2.3. Unsolved Examples

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

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.

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.

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 + 1.50.90)*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

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.

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.

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

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.

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.

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

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.

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!

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

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.

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.

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