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 separation | 3 m |
| Wave velocity | 5300 m/s |
| Arrival time difference (t2 - t1) | 200 us |
| Distance from sensor 1 | 0.97 m |
| Distance from sensor 2 | 2.03 m |
| Offset from array mid-point | 0.53 m |
| Saturation time difference | 566 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.