Survey instruments: magnetic, gyroscopic and inertial

Quick refresher

The tool finds the inclination and the toolface from the three components of gravity, that is, its parts along the three axes of the tool, which the accelerometers measure. It finds the azimuth from the three components of the magnetic field, together with the gravity components. So an error in the accelerometers spoils all three angles.
We convert the magnetic azimuth to grid north, the direction of the vertical lines of the map grid. To do this, we add the declination (the angle between true north and magnetic north) for the place and the date of the survey. Then we subtract the grid convergence (the angle between true north and grid north). The declination changes from year to year. Near Atyrau it grew by almost 2° in 30 years.
Magnetised steel in the drill string adds a field along the axis of the tool. The azimuth error that it causes is proportional to $\sin\theta\,\sin\varphi$ and inversely proportional to the horizontal component of the field. It is worst in steeply inclined wells that run east or west, and at northern latitudes.
Gyroscopes do not notice steel. So they work inside casing and near neighbouring wells. A gyrocompass finds north from the horizontal component of the rotation of the Earth, $\Omega\cos\Phi$ , which gets weaker towards the poles.
A check against the total gravity, the total field and the dip angle rejects gross errors. But it misses a wrong declination and a moderate axial interference.
Position uncertainty grows with depth. In an inclined well it is largest across the direction of the well, because of the azimuth error.

Key relationships

\tan\theta = \dfrac{\sqrt{G_x^2 + G_y^2}}{G_z}

inclination from the accelerometers

\tan\varphi_{\text{m}} = \dfrac{(G_xB_y - G_yB_x)\,G}{B_z(G_x^2 + G_y^2) - G_z(G_xB_x + G_yB_y)}

magnetic azimuth; the signs of the numerator and the denominator show the quadrant (the quarter of the circle) where the angle lies

\Delta\varphi \approx -\dfrac{\Delta B_z\sin\theta\sin\varphi}{B\cos I}

azimuth error (in radians) from the axial interference $\Delta B_z$

\Delta B_z = \dfrac{P}{4\pi d^2}

field of the pole of magnetised steel at distance $d$ along the axis

\Omega_h = \Omega\cos\Phi, \qquad \Omega = 15.041^\circ/\text{h}

horizontal component of the rotation of the Earth, from which the gyrocompass finds north

Common pitfalls

Recalculating old magnetic surveys with today's declination.
Assuming that one non-magnetic drill collar removes the interference for any direction of the well and at any latitude.
Taking a survey that passed the check against the field of the Earth as free of errors.
Taking magnetic surveys inside casing or near the casing strings of neighbouring wells.
Mixing up the gravity toolface (measured from the high side of the hole, its upper wall) with the magnetic toolface (measured from north).

After this lesson you can

explain how to calculate the inclination, the toolface and the magnetic azimuth from the readings of three accelerometers (gravity sensors) and three magnetometers (sensors of the magnetic field)
choose a survey instrument (magnetic single-shot or multi-shot, MWD, gyroscopic or inertial) for open hole, for a cased well and for a well pad (a site where several wells stand close together)
find the grid convergence (the angle between true north and the north of the map grid) from coordinates, and explain why we recalculate old surveys with the declination of their own year
estimate the axial interference (the extra magnetic field along the tool) from the steel of the drill string, and the distance needed between the sensors and the steel parts of the assembly
check a survey against gravity, the total magnetic field and the magnetic dip angle, and understand which errors such a check cannot see
explain where the uncertainty region of the position of the well (the area in which the well may really lie) comes from and how it is used to keep wells from colliding

What a survey instrument measures and what types there are

Two angles give the direction of the axis of the well at a point (what one survey reading gives). The inclination $\theta$ is the tilt of the axis from the vertical. The azimuth $\varphi$ is the compass direction of the axis as seen from above (its horizontal projection). To measure the two angles, the tool needs two references. Gravity gives the vertical. A pendulum or an accelerometer (a sensor that measures acceleration, and so gravity) senses it. To find north, the tool uses either the magnetic field of the Earth or the rotation of the Earth. A gyroscope, a fast spinning wheel that keeps its direction, senses the rotation. So there are two large families of instruments: magnetic and gyroscopic.

The third quantity is the toolface $\omega$ . We need it when we bend the well on purpose (how it is used). It is the rotation of the tool, and with it of the deflection tool, around the axis of the well. We measure it clockwise from the high side of the hole (its upper wall), looking down the well. In an almost vertical well there is no high side, so we measure the angle from magnetic north instead. The first angle is called the gravity toolface, the second the magnetic toolface.

The main types of survey instruments

Instrument: Magnetic single-shot or multi-shot
Finds north from: the magnetic field
Where it works: open hole, inside a non-magnetic drill collar
How it gets into the well: on a cable, or dropped into the drill pipe

Instrument: MWD sensors (measurement while drilling)
Finds north from: the magnetic field
Where it works: open hole, inside a non-magnetic drill collar
How it gets into the well: in the bottom hole assembly; the reading is sent to the surface while drilling (Measurement while drilling)

Instrument: Gyroscopic
Finds north from: the rotation of the Earth, or an initial setting
Where it works: inside casing, on a well pad, near steel
How it gets into the well: on a cable or dropped; some versions go in the bottom hole assembly

Instrument: Inertial
Finds north from: the rotation of the Earth
Where it works: a hole or a casing string that is wide enough
How it gets into the well: on a cable

Instrument: Steering tool (a wireline telemetry system: the readings go up a cable)
Finds north from: the magnetic field
Where it works: open hole while drilling with a motor
How it gets into the well: on a cable inside the drill pipe

Instrument	Finds north from	Where it works	How it gets into the well
Magnetic single-shot or multi-shot	the magnetic field	open hole, inside a non-magnetic drill collar	on a cable, or dropped into the drill pipe
MWD sensors (measurement while drilling)	the magnetic field	open hole, inside a non-magnetic drill collar	in the bottom hole assembly; the reading is sent to the surface while drilling (Measurement while drilling)
Gyroscopic	the rotation of the Earth, or an initial setting	inside casing, on a well pad, near steel	on a cable or dropped; some versions go in the bottom hole assembly
Inertial	the rotation of the Earth	a hole or a casing string that is wide enough	on a cable
Steering tool (a wireline telemetry system: the readings go up a cable)	the magnetic field	open hole while drilling with a motor	on a cable inside the drill pipe

A single-shot instrument takes one reading in each run. A multi-shot instrument records a series of readings in its memory, one after another in time. Later we match the depth of each reading with the record of how the instrument was moved in the well.

Magnetic instruments are simpler, cheaper and tougher. But they need a “clean” section of the hole, because steel distorts the field: the steel of their own drill string, casing pipes, and the casing strings of neighbouring wells. Gyroscopic instruments are not afraid of steel. So we cannot do without them in cased holes, when we sidetrack from old wells (drill a new branch out of an existing hole), and in the upper part of wells on a crowded well pad, where wells stand close together. We also use them to check magnetic surveys independently. Today the magnetic single-shot and multi-shot instruments with memory mostly serve to check and back up MWD. When we drill an inclined well, the main source of readings is the MWD sensors in the assembly.

