Deflection tools and downhole motors

Quick refresher

The bit goes to the side if its axis is tilted to the axis of the hole (a tilt) or if a side force presses it against the wall. Every deflection tool uses one of these ways, or both.
A motor with a tilt bends the well only while the string does not rotate, that is, in sliding drilling (the string only slides down, and the motor alone turns the bit). If the string rotates, the tilt turns with it, its effect averages out, and the well goes almost straight.
In a PDM the rotational speed is proportional to the flow rate and the torque is proportional to the pressure drop across the motor, so the standpipe pressure (the pressure on the gauge at the rig) shows the load. A turbodrill is different: its torque falls as the speed grows, and its pressure drop hardly depends on the load. On the other hand, it has no rubber parts, so it works at high temperature.
The toolface sets the direction of the bend relative to the high side of the hole (its upper wall): 0° builds the angle, 180° drops it, 90° turns the well to the right, and 270° to the left. The reaction torque of the motor turns the tool counter-clockwise.
Assemblies without a motor change the angle by the bending of the drill collars (heavy, thick-walled pipes). A near-bit stabiliser (a short section with blades that touch the wall) acting as the fulcrum of a lever builds the angle, several stabilisers placed close together hold it, and a long free section of collars above the bit (a pendulum) drops it.
A rotary steerable system deflects the bit while the string keeps rotating. The hole is cleaner and smoother, the string does not “hang up”, and we can drill farther. The price is the cost, the complexity and the temperature limits of the electronics.

Key relationships

i = \dfrac{2\,\delta\,l_0}{L}

three-point dogleg severity of a deflecting motor: $\delta$ is the bend angle, $L$ is the distance from the bit to the upper stabiliser; for $l_0 = 30$ m, $i = 60\,\delta/L$ in °/30 m

f = \dfrac{i_{\text{plan}}}{i_{\text{slide}}}

share of sliding drilling for the planned rate (rotary drilling goes straight)

n = \dfrac{Q}{q}, \qquad M = \dfrac{\Delta p\,q}{2\pi}

an ideal PDM: the speed from the flow rate, the torque from the pressure drop

\Delta\theta \approx \beta\cos\omega, \qquad \Delta\varphi \approx \dfrac{\beta\sin\omega}{\sin\bar\theta}

how the toolface $\omega$ divides the dogleg $\beta$ between the inclination and the azimuth

F = \tfrac12\,w\,L_{\text{p}}\sin\theta

pendulum force at the bit; $w$ is the weight of a metre of collars in the mud

Common pitfalls

Expecting a motor with a tilt to bend the well while the string rotates.
Taking the three-point rate as a promise. The measurements give the actual rate, and it is usually smaller.
Orienting the deflection tool without leaving an allowance for the reaction torque.
Judging the load on a turbodrill by the standpipe pressure.
Putting a stabiliser right above the bit to drop the angle. It works as the fulcrum of a lever and builds the angle.

After this lesson you can

explain the two ways to deflect the bit, a tilt of its axis and a side force, and recognise them in each deflection tool
describe the construction and the characteristics of a positive displacement motor and of a turbodrill, and calculate the rotational speed and the torque (the turning force) of a PDM from the flow rate and the pressure drop
estimate the dogleg severity (how sharply the hole bends) of a deflecting motor with the three-point method, and calculate what part of an interval must be drilled in the sliding mode (with the string not rotating) to get the planned rate
choose the toolface (the angle that shows which way the bend points) to change the inclination and the azimuth at the same time, and allow for the reaction torque of the motor (the twist that the turning motor puts on the string)
explain how assemblies without a motor build, hold and drop the angle, and estimate the pendulum force (the sideways force that the weight of the free collars puts on the bit)
compare a steerable motor with a rotary steerable system, and list what measurement while drilling and logging while drilling (MWD and LWD) tell the driller

How to make the bit turn

The bit deepens the hole in the direction in which the assembly above it points it. A straight assembly that the stabilisers (short sections with blades that centre the string in the hole) hold in the middle of the hole leads the bit along the axis of the hole that is already drilled. To make the well turn, this symmetry must be broken, and there are two ways to do it.

Tilt: the axis of the bit is tilted by a small angle to the axis of the hole. The bit breaks the rock mostly with its face, so the new hole goes along the axis of the bit and departs gradually from the old hole. This is how a bent sub and a bent motor housing work, and rotary steerable systems that tilt the shaft. In assemblies without a motor, the same thing is done by the bending of the drill collars (heavy, thick-walled pipes) under load.
Side force: the bit is pressed against one side of the hole. Its side cutters and its gauge surface (the part that sets the diameter of the hole) cut into the wall, and the hole shifts in the direction of the force. A side force is made by a whipstock, by the weight of a free section of collars in a pendulum assembly, by the lever action of a near-bit stabiliser, and by the extendable pads of a rotary steerable system.

