Resolution and the wavelength problem

Vertical resolution: the $\lambda/4$ rule

Two reflectors closer in time than about a quarter of a wavelength cannot be resolved: their wavelets overlap and appear as a single event.

The wavelength in the rock is $\lambda = V / f_{\mathrm d}$, where $V$ is the interval velocity and $f_{\mathrm d}$ the dominant frequency, the inverse of the wavelet's dominant period $T_{\mathrm d}$ (for a Ricker wavelet, the time between its two troughs). The resolution threshold is:

For a 30 Hz dominant frequency in 3000 m/s rock, $\lambda = 100$ m and $\lambda/4 = 25$ m. Two reflectors closer than 25 m (about 17 ms of two-way time in this rock) appear as one event. Farther apart than that, they progressively separate.

Two cautions. First, the frequency in the rule is the dominant one. A Ricker wavelet is usually labelled by its peak frequency $f_{\mathrm p}$, and its dominant frequency is about 1.28 times higher, so $\lambda/4$ worked out with $f_{\mathrm p}$ overstates the limit by 28%. Second, the word "about" does real work. The precise value depends on wavelet shape, phase and noise: $\lambda/4$ is the classical (Rayleigh) rule of thumb, Widess put the limit for noise-free data at $\lambda/8$, and noisy data do worse than $\lambda/4$. Treat it as an order-of-magnitude constraint, not a sharp cutoff.

Tuning: resolution's twin

Below the resolution threshold, reflectors do not simply vanish: their wavelets interfere. Take a thin bed with a reflection coefficient $+R$ at its top and $-R$ at its base, a sand harder than the shales around it (a softer sand flips both signs and behaves the same way). As the bed thins, the side lobe of the base's wavelet lands on the main lobe of the top's, and the composite event grows brighter than either reflection alone. The amplitude peaks at the tuning thickness, where a Ricker wavelet gives about 1.45 times the amplitude of one interface. That thickness is half the wavelet's dominant period, $T_{\mathrm d}/2$ of two-way time, which is exactly $\lambda/4$ at the dominant frequency. Thinner still, the top and base wavelets cancel more and more, and the amplitude falls in proportion to the thickness.

This creates a trap. A thin sand at the tuning thickness produces a bright reflection that can be mistaken for a thicker, more significant body, or for a gas accumulation. Conversely, a much thinner bed can sink into the noise, and its absence on seismic does not mean the bed is absent.

Well above tuning, amplitude reflects impedance; below tuning, it reflects thickness

This is the most important sentence in Section 1.7.

For beds thicker than about twice the tuning thickness, peak amplitude approximates the true reflection coefficient of the top interface, and you read amplitude as a proxy for impedance contrast. Between tuning and twice tuning the side lobes still add: at 25 Hz a 20 ms bed reads 1.35 times as bright as one interface, and a 26 ms bed 1.12 times.

For beds thinner than the tuning thickness, peak amplitude varies mainly with bed thickness, not with impedance contrast, and you read amplitude as a proxy for thin-bed geometry. Two beds with the same impedance contrast but different thicknesses can have very different amplitudes, and two beds with different contrasts can have the same amplitude if their thicknesses differ in the right way. Any amplitude-to-rock-property interpretation in the thin-bed regime has to account for this.

Lateral resolution: the Fresnel zone

Vertical resolution gets most of the attention because the $\lambda/4$ rule is simple. But lateral resolution, the smallest feature that can be distinguished in map view, is just as important and is controlled by a different mechanism.

On unmigrated data, the Fresnel zone is the patch of the reflector whose echoes arrive within half a period of the first one, so that they interfere constructively at the receiver. Its radius is approximately

where $z$ is the depth to the reflector. For 30 Hz data at 2000 m depth in 3000 m/s rock, $\lambda = 100$ m and $r_F \approx \sqrt{100 \cdot 2000 / 2} = 316$ m. Anything smaller than the Fresnel-zone diameter (about 630 m) blurs together on unmigrated data.

Migration improves this dramatically. It collapses the Fresnel zone toward about half a wavelength, 50 m in this example, limited in practice by the migration aperture, the accuracy of the velocities and the noise. The survey bins must be small enough to sample that, which is one reason typical 3D bins are 25 × 25 m or smaller. A 2D line is migrated only along its own direction, so across the line the Fresnel zone stays. This is one of the strongest arguments for 3D surveys and for treating migration as non-negotiable in modern interpretation.

Detection vs. resolution

A subtle but important distinction. Resolution is the ability to see two features as separate. Detection is the ability to see that a feature is present at all. A thin bed below the tuning thickness is not resolved: you cannot see its top and base as separate events. But it can still be detected as an amplitude anomaly, as long as that amplitude stands above the noise. The wavelet sets the resolution; the noise sets the detection limit, which is often a small fraction of it. In the figure, a 25 Hz wavelet with noise at 10% of one interface detects beds down to 2 ms (3 m), while only beds thicker than about 16 ms are resolved.

This matters for reservoir characterization. A 5 m gas sand below the 25 m resolution limit will not show a clear top-and-base pair on seismic. But it will show an amplitude anomaly whose brightness depends on its thickness (through tuning) and on its impedance contrast (through the fluid). With calibration from well data, that anomaly can be inverted for sand thickness. The anomaly is detectable; the sand is not resolvable. Both facts are useful, and the interpreter needs to know which is which.