Think about building a hill out of sand in your backyard. You could estimate how much sand you'll need by measuring the width and height at various points, similar to gathering data at levels of the embankment. The trapezoidal method is like saying 'I just want to find a flat area on average,' while the prismoidal method incorporates how the hill slopes and curves, giving you a better idea of that sand scoop you’ll need.

Example 1.21: Volume of Water Calculation Using Contours

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The areas enclosed by the contours in the lake are as follows: Contour (m) Area (m²) 270 2050 275 8400 280 16300 285 24600 290 31500 Calculate the volume of water between the contours 270 m and 290 m by Trapezoidal formula.

Solution:
Volume according to trapezoidal formula:
= 5 / 2{2050 + 31500 + 2(8400 + 16300 + 24600)}
= 330,250 m³

Detailed Explanation

The example illustrates the calculation of the volume of water between various contour levels in a lake. By using the trapezoidal formula, the volume is computed from the areas specified at each contour. The formula essentially averages the areas at both contour lines and sums the contributions from additional contours between them to determine the total volume of water.

Examples & Analogies

Picture a swimming pool with varying depths. Using contours is like mapping out the depths at different points to get an average understanding of how deep the water is overall. By averaging the depths (areas), we can calculate how much water is actually in the pool—a practical approach used in designing everything from irrigation systems to flood control.

Example 1.22: Tacheometric Constants Calculation

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The following readings were taken with a tacheometer on to a vertical staff, calculate the tacheometric constants.
Horizontal Distance (m) Stadia Readings (m) 45.00 0.885 1.110 1.335 60.00 1.860 2.160 2.460

Solution:
D = KS + C
1 1
D = KS + C
2 2
... K = 100
Now put the value of K in either equation 3 or 4, we get C =0

Detailed Explanation

In this example, the process of determining the tacheometric constants K and C from measurements taken with a tacheometer is explained. By setting up the equations based on two sets of observations, students can understand how to derive these constants, which are critical for distance measurement in surveying. Finding K and C allows the surveyor to convert stadia readings directly into horizontal distances.

Examples & Analogies

Imagine using a laser measuring tool to gauge distances to a distant tree. If you find that lengths measured at specific angles consistently yield a formula (like K and C), you can use that to simplify future measurements. This consistency leads to speedy and reliable measurements, crucial for any construction or land surveying project!

Example 1.23: Horizontal Distance and Reduced Level Calculation

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The tacheometer instrument was setup over a station P of RL 1850.95 m and the height of instrument above P was 1.475 m. The staff was held vertical at a station Q and the three readings were 1.050, 1.900 and 2.750 m with the line of sight horizontal. Calculate the horizontal distance of PQ and RL of Q point.

Solution:
D = KS + C
Now, S = 2.750 – 1.050 = 1.700 m
D = 100 (1.700) + 0 = 170 m
RL of Q = 1850.95 + 1.475 – 1.900
= 1850.525 m

Detailed Explanation

This example shows how to calculate both the horizontal distance and the reduced level at a point when using a tacheometer. By recording the staff readings and calculating the difference between those readings, the surveyor can find the horizontal distance to the observed point. The process concludes with the calculation of the reduced level at point Q using the readings and the known height of the instrument.

Examples & Analogies

If you're navigating in a dense forest, having a map and knowing the height of trees (like the staff readings) helps you estimate how far away certain points are without having a clear line of sight. When you know how high your view point is (similar to the instrument's height), you can piece together the vertical difference and get a clearer picture of how far your destination truly is.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Area Computation: Understanding the use of the Trapezoidal and Simpson’s rules in calculating the area between a chain line and an irregular boundary. The section begins with offsets taken from a chain line to an irregular boundary line, leading to practical computations of areas, such as 188 m² using the Trapezoidal rule and 196.66 m² using Simpson's rule.

  • Volume of Earthwork: The section continues to demonstrate how to compute the volume of earthwork needed for an embankment using both the Trapezoidal and Prismoidal formulas, highlighting the significance of cross-sectional areas in volume calculations.

  • Tacheometric Measurements: The examples also cover how to calculate horizontal distances and reduced levels using tacheometric readings, demonstrating the concepts of tacheometric constants effectively.

  • Trigonometric Applications: Lastly, it includes practical applications involving angles of depression and elevation, showcasing real-world scenarios that students may encounter in surveying practice.

  • Overall, this section emphasizes practical problem-solving skills by illustrating various unsolved examples that require students to think critically about the principles and methods discussed in preceding chapters.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • Example 1: Calculate the area between a chain line and an irregular boundary line using the Trapezoidal rule.

  • Example 2: Use the Prismoidal formula to compute the volume of earthwork for an embankment.

  • Example 3: Determine horizontal distance and reduced level using tacheometric readings.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎵 Rhymes Time

  • To find the area, we take a slice, with trapezoids or parabolas, which is nice!

📖 Fascinating Stories

  • Imagine a farmer measuring his field. He uses the Trapezoidal Rule to find out how much space he has for crops, and later updates his plan using Simpson's more precise method.

🧠 Other Memory Gems

  • TAPE: Trapezoidal Area for Precision Estimation.

🎯 Super Acronyms

T.E.A

  • Tacheometry
  • Elevation
  • Area - remember these surveying essentials!

Flash Cards

Review key concepts with flashcards.

Glossary of Terms

Review the Definitions for terms.

  • Term: Trapezoidal Rule

    Definition:

    A numerical method for estimating the area under a curve using trapezoidal segments.

  • Term: Simpson’s Rule

    Definition:

    A numerical method for estimating the area under a curve using parabolic segments.

  • Term: Tacheometry

    Definition:

    A surveying technique to determine distances and elevations using angular measurements.

  • Term: Earthwork Volume

    Definition:

    The quantity of soil or rock that needs to be excavated or filled for construction purposes.