In a real assembly both ways usually act together: an assembly with a tilt, squeezed into the hole, also presses the bit against the wall. A third, auxiliary way is to remove the rock on one side. In soft rocks a jet of mud from one large nozzle of the bit washes out a hollow, and the bit goes into it (this is called jetting). It is rarely used today.

SchematicFour ways to deflect the bit (a sketch, not to scale). A whipstock in a casing string pushes the mill to the side. A motor with a bent housing tilts the axis of the bit in sliding drilling. A rotary steerable system with extendable pads pushes off the wall and presses the bit to the opposite side. A system that tilts the shaft tilts the axis of the bit while the string rotates. The dashed line shows the direction of the new hole.

Deflection tools and where they fit

Tool: Whipstock
How it deflects: side force: the wedge pushes the mill or the bit to the side
Where it is used: starting a sidetrack out of a cased well
Main limit: one fixed deflection, separate trips

Tool: Motor with a bent sub
How it deflects: tilt above the motor
Where it is used: in the past, for the kick-off and for corrections; rarely today
Main limit: the string cannot be rotated: after the build, the assembly is changed

Tool: Steerable motor (a bend in the housing)
How it deflects: tilt right at the bit
Where it is used: build, turn and hold with one assembly
Main limit: sliding: friction, hole cleaning, a stepped hole

Tool: Bottom hole assembly (BHA) without a motor
How it deflects: bending and weight of the collars, lever action of a stabiliser
Where it is used: vertical holes, tangent sections
Main limit: the rate is hard to predict; it does not control the azimuth

Tool: Rotary steerable system
How it deflects: pads or a tilt of the shaft while rotating
Where it is used: complex, extended-reach and horizontal wells
Main limit: cost, complexity, temperature limits

Tool	How it deflects	Where it is used	Main limit
Whipstock	side force: the wedge pushes the mill or the bit to the side	starting a sidetrack out of a cased well	one fixed deflection, separate trips
Motor with a bent sub	tilt above the motor	in the past, for the kick-off and for corrections; rarely today	the string cannot be rotated: after the build, the assembly is changed
Steerable motor (a bend in the housing)	tilt right at the bit	build, turn and hold with one assembly	sliding: friction, hole cleaning, a stepped hole
Bottom hole assembly (BHA) without a motor	bending and weight of the collars, lever action of a stabiliser	vertical holes, tangent sections	the rate is hard to predict; it does not control the azimuth
Rotary steerable system	pads or a tilt of the shaft while rotating	complex, extended-reach and horizontal wells	cost, complexity, temperature limits

The whipstock and sidetracking

A whipstock is a long steel wedge with a concave inclined surface, which is set in the hole at the kick-off depth. It works as a guide. A tool run from above comes up against the inclined surface and is pushed to the side, and this is the side force. The slope of the wedge is small, a few degrees, so the new hole leaves the old one gradually.

In the past, before downhole motors became an ordinary tool of directional drilling, whipstocks were also used to deflect an open hole. The whipstock was run together with a small-diameter bit, a short pilot hole was drilled with it, the whipstock was pulled out, and the hole was reamed out to the required diameter. Today a whipstock is needed first of all to start a sidetrack out of a cased well. In mature fields this is how a new hole leaves an old well to reach parts of the formation that the old well does not drain, or to bypass a tool left in the hole.

Leaving the well through a window in the casing

1
A base
Below the kick-off depth we set a packer or an anchor on which the whipstock will sit. Before this, the casing is drifted (a gauge tool is run through it to check that it is clear).
2
Orientation
The whipstock is run on pipe, turned with its inclined surface to the required side by the readings of direction sensors, and fixed on the base. Often the whipstock and the mill are run in one trip: the mill is held on the whipstock by a shear bolt.
3
Milling the window
The mills, pushed aside by the whipstock, cut through the wall of the casing, and a window a few metres long is made. The mud carries the metal chips out, and at the surface they are caught by magnets and screens.
4
The new hole
A bottom hole assembly with a deflecting motor is run through the window, and the angle is built in the new direction.

