Sonic log · MunaiHub

Quick refresher

The first wave to reach the receiver is the compressional wave that has travelled through the rock along the borehole wall, because the rock is faster than the mud. The shear wave, the mud wave and the Stoneley wave follow it.
The interval transit time is the reciprocal of the velocity: 250 µs/m corresponds to 4 km/s. To go from µs/ft to µs/m, divide the value by 0.3048: 55.5 µs/ft is 182 µs/m.
Two receivers cancel the path of the wave through the mud: $\Delta t$ is the difference in arrival times divided by the span (the distance between the receivers). Two transmitters, one above and one below the receivers, cancel the effect of hole enlargements and of tool tilt. This is the borehole-compensated tool.
A weakened signal triggers the far receiver one period of the wave late. This is cycle skipping, and it makes a sharp spike of $\Delta t$ towards larger values. Gas, fractures, loose rocks and a bad hole cause it.
In rocks of the same composition and the same degree of compaction, $\Delta t$ grows almost linearly with porosity. This is why the sonic log is one of the three porosity methods. In loose sands, shaly beds and gas-bearing beds, the calculation needs corrections. Vugs hardly show in $\Delta t$ .
Interval transit times add up: their sum over the section is the travel time we need to tie seismic data to depths. We calculate dynamic elastic moduli from the $\Delta t$ of the compressional and shear waves and the density. In a cased well, the same method evaluates the cement.

Key relationships

\Delta t = \dfrac{10^6}{V}

interval transit time in µs/m for a velocity in m/s; in µs/ft for a velocity in feet per second

\Delta t = \dfrac{t_2 - t_1}{s}

from the arrival times at two receivers separated by the span $s$ ; in a compensated tool, the mean of two opposite measurements

\phi_S = \dfrac{\Delta t - \Delta t_{ma}}{\Delta t_f - \Delta t_{ma}}

the time-average equation, a first estimate of porosity in clean compacted rocks

t = \sum_i \Delta t_i\,h_i

the travel time through an interval is the sum over the layers; the average velocity equals the thickness of the interval divided by this time

G = \dfrac{\rho_b}{\Delta t_s^2}

dynamic shear modulus from the shear-wave interval transit time and the density, in SI units

Common pitfalls

Reading a scale in µs/ft as µs/m, or the other way round. The values differ by a factor of 3.28.
Taking a $\Delta t$ spike caused by cycle skipping for a thin high-porosity layer.
Believing that the compensated tool corrects everything. It does not remove cycle skipping, the arrival of the mud wave or the altered zone.
Calculating the average velocity of a section as the mean of the layer velocities. The times add up, not the velocities.

After this lesson you can

explain which elastic waves arise in the borehole and why the first to reach the receiver is the compressional wave that has travelled through the rock
convert the interval transit time to velocity and from µs/ft to µs/m, and recognise the main rocks by typical values of $\Delta t$
explain how two receivers remove the path of the wave through the mud and why a compensated tool needs two transmitters
recognise cycle skipping, the arrival of the mud wave and the effect of the altered zone on a log, and choose a tool that suits the conditions in the well
explain how the interval transit time is related to porosity and why we do not convert it to porosity without checking the quality of the record
calculate the travel time of a wave through a section for the tie to seismic, and the dynamic elastic moduli from the interval transit times and density

Elastic waves in the borehole

The transmitter of a sonic tool is a piezoceramic element (a ceramic that changes size when an electric signal reaches it). The signal makes it change size abruptly and push the surrounding liquid. In this way a short pulse goes into the drilling mud. Its frequency is about 10–20 kHz (10,000–20,000 oscillations per second). In the mud it is a pressure wave. When it reaches the borehole wall, it sets up elastic waves of two kinds in the rock. Along the hole it also sets up waves that belong to the borehole itself.

Compressional wave (P): The particles oscillate along the direction in which the wave travels, so the rock is alternately squeezed and stretched. It is the fastest of the elastic waves. It passes through a solid, a liquid and a gas.
Shear wave (S): The particles oscillate across the direction in which the wave travels, so the rock is sheared (its layers slide sideways). It is one and a half to two times slower than the compressional wave. It travels only through a solid, because liquids and gases do not resist shear.
Mud wave: A compressional wave that goes from the transmitter to the receiver directly through the column of mud. It travels at the speed of sound in the mud, about 1.5–1.6 km/s. In Russian-language literature it is also called the hydrowave.
Stoneley wave: A wave that travels along the boundary between the mud and the borehole wall. It is also called the tube wave. It arrives last, a little after the mud wave, but with the largest amplitude (the biggest swing). Its velocity depends not only on the rock. The frequency, the hole diameter and the properties of the mud also affect it. On the rock side, the shear modulus matters most.

How does the wave get from the mud into the rock and back to the tool? At the boundary of two media a wave is refracted: it changes direction, like a ray of light at the surface of water. The rock is faster than the mud. So a ray that hits the wall at one particular angle, the critical angle, does not go deeper. It glides along the wall through the rock at the velocity of the rock. All along the way it sends energy back into the mud at the same angle. Such a wave is called a head wave, or refracted wave. The receivers record it.

The velocities of the waves set the order in which they reach the receiver. The compressional head wave arrives first. Its path is longer than the direct one, but it travels most of the way through fast rock. The shear head wave follows, if the rock is so stiff that the shear wave in it is faster than sound in the mud. Then comes the mud wave, and last the Stoneley wave. The record of the whole oscillation at a receiver is called the waveform.

The body of the tool also conducts sound, and steel is faster than most rocks. The signal along the body must not overtake the signal through the rock. So the section of the body between the transmitter and the receivers has slots and inserts. They lengthen the path of the wave and damp it.

Interval transit time: units and what it depends on

The tool measures time. So we record the result not as a velocity but as the interval transit time $\Delta t$ , the travel time of the wave per unit length. On logs the curve is labelled DT, and the quantity itself is also called slowness. Time has a property that velocity lacks: the travel times over neighbouring intervals simply add up. Also, in compacted rocks $\Delta t$ is almost linearly related to porosity.

Interval transit time and wave velocity

\Delta t = \frac{10^6}{V}

where:

$\Delta t$ interval transit time, µs/m
$V$ velocity of the wave in the rock, m/s

Units: $\Delta t$ in µs/m, $V$ in m/s

Holds when

The factor $10^6$ converts seconds to microseconds.
The velocity and the interval transit time refer to the same wave, the compressional one unless stated otherwise.

The faster the rock, the smaller $\Delta t$ . Dense dolomite has about 143 µs/m (7 km/s) and water about 620 µs/m (1.6 km/s). The readings in porous rocks filled with liquid lie between the values of the matrix and of the fluid.

In other unit systems (2)

Oilfield units

\Delta t\,[\text{$\mu$s/ft}] = \frac{10^6}{V\,[\text{ft/s}]}

$\Delta t$ in microseconds per foot (µs/ft), $V$ in feet per second.

The same factor $10^6$ : seconds to microseconds.

Conversion between the systems

\Delta t\,[\text{$\mu$s/m}] = \frac{\Delta t\,[\text{$\mu$s/ft}]}{0.3048}

Dividing by 0.3048 is equivalent to multiplying by 3.281.

A foot is 0.3048 m. So a path of one metre is 3.281 times longer than a path of one foot, and the wave needs 3.281 times as long to cover it.

