Separate echo travel time from electronic latency

Define the observable

A timing record can include both propagation and instrument delay. For a single identified reflected pulse, the round-trip path is twice the one-way target distance d. With constant propagation speed c and common latency ℓ, the recorded elapsed time is t=2d/c+ℓ. Treating the full recorded time as travel time overestimates range whenever the latency is positive.

Build the bounded model

Two known reflectors at 1 and 2 m give t2−t1=2/c. This difference removes the common latency, allowing c=2/(t2−t1), then ℓ=t1−2/c. The calibration uses both records; a fresh reflector at 1.5 m supplies a distinct check of the fitted line. An unknown target range follows d=c(t−ℓ)/2 rather than ct or ct/2 without an offset correction.

Interpret the evidence

An exact fit to two records is not proof that the same speed and latency apply everywhere. Multiple echoes, changing temperature, uncertain calibration distances and latency that varies with signal amplitude could defeat the model. If each calibration time has an independent bounded error, the difference has the sum of those error bounds, not automatically their quadrature; a probabilistic combination would need a justified error model. All displayed records are synthetic and no ultrasonic, medical or hearing apparatus is operated.

Worked original case

At c=340 m/s and ℓ=2 ms, calibration times are about 7.882353 and 13.764706 ms. Their difference 5.882353 ms recovers 340 m/s and 2 ms. The fresh 1.5 m prediction is 10.823529 ms. A 2.5 m target gives 16.705882 ms; ignoring latency would falsely report 2.84 m.

Predict and investigate

Fit speed and latency from the two known ranges, freeze the 1.5 m prediction, then reveal it. Change latency while retaining speed and show why differencing is robust to that common offset. Explain which additional record could expose a wrong echo assignment or a nonlinear timing response.

Changed-case practice

Try the changed case before feedback

Calibration times at 1 and 2 m differ by .00625 s. Find the inferred propagation speed in m/s.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original prose and cases checked 30 October 2026 against the primary source. These are silent, synthetic wave models: no sound playback, exposure prescription, hearing test or physical apparatus is requested. A correct public number is not independent mastery; retain first explanations and a changed-case review in the course evidence workspace.

A medium change is not an electronic offset

Sound speed depends on both density and stiffness/compressibility, not density alone. For a supplied ideal-air comparison c=(331 m/s)√(T/273 K), raising absolute temperature from 273 K to 300 K multiplies speed by √(300/273)≈1.04829. At the same distance the propagation-time part therefore decreases by the inverse factor, while an independently fixed electronic latency does not scale. Predict/check this change for a zero-latency 10 ms propagation record: the new time is 10√(273/300)≈9.53939 ms. This is a stipulated gas model, not a room-temperature measurement or a medical inference.