From sensor readings to angles

A modern survey instrument, either a magnetic tool with memory or MWD sensors, contains three accelerometers and three magnetometers (sensors of the magnetic field). Their axes are at right angles to each other. The $z$ axis points along the tool, down the hole. The $x$ and $y$ axes lie in the cross-section of the tool and turn with it. An accelerometer in a tool that is not moving measures the component of gravity along its own axis, in units of $g$ (the acceleration of gravity). A magnetometer measures the component of the magnetic field of the Earth, in nanotesla (nT).

Accelerometers are usually of the force-balance type. A position sensor notices the smallest movement of a small suspended weight (the test mass), and an electromagnet pushes the weight back into place. The current in the electromagnet is the measure of the acceleration along the axis. Magnetometers are usually of the fluxgate type. An alternating current drives a core of soft magnetic alloy to saturation (the point where it cannot be magnetised any further), first in one direction and then in the other. An outside field breaks the symmetry of this cycle. The size of the break tells us the field along the axis.

SchematicGravity in the axes of the tool (diagram). Left: the vertical plane of the well. Along the axis of the tool gravity gives

g\cos\theta

, and across it

g\sin\theta

, which points to the low side (the lower wall of the hole). Right: the cross-section of the hole, looking down the hole. The

x

axis is turned clockwise from the high side by the toolface angle

\omega

, and the cross component is shared between the

x

and

y

axes.

Inclination and toolface from the accelerometers

\tan\theta = \frac{\sqrt{G_x^2 + G_y^2}}{G_z}, \qquad \tan\omega = \frac{G_y}{-G_x}

where:

$G_x,\ G_y,\ G_z$ components of gravity along the tool axes: $G_z = g\cos\theta$ , $G_x = -g\sin\theta\cos\omega$ , $G_y = g\sin\theta\sin\omega$ , g
$\theta$ inclination, degrees
$\omega$ gravity toolface, measured clockwise from the high side of the hole, looking down the hole; the quadrant follows from the signs of $G_y$ and $-G_x$ , degrees

Units: the components are in units of $g$ or in m/s²: only their ratios enter the formulas

Holds when

The tool is at rest. When it moves, the accelerometers also feel the accelerations of the string and its vibration.
The sensor axes coincide with the tool axes, and the tool axis with the axis of the hole. If the tool sits crooked in the hole, or the drill collar sags, the inclination is in error.
The signs follow the convention used here. At $\omega = 0$ the $x$ axis points to the high side and the $y$ axis points to the right. Manufacturers orient the axes differently, and then the signs in the formulas change.

The inclination is the angle between the axis of the tool and the direction of gravity. When the well is almost vertical, the cross component is small and the toolface from the accelerometers becomes undefined (it can no longer be found). Then we orient the deflection tool by the magnetic toolface, $\omega_{\text{m}} = \varphi_{\text{m}} + \omega$ , measured from magnetic north.

Magnetic azimuth from the accelerometers and magnetometers

\tan\varphi_{\text{m}} = \frac{(G_xB_y - G_yB_x)\,G}{B_z\,(G_x^2 + G_y^2) - G_z\,(G_xB_x + G_yB_y)}

where:

$G_x,\ G_y,\ G_z$ components of gravity along the tool axes (the formula above), g
$B_x,\ B_y,\ B_z$ components of the magnetic field along the tool axes, nT
$G$ total gravity from the accelerometers, $\sqrt{G_x^2 + G_y^2 + G_z^2}$ , g
$\varphi_{\text{m}}$ magnetic azimuth, measured from the direction to magnetic north, degrees

Units: $B$ is in nT and $G$ is in $g$ . The formula uses a ratio, so the unit of the field does not matter

Holds when

The magnetometers see only the field of the Earth: there is no steel nearby, or its effect has been allowed for.
The quadrant follows from the signs: the numerator plays the role of a sine and the denominator the role of a cosine (like the atan2 function in computer programs).
As $\theta \to 0$ the numerator and the denominator both go to zero, and the azimuth loses its meaning, as it does for a truly vertical well.

The formula moves the magnetic field from the axes of the tool into the horizontal plane. The accelerometers tell where up is, and the magnetometers tell where magnetic north is. So an error in the accelerometers distorts not only the inclination but also the azimuth. The derivation is in the advanced section.

Worked exampleFrom six readings to three angles

Given (assumed data)

Accelerometers $G_x$ , $G_y$ , $G_z$: −0.4396; 0.3688; 0.8195 g
Magnetometers $B_x$ , $B_y$ , $B_z$: −39,059; 6,955; 35,251 nT

Solution

Cross component of gravity: $G_x^2 + G_y^2 = 0.4396^2 + 0.3688^2 = 0.32926$ . Total gravity: $G = \sqrt{0.32926 + 0.8195^2} = 1.00042$ g.
Inclination: $\tan\theta = \sqrt{0.32926}/0.8195 = 0.70020$ , so $\theta = 35.00^\circ$ .
Toolface: $\tan\omega = 0.3688/0.4396 = 0.83894$ . Both $G_y > 0$ and $-G_x > 0$ , so this is the first quadrant, and $\omega = 39.99^\circ$ .
Numerator: $G_xB_y = -3057.4$ and $G_yB_x = -14{,}405.0$ . So $(G_xB_y - G_yB_x)\,G = (-3057.4 + 14{,}405.0)\cdot 1.00042 = 11{,}352$ .
Denominator: $G_xB_x + G_yB_y = 17{,}170.3 + 2565.0 = 19{,}735.3$ . Then $35{,}251\cdot 0.32926 - 0.8195\cdot 19{,}735.3 = -4566$ .
The numerator is positive and the denominator is negative, so this is the second quadrant: $\varphi_{\text{m}} = 180^\circ - \arctan(11{,}352/4566) = 111.91^\circ$ .

Answer: The inclination is 35.00°, the magnetic azimuth is 111.91°, and the deflection tool is turned 39.99° to the right of the high side.

What it means. The well runs east-south-east. The azimuth is still magnetic. To calculate coordinates, we convert it to grid north (next section). The total gravity of 1.00042 g is the first sign of quality: for a tool at rest that works properly, it must be close to 1. The data in this example are assumed for illustration: the readings were calculated for a chosen direction of the well, with small sensor errors, and rounded as the tool reports them.

The magnetic field of the Earth, declination and grid convergence

At every point the magnetic field of the Earth is a vector, an arrow with a size and a direction. Its size $B$ is about 51–59 thousand nT in Kazakhstan. The arrow points at the magnetic dip angle $I$ below the horizontal (in the northern hemisphere the field points down). Its horizontal component, $B_h = B\cos I$ , points to magnetic north. Magnetic north differs from true north (the direction to the geographic North Pole) by the magnetic declination $\delta$ . A magnetic tool sees the whole vector, but only the horizontal component gives it the direction to north. At a dip angle of 63–70°, as in the areas of the table below, that is only 35–45% of the field. The smaller this component is, the more sensitive the azimuth is to interference.