There are other ways to leave an old hole. We can mill out a section of the casing around its whole circumference (section milling), set a cement plug in this interval, and kick off from it with a deflecting motor. In an open hole we kick off from a cement plug, or directly from the wall: a deflecting motor slowly cuts into the wall with a small weight on the bit until the bit goes into new rock.

Downhole motors: the PDM and the turbodrill

For a tilt to work, the bit must rotate and the string above the tilt must stand still. Otherwise the tilt turns together with the string. A downhole motor solves this contradiction. Its housing is part of the stationary string, and its shaft turns the bit, taking energy from the flow of drilling mud, which has to be pumped anyway to clean the bottom of the hole.

A positive displacement motor (PDM, also called a mud motor) is built like a screw pump that works in reverse. In a steel housing there is a stator lined with an elastomer (rubber) in which helical (spiral) lobes are formed. Inside it a steel rotor with helical lobes turns, and it has one lobe fewer than the stator. Closed cavities form between them. The mud, under pressure, pushes the cavities along the motor and turns the rotor. The axis of the rotor moves around a circle as it does so. So the rotation is passed through a jointed or flexible shaft to the shaft of the spindle (the bearing section), which is in line with the string. The bearings of the spindle carry the axial load on the bit and the radial forces. At the top of the motor there is a bypass valve, through which the string fills with mud and empties when it is run in and pulled out.

In one revolution of the rotor a certain volume of mud passes through the motor. This is the displacement per revolution $q$ . From this come the two main properties of a PDM. The rotational speed is set by the flow rate, and the torque is set by the pressure drop across the motor, because the hydraulic power of the flow, $\Delta p\,Q$ , turns into the mechanical power of the shaft, $M\omega$ .

An ideal positive displacement motor

n = \frac{Q}{q}, \qquad M = \frac{\Delta p\,q}{2\pi}

where:

$n$ rotational speed of the shaft, rev/s (multiply by 60 for rpm)
$Q$ flow rate of the mud through the motor, m³/s
$q$ displacement per revolution: the volume of mud that passes through the motor in one revolution of the rotor, m³/rev
$\Delta p$ pressure drop across the motor, the difference between the pressures at the inlet and the outlet, Pa
$M$ torque on the shaft, N·m

Units: SI

Holds when

There are no leaks and no friction: the power of the flow $\Delta p\,Q$ turns entirely into the mechanical power $M\omega$ , where $\omega = 2\pi n$ .
A real motor turns more slowly, because part of the mud leaks through the seal between the rotor and the stator, the more so the higher the pressure drop (the volumetric efficiency). It also gives a smaller torque because of friction (the mechanical efficiency). Real values are taken from the characteristic on the data sheet.
The motor works within the range of flow rate and pressure drop of its data sheet.

For the same diameter, motors with more lobes on the rotor are slower and develop more torque. Such motors are needed for PDC bits and for steerable drilling. If the torque is not enough for the bit, the rotor stops and the standpipe pressure rises sharply. This is a stall of the motor. Working for long in a stall destroys the elastomer of the stator.

In other unit systems (2)

Metric oilfield units

n\,[\text{rpm}] = \frac{60\,Q\,[\text{L/s}]}{q\,[\text{L/rev}]}, \qquad M\,[\text{N·m}] = 159.2\,\Delta p\,[\text{MPa}]\,q\,[\text{L/rev}]

$Q$ in L/s, $q$ in litres per revolution, $\Delta p$ in MPa, $n$ in rpm, $M$ in N·m.

$159.2 = 10^6\cdot 10^{-3}/(2\pi)$ : MPa to Pa and litres to m³.

Oilfield units

n\,[\text{rpm}] = \frac{Q\,[\text{gal/min}]}{q\,[\text{gal/rev}]}, \qquad M\,[\text{ft·lbf}] = 3.064\,\Delta p\,[\text{psi}]\,q\,[\text{gal/rev}]

$Q$ in US gallons per minute, $q$ in gallons per revolution, $\Delta p$ in psi, $M$ in pound-force feet.

1 psi · 1 gal $= 6894.76\cdot 3.7854\cdot10^{-3} = 26.10$ J; $26.10/(2\pi) = 4.154$ N·m $= 3.064$ ft·lbf.