Interval transit time of the compressional wave in rocks and fluids

Substance: Sandstone, quartz matrix
$\Delta t$ , µs/m: 182–167
$\Delta t$ , µs/ft: 55.5–51.0
Velocity, km/s: 5.5–6.0

Substance: Limestone, calcite
$\Delta t$ , µs/m: 156–143
$\Delta t$ , µs/ft: 47.6–43.5
Velocity, km/s: 6.4–7.0

Substance: Dolomite
$\Delta t$ , µs/m: 143
$\Delta t$ , µs/ft: 43.5
Velocity, km/s: 7.0

Substance: Anhydrite
$\Delta t$ , µs/m: 164
$\Delta t$ , µs/ft: 50.0
Velocity, km/s: 6.1

Substance: Rock salt
$\Delta t$ , µs/m: 220
$\Delta t$ , µs/ft: 67.0
Velocity, km/s: 4.5

Substance: Casing steel
$\Delta t$ , µs/m: 187
$\Delta t$ , µs/ft: 57.0
Velocity, km/s: 5.3

Substance: Water, fresh filtrate
$\Delta t$ , µs/m: 620
$\Delta t$ , µs/ft: 189
Velocity, km/s: 1.6

Substance: Drilling mud
$\Delta t$ , µs/m: 620–660
$\Delta t$ , µs/ft: 190–200
Velocity, km/s: 1.5–1.6

Substance: Shales
$\Delta t$ , µs/m: 230–500
$\Delta t$ , µs/ft: 70–150
Velocity, km/s: 2.0–4.3

Substance: Coal
$\Delta t$ , µs/m: 330–460
$\Delta t$ , µs/ft: 100–140
Velocity, km/s: 2.2–3.0

Substance	$\Delta t$ , µs/m	$\Delta t$ , µs/ft	Velocity, km/s
Sandstone, quartz matrix	182–167	55.5–51.0	5.5–6.0
Limestone, calcite	156–143	47.6–43.5	6.4–7.0
Dolomite	143	43.5	7.0
Anhydrite	164	50.0	6.1
Rock salt	220	67.0	4.5
Casing steel	187	57.0	5.3
Water, fresh filtrate	620	189	1.6
Drilling mud	620–660	190–200	1.5–1.6
Shales	230–500	70–150	2.0–4.3
Coal	330–460	100–140	2.2–3.0

For minerals and steel, the values are for the dense substance without pores. For sandstone, take the lower velocity for ordinary varieties and the higher one for well-cemented, quartzite-like varieties. For limestone, 156 µs/m (47.5 µs/ft) is the value used most often. Shales and coals are shown as logs usually see them. Compacted deep shales read 230–300 µs/m. Young undercompacted shales (not yet squeezed to their normal density) read more than 330. The values in µs/m for minerals, steel and water are rounded to the nearest whole number. The ranges for the mud, shales and coals are rounded to tens.

Worked exampleFrom µs/ft to µs/m and to velocity

Given (assumed data)

Reading on a log with an oilfield scale: $\Delta t = 72$ µs/ft
To find: the value in µs/m and the wave velocity

Solution

In metric units: $72/0.3048 = 236.2$ µs/m.
Velocity: $V = 10^6/236.2$ m/s, that is, 4.23 km/s. In feet per second: $10^6/72 = 13{,}889$ .
Plausibility check: 236.2 µs/m lies between the sandstone matrix (182) and water (620), closer to the matrix: this is what a porous compacted rock looks like.

Answer: 72 µs/ft is 236.2 µs/m; the velocity of the compressional wave is about 4.23 km/s.

What it means. It is easy to mix up the scales, because the numbers differ by a factor of 3.28. But the physics gives the mistake away. No rock has a value of 72 µs/m: even dense dolomite gives 143. And 236.2 µs/ft would mean a velocity of 1.3 km/s, which is less than in water. The simplest way to check the scale is to look at the readings in known rocks: anhydrite gives 164 µs/m (50 µs/ft) and casing 187 (57). The log heading (the block of data at the top of the log) also shows the direction of the scale. Most often $\Delta t$ increases to the left, like porosity on the neighbouring curves. The data in this example are assumed for illustration.

Why are some rocks fast and others slow? Two things set the velocity of an elastic wave: the stiffness of the medium and its density.

Velocities of the compressional and shear waves in an elastic medium

V_p = \sqrt{\frac{K + \tfrac{4}{3}G}{\rho_b}}, \qquad V_s = \sqrt{\frac{G}{\rho_b}}

where:

$V_p,\ V_s$ velocities of the compressional and shear waves, m/s
$K$ bulk modulus of the rock together with the fluid in the pores: the resistance to a change of volume, Pa
$G$ shear modulus: the resistance to a change of shape, Pa
$\rho_b$ bulk density of the rock, kg/m³

Units: SI: moduli in Pa, density in kg/m³, velocities in m/s

Holds when

A uniform, isotropic medium (the same in all directions) that behaves elastically (it springs back after it is deformed).
The wavelength is much greater than the grains and pores, and the rock behaves as a continuous medium with averaged properties.
For a liquid and a gas $G = 0$ : there is no shear wave in them, and $V_p = \sqrt{K/\rho}$ , where $K$ and $\rho$ are the bulk modulus and the density of the fluid itself.

A stiff rock is fast, and a soft (compliant) one is slow. Density is in the denominator, but dense rocks are usually also the stiffest. So in practice velocity increases with density. The compressional wave is always faster than the shear wave. For rocks the ratio $V_p/V_s$ is from 1.5 to 2 or more.

Mineral matrix. Calcite and dolomite resist compression more strongly than quartz. So carbonates are faster than sandstones of the same porosity. Rock salt is slower, and shales are slower still.
Porosity. The fluid in the pores is several times slower than the matrix, and the pores themselves weaken the framework of the rock. The more intergranular pores there are, the greater $\Delta t$ is. This is the basis of the porosity calculation (Porosity from the sonic log).
Compaction and cementation. In a loose sand, the grains touch each other at separate points, and the framework is soft. With depth the contacts widen, cement joins the grains, and the velocity grows even at the same porosity. So the same porosity gives a different $\Delta t$ in a young loose sand and in an ancient sandstone.
Effective stress. The effective stress squeezes the framework of the rock. It is the difference between the overburden pressure (the weight of all the rock above) and the pore pressure (the pressure of the fluid in the pores). At abnormally high pore pressure, the rock is undercompacted (it has been squeezed less than is normal for its depth) and slower than it should be at that depth (Abnormal formation pressures: origin, prediction and recognition).
Fluid in the pores. In a stiff cemented rock, the framework carries the wave, and replacing water by oil or gas changes $\Delta t$ little. In a loose rock the fluid matters a lot: gas is tens or hundreds of times easier to compress than water, so it slows the compressional wave noticeably.
Shape of the voids. Rounded vugs weaken the framework far less than the same volume of intergranular pores and thin gaps between grains. So vuggy porosity raises $\Delta t$ much less than intergranular porosity.
Shales. Clay minerals are soft and hold bound water (water stuck to the clay). Shales are slow, so shaly streaks (thin shale layers) usually increase the $\Delta t$ of a sandstone.

How the tool measures interval transit time

The receivers sit in the tool at known distances from the transmitter. The electronics note the moment of the first arrival, when the signal at the receiver first exceeds a set threshold. This gives the travel time from the transmitter to the receiver. Centralisers (spring arms that keep the tool in the middle of the hole) hold the tool in the centre, so that the path of the wave through the mud is the same all round.

