An eddy current instrument samples at a fixed rate while the probe moves at some speed, so the data is a series of points spaced v / rate apart along the scan. A flaw of length L is therefore represented by L x rate / v samples. Below about three samples across a flaw the peak is systematically under-recorded: the sampling can miss the maximum, the indication looks smaller than it is, and a flaw at the reporting threshold drops below it. Requiring a minimum number of samples across the shortest reportable flaw and solving for speed gives v_max = rate x L / n.

This is the arithmetic behind the scan speed limit in a procedure. It is also why raising the scan speed is not a free productivity gain: doubling the speed halves the samples across every flaw, and the flaws that disappear first are the small ones near the acceptance limit, which are exactly the ones the examination exists to find.

Two practical qualifications. First, the rate that matters is the rate each channel actually receives – on a multiplexed array that is the instrument rate divided by the number of firing groups, and on a rotating probe it is set by the rotation speed and the helical advance. Second, the instrument low-pass filter must pass the flaw frequency: a flaw of length L passing at speed v produces a signal at roughly v / L hertz, and a filter set for a slower scan will attenuate it regardless of how densely the data is sampled.

Use this before mobilising to check that the planned production rate is compatible with the procedure, and afterwards to defend or challenge a data set whose encoder record shows speeds above the qualified limit.

Worked example

Acquisition rate per channel	500 Hz
Smallest reportable flaw length	5 mm
Samples required across the flaw	3
Planned scan speed	500 mm/s
Scan length	6000 mm
Maximum scan speed	833.3 mm/s
Maximum sample spacing	1.667 mm
Sample spacing at planned speed	1 mm
Samples across flaw at planned speed	5 samples
Flaw signal frequency	100 Hz
Time per pass	12 s

500 samples per second per channel, a 5 mm smallest reportable flaw and a requirement for at least 3 samples across it. v_max = 500 x 5 / 3 = 833.3 mm/s, corresponding to a maximum sample spacing of 5/3 = 1.667 mm. At the planned 500 mm/s the spacing is 500/500 = 1.000 mm, so the 5 mm flaw receives 5.00 samples - comfortably above the 3 required. The flaw signal envelope is about 500/5 = 100 Hz, so the low-pass filter must be set above that. A 6000 mm tube takes 6000/500 = 12.0 s per pass.