Worked exampleSpeed and torque of a PDM from the flow rate and the standpipe pressure

Given (assumed data)

Flow rate of the mud $Q$: 30 L/s (1800 L/min)
Displacement of the motor $q$: 15 L/rev
Standpipe pressure: 17.0 MPa off bottom, 20.5 MPa while drilling
Volumetric and mechanical efficiency at this pressure drop: 0.90 and 0.85 (from the data sheet)
Limit of the pressure drop on the data sheet: 5.5 MPa

Solution

The ideal speed: $n = 60\cdot 30/15 = 120$ rpm. Leaks lower it to $120\cdot 0.90 = 108$ rpm.
The pressure drop across the motor is the rise of the standpipe pressure when the bit is put on the bottom: $20.5 - 17.0 = 3.5$ MPa.
The ideal torque: $M = 3.5\cdot 10^6\cdot 0.015/(2\pi) = 8355.6$ N·m. With the mechanical efficiency it is $8355.6\cdot 0.85 = 7102$ N·m.
The power on the shaft: $M\cdot 2\pi n/60 = 80.3$ kW, at a hydraulic power of $\Delta p\,Q = 105$ kW. The overall efficiency is about 76.5%.
At the limit on the data sheet, 5.5 MPa, the torque will reach 11.2 kN·m, and the standpipe pressure will be $17.0 + 5.5 = 22.5$ MPa. If the pressure rises above this, the motor is close to a stall, and the weight on the bit is reduced.

Answer: The motor turns the bit at about 108 rpm and develops a torque of about 7.1 kN·m.

What it means. The rise of the standpipe pressure is a direct indicator of how a PDM works. From the gauge the driller sees the torque at the bit and notices in time that a stall is near. If the string is also rotated from the surface at 40 rpm, the two rotations add up, and the bit turns at 148 rpm. The data in this example are assumed for illustration.

A turbodrill is a multistage turbine, that is, a turbine made of many stages. Rotors with blades are fixed on the shaft, and stators in the housing swirl the flow and direct it on to the blades of the rotors. A turbodrill has tens or hundreds of stages, each adding its share of torque and of pressure drop, so turbodrills are long and are often assembled from several sections.

The characteristic of a turbine (how its torque depends on its speed) is different from that of a PDM. At a constant flow rate its torque is greatest when the shaft is stopped, and it falls as the speed grows, to zero when the turbine runs with no load. The power, which is the torque times the speed, is greatest in the middle. Under load a turbodrill does not hold its speed but loses it, and with too much weight on the bit it stops. The pressure drop across a turbine hardly depends on the load, so from the standpipe gauge the driller cannot see what regime the turbodrill is working in.

Characteristic of a turbodrill at a constant flow rate

M = M_{\text{stall}}\left(1 - \frac{n}{n_{\text{run}}}\right), \qquad N_{\text{sh}} = 2\pi n M, \qquad N_{\text{sh}\,\max} = \frac{\pi}{2}\,M_{\text{stall}}\,n_{\text{run}}

where:

$M$ torque on the shaft, N·m
$M_{\text{stall}}$ stall torque, the torque of a stopped shaft, N·m
$n$ rotational speed of the shaft, rev/s
$n_{\text{run}}$ runaway speed, the speed of rotation with no load, rev/s
$N_{\text{sh}}$ power on the shaft; the greatest value $N_{\text{sh}\,\max}$ is at $n = n_{\text{run}}/2$ , W

Units: SI; with the speed in rpm the power is $N_{\text{sh}} = 2\pi n M/60$

Holds when

The flow rate and the density of the mud are constant.
The linear dependence of the torque on the speed follows from the turbine equation for blades with fixed angles. In real turbodrills it is close to linear in the working range.
When the flow rate changes (with the same blades), the runaway speed changes in proportion to $Q$ , the stall torque as $Q^2$ , and the power as $Q^3$ .

A turbodrill is a good choice when the bit needs high speed at a moderate torque, for example diamond and impregnated diamond bits (bits with diamonds set in the body of the matrix) in hard rocks. It has no elastomer, and it works at temperatures at which the stator of a PDM would be destroyed. The weight on the bit is chosen for a turbodrill so as to stay near the regime of the greatest power.

Worked exampleA turbodrill under load

Given (assumed data)

A turbodrill at a flow rate of 30 L/s: runaway speed 900 rpm, stall torque 3200 N·m
The torque that the bit demands: first the regime of the greatest power, then 2400 N·m

Solution

The greatest power is at $n = 900/2 = 450$ rpm. The torque there is $M = 3200\cdot(1 - 0.5) = 1600$ N·m, and the power is $1600\cdot 2\pi\cdot 450/60 = 75.4$ kW.
The bit demands 2400 N·m: the speed falls to $900\cdot(1 - 2400/3200) = 225$ rpm, and the power to 56.5 kW.
At 3200 N·m the turbodrill stalls, although the standpipe pressure hardly changes.

