AE linear source location

Linear location is the simplest acoustic emission location method. Two sensors are placed a known distance L apart on the item — along a pipe run, a weld seam, a bolt line — and the only measurement needed is the difference in first-threshold-crossing time between them, Δt. Because the wave travels at the same speed in both directions, that time difference converts directly into a path-length difference Δx = c·Δt, and the source must sit half of that difference away from the mid-point, toward whichever sensor triggered first.

The distance from the first-hit sensor is d₁ = (L − c·Δt)/2. Note the whole method rests on one number the technician has to earn: the velocity. AE in plate and pipe wall does not travel at the bulk longitudinal velocity — it propagates as guided (Lamb) modes whose speed depends on wall thickness, frequency and mode. The extensional (S₀) mode in thin carbon steel runs at roughly 5300 m/s, the flexural (A₀) mode is slower and strongly dispersive, and a surface (Rayleigh) wave sits near 3000 m/s. Always establish the velocity on the actual structure with Hsu–Nielsen pencil-lead breaks before trusting any location.

Two limits bound the result. If |c·Δt| equals L, the source is at (or beyond) one of the sensors and the arrival difference saturates — anything outside the array collapses onto the end sensor and cannot be located. And the location error is roughly c·δt/2, so a 10 µs timing uncertainty on steel at 5300 m/s is already ±27 mm of position error before geometry is considered. Threshold crossing on a slow-rising flexural arrival can easily cost far more than that, which is why velocity checks and consistent threshold settings matter more than arithmetic precision here.

Worked example

Sensor separation3 m
Wave velocity5300 m/s
Arrival time difference (t2 - t1)200 us
Distance from sensor 10.97 m
Distance from sensor 22.03 m
Offset from array mid-point0.53 m
Saturation time difference566 us

c·Δt = 5300 m/s x 200e-6 s = 1.060 m path difference. d1 = (3.000 − 1.060)/2 = 0.970 m from the first-hit sensor, so d2 = 3.000 − 0.970 = 2.030 m, i.e. 0.530 m off the mid-point. Saturation occurs at Δt = L/c = 3/5300 = 566.0 µs.

Use at your own risk — verify before you act

These calculators support, and never replace, the judgement of qualified NDT and engineering personnel. Results are provided as is, without warranty of any kind, express or implied, and must be independently verified against the governing code edition named in your contract before being used in any inspection, acceptance, rejection, radiation-safety or fitness-for-service decision. By using them you accept full responsibility for how the results are applied; NDT Inspect, its owners and contributors accept no liability for any loss, damage, injury or death arising from their use or from reliance on them. If a result matters to safety, check it by hand and have it reviewed by a competent person.

Privacy Overview

This website uses cookies so that we can provide you with the best user experience possible. Cookie information is stored in your browser and performs functions such as recognising you when you return to our website and helping our team to understand which sections of the website you find most interesting and useful.