But the time to one receiver is not just the time for the path through the rock. The wave crosses the mud twice: from the transmitter to the wall and from the wall to the receiver. Think of the wave as a ray. Then the travel time of the head wave has two parts.

Travel time of the head wave from the transmitter to the receiver

t = L\,\Delta t + (a_T + a_R)\,\Delta t_m\cos\theta_c, \qquad \sin\theta_c = \frac{\Delta t}{\Delta t_m}

where:

$t$ travel time from the transmitter to the receiver, µs
$L$ distance between the transmitter and the receiver along the tool, m
$\Delta t$ interval transit time of the rock at the borehole wall, µs/m
$a_T,\ a_R$ gaps between the tool and the wall at the transmitter and at the receiver, m
$\Delta t_m$ interval transit time of the drilling mud, µs/m
$\theta_c$ critical angle between the ray in the mud and the perpendicular to the wall, —

Units: times in µs, distances in m, interval transit times in µs/m

Holds when

The ray approximation: the wave travels along straight segments, from the transmitter to the wall, along the wall and from the wall to the receiver.
The rock is faster than the mud ( $\Delta t < \Delta t_m$ ); otherwise there is no head wave.
The wall in the path of the wave is smooth and parallel to the tool, and the rock along it is uniform.

The first term is the time in which the wave would cover the distance $L$ through the rock. The second is the addition for the two oblique segments in the mud. It does not depend on $L$ : only the gaps and the velocities set it. The derivation of the formula is in the advanced section.

SchematicA schematic drawing of the borehole-compensated tool in the hole (not to scale). The hole is drawn horizontally, with the surface on the left. Solid lines are the wave paths from the upper transmitter T₁ to the receivers R₁ and R₂. Dashed lines are the paths from the lower transmitter T₂. The hole is enlarged opposite receiver R₂, so the path segments in the mud that lead to it are longer.

The addition for the mud is the same for receivers that have the same gap in front of them. So we only need the difference of the arrival times at two receivers, and the path through the mud cancels. What remains is the travel time through the rock between the points of the wall from which the wave goes up to the receivers. The distance between the receivers is called the span of the tool.

Interval transit time from two receivers

\Delta t = \frac{t_2 - t_1}{s}

where:

$t_1,\ t_2$ first-arrival times at the near and far receivers, µs
$s$ span of the tool, the distance between the receivers, m
$\Delta t$ interval transit time of the rock, averaged over the span, µs/m

Units: times in µs, span in m, $\Delta t$ in µs/m

Holds when

Both receivers trigger on the same oscillation of the same wave, the first arrival of the compressional head wave.
The gaps between the tool and the wall at both receivers are equal.
The rock is uniform within the span: the measured $\Delta t$ is the average over an interval of length $s$ .

The span sets the vertical resolution: with a span of 0.61 m (2 ft) the tool averages the rock over an interval of about 0.6 m. The depth of investigation is small. The wave glides through only the first centimetres or decimetres of rock behind the wall. So the sonic log, like the other porosity methods, measures the flushed zone (the zone next to the wall where the mud filtrate has pushed out the original fluids) (Borehole environment: invaded zone and mudcake).

Worked exampleTwo receivers: the path through the mud cancels

Given (assumed data)

Rock: compacted sandstone, $\Delta t = 250$ µs/m (4.0 km/s)
Drilling mud: $\Delta t_m = 650$ µs/m (about 1.54 km/s)
Gap between the tool and the wall: $a = 6$ cm at the transmitter and at both receivers
Distances from the transmitter to the receivers: 0.9144 m (3 ft) and 1.524 m (5 ft)
Span: $s = 0.6096$ m (2 ft)

Solution

Critical angle: $\sin\theta_c = 250/650 = 0.385$ , so $\cos\theta_c = 0.923$ (the angle is about 23°).
Addition for the two segments in the mud: $2\cdot 0.06\cdot 650\cdot 0.923 = 72.0$ µs.
Near receiver: $t_1 = 0.9144\cdot 250 + 72.0 = 300.6$ µs. Far receiver: $t_2 = 1.524\cdot 250 + 72.0 = 453.0$ µs.
From the two receivers: $\Delta t = \dfrac{453.0 - 300.6}{0.6096} = 250$ µs/m.
With a single receiver: $300.6/0.9144 = 329$ µs/m, which is 31% more than the true value.
The mud wave would reach the near receiver in $0.9144\cdot 650 = 594$ µs, almost twice as late as the head wave.

Answer: The difference of the times gives exactly 250 µs/m. The addition of 72.0 µs is in both times, so it cancels when we subtract.

What it means. A single receiver would give an interval transit time that is almost a third too high, and the error would change with the hole diameter. Two receivers remove the error as long as the gaps in front of them are equal. The next paragraphs show what happens when they are not. The data in this example are assumed for illustration.

The rule that the gaps are equal breaks in two ordinary cases. The first is a hole enlargement, a washout in the wall. One of the receivers is opposite it, and the layer of mud in front of it is thicker. The second is tilt. The tool stands in the hole at an angle, and the gap changes smoothly along it. Then the difference $t_2 - t_1$ still contains some extra time in the mud. A tool with a single transmitter then shows a wrong $\Delta t$ , too high or too low, depending on which receiver is farther from the wall.

The solution is to measure $\Delta t$ twice, in opposite directions. The compensated tool (borehole-compensated, BHC) has two transmitters, one above the receivers and one below. They fire in turn. For the upper transmitter the far receiver is the lower one. For the lower transmitter the far receiver is the upper one. So the extra path in the mud enters the two measurements with opposite signs. The mean $\Delta t = (\Delta t_{\downarrow} + \Delta t_{\uparrow})/2$ is free of it. In a common design, the distances from the transmitter to the receivers are 0.9 and 1.5 m, and the span is 0.61 m. Each transmitter has its own pair of receivers, but the principle is the same.

Worked exampleA hole enlargement opposite one receiver

Given (assumed data)

Rock, mud and tool: as in the previous example: $\Delta t = 250$ µs/m, $\Delta t_m = 650$ µs/m, span 0.6096 m
Gap at the lower receiver: 11 cm instead of 6 cm: the receiver stands opposite a hole enlargement
Gaps at the upper receiver and at the transmitters: 6 cm

Solution

Difference of the travel times through the rock within the span: $0.6096\cdot 250 = 152.4$ µs.
The gap at the lower receiver is 0.05 m greater. The extra time in the mud: $0.05\cdot 650\cdot 0.923 = 30.0$ µs.
The upper transmitter fires: the lower receiver is the far one, and its time is greater by 30.0 µs. $\Delta t_{\downarrow} = \dfrac{152.4 + 30.0}{0.6096} = 299.2$ µs/m.
The lower transmitter fires: the lower receiver is the near one, and the time subtracted is already greater by 30.0 µs. $\Delta t_{\uparrow} = \dfrac{152.4 - 30.0}{0.6096} = 200.8$ µs/m.
Mean: $(299.2 + 200.8)/2 = 250$ µs/m.
What a single measurement costs: by the time-average equation for sandstone (Porosity from the sonic log), 250 µs/m is a porosity of 0.155, but 299.2 µs/m already gives 0.268.

Answer: Each of the opposite measurements is wrong by 49 µs/m, and their mean equals the true 250 µs/m.

