To find the volume corresponding to a sounding of 4.13 m, we can use linear interpolation between the two given values:

• Sounding = 4.10 m, Volume = 120.91 m³
• Sounding = 4.15 m, Volume = 124.27 m³

Given:

• You want to find the volume at Sounding = 4.13 m.

The formula for linear interpolation is:

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

Where:

• $S$ = 4.13 m (the sounding for which we need to find the volume)
• $S_1 = 4.10$ m, $V_1 = 120.91$ m³
• $S_2 = 4.15$ m, $V_2 = 124.27$ m³

Now, we can calculate the volume $V$. I'll compute it.The volume corresponding to a sounding of 4.13 meters is approximately 122.93 m³.

Would you like further details or clarifications? Here are some related questions to expand your understanding:

1. How would the volume change for a sounding of 4.12 meters using the same method?
2. Can you apply linear interpolation in other situations, like temperature or pressure tables?
3. What assumptions do we make when using linear interpolation in this context?
4. How would the result differ if the values were not linearly related?
5. Can we calculate this volume using a different interpolation method, like quadratic interpolation?

Tip: When using interpolation, make sure the relationship between the variables is approximately linear, otherwise the result might not be accurate.