Answer: In the best regime the turbodrill gives about 75.4 kW at 450 rpm. Under a load of 2400 N·m it gives only 56.5 kW at 225 rpm.

What it means. Compare this with the PDM of the previous example. At the same flow rate it develops 7.1 kN·m at 108 rpm, and loses only a little speed as the load grows. A turbodrill is a machine of high speed and moderate torque. Extra weight on the bit does not speed up drilling but “chokes” it. The data in this example are assumed for illustration.

CalculatedThe calculated characteristics of the two assumed motors of the examples at a flow rate of 30 L/s: rotational speed against torque. The PDM (displacement 15 L/rev, leaks grow with the pressure drop) almost holds its speed up to the data sheet limit of the pressure drop, 5.5 MPa. For the turbodrill (runaway speed 900 rpm, stall torque 3.2 kN·m) the speed falls with the torque along a straight line. The dots are the regimes of the examples.

PDM and turbodrill

Feature: Rotational speed
PDM: low to medium, set by the flow rate
Turbodrill: high, falls with the load

Feature: Torque
PDM: large, grows with the pressure drop
Turbodrill: moderate, greatest with a stopped shaft

Feature: Load on the standpipe gauge
PDM: visible: the pressure drop grows with the torque
Turbodrill: almost invisible

Feature: Weak point
PDM: the elastomer of the stator: temperature, oil-based muds, wear
Turbodrill: low torque, hard to control the regime

Feature: Where it is strong
PDM: steerable drilling, PDC bits
Turbodrill: hard rocks, high temperatures, diamond bits

Feature	PDM	Turbodrill
Rotational speed	low to medium, set by the flow rate	high, falls with the load
Torque	large, grows with the pressure drop	moderate, greatest with a stopped shaft
Load on the standpipe gauge	visible: the pressure drop grows with the torque	almost invisible
Weak point	the elastomer of the stator: temperature, oil-based muds, wear	low torque, hard to control the regime
Where it is strong	steerable drilling, PDC bits	hard rocks, high temperatures, diamond bits

Oil-based muds (Drilling fluids: functions and properties) can make the elastomer swell. For them, resistant grades of rubber are chosen.

A deflecting motor: a bent sub and a bent housing

The classic way to build the angle is to put a bent sub above the downhole motor. This is a short connecting piece in which the axes of the upper and lower threads cross at an angle of one to three degrees. Everything below it, the motor and the bit, is tilted relative to the string. The side of the bend is marked on the sub, and the assembly is oriented by it. The motor turns the bit, and the string stands still. In Russian-language practice such a set is called a deflecting motor (also a turbine deflector). The same Russian word is also used for a deflection tool in general.

The weak point of a bent sub is the distance. The bend is above the motor, 8–10 m from the bit, and the centre of the bit is displaced from the extension of the axis of the string by $e = l_{\text{b}}\sin\delta$ , where $l_{\text{b}}$ is the distance from the bend to the bit and $\delta$ is the bend angle. This is tens of centimetres, comparable with the radius of the hole or more. Such an assembly cannot be rotated: to fit in the hole it would have to bend strongly at every revolution. So after the angle is built, it is pulled out and another assembly, for holding the angle, is run in, and for the next correction a deflecting motor again. Each change is a trip, and in deep wells it takes many hours.

A steerable motor moves the bend into the housing of the motor itself, between the power section (the part with the rotor and the stator) and the spindle, one or two metres from the bit. The bend angle is usually smaller, and it can often be adjusted on the rig (an adjustable bent housing). The displacement of the bit becomes small, and the assembly can not only be oriented but also rotated together with the string. In this way one assembly drills both curved and straight sections, without extra trips. At large bend angles rotation is limited or forbidden, because the bending stresses, the vibration and the wear grow. The limits are set by the manufacturer of the motor.

What dogleg severity will a motor give in sliding drilling? The simplest estimate is the three-point method. The assembly touches the walls of the hole at three points: the bit, the lower stabiliser on the housing of the spindle, and the upper stabiliser above the motor. Only one circle passes through three points, and its radius sets the curvature.

SchematicThe three-point method for a deflecting motor (a sketch; the bend angle and the curvature are greatly exaggerated). The bit B, the lower stabiliser S and the upper stabiliser U touch the walls and lie on one circle of radius

R

with its centre O. The chords B–S and S–U diverge by the bend angle

\delta

;

L_1

and

L_2

are the distances between the points of support.