What it means. A hole enlargement of only 5 cm in radius would shift the reading of a single-transmitter tool by 20%. It would shift the calculated porosity by 11 porosity units (p.u.; a porosity unit is 0.01, or one percentage point). The mean of the two opposite measurements does not depend on the enlargement. The compensation is exact as long as both measurements refer to the same interval of rock and the tilt of the tool does not change between the firings of the transmitters. The calculation is simplified: it accounts only for the different thickness of the mud layer in front of the receivers. The data in this example are assumed for illustration.

CalculatedA calculated record opposite a hole enlargement, from the ray model and the data of the example. The rock is uniform (250 µs/m), and the hole is enlarged by 5 cm in radius over an interval of 1 m. Each of the two opposite measurements gives false spikes at the boundaries of the enlargement. The spikes are as wide as the span of the tool, and they have opposite signs in the two measurements. Their mean is a straight line.

The surface equipment (the equipment of the logging unit) not only averages the opposite measurements but also adds up $\Delta t$ over depth. The result is the integrated travel time: the time the wave would need to travel from one depth to another. On the log it is shown by marks every millisecond. We will need it for the tie to seismic.

Long-spaced tools, receiver arrays and dipoles

A short compensated tool is reliable in a smooth hole of ordinary diameter and in rock that drilling has not altered. There are two cases where its first arrival is not the wave we need.

Altered zone. In some shales, the rock right at the wall loosens after the hole is drilled. The velocity in it is then lower than it is a few decimetres away from the hole. The same happens in loose rocks, and in frozen ground that has thawed around the hole. The wave that glides along the wall travels through this slow zone, so the short tool shows a $\Delta t$ that is too high. A wave that goes deeper, into unaltered rock, travels faster, but its path there and back is longer. It can overtake the first wave only if the distance from the transmitter to the receiver is large enough.

Large hole diameter. The thicker the layer of mud, the larger the addition for the two segments in the mud for the head wave. The time of the direct wave through the mud does not depend on the thickness of the layer. Suppose the rock is not much faster than the mud and the hole is wide. Then the mud wave arrives at the near receiver first. We find the distance beyond which the head wave arrives first by setting the travel times of the two waves equal.

Distance from which the head wave overtakes the mud wave

L_c = 2a\,\sqrt{\frac{\Delta t_m + \Delta t}{\Delta t_m - \Delta t}}

where:

$L_c$ the smallest distance from the transmitter to the receiver at which the head wave arrives before the mud wave, m
$a$ gap between the tool and the wall, m
$\Delta t,\ \Delta t_m$ interval transit times of the rock and of the mud, µs/m

Units: distances in m; interval transit times in the same units

Holds when

The ray approximation; the gap is the same at the transmitter and at the receiver.
The rock is faster than the mud: $\Delta t < \Delta t_m$ .

The distance is proportional to the gap. It grows sharply when the velocity in the rock comes close to the velocity in the mud. For dense rocks in a hole of ordinary diameter it is 15–25 cm, so the short tool works with a large margin.

Worked exampleWide hole and slow rock: the mud wave arrives first

Given (assumed data)

Rock: weakly compacted shallow deposits, $\Delta t = 540$ µs/m (1.85 km/s)
Drilling mud: $\Delta t_m = 650$ µs/m
Hole: 445 mm (17½ in), tool in the centre, gap $a = 17.6$ cm
Short tool: receivers 0.9144 and 1.524 m from the transmitter
Long tool: receivers 2.438 and 3.048 m from the transmitter (8 and 10 ft)

Solution

Twice the gap is $2a = 0.352$ m. Critical distance: $L_c = 0.352\cdot\sqrt{\dfrac{650 + 540}{650 - 540}} = 1.16$ m. The near receiver of the short tool (0.91 m) is closer than that.
Addition for the two segments in the mud for the head wave: $0.352\cdot 650\cdot 0.557 = 127$ µs (here $\cos\theta_c = 0.557$ , because $\sin\theta_c = 540/650$ ).
Near receiver. Mud wave: $0.9144\cdot 650 = 594$ µs. Head wave: $0.9144\cdot 540 + 127 = 621$ µs. The mud wave arrives first.
Far receiver. Mud wave: 991 µs, head wave: 950 µs. The head wave arrives first.
The tool subtracts the times of different waves: $\dfrac{950 - 594}{0.6096} = 584$ µs/m instead of 540.
Long tool: both receivers are farther than 1.16 m, both trigger on the head wave, and the tool shows 540 µs/m.
In a 216 mm hole (gap 6.2 cm) $L_c = 0.41$ m, and the short tool would work without distortion.

Answer: In a wide hole the short tool shows 584 µs/m, which is 8% more than the true value. The long tool gives the correct 540 µs/m.

What it means. Compensation does not help here. Both opposite measurements are wrong in the same way, and averaging does not remove the error. For the tie to seismic the error matters: over 500 m of section, 22 ms builds up in one direction. That is why the upper large-diameter intervals, which are logged precisely for the sake of seismic, are surveyed with long-spaced tools. The data in this example are assumed for illustration.

In long-spaced tools, the distances from the transmitter to the receivers are 2.4–3.7 m (8–12 ft). Their first arrival is the wave that has travelled beyond the altered zone, and the mud wave arrives later even in a wide hole. The depth of investigation of such a tool is about 30 cm, against 10–15 cm for a short one. At such distances the symmetrical arrangement, with transmitters on both sides of the receivers, is not used: the transmitters would be about 6 m apart. The opposite measurement is obtained in a different way.

Two transmitters stand next to each other, a span apart, and so do two receivers. The tool moves, so the same interval of rock is first between the receivers and later between the transmitters. In the first position, we measure the difference of the arrival times at two receivers from one transmitter, $\delta t_1$ , and store it. In the second position, we measure the difference of the arrival times at one receiver from two transmitters, $\delta t_2$ . In this second measurement the opposite end of the interval is the far one, so the extra time in the mud enters with the opposite sign. Then $\Delta t = (\delta t_1 + \delta t_2)/(2s)$ . This is depth-derived compensation. Its weak point is that the two measurements are made at different times. Suppose that in the meantime the tool has moved differently from what the depth counter showed, or has changed its tilt. Then the extra time in the mud is not fully subtracted.

Array sonic tools carry eight or more receivers, spaced 15 cm apart, at distances of 2.4–3.5 m from the transmitter. They record the waveform digitally at each receiver. The processing looks for waves that pass all the receivers with the same delay from one receiver to the next. From this delay it finds the interval transit time of each wave: the compressional $\Delta t_p$ , the shear $\Delta t_s$ and that of the Stoneley wave. With a separate short transmitter–receiver pair, such a tool can also measure the interval transit time of the mud itself. In Russian-language practice, the recording of waveforms is called waveform acoustic logging.

The ordinary transmitter is a monopole: it expands and contracts equally in all directions. It excites a shear head wave only if the shear wave in the rock is faster than sound in the mud. This is true in stiff, “fast” rocks. In “slow” rocks (loose sands, weak shales) there is no critical angle for the shear wave, and it does not reach the receivers. A dipole transmitter pushes the borehole wall to one side. It excites a flexural wave, and the hole oscillates like a string. At low frequencies this wave travels along the hole at practically the velocity of the shear wave in the rock. So a dipole tool measures $\Delta t_s$ in any rock. Two dipoles turned at a right angle show whether the shear velocity depends on direction (anisotropy). This is a sign of fractures or of unequal horizontal stresses.

