ToFD depth from time of flight

In time-of-flight diffraction the transmitter and receiver sit on opposite sides of the weld at a fixed probe centre separation (PCS). Energy diffracted from a crack tip travels from the transmitter index point down to the tip and back up to the receiver index point. With the tip on the centreline between the probes the two legs are equal, so the total path is 2 x sqrt(S squared + d squared), where S is half the PCS and d is the depth of the tip below the scanning surface.

Inverting that path for depth gives d = sqrt((c*t/2)^2 - S^2). The time must first be corrected for the delay through the two wedges, because the instrument time zero is the transmit pulse and not the entry point. ToFD works on the compression (longitudinal) wave: the diffracted longitudinal signal is the first arrival at the receiver, which is what makes the timing unambiguous. Entering a shear velocity here shortens the calculated half path until it is less than the half separation, so the calculation returns no solution at all for any ordinary geometry, and a grossly under-estimated depth in the few cases where it does resolve. It is always wrong for ToFD.

The relationship is strongly non-linear. Near the scanning surface a large change in depth produces only a small change in arrival time, so near-surface depth accuracy is poor and very sensitive to PCS error. Deeper in the wall the curve straightens and depth resolution improves. That is why the PCS is normally set so the beams cross at about two-thirds of the wall, and why the lateral wave and backwall are used to prove velocity and delay before any depth is reported.

Depths calculated this way assume the diffractor lies on the centreline between the probes. A tip offset across the weld has a longer path and therefore always reads deeper than it really is.

Compression wave

Worked example

Probe centre separation70 mm
Measured arrival time15.1 µs
Total probe delay2.5 µs
Compression velocity5900 m/s
Half separation35 mm
Delay-corrected time of flight12.6 µs
Half sound path37.17 mm
Depth below scanning surface12.51 mm

S = 70/2 = 35 mm. c = 5900 m/s = 5.9 mm/us. t_eff = 15.10 - 2.50 = 12.60 us. Half path = 5.9 x 12.60 / 2 = 37.17 mm. d = sqrt(37.17^2 - 35^2) = sqrt(1381.6089 - 1225) = sqrt(156.6089) = 12.5143 mm, so 12.51 mm to 2 dp.

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