SchematicThe elements of the magnetic field and the three norths (diagram; the angles are exaggerated). Left: the vertical plane that runs towards magnetic north (the magnetic meridian), with the field vector

B

, its horizontal component

B_h

, its vertical component

B_v

and the dip angle

I

. Right: view from above. Magnetic north is turned east from true north by the declination

\delta

. Grid north is turned from true north by the grid convergence

\gamma

. Here the point lies west of the central meridian of the zone (the line of longitude along the middle of the zone), so grid north is turned to the west. The azimuth of the well

\varphi_{\text{m}}

is measured from magnetic north. From true north or from grid north it is different.

The geomagnetic field in areas of Kazakhstan (IGRF-14 model, October 2026, near the surface)

Area: Atyrau
$B$ , nT: 53,090
$I$ , °: 66.3
$\delta$ , °: +9.3
Change of $\delta$ , °/year: +0.03

Area: Aktau
$B$ , nT: 51,880
$I$ , °: 63.3
$\delta$ , °: +7.8
Change of $\delta$ , °/year: +0.03

Area: Aksai
$B$ , nT: 54,430
$I$ , °: 69.6
$\delta$ , °: +11.6
Change of $\delta$ , °/year: +0.04

Area: Kyzylorda
$B$ , nT: 54,750
$I$ , °: 65.2
$\delta$ , °: +7.7
Change of $\delta$ , °/year: +0.02

Area: Almaty
$B$ , nT: 56,020
$I$ , °: 64.1
$\delta$ , °: +5.1
Change of $\delta$ , °/year: 0.00

Area	$B$ , nT	$I$ , °	$\delta$ , °	Change of $\delta$ , °/year
Atyrau	53,090	66.3	+9.3	+0.03
Aktau	51,880	63.3	+7.8	+0.03
Aksai	54,430	69.6	+11.6	+0.04
Kyzylorda	54,750	65.2	+7.7	+0.02
Almaty	56,020	64.1	+5.1	0.00

Calculated from the coefficients of the 14th generation of the International Geomagnetic Reference Field model (IGRF) for these points: Atyrau 47.1° N and 51.9° E, Aktau 43.7 and 51.2, Aksai 51.2 and 53.0, Kyzylorda 44.9 and 65.5, Almaty 43.2 and 77.0. The declination is east everywhere. These values are only a guide: in real work we take the field for the wellhead and the date of the survey.

The field changes with time. Slow secular variation shifts the declination by hundredths of a degree per year. Over decades this adds up to whole degrees. Near Atyrau, the same model gives a declination of about +7.4° in 1996 and about +9.3° in 2026. So we recalculate every magnetic survey with the declination of its own date. Fast changes come from the daily swing of the field and from magnetic storms. At middle latitudes the daily swing of the declination is about a tenth of a degree. During strong storms, especially at high latitudes, the deviations can be several times larger, and at such times we repeat the surveys. Finally, global models describe the field that comes from the core of the Earth. They do not see local anomalies from magnetised rocks of the Earth's crust. Where high accuracy is needed, we use high-resolution models or measure the field in the oil field itself (in-field referencing, IFR).

The second conversion goes from true north to grid north (the rule for the conversion). Grid north is the direction of the vertical lines of the map coordinate grid. The grid convergence $\gamma$ depends only on the position of the point in the zone of the coordinate grid, and we calculate it from the coordinates:

Grid convergence from the coordinates of the point

\tan\gamma = \tan\Delta\lambda\cdot\sin\Phi, \qquad \gamma \approx \Delta\lambda\,\sin\Phi

where:

$\gamma$ grid convergence: the angle from true north to grid north, positive if grid north lies to the east, degrees
$\Delta\lambda$ difference between the longitude of the point and that of the central meridian of the zone; positive east of it, degrees
$\Phi$ latitude of the point (surveyors often write it as $B$ ), degrees

Units: degrees; in the approximate form $\gamma$ and $\Delta\lambda$ are in the same unit of angle

Holds when

A transverse Mercator projection (Gauss–Krüger or UTM, the usual ways to draw the Earth on a flat map) and a spherical Earth. The flattening of the Earth changes $\gamma$ in a zone 6° wide by only fractions of an arc second (an arc second is 1/3600 of a degree).
The approximate form differs from the exact one by thousandths of a degree within the zone.
The central meridians of the 6° zones are 3, 9, 15, … °E. For Gauss–Krüger zone number $n$ the central meridian is $6n - 3$ . In UTM the number of the same zone is 30 higher.

On the central meridian, grid north and true north are the same. The convergence grows towards the edges of the zone and towards the pole. In the northern hemisphere it is positive east of the central meridian, where grid north is turned to the east of true north, and negative west of it.

Worked exampleGrid convergence and declination on the date of the survey near Atyrau

Given (assumed data)

Wellhead: 47.1° N, 51.9° E. The coordinates are in the zone with the central meridian at 51° E (Gauss–Krüger zone 9, which is UTM zone 39)
Survey of 2026: magnetic azimuth 111.91° (the example above)
Declination from IGRF-14: +9.3° in 2026, +7.4° in 1996
Old well on the same pad: surveyed with a magnetic multi-shot instrument in 1996; the horizontal departure of its bottom (its distance sideways from the wellhead) is 900 m

Solution

Difference in longitude from the central meridian: $\Delta\lambda = 51.9 - 51 = 0.9^\circ$ . Grid convergence: $\gamma \approx 0.9\cdot\sin 47.1^\circ = 0.659^\circ$ . The exact formula gives 0.659°.
Grid azimuth of the 2026 survey: $111.91 + 9.3 - 0.66 = 120.55^\circ$ .
We recalculate the surveys of 1996 with the declination of their own year, +7.4°. If we take today's +9.3° instead, all azimuths of the old well turn by $9.3 - 7.4 = 1.9^\circ$ .
Then the bottom of the old well shifts sideways by $900\cdot\sin 1.9^\circ = 29.8$ m.

Answer: The grid convergence is 0.66°, and the grid azimuth of the survey is 120.55°. A declination of the wrong year would move the bottom of the old well by 29.8 m.

What it means. Thirty metres is more than the error of a good survey. An error of this size is enough to misjudge the distance to the old well when we drill a new one beside it. So in the well records, next to each set of surveys, we write down the field model, the date, and the declination and grid convergence that were used. Before we design a new well beside old ones, we calculate the old data again. The declination values come from the IGRF-14 model. The other data in this example are assumed for illustration.

Interference from the steel of the string and non-magnetic drill collars

Steel drill collars, pipes, the downhole motor (the motor that turns the bit, driven by the flow of drilling mud) and the bit become magnetised. The causes are the field of the Earth, knocks and vibration while drilling, and magnetic particle inspection in the workshop (a test that finds cracks with a magnet). A long, thin piece of steel is magnetised mainly along its axis. It behaves like a magnet with poles at its ends. So we put the magnetic sensors inside non-magnetic drill collars (what they are made of), as far as possible both from the steel above and from the motor and the bit below. The interference that reaches the sensors points mainly along the axis of the tool. It adds an amount $\Delta B_z$ to the reading $B_z$ .