The three-point estimate of the dogleg severity of a deflecting motor

i = \frac{2\,\delta\,l_0}{L}, \qquad L = L_1 + L_2

where:

$i$ dogleg severity in sliding drilling: degrees per the reference length $l_0$ , °/30 m
$\delta$ bend angle (the angle of the tilt) between the axis of the lower part of the assembly and the axis of the motor, degrees
$L_1$ distance from the bit to the lower stabiliser, m
$L_2$ distance from the lower to the upper stabiliser, m
$L$ distance from the bit to the upper stabiliser, m
$l_0$ reference length to which the rate is related: 30 m, 10 m or 100 ft, m

Units: $\delta$ and $i$ in degrees; $L$ and $l_0$ in the same units of length

Holds when

The bit and both stabilisers touch the walls and lie on one circle; the stabilisers are full gauge.
All the tilt is concentrated at the lower stabiliser. If the bend is above it, the effective angle is a little smaller than $\delta$ .
The angles are small. The formula follows from the fact that the angle between two chords of a circle equals half the sum of the arcs on which they rest (the derivation is in the advanced section).
The clearances, the washing out of the hole in soft rock, the weight on the bit, the hardness and bedding of the rocks, and the ability of the bit to cut the wall are not taken into account. So this is an upper, geometric estimate.

The rate grows with the bend angle and is inversely proportional to the length of the assembly between the end points of support: with a greater angle, or with the upper stabiliser put lower, we get a steeper build. The actual rate that the assembly gives in a given section (the motor yield) is found from survey measurements (readings of the inclination and azimuth taken along the hole).

In other unit systems (3)

Degrees per 30 m

i\,[^\circ/30\ \text{m}] = \frac{60\,\delta}{L\,[\text{m}]}

$\delta$ in degrees, $L$ in metres.

$60 = 2\cdot 30$ .

Degrees per 100 ft (oilfield units)

i\,[^\circ/100\ \text{ft}] = \frac{200\,\delta}{L\,[\text{ft}]}

$\delta$ in degrees, $L$ in feet.

$200 = 2\cdot 100$ .

Degrees per 10 m

i\,[^\circ/10\ \text{m}] = \frac{20\,\delta}{L\,[\text{m}]}

$\delta$ in degrees, $L$ in metres.

$20 = 2\cdot 10$ .

Worked exampleDogleg severity and the bit offset of a steerable motor

Given (assumed data)

Bend angle $\delta$: 1.25°
From the bit to the lower stabiliser $L_1$ (the bend is at the level of the stabiliser): 1.5 m
From the lower to the upper stabiliser $L_2$: 7.5 m
Hole diameter: 215.9 mm
For comparison: a bent sub: 2° at 8.5 m above the bit

Solution

$L = 1.5 + 7.5 = 9.0$ m; $i = 60\cdot 1.25/9.0 = 8.33^\circ$ per 30 m.
The radius of curvature: $R = 1718.9/8.33 = 206$ m, by the formula from the lesson on the well profile.
The offset of the bit of a steerable motor: $e = 1500\cdot\sin 1.25^\circ = 32.7$ mm, which is 3.3 times smaller than the radius of the hole, 108 mm.
A bent sub: $e = 8500\cdot\sin 2^\circ = 297$ mm, almost three times the radius of the hole.

Answer: The geometric dogleg severity is about 8.33°/30 m (a radius of 206 m), and the bit offset of 32.7 mm makes it possible to rotate the assembly.

What it means. In neighbouring wells such motors gave about 7.5°/30 m in this section. This is less than the geometric estimate: the stabilisers are a little smaller than the hole, the rock is washed out in places, and the bit does not cut into the wall at once. This actual number is the one used in planning the sliding. The data in this example are assumed for illustration.

Sliding, rotating and the toolface

A steerable motor works in two modes. In sliding (oriented drilling) the string is not rotated but only fed downwards. The motor turns the bit, the bend points to one chosen side, and the well curves. In rotary mode the string is rotated from the surface, usually at a few tens of revolutions per minute. The bend goes round the whole circle, its effect averages out, and the well goes almost straight, a little wider than the bit, and with the slight tendency to build or drop the angle that is set by the placing of the stabilisers.

By alternating the modes, the driller gets any rate smaller than the rate that the assembly gives in sliding. If rotary drilling is taken as straight, the average rate over an interval is proportional to the share of sliding.