Which tool measures what

Tool: Short compensated
Distances: 0.9 and 1.5 m; span 0.61 m
What it gives: $\Delta t_p$ at the wall
When it is needed: smooth hole, compacted rocks

Tool: Long-spaced
Distances: 2.4–3.7 m; span 0.61 m
What it gives: $\Delta t_p$ beyond the altered zone
When it is needed: shales, loose rocks, wide hole

Tool: Array
Distances: 8 receivers 15 cm apart
What it gives: $\Delta t_p$ , $\Delta t_s$ , the Stoneley wave
When it is needed: elastic properties, fractures

Tool: Dipole
Distances: array, low frequency
What it gives: $\Delta t_s$ in any rocks
When it is needed: loose rocks, anisotropy

Tool	Distances	What it gives	When it is needed
Short compensated	0.9 and 1.5 m; span 0.61 m	$\Delta t_p$ at the wall	smooth hole, compacted rocks
Long-spaced	2.4–3.7 m; span 0.61 m	$\Delta t_p$ beyond the altered zone	shales, loose rocks, wide hole
Array	8 receivers 15 cm apart	$\Delta t_p$ , $\Delta t_s$ , the Stoneley wave	elastic properties, fractures
Dipole	array, low frequency	$\Delta t_s$ in any rocks	loose rocks, anisotropy

The distances are typical; specific tools differ. A monopole array tool gives $\Delta t_s$ only in stiff rocks. Two dipoles turned at a right angle measure anisotropy. The vertical resolution is about 0.6 m with a span of 0.61 m, about 0.9 m when the processing uses the whole array, and down to 15 cm with special high-resolution processing.

Cycle skipping and quality control

A receiver triggers when the signal first exceeds a threshold. Usually this is the very first oscillation of the compressional wave. But the signal weakens on the way to the far receiver. If the first oscillation does not reach the threshold, the receiver triggers on the next one, a period later. This is cycle skipping. The time $t_2$ is too large by one period of the oscillation. At a frequency of 20 kHz the period is 50 µs, so a measurement over a span of 0.61 m gives an extra 82 µs/m (25 µs/ft). The compensated curve is the mean of the two opposite measurements. A skip in one of them raises it by half of this amount, about 41 µs/m, and a skip in both raises it by the full 82. Two or three cycles can also be skipped.

CalculatedCalculated model signals of one of the two opposite measurements, at the near and far receivers, for the rock of the examples (250 µs/m). The pulse shape is assumed, and the frequency is 20 kHz. Triangles mark the arrival of the wave (301 and 453 µs). Dashed lines mark the threshold. Dots with labels mark the moments of triggering. The weakened signal does not reach the threshold with its first oscillation, so the far receiver triggers on the second one, 44 µs later. This is slightly less than a period, because the second oscillation is larger and crosses the threshold earlier in its phase. The measured

\Delta t

is about 325 µs/m instead of 250.

Anything that weakens the signal strongly causes cycle skipping: gas in the pores or in the mud, open fractures, loose uncompacted rocks, hole enlargements and an uneven hole. So cycle skipping is not only a nuisance. It is also a qualitative sign. A cluster of such spikes in a dense carbonate may point to fractures, and in a sandstone to gas. An uneven hole, however, gives the same picture. We cannot use $\Delta t$ at these points quantitatively.

Even without cycle skipping, a weakened signal crosses the threshold slightly later than a strong one, so $\Delta t$ comes out slightly too high (by 1% in the figure). Array tools have given up the threshold method. They find $\Delta t$ from how similar the shapes of the signals are at all the receivers, and this depends far less on the amplitude.

Signs of an unreliable interval transit time

Sign on the log: Sharp spikes towards larger $\Delta t$
What is happening: cycle skipping: the far receiver triggered a period late
What to do: exclude the points; check for gas, fractures and the condition of the hole

Sign on the log: Sharp spikes towards smaller $\Delta t$
What is happening: noise: a knock or friction of the tool against the wall triggered the far receiver before the signal
What to do: exclude the points; compare with the repeat section (a second pass over the same interval)

Sign on the log: In a wide hole $\Delta t$ is too high, up to the value for the mud
What is happening: the first wave to arrive is the mud wave, not the head wave
What to do: use a long-spaced tool

Sign on the log: In shales the $\Delta t$ of the short tool is greater than that of the long one
What is happening: altered zone at the wall
What to do: use the readings of the long-spaced tool

Sign on the log: A constant 187 µs/m (57 µs/ft)
What is happening: the signal travels along the casing
What to do: this is not a formation but the steel of the casing

Sign on the log	What is happening	What to do
Sharp spikes towards larger $\Delta t$	cycle skipping: the far receiver triggered a period late	exclude the points; check for gas, fractures and the condition of the hole
Sharp spikes towards smaller $\Delta t$	noise: a knock or friction of the tool against the wall triggered the far receiver before the signal	exclude the points; compare with the repeat section (a second pass over the same interval)
In a wide hole $\Delta t$ is too high, up to the value for the mud	the first wave to arrive is the mud wave, not the head wave	use a long-spaced tool
In shales the $\Delta t$ of the short tool is greater than that of the long one	altered zone at the wall	use the readings of the long-spaced tool
A constant 187 µs/m (57 µs/ft)	the signal travels along the casing	this is not a formation but the steel of the casing

From interval transit time to porosity, briefly

The matrix of the rock is fast, and the fluid in the pores is three to four times slower. So in rocks of the same composition and the same degree of compaction, $\Delta t$ grows with porosity. The simplest form of this relationship is the time-average equation. The travel times in the matrix and in the fluid add up in proportion to their volume fractions.

Time-average equation

\phi_S = \frac{\Delta t - \Delta t_{ma}}{\Delta t_f - \Delta t_{ma}}

where:

$\phi_S$ sonic porosity, the porosity from the sonic log, fraction
$\Delta t$ interval transit time from the log, µs/m
$\Delta t_{ma}$ interval transit time of the matrix: 182 µs/m for sandstone, 156 for limestone, 143 for dolomite, µs/m
$\Delta t_f$ interval transit time of the fluid in the pores at the borehole wall, the filtrate, about 620 µs/m, µs/m

Units: interval transit times in the same units, µs/m or µs/ft; porosity as a fraction

Holds when

A clean compacted rock with intergranular pores filled with liquid.
The relationship is empirical (it comes from experience and is not derived from the mechanics of elastic waves). The topic Porosity from the sonic log discusses its assumptions, corrections and calibration.

At $\Delta t = \Delta t_{ma}$ the porosity is zero, at $\Delta t = \Delta t_f$ it is one. For sandstone every 4–5 µs/m is one porosity unit, so a measurement error in $\Delta t$ passes straight into the porosity.

Outside these conditions the simple formula gives wrong answers, and the error always goes in a known direction. In loose undercompacted sands it gives a porosity that is too high. Then we divide the result by a correction factor, or we replace the straight line with an empirical curve. Shaly streaks and gas in high-porosity sands increase $\Delta t$ , so the porosity comes out too high there as well. Vugs, on the contrary, hardly show in $\Delta t$ . The sonic porosity of a vuggy carbonate is smaller than its total porosity. The difference between it and the density and neutron porosity gives an estimate of the secondary porosity (the porosity that formed after the rock was deposited, for example vugs and fractures).

The topic Porosity from the sonic log covers the choice of the matrix and fluid interval transit times, the corrections for undercompaction, shale, oil and gas, the estimate of secondary porosity and the calibration against core. For the measurement itself, something else matters: any error in it passes into the porosity. So we calculate porosity only over the parts of the curve that have passed the quality check, with no cycle skipping, no mud-wave arrival and no effect of the altered zone.