Geometry shows why such an addition spoils the azimuth. The tool finds north from the horizontal component of the field. The axial addition, seen from above (projected onto the horizontal plane), lies along the direction of the well. While the well runs north or south, the addition only makes the horizontal vector longer or shorter and does not turn it. When the well runs east or west, the addition points across this vector and turns it, so the azimuth goes wrong.

Azimuth error from axial magnetic interference

\Delta\varphi \approx -\frac{\Delta B_z\,\sin\theta\,\sin\varphi_{\text{m}}}{B_h}, \qquad B_h = B\cos I

where:

$\Delta\varphi$ error of the magnetic azimuth: measured minus true, rad (× 57.3 for degrees)
$\Delta B_z$ axial interference: an addition to the field along the axis of the tool; positive down the hole, nT
$\theta,\ \varphi_{\text{m}}$ inclination and magnetic azimuth of the well, degrees
$B_h$ horizontal component of the field of the Earth, nT

Units: $\Delta B_z$ and $B_h$ are in the same units (nT); $\Delta\varphi$ is in radians

Holds when

The interference acts along the axis of the tool. The formula does not describe cross interference (from neighbouring casing strings, or from magnetised spots in the non-magnetic drill collar itself).
The formula is a first-order approximation (it keeps only the main term). It holds when $\Delta B_z$ is much smaller than $B_h$ .
The error is systematic: it is the same in all surveys as long as the assembly and its magnetisation do not change.

The error grows with the inclination. It is largest at azimuths of 90° and 270°, vanishes at 0° and 180°, and is inversely proportional to the horizontal component of the field. With the same assembly, surveys in a well that runs north are almost exact, and in a well that runs east they are shifted. In the north of Kazakhstan, where $B_h$ is smaller, they are shifted more than in the south.

CalculatedCalculated error of the magnetic azimuth from an axial interference of 200 nT, against the azimuth of the well, for inclinations of 20, 45 and 90°. The field is that of Atyrau (

B_h

about 21,300 nT). The calculation uses the full sensor formulas, without the simplification (linearisation) of the first-order formula.

The pole model estimates how strong the interference is at distance $d$ from the end of a steel part. A magnetised part behaves as if a magnetic “charge” were concentrated at its end: a pole of strength $P$ , measured in microwebers (µWb). When the string is magnetised in the usual way, by the field of the Earth, the fields of the poles of the steel above and below the sensors add up. They do not cancel each other.

Axial interference from the poles of steel above and below the sensors

\Delta B_z = \frac{1}{4\pi}\left(\frac{P_{\text{above}}}{d_{\text{above}}^2} + \frac{P_{\text{below}}}{d_{\text{below}}^2}\right)

where:

$P_{\text{above}},\ P_{\text{below}}$ strengths of the poles of the steel above and below the sensors, Wb
$d_{\text{above}},\ d_{\text{below}}$ distances along the axis from the sensors to these poles, m
$\Delta B_z$ axial interference at the sensors, T

Units: SI: $P$ in Wb, $d$ in m, $\Delta B_z$ in T

Holds when

The poles are points and lie on the axis of the tool. The field of the far ends of the steel parts, which has the opposite sign, is not included, so the estimate is slightly too high.
When the steel is magnetised by the field of the Earth, the lower end of the steel above the sensors and the upper end of the steel below them give a field in the same direction. Magnetisation from knocks, from magnetic particle inspection and from earlier jobs cannot be predicted in strength or sign. So we take the pole strengths from experience with surveys in the same kinds of assemblies.

The field of a pole falls with the square of the distance: twice as far means four times weaker. We choose the length of the non-magnetic section so that the total interference stays within the azimuth error that is allowed for this well.

In other unit systems (1)

P

in µWb,

\Delta B_z

in nT

\Delta B_z = 79.6\left(\frac{P_{\text{above}}}{d_{\text{above}}^2} + \frac{P_{\text{below}}}{d_{\text{below}}^2}\right)

$P$ in µWb, $d$ in m, $\Delta B_z$ in nT.

$79.6 = 10^{-6}\cdot 10^{9}/(4\pi)$ : it converts µWb to Wb and T to nT.

Worked exampleHow much non-magnetic drill collar we need above the sensors

Given (assumed data)

Field (Atyrau area): $B = 53{,}090$ nT, $I = 66.3^\circ$
Well: inclination 60°, magnetic azimuth 90° (east)
Allowed azimuth error from interference: 0.5°
Below the sensors: steel of the motor: a pole of 150 µWb at 12 m from the sensors
Above the sensors: steel drill collars: a pole of 800 µWb

Solution

Horizontal component of the field: $B_h = 53{,}090\cdot\cos 66.3^\circ = 21{,}339$ nT.
Allowed axial interference: $\Delta B_z = \dfrac{0.5\cdot\pi/180\cdot 21{,}339}{\sin 60^\circ\cdot\sin 90^\circ} = 215.03$ nT.
Contribution of the steel below: $79.6\cdot 150/12^2 = 82.9$ nT. What is left for the steel above is $215.03 - 82.9 = 132.1$ nT.
Distance to the pole of the steel above: $d_{\text{above}} = \sqrt{79.6\cdot 800/132.1} \approx 22$ m.
The same well at azimuth 30°: $\sin 30^\circ = 0.5$ , so the allowance is twice as large, 430.06 nT, and for the steel above $\sqrt{79.6\cdot 800/347.2} \approx 14$ m is enough.
A well that runs east in the Aksai area, where $B_h = 54{,}430\cdot\cos 69.6^\circ = 18{,}973$ nT: the allowance is 191 nT, and about 24 m is needed above.

Answer: The steel above must start no closer than 22 m from the sensors. A well that runs north-north-east would need 14 m, and a well that runs east near Aksai would need 24 m.

What it means. We choose the length of the non-magnetic section for the worst combination of direction and inclination that will occur in the interval, and for the latitude of the area. If the section cannot be made longer, we correct for the axial interference by calculation from the known field (next section), or we switch to gyroscopic surveys. The pole strengths are assumed, at about the size met in practice. The real values are found from surveys. The data in this example are assumed for illustration.

Checking a survey against the field of the Earth

A good survey must “see” the Earth as it really is: the total gravity is 1 g, and the total field and the dip angle are as in the model for this place and date. We calculate these three quantities from the six readings, and the azimuth is not needed for them. So comparing them with the reference is a simple and strong check of every survey:

Check quantities of a survey

G = \sqrt{G_x^2 + G_y^2 + G_z^2}, \qquad B = \sqrt{B_x^2 + B_y^2 + B_z^2}, \qquad \sin I = \frac{G_xB_x + G_yB_y + G_zB_z}{G\,B}

where:

$G$ total gravity from the accelerometers, g
$B$ total field from the magnetometers, nT
$I$ magnetic dip angle from the survey: the angle between the field vector and the horizontal, degrees

Units: $G$ in $g$ , $B$ in nT

Holds when

The tool is at rest.
The reference $B$ and $I$ come from a field model for the place and the date, or from local measurements.
The tolerances come from the error model of the particular instrument. Different companies use different ones.

