An HMX explosive having Velocity of Detonation (VOD) of 10500 m/s is tested by D'Auriche method with a detonating fuse of VOD 7000 m/s, as shown in the figure. The impression mark on the lead plate will be obtained at a distance \(L\), in m, from the midpoint of the fuse, is:
Show Hint
In D'Auriche method, the distance \( L \) is calculated by taking half the difference in the VOD of the explosive and the fuse, normalized by the VOD of the explosive.
In the D'Auriche method, the distance \( L \) is given by the following relationship:
\[
L = \frac{1}{2} \times \left( \frac{{VOD of explosive} - {VOD of fuse}}{{VOD of explosive}} \right) \times 1
\]
Step 1: Applying the given values.
The Velocity of Detonation (VOD) of the explosive is \( 10500 \, {m/s} \), and the VOD of the fuse is \( 7000 \, {m/s} \). Substituting these values into the equation:
\[
L = \frac{1}{2} \times \left( \frac{10500 - 7000}{10500} \right) \times 1
\]
Step 2: Calculation.
Now, simplify the equation:
\[
L = \frac{1}{2} \times \left( \frac{3500}{10500} \right) = \frac{1}{2} \times 0.3333 = 0.35 \, {m}
\]
Conclusion: The distance \( L \) from the midpoint of the fuse to the impression mark on the lead plate is \( \mathbf{0.35} \, {m} \).
