To find the volume ($V$) corresponding to a sounding of 4.13 meters, we can use linear interpolation since the table gives us the values for soundings at 4.10 meters and 4.15 meters.

Given dataTo find the volume ($V$) corresponding to a sounding of 4.13 meters, we can use linear interpolation since the table gives us the values for soundings at 4.10 meters and 4.15 meters.

Given

• Sounding at 4.10 meters corresponds to a volume of 120.91 cubic meters.
• Sounding at 4.15 meters corresponds to a volume of 124.27 cubic meters.

The formula for linear interpolation is:

$V = V_1 + \frac{(V_2 - V_1)}{(S_2 - S_1)} \times (S - S_1)$

Where:

• $S_1 = 4.10 \, \text{m}$, $V_1 = 120.91 \, \text{m}^3$
• $S_2 = 4.15 \, \text{m}$, $V_2 = 124.27 \, \text{m}^3$
• $S = 4.13 \, \text{m}$ (the value for which we want to find the volume)

Let me calculate this for you.The volume corresponding to a sounding of 4.13 meters is approximately 122.93 cubic meters.

Would you like more details on how the interpolation works, or do you have any further questions?

Here are 5 related questions that can expand on this topic:

1. What is the importance of interpolation in engineering calculations?
2. How can interpolation be extended to multiple dimensions (e.g., bilinear interpolation)?
3. Can interpolation be used for non-linear data? If so, how?
4. What is the difference between interpolation and extrapolation?
5. How would this method change if we had a larger data set?

Tip: Interpolation assumes the data between two known points follows a linear trend. Always check the appropriateness of this assumption for your data.