The scalar product (dot product) of the gravity vector and the field vector is one number that shows how closely the two arrows line up. It does not depend on how the tool is turned. So the dip angle is an independent check of the magnetometers together with the accelerometers. A deviation of $G$ points to the accelerometers, or to movement of the tool during the survey. Deviations of $B$ and $I$ point to the magnetometers or to magnetic interference.

What the check does not see: errors that hardly change the length of the field vector and its angle with the vertical. These are a wrong declination, cross interference and, for some directions of the well, a moderate axial interference (example below).

Worked exampleTwo surveys: a good one and one with interference

Given (assumed data)

Reference from the field model: $B = 53{,}090$ nT, $I = 66.3^\circ$
Tolerances of the example: $G$ ±0.003 g, $B$ ±300 nT, $I$ ±0.3°
Survey A (the example above): −0.4396; 0.3688; 0.8195 g and −39,059; 6,955; 35,251 nT
Survey B (the next stand of pipe, two or three pipes screwed together): −0.1560; 0.5814; 0.7990 g and −28,234; 29,587; 34,101 nT

Solution

Survey A: $G = 1.00042$ g. $B = 53{,}071.7$ nT, which is 18.3 nT below the reference. $G_xB_x + G_yB_y + G_zB_z = 48{,}623.5$ , so $\sin I = 48{,}623.5/(1.00042\cdot 53{,}071.7) = 0.91580$ and $I = 66.32^\circ$ . Everything is within tolerance.
Survey B: $G = 1.00038$ g. $B = 53{,}248.7$ nT, which is 158.7 nT above the reference. $\sin I = 48{,}853.1/(1.00038\cdot 53{,}248.7) = 0.91710$ and $I = 66.51^\circ$ , which is 0.21° more. This is also within tolerance.
The angles of survey B from the formulas of the section on sensors: $\theta = 36.99^\circ$ , $\varphi_{\text{m}} = 112.57^\circ$ .
Correction for the axial interference, using the known vertical component of the field, $B\sin I = 53{,}090\cdot\sin 66.3^\circ = 48{,}612.5$ nT. The readings must give $G_xB_x + G_yB_y + G_zB_z = G\cdot 48{,}612.5$ . From this, $B_z = (1.00038\cdot 48{,}612.5 - 4404.5 - 17{,}201.9)/0.7990 = 33{,}823$ nT, which is 278 nT less than the measured value.
Azimuth with the corrected $B_z$ : 112.99°, which is 0.42° more than without the correction.

Answer: Both surveys passed the check, but in survey B an axial interference of about 278 nT shifted the azimuth by 0.42°.

What it means. The check against the field of the Earth is a sieve for gross errors. It is not proof of accuracy. With an interference of almost 280 nT, the total field deviated from the reference by only 158.7 nT, and the dip angle by only 0.21°. The reason is that only a part of the field vector lies along the axis of the tool. An axial addition changes the length and the tilt of the vector much less than it changes its direction seen from above. So when the non-magnetic section is shorter than it should be, we recalculate the axial component for every survey from the known field. The method shown rests on the vertical component. It works as long as $G_z$ is clearly larger than zero. It is only as accurate as the reference $B_v$ . An error in $B_v$ passes into $B_z$ multiplied by $G/G_z$ . For survey B, an error of 100 nT in the field model would give an error in $B_z$ of about 125 nT and an error in the azimuth of about 0.19°. In a horizontal well that runs east or west, neither $B$ nor $I$ depends much on the axial component. So any such correction loses accuracy exactly where the interference spoils the azimuth most. The data in this example are assumed for illustration.

Gyroscopic survey instruments

Gyroscopic instruments do not notice steel. So they work where magnetic instruments are useless: inside casing, when we sidetrack from old wells, and in the upper part of a well on a pad, next to the casing strings of neighbouring wells. They also give an independent check of magnetic surveys. They are built in different ways.

Free gyro (a positional gyroscope): a rotor that spins fast in gimbals (a ring mount that lets it turn freely) keeps the direction of its axis in space. At the surface, the axis is pointed at a reference mark with a known azimuth, and the tool measures the azimuth from it. The Earth turns under the fixed axis. We compensate this apparent rotation using the latitude of the place. The drift that remains (the slow wandering of the reading) is found by comparing the reading with the same reference mark after the tool comes back up.
Gyrocompass (a north-seeking gyro): rate sensors (sensors of angular speed) measure the components of the rotation of the Earth and use them to find true north by themselves, with no initial setting. The horizontal component of the rotation, $\Omega\cos\Phi$ , sets the direction of north.
Continuous gyro survey instrument: after it has found north, the tool moves along the well without stopping. It adds up (integrates) the angular speeds to follow its own turns. This is a fast way to survey the whole path inside casing. There are also gyroscopic modules in the bottom hole assembly for the first sections of a well, where magnetic surveys are not possible.

Components of the rotation of the Earth

\Omega_h = \Omega\cos\Phi, \qquad \Omega_v = \Omega\sin\Phi, \qquad \Omega = 15.041^\circ/\text{h}

where:

$\Omega$ angular speed of the rotation of the Earth relative to the stars: one turn in 23 h 56 min, °/h
$\Omega_h$ horizontal component, which points to true north, °/h
$\Omega_v$ vertical component: the turning of the local meridian (the north–south line) around the vertical, °/h
$\Phi$ latitude of the place, degrees

Units: °/h; latitude in degrees

Holds when

The tool is at rest relative to the Earth. The latitude is the geographic latitude.

The gyrocompass looks for north using $\Omega_h$ . At the latitude of Atyrau this is about 10.2°/h. At the pole it is zero, so gyrocompassing (finding north with a gyrocompass) loses accuracy towards high latitudes. A free gyro, by contrast, “sees” the rotation of the Earth around the vertical. Without compensation its axis would drift relative to the meridian at $\Omega_v$ , which is about 11°/h at the same latitude.

Worked exampleCorrection for the drift of a free gyro

Given (assumed data)

Latitude of the wellhead: 47.1° N
Instrument: free gyro; the apparent rotation of the Earth is compensated using the latitude
Run: 2.5 h; after the tool is pulled up, it reads −1.25° on the same reference mark instead of 0°
Readings: at 0.6, 1.0 and 1.4 h after setting on the reference mark

Solution

Without compensation the axis would drift by $\Omega_v = 15.041\cdot\sin 47.1^\circ = 11.02$ °/h, which is about 28° over the run.
Residual drift: by the end of the run the readings had fallen behind by 1.25°, which is $1.25/2.5 = 0.50$ ° per hour.
If we take the drift as steady, we add $0.50\cdot 0.6 = 0.30^\circ$ , $0.50\cdot 1.0 = 0.50^\circ$ and $0.50\cdot 1.4 = 0.70^\circ$ to the azimuths.