Not only porosity: seismic, elastic moduli, cement

Tie to seismic. A seismic section is recorded in time. Its vertical axis shows the travel time of the wave from the surface to a reflecting boundary and back. To find which bed gives which reflection, we need the travel time to each depth in the well. The sum of the interval transit times gives it.

Travel time through an interval and the average velocity

t = \sum_i \Delta t_i\,h_i, \qquad \bar V = \frac{10^6\sum_i h_i}{t}

where:

$t$ one-way travel time of the wave through the interval; the two-way time is $2t$ , µs
$\Delta t_i$ interval transit time of the $i$ -th layer, µs/m
$h_i$ vertical thickness of the $i$ -th layer, m
$\bar V$ average velocity in the interval, m/s

Units: $\Delta t$ in µs/m, $h$ in m, $t$ in µs, $\bar V$ in m/s (the factor $10^6$ converts microseconds to seconds)

Holds when

The wave travels vertically; in a deviated well, vertical thicknesses are used.
The velocity at logging frequencies (kilohertz) is usually somewhat higher than at seismic frequencies (tens of hertz), and the tool sees only the rock at the wall. So we calibrate the integrated time against a check-shot survey or vertical seismic profiling (VSP). Both are direct measurements of the travel time from the surface to a receiver in the well.

Times add up. So the average velocity is the thickness divided by the total time, not the mean of the layer velocities. The product of velocity and density is the acoustic impedance. From the $\Delta t$ and $\rho_b$ curves we calculate a synthetic seismogram (a seismic trace computed from the logs) and compare it with the real seismic trace at the well (Seismic waves, acoustic impedance and reflections).

Worked exampleTravel time through 300 m of section

Given (assumed data)

Shale, 120 m: $\Delta t = 330$ µs/m
Sandstone, 40 m: $\Delta t = 265$ µs/m
Shale, 70 m: $\Delta t = 320$ µs/m
Limestone, 70 m: $\Delta t = 175$ µs/m
Travel time between the same depths from the check-shot survey: 86.5 ms one way

Solution

Times for the layers: $120\cdot 330 = 39{,}600$ ; $40\cdot 265 = 10{,}600$ ; $70\cdot 320 = 22{,}400$ ; $70\cdot 175 = 12{,}250$ µs.
Sum: 84,850 µs, that is, 84.85 ms one way and 169.7 ms two-way time.
Average velocity: $300/84.85\cdot 1000 = 3536$ m/s. The thickness-weighted mean of the layer velocities would give 3778 m/s, which is 7% more.
Discrepancy with the check-shot survey: $86.5 - 84.85 = 1.65$ ms, or 1.9%.

Answer: From the sonic log the interval gives 84.85 ms one way (169.7 ms two-way time); the average velocity is 3536 m/s.

What it means. A discrepancy of 1.9% is a normal size: sound at logging frequencies travels slightly faster than a seismic wave. We correct the integrated time so that it matches the check-shot survey. This means stretching it by a factor of 1.019. Only after that do we build the time–depth relationship. The fast limestone is almost a quarter of the thickness but only 14% of the time. In an average of velocities, its weight would be too large. The data in this example are assumed for illustration.

Elastic properties of rocks. Suppose we know the velocities of both waves and the density. Then we can turn the equations for $V_p$ and $V_s$ around and find the elastic moduli of the rock directly in the well.

Dynamic shear modulus and Poisson's ratio from the sonic log

G = \frac{\rho_b}{\Delta t_s^2}, \qquad \nu = \frac{\tfrac{1}{2}R^2 - 1}{R^2 - 1}, \qquad R = \frac{\Delta t_s}{\Delta t_p}

where:

$G$ dynamic shear modulus, Pa
$\nu$ dynamic Poisson's ratio (the ratio of the sideways shortening to the lengthening when the rock is stretched), —
$\rho_b$ bulk density from the density log, kg/m³
$\Delta t_p,\ \Delta t_s$ interval transit times of the compressional and shear waves, s/m
$R$ ratio of the interval transit times, equal to $V_p/V_s$ , —

Units: SI: density in kg/m³, interval transit time in s/m, modulus in Pa

Holds when

A uniform isotropic elastic rock.
The moduli are dynamic: they describe the small, fast deformations in a sound wave. Static moduli, which are measured by loading core slowly, are usually lower. For engineering calculations we obtain them from relationships checked on core.
$\Delta t_s$ is measured, not estimated from $\Delta t_p$ .

Young's modulus (it says how stiff the rock is when stretched) is $E = 2G(1 + \nu)$ , and the bulk modulus is $K = \rho_b\,(V_p^2 - \tfrac{4}{3}V_s^2)$ . Poisson's ratio depends only on the ratio $R$ : at $R = 1.5$ it is 0.10, and at $R = 2$ it is 0.33. We need these quantities to estimate the stability of the borehole walls, to predict sand production (sand carried into the well with the oil) and to design hydraulic fracturing (Fracture gradient and leak-off testing, Rock mechanics for hydraulic fracturing).

In other unit systems (2)

Metric oilfield units

G\,[\text{GPa}] = \frac{10^6\,\rho_b}{\Delta t_s^2}

$\rho_b$ in g/cm³, $\Delta t_s$ in µs/m, $G$ in GPa.

$10^6 = 10^3\cdot 10^{12}\cdot 10^{-9}$ : conversion of g/cm³ to kg/m³, of µs/m to s/m (squared) and of Pa to GPa.

Oilfield units

G\,[\text{psi}] = \frac{1.347\cdot 10^{10}\,\rho_b}{\Delta t_s^2}

$\rho_b$ in g/cm³, $\Delta t_s$ in µs/ft, $G$ in psi.

The constant is $10^{15}\cdot 0.3048^2/6895$ : conversion of g/cm³ to kg/m³ and of µs to seconds, of feet to metres, and 6895 Pa per psi. In reference books it is usually given as $1.34\cdot 10^{10}$ .

Worked exampleElastic moduli of a sandstone from two interval transit times and the density

Given (assumed data)

Compressional wave: $\Delta t_p = 250$ µs/m
Shear wave: $\Delta t_s = 425$ µs/m
Bulk density: $\rho_b = 2.39$ g/cm³

Solution

Velocities: $V_p = 10^6/250 = 4000$ m/s, $V_s = 10^6/425 = 2353$ m/s; ratio $R = 425/250 = 1.70$ .
Shear modulus: $G = 10^6\cdot 2.39/425^2 = 13.2$ GPa.
Poisson's ratio: $\nu = \dfrac{0.5\cdot 1.70^2 - 1}{1.70^2 - 1} = 0.24$ .
Young's modulus: $E = 2\cdot 13.2\cdot(1 + 0.24) = 32.7$ GPa. Bulk modulus: $K = 20.6$ GPa.

Answer: Dynamic moduli:

G \approx 13.2

GPa,

E \approx 32.7

GPa, Poisson's ratio 0.24.

What it means. The values are typical of a compacted sandstone. $\Delta t_s$ enters the modulus squared, so an error of 2% in the interval transit time gives an error of 4% in the shear modulus. Before calculating stresses and hydraulic fracturing pressures, we convert the dynamic moduli to static ones. The data in this example are assumed for illustration.