The share of sliding drilling

f = \frac{l_{\text{slide}}}{l} = \frac{i_{\text{plan}}}{i_{\text{slide}}}

where:

$f$ share of the interval drilled in sliding, —
$l_{\text{slide}},\ l$ length of sliding in the interval and the length of the interval (for example, a stand, which is two or three joints of pipe screwed together), m
$i_{\text{plan}}$ the required, planned dogleg severity, °/30 m
$i_{\text{slide}}$ the rate that the assembly gives in sliding, °/30 m

Units: rates in the same units, lengths in metres

Holds when

In rotation the assembly drills straight. If it has a tendency of its own, $i_{\text{rot}}$ (a build is positive, a drop negative), then $f = (i_{\text{plan}} - i_{\text{rot}})/(i_{\text{slide}} - i_{\text{rot}})$ .
During sliding the toolface is held in one position.
The rate is averaged over a stand or two. Within the interval the hole is stepped.

Sliding is slower and harder than rotating, so the assembly is chosen with a margin of rate in case the rock “does not give” the angle, but not too much of it: the bigger the margin, the sharper the local bends of the hole.

The direction in which the well curves in sliding is decided by the toolface $\omega$ . This is the direction of the bend when we look down along the hole. In an inclined hole it is measured clockwise from the high side of the hole, that is, from the line along the top of the wall, where the vertical plane through the axis of the hole (called the apsidal plane in Russian-language literature) crosses the wall. This is the gravity toolface: the sensors find the high side from the direction of gravity. In an almost vertical hole there is no high side, and the angle is measured from north. This is the magnetic toolface. We change to the gravity toolface when the inclination exceeds a few degrees.

SchematicThe toolface (a sketch): a view looking down along the hole. The angle is measured clockwise from the high side of the hole: 0° builds the inclination, 90° turns the well to the right (the azimuth grows), 180° drops the angle, and 270° turns the well to the left. A bend set at the angle

\omega

divides the dogleg

\beta

into a build

\beta\cos\omega

and a turn

\beta\sin\omega

How the toolface divides the dogleg between the inclination and the azimuth

\begin{aligned} \Delta\theta &\approx \beta\cos\omega, & \Delta\varphi &\approx \frac{\beta\sin\omega}{\sin\bar\theta},\\ \beta &\approx \sqrt{\Delta\theta^2 + (\Delta\varphi\sin\bar\theta)^2}, & \tan\omega &\approx \frac{\Delta\varphi\sin\bar\theta}{\Delta\theta} \end{aligned}

where:

$\beta$ dogleg over the sliding interval, degrees
$\omega$ toolface from the high side, clockwise, degrees
$\Delta\theta,\ \Delta\varphi$ changes of the inclination and the azimuth over the interval, degrees
$\bar\theta$ mean inclination of the interval: the initial value plus half of $\Delta\theta$ , degrees

Units: angles in degrees

Holds when

The dogleg is small (a few degrees) and the toolface is held in one position all the time.
The hole is not close to vertical. At an inclination of a few degrees the formula for the azimuth becomes crude, and in a vertical hole it loses its meaning (the exact formulas are in the advanced section).
In the inverse problem the quadrant of the angle $\omega$ is chosen by the signs of the numerator and the denominator, as when calculating the azimuth. A turn to the left is a negative angle (or 270–360°), and a drop is more than 90° in absolute value.

For $\omega = 0$ all the dogleg goes into the build of the angle, for 180° into the drop, and for 90° and 270° into the turn. The factor $1/\sin\bar\theta$ explains why the azimuth of an almost vertical hole “wanders” so easily: a small shift to the side turns it by tens of degrees.

The deflection tool is oriented by turning the string from the surface, and the measurement while drilling (MWD) system reports its position. The reaction torque gets in the way. The rotor of the motor turns the bit clockwise (seen from above), and the stator, and with it the housing and the string, receives an equal torque in the opposite direction. When the bit is put on the bottom and the torque grows, the string above the motor is twisted, and the toolface turns to the left. The turn is greater the greater the pressure drop across the motor and the longer the string, and the friction of the string against the walls reduces it. At depth this is usually tens of degrees, and with a large torque and a long string, even more. So before putting the bit on the bottom, the toolface is set to the right of the required position by an amount known from experience, and then it is corrected by the readings of the MWD.

