ToFD near-surface dead zone
The lateral wave is not an instantaneous spike; it occupies a time equal to the duration of the transmitted pulse. Any diffracted signal arriving inside that interval is buried in it. Setting the arrival time of a tip at depth d equal to the end of the lateral wave pulse and solving for d gives the shallowest depth at which a diffractor can be resolved: d = sqrt(S*c*tau + (c*tau/2)^2), where tau is the pulse duration and S is half the probe centre separation.
Everything shallower than that is the near-surface dead zone, and it is usually the limiting factor of a ToFD examination. It sits exactly where many fabrication and service defects occur, including lack of side-wall fusion at the cap, undercut and toe cracking. Its depth grows with both the probe centre separation and the pulse duration, so a heavily damped higher-frequency pair at the smallest workable PCS gives the smallest dead zone.
Pulse duration is taken as the number of cycles in the transmitted pulse divided by the centre frequency. Two cycles at 5 MHz is 0.4 microseconds. Measure it on the actual A-scan rather than assuming it: wedge damping, cabling and receiver bandwidth all change the effective pulse length, and because the dead zone grows with the square root of the pulse duration, a probe that rings for four cycles instead of two makes it about 40 percent deeper.
The dead zone cannot be removed by processing, so ISO 10863 and ASME require it to be covered by a complementary technique, normally a pulse-echo or phased-array scan of the near surface, or a second ToFD setup dedicated to the upper zone. Calculate it, state it in the procedure and record it on the report.
Compression wave
Worked example
| Probe centre separation | 70 mm |
| Probe centre frequency | 5 MHz |
| Cycles in the pulse | 2 |
| Compression velocity | 5900 m/s |
| Half separation | 35 mm |
| Pulse duration | 0.4 µs |
| Pulse length in material | 2.36 mm |
| Near-surface dead zone depth | 9.16 mm |
S = 35 mm and c = 5.9 mm/us. Pulse duration tau = 2 cycles / 5 MHz = 0.400 us, which is 5.9 x 0.4 = 2.36 mm of material. d_dead = sqrt(35 x 2.36 + (2.36/2)^2) = sqrt(82.60 + 1.3924) = sqrt(83.9924) = 9.1647 mm, so 9.16 mm to 2 dp.