Lithology and correlation. Rock salt and anhydrite give constant, characteristic values, 220 and 164 µs/m. Together with density, they are easy to identify. Coals stand out by a very large $\Delta t$ . In thick shale and salt-bearing sequences, the other curves are monotonous (they hardly change). There the pattern of the $\Delta t$ curve helps to correlate the sections of neighbouring wells. Combining $\Delta t$ with density and neutron porosity gives a better picture of the mineral composition (Lithology and porosity from crossplots).
Pore pressure. We use how far the $\Delta t$ of shales departs from the normal compaction trend to predict and estimate abnormally high pressures (Abnormal formation pressures: origin, prediction and recognition).
Fractures and permeable intervals. Open fractures weaken the waves and cause cycle skipping. The Stoneley wave is reflected from them and loses strength, and in permeable beds it is slowed down. These are qualitative signs. We check them with image logs (pictures of the borehole wall) and well tests.
Cementing. In a cased well the signal travels along the steel casing. Free casing “rings”, and the signal amplitude is large. Cement bonded to the pipe damps the oscillations, and the amplitude falls many times over. A receiver about 0.9 m from the transmitter measures the amplitude. The full waveform recorded by the receiver at 1.5 m shows whether there are arrivals from the formation behind the casing. That tells us whether the cement is also bonded to the formation. This is the cement bond log (CBL), discussed in the topic Remedial cementing, cement plugs and cement evaluation.
Organic matter. Kerogen (the solid organic matter in source rocks) is slower than the mineral matrix, so organic-rich shales stand out by a higher $\Delta t$ (Source rocks, maturation and migration).

Terms and symbols

Interval transit time $\Delta t$ (DT): The travel time of a wave per unit length of rock, in µs/m or µs/ft. It is the reciprocal of velocity. It is also called slowness.
Head (refracted) wave: A wave that glides through the rock along the borehole wall and radiates energy back into the mud. The receivers record its first arrival.
Span of the tool $s$: The distance between the two receivers. It sets the vertical resolution.
Borehole-compensated tool (BHC): A tool with transmitters above and below the receivers. The mean of two opposite measurements does not depend on hole enlargements or tool tilt.
Cycle skipping: The far receiver triggers one period late because the signal is weakened. It gives a sharp spike of $\Delta t$ towards larger values.
Waveform: The record of the whole oscillation at a receiver: the compressional and shear waves, the mud wave and the Stoneley wave.
Integrated travel time: The sum of the interval transit times over depth, that is, the time a wave needs to pass an interval of the section. It is the basis for tying seismic to the depths of the well.
Dynamic elastic moduli: The shear, Young's and bulk moduli and Poisson's ratio, calculated from the interval transit times of two waves and the density. They describe small, fast deformations.

Common mistakes and misconceptions

Wrong

“The interval transit time is the time from the transmitter to the receiver divided by the distance between them.”

In fact

That time includes the path through the mud. $\Delta t$ is the difference of the arrival times at two receivers divided by the span. Only this removes the mud.

Wrong

“The compensated tool corrects any distortion.”

In fact

Two transmitters remove the effect of hole enlargements and tilt. The arrival of the mud wave and the altered zone distort both opposite measurements in the same way, so averaging does not remove them. And a cycle skipped in even one of the measurements stays in the mean at half its size.

Wrong

“A $\Delta t$ peak in a gas-bearing sandstone is a thin high-porosity layer.”

In fact

A sharp single spike to values that this rock cannot have is cycle skipping: the gas has weakened the signal. We do not calculate porosity at such points.

Wrong

“The average velocity over an interval is the mean of the velocities of its layers.”

In fact

The travel times add up, not the velocities. To get the average velocity, divide the thickness of the interval by the sum of the times. The mean of the velocities comes out too high, because the fast layers get too much weight.

Wrong

“The moduli from the sonic log can be put straight into the calculation of wellbore stability and hydraulic fracturing.”

In fact

These are dynamic moduli: they describe small, fast deformations in a sound wave. The static moduli that engineering calculations need are usually lower. We obtain them from relationships checked on core.

In depth: advanced detail

Derivation of the head-wave travel-time formula

Let the transmitter and the receiver be at distances $a_T$ and $a_R$ from the wall, and let the distance between them along the tool be $L$ . The ray leaves the transmitter at the critical angle $\theta_c$ to the perpendicular. It travels a segment of length $a_T/\cos\theta_c$ in the mud and shifts along the hole by $a_T\tan\theta_c$ . The same happens at the receiver: a segment $a_R/\cos\theta_c$ and a shift $a_R\tan\theta_c$ . This leaves a path of $L - (a_T + a_R)\tan\theta_c$ along the wall through the rock.

The time in the mud is $(a_T + a_R)\,\Delta t_m/\cos\theta_c$ . The time in the rock is $[L - (a_T + a_R)\tan\theta_c]\,\Delta t$ . The law of refraction gives the critical angle: $\sin\theta_c = V_m/V = \Delta t/\Delta t_m$ ( $V_m$ is the speed of sound in the mud). So $\Delta t = \Delta t_m\sin\theta_c$ . We put this into the second term and add the terms with the gaps:

Adding the times in the mud and in the rock

\frac{\Delta t_m}{\cos\theta_c} - \Delta t_m\sin\theta_c\tan\theta_c = \Delta t_m\,\frac{1 - \sin^2\theta_c}{\cos\theta_c} = \Delta t_m\cos\theta_c

where:

$\Delta t_m$ interval transit time of the mud, µs/m
$\theta_c$ critical angle, —

Units: µs/m

Holds when

The same as for the head-wave travel-time formula: the ray approximation, a smooth wall parallel to the tool.

The factor of $(a_T + a_R)$ turned out to be $\Delta t_m\cos\theta_c$ , and the factor of $L$ stayed $\Delta t$ . This gives the formula of the main section. The oblique path in the mud is longer than the perpendicular one, but it shortens the path through the rock. In total, the addition is smaller than a perpendicular path there and back would give.

Set the head-wave time at equal gaps, $L\,\Delta t + 2a\,\Delta t_m\cos\theta_c$ , equal to the time of the direct wave through the mud, $L\,\Delta t_m$ . Write the cosine in terms of the interval transit times. This gives the critical distance $L_c$ . The ray model is crude, because the wavelength in a borehole is comparable to the hole diameter. But it describes the first-arrival times well enough, and these are exactly what we need.

The shear wave: lithology and gas

The interval transit time of the shear wave carries information that $\Delta t_p$ does not have. The ratio $\Delta t_s/\Delta t_p$ is different for dense rocks of different composition: 1.5–1.7 for quartz sandstone, about 1.9 for limestone and about 1.75 for dolomite. On a plot of $\Delta t_s$ against $\Delta t_p$ , the points of different minerals lie on different lines. We use this plot to find the lithology, in the same way as the other crossplots of porosity methods. In a water-bearing rock the ratio grows with porosity, but gas reduces it, because the compressional wave slows down and the shear wave does not. A low ratio in a porous bed is one of the signs of gas.

Approximate matrix parameters for the shear wave

Rock: Sandstone
$\Delta t_s$ , µs/m: 282
$\Delta t_s$ , µs/ft: 86
$\Delta t_s/\Delta t_p$: 1.55

Rock: Limestone
$\Delta t_s$ , µs/m: 295
$\Delta t_s$ , µs/ft: 90
$\Delta t_s/\Delta t_p$: 1.89

Rock: Dolomite
$\Delta t_s$ , µs/m: 249
$\Delta t_s$ , µs/ft: 76
$\Delta t_s/\Delta t_p$: 1.75

Rock: Anhydrite
$\Delta t_s$ , µs/m: 328
$\Delta t_s$ , µs/ft: 100
$\Delta t_s/\Delta t_p$: 2.00

Rock	$\Delta t_s$ , µs/m	$\Delta t_s$ , µs/ft	$\Delta t_s/\Delta t_p$
Sandstone	282	86	1.55
Limestone	295	90	1.89
Dolomite	249	76	1.75
Anhydrite	328	100	2.00

The ratios use matrix $\Delta t_p$ values of 55.5, 47.5, 43.5 and 50 µs/ft. The values are approximate, and core data refine them. The topic Porosity from the sonic log explains how to estimate porosity from $\Delta t_s$ .