Answer: The azimuths of the three readings are increased by 0.30, 0.50 and 0.70°.

What it means. Compensating for the rotation of the Earth is essential. Without it, the error over the run would be tens of degrees. A straight-line correction for the residual drift is right only if the drift was steady. A knock while the tool is run in, or sudden heating, makes the drift change suddenly. Readings after such an event are wrong in a different way than the correction assumes. So we run a free gyro in smoothly, check it against the reference mark before and after the run, and choose gyrocompasses for long runs. The data in this example are assumed for illustration.

Worked exampleAccuracy of a gyrocompass: latitude and inclination of the well

Given (assumed data)

Zero offset (bias, a constant error of the reading) of each of the two rate sensors across the axis: 0.05°/h
Latitudes: 47.1° (Atyrau) and 70°
Well: vertical; inclined: 60° towards the east

Solution

Horizontal component of the rotation of the Earth: $15.041\cdot\cos 47.1^\circ = 10.24$ °/h. At latitude 70° it is 5.14 °/h.
In a vertical well the azimuth error is $\varepsilon/\Omega_h$ : $0.05/10.24 = 0.00488$ rad, which is 0.28°. At latitude 70° it is 0.56°.
In a well with an inclination of 60° that runs east, the error is $1/\cos 60^\circ = 2$ times larger (advanced section): 0.56°.

Answer: Near Atyrau, such a gyrocompass finds north with an error of about 0.28° in a vertical well and 0.56° in an inclined well that runs east. At latitude 70° the error is already 0.56° in a vertical well.

What it means. The gyrocompass takes north from a signal of about 10°/h, and a sensor zero offset of 0.05°/h is half a per cent of the signal. That is why the error is tenths of a degree. So many gyroscopic instruments find north where the well is close to vertical. Further along the well they continuously track their own turn from the direction already found. The calculation includes only the zero offset. The error models of gyroscopic instruments include other sources too. The data in this example are assumed for illustration.

Inertial systems and wireline steering tools

An inertial navigation system is made of three gyroscopes and three accelerometers. They measure all the turns and all the accelerations of the tool. Adding up the accelerations twice (double integration) gives the distance moved, that is, the coordinates of the tool directly, and not only the angles. The price is a fast growth of the error. A constant offset (bias) $b$ of an accelerometer gives, after double integration, a position error of $b\,t^2/2$ , which grows with the square of time. An offset of only $10^{-5}$ g gives $0.5\cdot 9.81\cdot 10^{-5}\cdot 60^2 \approx 0.18$ m in one minute of movement, and already about 18 m in ten minutes. So the tool is often stopped. At rest its speed is exactly zero, and the speed error that has built up is reset (a zero-velocity update).

Inertial systems are among the most accurate downhole instruments, but for a long time they were large and expensive. They were used for reference surveys of well paths in cased wells of large diameter, above all on offshore platforms with closely spaced wells. The same principles, rate sensors and integration of turns, are also used by modern continuous gyro instruments of small diameter.

A steering tool (a wireline telemetry system, that is, one that sends its readings to the surface along a cable) is a magnetic survey instrument that is lowered inside the drill pipe on a single-conductor logging cable. It lands in an orienting sub (a pipe fitting that fixes the position of the tool) above the deflection tool. Through the cable it sends the inclination, the azimuth and the toolface all the time, and the driller sees how the deflection tool works while drilling goes on. The limit is the cable itself. The string cannot be rotated, so the system suits only drilling with a downhole motor in sliding mode (the string does not turn, and the motor alone turns the bit). At every connection (when a new pipe is added) the cable must be pulled up or taken out of the pipes through a side-entry sub (a short pipe with a side opening for the cable). Since MWD (see Measurement while drilling) appeared, such systems are rarely used.

Errors, the uncertainty region and wells that come close together

Every survey contains errors: zero offsets and scale errors of the sensors, a tilt of the tool against the axis of the hole, sag of the drill collars in an inclined hole, a depth error from stretch of the pipe, inaccuracy of the field model, and magnetic interference. Some of these errors are random and partly average out from one survey to the next. But many are systematic: the same declination error, or the interference of the same assembly, acts on all surveys and builds up along the well. So the position uncertainty grows with depth and depends on the path of the well.

The approach of the industry is an error model: a list of the sources of error with their sizes for each type of instrument, and the rules by which they add up along the well. A widely used model is that of the industry committee ISCWSA (Industry Steering Committee on Wellbore Survey Accuracy) for MWD, with related models for gyroscopic instruments. The result is a region around each point of the well in which the true point lies with a given probability. In space it is an ellipsoid (a stretched ball). Seen from above, and in the vertical plane, it is an ellipse of uncertainty (a stretched circle). Below is a simplified hand calculation of the same ideas for three systematic errors.

Position uncertainty from three systematic errors

\begin{aligned} \sigma_{\text{lat}} &= A\,\sigma_\varphi,\\ \sigma_{\text{along}} &= \sqrt{(\sigma_\theta\,(H - H_{\text{KOP}}))^2 + (\varepsilon_L A)^2},\\ \sigma_H &= \sqrt{(\sigma_\theta A)^2 + (\varepsilon_L H)^2} \end{aligned}

where:

$\sigma_{\text{lat}}$ standard deviation (the typical size of the error) of the position sideways, across the vertical plane of the well, m
$\sigma_{\text{along}},\ \sigma_H$ the same along the plane of the well, horizontally (along the departure) and vertically, m
$A,\ H$ horizontal departure and true vertical depth of the point, m
$H_{\text{KOP}}$ depth of the kick-off point (where the well starts to turn away from the vertical), m
$\sigma_\varphi,\ \sigma_\theta$ systematic errors of the azimuth and of the inclination, rad
$\varepsilon_L$ relative error of the depth along the hole, —

Units: lengths in metres, angles in radians

Holds when

The well is vertical down to the kick-off point and below it runs in one vertical plane.
The three errors are independent and constant along the whole well. The azimuth error turns the whole path, seen from above, around the wellhead. The inclination error turns the inclined part around the kick-off point. The depth error stretches the whole path from the rotary table.
Random errors of single surveys, tilts, sag and interference are not included. A real error model has dozens of terms and rules for adding them from survey to survey.

The azimuth error acts like a lever as long as the departure: the farther the target lies sideways, the larger the error. In inclined wells it usually sets the cross size of the ellipse. The errors of inclination and depth set its size along the plane of the well and vertically.

