The slope correction for a length of 30 m along a gradient of 1 in 20 is

📈 Understanding the Problem

This problem requires calculating the correction needed to convert a distance measured along a slope to its true horizontal equivalent. We are given:

• Slope Length (L): 30 m
• Gradient: 1 in 20. This means for every 20 units of horizontal distance, there is 1 unit of vertical rise. So, \( n = 20 \).

🔬 Detailed Calculation: Two Methods

We can solve this problem in two ways, both of which give the same result.

Method 1: Direct Formula Application

From the previous question, we derived the slope correction formula for a gradient of 1 in n:

$$ C \approx \frac{L}{2n^2} $$

Substituting the given values:

Method 2: Calculating from First Principles

This method involves finding the difference in height (h) first.

Step 1: Find the height difference (h)
Using the Pythagorean theorem, \( L^2 = H^2 + h^2 \). We also know the gradient relates H and h: \( H = 20h \). $$ 30^2 = (20h)^2 + h^2 $$ $$ 900 = 400h^2 + h^2 = 401h^2 $$ $$ h^2 = \frac{900}{401} \approx 2.244 $$

Step 2: Apply the fundamental correction formula
The most common formula for slope correction is \( C \approx h^2 / 2L \).

💡 Final Conversion and Conclusion

Both methods yield a result of approximately 0.0375 meters. The question asks for the answer in centimeters. To convert from meters to centimeters, we multiply by 100.

This matches option A. It's crucial to remember both the direct formula (\(L/2n^2\)) for speed and the fundamental formula (\(h^2/2L\)) for a deeper understanding.