How interval transit times are obtained from waveforms

The receiver array records the same oscillation eight times, each time a little later. A wave with interval transit time $\Delta t$ reaches the next receiver, 15 cm farther, with a delay of $0.15\,\Delta t$ . The processing tries many trial values of the interval transit time and of the arrival time. For each pair of trial values, it shifts the records of the receivers by the matching delays. Then it checks how similar the shifted records are to each other in a short time window. The measure of similarity is coherence. It is close to one when the trial interval transit time equals the true one and a real wave falls in the window. Otherwise it is small.

Plot the coherence on a plane of arrival time against interval transit time. Each wave then shows up as a separate maximum. The slower the wave, the later it arrives, and the higher its maximum lies on the interval-transit-time axis. The position of the maximum is the interval transit time we are looking for. In this way, depth by depth, we get continuous curves of $\Delta t_p$ , $\Delta t_s$ and the interval transit time of the Stoneley wave. The method is called slowness–time coherence. Seismic exploration analyses velocities in a similar way. The method compares the shapes of the signals, not the moment when a signal crosses a threshold. So it is not sensitive to a weak signal, and cycle skipping does not occur in it. The price is a coarser vertical resolution, because the result refers to the whole length of the array.

In a “slow” rock a monopole tool does not receive a shear head wave. Then we either measure $\Delta t_s$ with a dipole transmitter, or we estimate it from the velocity of the Stoneley wave, which depends on the shear modulus of the rock. The second estimate is less reliable.

Connections to other topics

Overview of logging tools — the place of the sonic log among the other methods, depth of investigation and resolution.
Density log and photoelectric factor (Pe) — another porosity method; density together with the interval transit time gives acoustic impedance and elastic moduli.
Neutron log — another porosity method; together with the density log it gives the total porosity for comparison with the sonic one.
Porosity — the types of porosity and why different methods see secondary porosity differently.
Porosity from the sonic log — calculation of porosity from the interval transit time: the time-average equation, corrections, calibration and secondary porosity.
Borehole environment: invaded zone and mudcake — the flushed zone and the condition of the hole, which is what a tool of small depth of investigation sees.
Abnormal formation pressures: origin, prediction and recognition — the interval transit time of shales as an indicator of undercompaction and abnormally high pressure.
Lithology and porosity from crossplots — lithology and porosity from the combination of the sonic, density and neutron logs.
Seismic waves, acoustic impedance and reflections — velocities, impedance and reflections: where the sonic log data go in seismic exploration.

Check yourself

1
Why does the wave that has travelled through the rock arrive at the receiver first, although its path is longer than the direct path through the mud?
1. A.Because the receivers are sensitive only to waves from the rock
2. B.Because the mud does not conduct sound
3. C.Because the mud wave dies out before reaching the receiver
4. D.Because the rock is faster than the mud, and the time gained along the wall is greater than the time lost on the two segments in the mud
Show the answer
Correct answer: D. Because the rock is faster than the mud, and the time gained along the wall is greater than the time lost on the two segments in the mud
The head wave travels most of its path through the rock, at the velocity of the rock. This is usually 2–4 times the velocity in the mud. The mud wave also reaches the receiver, but later. The exception is a wide hole with a slow rock. There the near receiver can receive the mud wave first, and a long-spaced tool is needed.
2
On a log with an oilfield scale, in a clean compacted sandstone $\Delta t = 85$ µs/ft. Convert the reading to µs/m and find the wave velocity. How can you tell that the scale has been read correctly?
Show the answer
$85/0.3048 \approx 279$ µs/m; the velocity is $10^6/279 \approx 3580$ m/s. The value lies between the sandstone matrix (182 µs/m) and water (620 µs/m), closer to the matrix. This is just what a porous compacted rock looks like. If the same 85 were in µs/m, the rock would be faster than any matrix mineral, because even dense dolomite gives 143 µs/m.
3
The signal frequency is 20 kHz and the span of the tool is 0.61 m. By how much does $\Delta t$ increase in one of the opposite measurements if the far receiver skips one cycle? How does this show on the compensated curve?
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The period of the oscillation is $1/20{,}000$ s $= 50$ µs. The time at the far receiver is too large by 50 µs, so this measurement gives a $\Delta t$ that is greater by $50/0.61 \approx 82$ µs/m (25 µs/ft). The compensated curve is the mean of the two opposite measurements. If the cycle is skipped in only one of them, the curve rises by 41 µs/m. If it is skipped in both, the curve rises by 82. On the log this is a sharp single spike towards larger values, with an equally sharp return. It is not a thin layer but a failure of the measurement: gas, fractures or a bad hole have weakened the signal.
4
A tool with one transmitter and two receivers passes a hole enlargement: the far receiver is opposite the washout, the near one opposite a smooth wall. What happens to the reading, and what does the compensated tool show at this point?
1. A.The reading will not change: two receivers eliminate the mud at any gaps
2. B.The reading is understated; the compensated tool will show the correct value
3. C.The reading is overstated; the compensated tool will show the correct value
4. D.The reading is overstated, and the compensated tool will be wrong in the same way
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Correct answer: C. The reading is overstated; the compensated tool will show the correct value
In front of the far receiver the layer of mud is thicker, so its time $t_2$ is larger, while $t_1$ stays the same. The difference $t_2 - t_1$ and $\Delta t$ are then too high. When the near receiver gets opposite the washout, the error changes sign. In a compensated tool, the opposite measurement is wrong by the same amount in the other direction, so the mean of the two measurements does not depend on the enlargement. Two receivers remove the mud only when the gaps are equal.
5
An interval consists of 200 m of shales with $\Delta t = 300$ µs/m and 100 m of limestone with $\Delta t = 200$ µs/m. Find the one-way travel time, the two-way time and the average velocity. What would the mean of the layer velocities give?
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Times for the layers: $200\cdot 300 = 60{,}000$ µs and $100\cdot 200 = 20{,}000$ µs. The sum is 80,000 µs, that is, 80 ms one way and 160 ms two-way time. The average velocity: $300/0.080 = 3750$ m/s. The layer velocities are 3333 and 5000 m/s. Their mean, weighted by thickness, is 3889 m/s, which is 4% more. The times add up, not the velocities.
6
We need to measure the interval transit time of the shear wave in a loose sand, where the shear wave is slower than sound in the mud. Which tool should we choose?
1. A.A monopole array tool
2. B.A long-spaced tool
3. C.A dipole tool
4. D.A short compensated tool
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Correct answer: C. A dipole tool
A monopole transmitter excites a shear head wave only if that wave is faster than sound in the mud. In a “slow” rock there is no critical angle for it. A dipole transmitter excites a flexural wave, which at low frequencies travels along the hole at the velocity of the shear wave in the rock. So it works in any rock. A longer tool or more receivers do not remove this limit.

Updated: 3 October 2026