Worked exampleUncertainty at the target of the well from the lesson on the well profile

Given (assumed data)

Well: a build-and-hold profile from the lesson Directional wells: applications and profile: kick-off at 500 m, the target at a true vertical depth of 2400 m with a horizontal departure of 1100 m, length of the hole to the target 2710.6 m
Systematic errors (1σ, assumed): azimuth 0.5°, inclination 0.1°, depth 1 m per 1000 m
Neighbouring well: 60 m from the target, across the plane of the well; the semi-axis (half-width) of its 95% ellipse towards us is 15 m

Solution

Sideways: $\sigma_{\text{lat}} = 1100\cdot 0.5\cdot\pi/180 = 9.60$ m.
Along the plane of the well: from the inclination, $0.1\cdot\pi/180\cdot(2400 - 500) = 3.316$ m; from the depth, $0.001\cdot 1100 = 1.100$ m. Together, $\sqrt{3.316^2 + 1.100^2} = 3.494$ m.
Vertically: from the inclination, $0.1\cdot\pi/180\cdot 1100 = 1.920$ m (the point is higher if the angle is too large); from the depth, $0.001\cdot 2400 = 2.400$ m. Together, 3.07 m.
The ellipse seen from above, for a probability of 95%: we multiply the semi-axes by $\sqrt{-2\ln 0.05} = 2.448$ . This gives 23.50 m across and 8.55 m along the plane of the well.
Separation factor with the neighbouring well: $60/(23.50 + 15) = 1.56$ .

Answer: Seen from above, the target is known to within ±23.50 m across and ±8.55 m along the plane of the well (95%). The standard deviation vertically is 3.07 m. The separation factor with the neighbouring well is 1.56.

What it means. The cross size of the ellipse is almost three times larger than the size along the well. In an inclined well, the azimuth above all decides how accurate the position is. So the most effective measures are to improve the field data (a high-resolution model, local measurements), to correct for the interference of the string, and to check the azimuth with a gyroscope. The factor 2.45 applies to a two-dimensional ellipse. For a single coordinate, 95% means ±1.96σ, and for a three-dimensional ellipsoid about 2.8σ. So reports always state which level was used. The data in this example are assumed for illustration.

CalculatedCalculated growth of the uncertainty (1σ) along the well of the example. Sideways, it comes from the azimuth error. Along the plane of the well and vertically, it comes from the errors of inclination and depth. The model is simplified and the errors are assumed.

We need uncertainty regions above all for collision avoidance (anti-collision), which keeps one well from running into another. We compare the distance between the wells with the sizes of their uncertainty regions in the direction towards each other:

Separation factor of two wells

SF = \frac{D}{R_1 + R_2}

where:

$SF$ separation factor, —
$D$ distance between the centres of the wells at the points considered, m
$R_1,\ R_2$ sizes of the uncertainty regions of the two wells in the direction towards each other, for the chosen probability, m

Units: metres; the result has no units

Holds when

The uncertainty regions come from the error models of both wells, for one probability level (often 2–3σ).
Company standards have variants: some subtract the radii of the wells from the distance, and some add the sizes in other ways. The standards themselves set the limits of allowed values.

At $SF = 1$ the regions touch. At $SF < 1$ they overlap, and we cannot say that the wells will not meet. The farther from the wellhead and the larger the ellipses, the larger the gap needed between neighbouring wells.

Terms and symbols

Directional survey, survey instrument: The measurement of the inclination and azimuth at a known depth, and the instrument that does it.
Accelerometer, magnetometer: Sensors of the components of gravity and of the magnetic field along their own axis. A tool has three of each.
Toolface $\omega$: The gravity toolface is measured from the high side of the hole. In an almost vertical hole, the magnetic toolface is measured from magnetic north.
Magnetic dip angle $I$: The angle between the field vector of the Earth and the horizontal. It sets the share of the horizontal component, $B\cos I$ .
Declination $\delta$ , secular variation: The angle from true north to magnetic north, and its slow change from year to year.
Grid convergence $\gamma$: The angle from true north to grid north. From the coordinates, $\tan\gamma = \tan\Delta\lambda\sin\Phi$ .
Non-magnetic drill collar: A drill collar made of a non-magnetic alloy, with the magnetic sensors inside it.
Axial interference $\Delta B_z$ , pole strength $P$: The extra field along the axis of the tool from magnetised steel, and its estimate from the pole model. The strength of a pole is measured in µWb.
Gyroscopic survey: Free gyro, gyrocompass and continuous gyro. They find the azimuth without the magnetic field.
Inertial system: Three gyroscopes and three accelerometers. It finds coordinates by integrating (adding up) the accelerations.
Ellipse of uncertainty, error model: The region of the probable position of the well, and a way to calculate it from the sources of error.
Separation factor $SF$: The distance between the wells divided by the sum of the sizes of their uncertainty regions.

Common mistakes and misconceptions

Wrong

“A non-magnetic drill collar solves the interference problem for any assembly.”

In fact

It only moves the sensors away from the steel. The length needed depends on the pole strengths, the inclination, the direction of the well and the latitude. If the length is not enough, we correct for the interference by calculation or switch to a gyroscope.

Wrong

“The survey passed the check on G, B and the dip angle, so the azimuth is right.”

In fact

The check catches gross errors, but not a wrong declination or a moderate axial interference. In the example of this lesson, an interference of about 280 nT shifted the azimuth by 0.4° and still stayed within the tolerances.

Wrong

“The declination is a constant of the oil field; we can take it from an old report.”

In fact

The declination changes with time. Near Atyrau it grew by almost 2° from 1996 to 2026. We recalculate every survey with the declination of its own date, and we write the value used next to the surveys.

Wrong

“A gyroscope shows true north equally accurately everywhere.”

In fact

A free gyro remembers only its initial setting, and it drifts. A gyrocompass looks for north using $\Omega\cos\Phi$ , and its accuracy falls with latitude and in inclined wells that run east or west.

Wrong

“Half a degree of azimuth is nothing.”

In fact

At a departure of 1100 m it is almost 10 m sideways at one standard deviation. In inclined wells it is the azimuth that sets the cross size of the ellipse of uncertainty and the margin to neighbouring wells.

In depth: advanced detail

Derivation of the azimuth formula and the formula for axial interference

Attach three unit vectors (arrows of length 1) to the well. $\mathbf t$ points along the axis, down the hole. $\mathbf h$ points across the axis to the high side. $\mathbf r$ points horizontally to the right, looking down the hole. The tool axis $x$ is turned from $\mathbf h$ towards $\mathbf r$ by the toolface $\omega$ , and the axis $y$ by 90° more. Gravity has the components $-g\sin\theta$ along $\mathbf h$ , zero along $\mathbf r$ and $g\cos\theta$ along $\mathbf t$ . From this, $G_x = -g\sin\theta\cos\omega$ , $G_y = g\sin\theta\sin\omega$ and $G_z = g\cos\theta$ , and the formulas for $\theta$ and $\omega$ follow.

A turn by $-\omega$ moves the magnetometer readings into the components along $\mathbf h$ and $\mathbf r$ : $B_{\mathbf h} = B_x\cos\omega - B_y\sin\omega$ and $B_{\mathbf r} = B_x\sin\omega + B_y\cos\omega$ . The horizontal direction of the well is $\cos\theta\,\mathbf h + \sin\theta\,\mathbf t$ , and the field of the Earth projects onto it as $B_h\cos\varphi$ . Onto $\mathbf r$ it projects as $-B_h\sin\varphi$ . So $\tan\varphi = -B_{\mathbf r}/(B_{\mathbf h}\cos\theta + B_z\sin\theta)$ .

