What is the formula for tofd depth resolution against depth?

S = PCS / 2; dt/dd = 2 × d / (c × √(S² + d²)); Δd = Δt × c × √(S² + d²) / (2 × d); τ = n_cycles / f; h_min = τ × c × √(S² + d²) / (2 × d). Converts a timing uncertainty into a depth uncertainty at a stated depth, using the gradient of the ToFD time-to-depth curve.

What should I check before relying on the result?

Timing contribution only. PCS error, velocity error and off-centreline position all add on top of this figure. Enter a realistic timing uncertainty: the digitisation interval plus operator cursor repeatability, not the digitisation interval alone. Deep in the wall the uncertainty tends to a floor of delta_t times c divided by two, independent of depth and PCS. Through-wall height is usually more accurate than absolute depth because part of the systematic error cancels between the two tip readings. The minimum resolvable tip separation is set by the pulse duration, not by the digitiser. A flaw of smaller through-wall height than h_min cannot be tip-sized at that depth, whatever the timing uncertainty. Verify against the governing edition: General engineering (differentiation of the ToFD depth equation); Charlesworth & Temple, Engineering Applications of Ultrasonic Time-of-Flight Diffract…

ToFD depth resolution against depth

Differentiating the ToFD depth relation shows how much arrival-time change a millimetre of depth actually produces: dt/dd = 2*d / (c*sqrt(S^2 + d^2)). Inverting it gives the depth uncertainty produced by a timing uncertainty, delta_d = delta_t * c * sqrt(S^2 + d^2) / (2*d).

The gradient goes to zero at the scanning surface, so near-surface depth uncertainty is unbounded: at a 70 mm PCS a diffractor at 1 mm and one at 3 mm differ by under 40 nanoseconds of arrival time. Deep in the wall the gradient tends to 2/c and the uncertainty tends to delta_t*c/2, which for steel and a 20 nanosecond timing uncertainty is about 0.06 mm. This asymmetry is why ToFD sizes mid-wall and lower-wall flaws far better than shallow ones, and why through-wall height of a deep flaw is more trustworthy than the absolute depth of a shallow one.

The timing uncertainty to enter is not only the digitiser step. It combines the digitisation interval, the operator repeatability in placing a cursor on a diffracted signal (typically a fraction of a cycle), and any jitter in the trigger. Using one digitisation interval alone gives an optimistic answer.

This figure is only the timing contribution. Errors in the PCS, in the assumed velocity, in probe alignment and in flaw position off the centreline all add to it, and near the surface the PCS term normally dominates. Quote depth accuracy from measurements on a reference block containing targets at known depths, and use this calculation to understand where the errors come from.

Timing uncertainty is not the only clock-related limit. The transmitted pulse has a finite duration tau = n_cycles/f, and two diffracted signals closer in time than that duration merge into one. Feeding tau through the same gradient gives the minimum resolvable separation of two tips at a given depth, h_min = tau*c*sqrt(S^2 + d^2)/(2*d). With a 70 mm PCS and a well damped two-cycle 5 MHz pulse that is about 3.6 mm at 12 mm depth: a flaw of smaller through-wall height there shows a single merged signal and cannot be sized from its tips at all, however fine the digitiser.

Compression wave

Worked example

Probe centre separation	70 mm
Depth of interest	12 mm
Timing uncertainty	0.02 µs
Compression velocity	5900 m/s
Probe centre frequency	5 MHz
Cycles in the pulse	2
Half separation	35 mm
Half sound path at that depth	37 mm
Time gradient dt/dd	0.1099
Depth uncertainty	0.182 mm
Uncertainty as percent of depth	1.52 %
Pulse duration	0.4 µs
Minimum resolvable tip separation	3.64 mm

S = 35 mm and c = 5.9 mm/us. R = sqrt(35^2 + 12^2) = sqrt(1225 + 144) = sqrt(1369) = 37.00 mm exactly. Gradient = 2 x 12 / (5.9 x 37) = 24 / 218.3 = 0.10994 us/mm, so 0.1099 to 4 dp. Depth uncertainty = 0.02 x 5.9 x 37 / 24 = 4.366 / 24 = 0.18192 mm, that is 0.182 mm, which is 100 x 0.18192 / 12 = 1.5160 percent, so 1.52 percent. Pulse-limited two-tip resolution: tau = 2/5 = 0.400 us, so h_min = 0.400 x 5.9 x 37 / (2 x 12) = 87.32 / 24 = 3.6383 mm, that is 3.64 mm — a flaw of smaller height at 12 mm depth shows one merged tip signal.

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