Substitute $\cos\omega = -G_x/(g\sin\theta)$ , $\sin\omega = G_y/(g\sin\theta)$ , $\cos\theta = G_z/g$ and $\sin^2\theta = (G_x^2 + G_y^2)/g^2$ , and multiply the numerator and the denominator by the positive $g^2\sin\theta$ . This gives the formula of the lesson, with $G = g$ . The formula for the axial interference follows from the same result. The addition $\Delta B_z$ enters only the denominator, with the factor $\sin\theta$ , and the denominator itself equals $B_h\cos\varphi$ while the numerator is $B_h\sin\varphi$ . The derivative of the arctangent (its rate of change) gives $\Delta\varphi \approx -\Delta B_z\sin\theta\sin\varphi/B_h$ .

Why a gyrocompass loses accuracy in an inclined well

Let two rate sensors measure the projections of the rotation of the Earth onto $\mathbf h$ and $\mathbf r$ (the accelerometers show where the high side is). In the axes “north, east, down” the rotation of the Earth is $(\Omega\cos\Phi,\ 0,\ -\Omega\sin\Phi)$ . So $\omega_{\mathbf r} = -\Omega\cos\Phi\sin\varphi$ and $\omega_{\mathbf h} = \Omega\cos\Phi\cos\theta\cos\varphi + \Omega\sin\Phi\sin\theta$ , where $\varphi$ is the true azimuth.

Azimuth error of a gyrocompass from the zero offsets of the sensors

\sigma_\varphi = \frac{\varepsilon}{\Omega\cos\Phi}\sqrt{\cos^2\varphi + \frac{\sin^2\varphi}{\cos^2\theta}}

where:

$\sigma_\varphi$ azimuth error, rad
$\varepsilon$ zero offset of each of the two rate sensors, °/h
$\Omega\cos\Phi$ horizontal component of the rotation of the Earth, °/h
$\theta,\ \varphi$ inclination and true azimuth of the well, degrees

Units: $\varepsilon$ and $\Omega$ are in the same unit of angular speed; the result is in radians

Holds when

The zero offsets of the two sensors are independent and equal in size.
The inclination and the toolface are known exactly, and the tool is at rest.

The sensor along $\mathbf r$ finds the deviation from the north–south direction, and this sensor is always horizontal. The deviation from the east–west direction has to be taken from $\omega_{\mathbf h}$ . But the axis $\mathbf h$ is tilted by $\theta$ , so it feels only the part $\cos\theta$ of the horizontal rotation. So in a well that runs east or west the error grows as $1/\cos\theta$ , and in a well that runs north or south it does not depend on the inclination.

Connections to other topics

Directional survey calculations — how the coordinates of the well are calculated from surveys, and how azimuths are brought to one north.
Measurement while drilling — how the direction sensors work in the assembly and send the survey to the surface while drilling.
Deflection tools and downhole motors — how the toolface is used to orient the deflection tool.
Directional wells: applications and profile — the planned profile, the tolerance circle and the distances to neighbouring wells.
Drill string: pipe, tool joints, drill collars and heavy-weight drill pipe — non-magnetic drill collars and other parts of the bottom hole assembly.
Horizontal wells — why the accuracy of the azimuth and of the true vertical depth is especially important in horizontal wells.

Check yourself

1
Before a sidetrack is cut, we must find out the path of the old well inside the casing. Which instrument should we choose?
1. A.Any of them: the casing does not affect the survey
2. B.A magnetic multi-shot instrument in a non-magnetic drill collar
3. C.MWD sensors
4. D.A gyroscopic instrument
Show the answer
Correct answer: D. A gyroscopic instrument
The steel casing distorts the magnetic field inside it so much that magnetic surveys are useless, both with a magnetic multi-shot instrument and with MWD. A gyroscope finds direction without the magnetic field, from its initial setting or from the rotation of the Earth, and it works inside casing.
2
A well with an inclination of 45° runs at a magnetic azimuth of 270°. The horizontal component of the field is 23,000 nT, and the axial interference is +150 nT. Estimate the azimuth error and its sign.
Show the answer
$\Delta\varphi \approx -150\cdot\sin 45^\circ\cdot\sin 270^\circ/23{,}000 = +150\cdot 0.7071/23{,}000 \approx +0.0046$ rad, which is about +0.26°. The tool will show about 270.26° instead of 270°. At an azimuth of 90° the same interference would give −0.26°, and at 0° or 180° almost zero.
3
Why is a gyrocompass at latitude 70° less accurate than at latitude 45°, with the same sensors?
1. A.The horizontal component of the rotation of the Earth is smaller there
2. B.Gravity is stronger there
3. C.The temperature of the rocks is lower there
4. D.The magnetic field is stronger there
Show the answer
Correct answer: A. The horizontal component of the rotation of the Earth is smaller there
A gyrocompass finds north from the horizontal component of the rotation of the Earth, $\Omega\cos\Phi$ . At 45° this is 10.6°/h, and at 70° it is 5.1°/h. The same sensor error is twice as large a share of the signal, and the azimuth error roughly doubles.
4
A survey gave $G = 1.0002$ g, $B = 52{,}760$ nT and a dip angle of 66.9°, against a reference of 53,090 nT and 66.3° (tolerances ±0.003 g, ±300 nT, ±0.3°). What does this mean, and what should we do?
Show the answer
Gravity is within tolerance, so the accelerometers work and the tool was at rest. The total field is 330 nT below the reference and the dip angle is 0.6° above it. Both are outside the tolerances, so the magnetic part of the survey cannot be trusted (interference from the string or from steel nearby, or faulty magnetometers). We cannot use the azimuth of this survey. We repeat the survey, check the distances to steel in the assembly and the neighbouring casing strings, apply a correction for the interference if its nature is clear, or check the direction with a gyroscope.
5
The wellhead is at 45.0° N, 2.0° west of the central meridian of the zone. Find the grid convergence and the grid azimuth for a true azimuth of 30.0°.
Show the answer
$\gamma \approx -2.0\cdot\sin 45^\circ \approx -1.41^\circ$ (the exact formula gives the same to hundredths). West of the central meridian, grid north is turned to the west of true north. The grid azimuth equals the true azimuth minus the convergence: $30.0 - (-1.41) = 31.41^\circ$ .
6
The same assembly, with a short non-magnetic section, gave good azimuths in a well that ran north and clearly shifted azimuths in a well that ran east, at the same inclination. Why?
Show the answer
The interference from the steel points along the axis of the tool. Its horizontal projection lies along the direction of the well. In a well that runs north, it is parallel to the horizontal component of the field and only changes its length. In a well that runs east, it is perpendicular to it and turns it. The error is proportional to $\sin\varphi$ : zero for north, and largest for east and west.

Updated: 19 October 2026