Worked exampleHow much to slide to hold a build of 3° per 30 m, and how to correct the azimuth

Given (assumed data)

Planned build-up rate: 3° per 30 m (the profile from the lesson on the well profile, a build section of 324 m along the hole)
Rate of the assembly in sliding: 7.5° per 30 m (from neighbouring wells)
Rate of penetration: 9 m/h in sliding, 24 m/h in rotating
Length of a stand (two or three joints of pipe screwed together): 30 m
Survey at 660 m: inclination 16.4°, azimuth 59.0°, to the right of the planned 53.1°

Solution

The share of sliding: $f = 3/7.5 = 0.40$ . On every stand we slide $30\cdot 0.40 = 12$ m and rotate 18 m.
The time per stand: $12/9 + 18/24 = 2.08$ h, against $30/24 = 1.25$ h with continuous rotation.
The whole build section is $324\cdot 0.40 = 130$ m of sliding and about 22.5 h of pure drilling, instead of 13.5 h.
The correction on the next stand: build 3° and turn to the left by 1.5°. The mean inclination is $\bar\theta = 16.4 + 1.5 = 17.9^\circ$ ; the turn is $\Delta\varphi\sin\bar\theta = -1.5\cdot\sin 17.9^\circ = -0.461^\circ$ .
The dogleg is $\beta = \sqrt{3^2 + (-0.461)^2} = 3.04^\circ$ . The toolface is $\omega = \arctan(-0.461/3) = -8.7^\circ$ , that is 8.7° to the left of the high side (on a 0–360° scale, 351.3°).
The length of sliding on this stand is $3.04/7.5\cdot 30 = 12$ m.

Answer: To build 3° per 30 m we slide 12 m out of every 30. For the correction we slide 12 m with the toolface 8.7° to the left of the high side.

What it means. Sliding makes the drilling of every stand 1.7 times longer, so an assembly with an excessive rate is not chosen. The allowance for the reaction torque is not included in this calculation: when the toolface is oriented before the bit is put on the bottom, it is set still further to the right. The small turn needed the toolface to be moved from the high side by only 8.7°. At an inclination of about 18° the azimuth is three times as sensitive to a sideways dogleg as the inclination is to a build. The data in this example are assumed for illustration.

Friction. A stationary string lies on the low side of the hole, and the force of static friction is large. In long inclined and horizontal holes the weight does not reach the bit, the string “hangs up”, and it has to be fed in jerks.
Hole cleaning. Without rotation the cuttings settle on the low side, and the risk of stuck pipe, including differential sticking (the pipe is held against the wall by the pressure difference), grows.
Speed. The rate of penetration in sliding is usually much lower than in rotating.
Quality of the hole. Alternating sliding and rotating gives a stepped, winding hole with local bends (tortuosity). The torque and the drag when strings are run in grow.

Assemblies without a motor: build, hold, drop

Without a downhole motor the angle is changed by a rotary assembly, by choosing the placing of the stabilisers (Russian-language practice calls them centralisers, and also gauge reamers and centring supports) in the bottom hole assembly (BHA, the lowest part of the drill string). Even thick drill collars are flexible. Under the weight on the bit and under their own weight they bend, and where the bit tilts and which wall it is pressed to depend on where the collars rest against the walls.

The build assembly (fulcrum assembly): a stabiliser stands right above the bit, and the next one 15–25 m higher. Under load the collars between them bend towards the low side, the near-bit stabiliser works as the fulcrum of a lever, and the bit turns towards the high side. The build gets stronger as the weight on the bit grows, as the stiffness of the collars above the near-bit stabiliser falls, and as the second stabiliser is put farther away.
The hold assembly (packed assembly): three or four stabilisers at short distances (at the bit, a few metres above it, and higher still) on stiff collars. There is almost no room to bend, and the bit goes along the axis of the hole. In practice even this assembly slowly builds or drops the angle a little, and it is chosen by experience.
The pendulum assembly: the first stabiliser stands 10–20 m above the bit, and below it the collars hang free, like a pendulum. The part of their weight that acts across the hole presses the bit to the low side, and the angle drops. The same assembly keeps a vertical hole vertical.

SchematicThree assemblies without a motor in an inclined hole (a sketch; the bending of the collars is exaggerated). At the top is the build assembly: the near-bit stabiliser serves as the fulcrum, the collars bend under load towards the low side, and the bit turns upwards. In the middle is the stiff assembly: the stabilisers hold the collars straight. At the bottom is the pendulum: the free section of collars presses the bit to the low side by its weight. The arrow at the bit is the side force, and the arrow on the right is the direction of gravity.

In both of these assemblies (the build assembly and the pendulum) the weight on the bit works towards building the angle. In the build assembly it bends the collars above the fulcrum and makes the build stronger. In the pendulum it also bends the collars towards the low side and so tilts the bit upwards, against the pendulum. This is why the angle is dropped with a moderate weight on the bit. We can estimate the pendulum force if we treat the free section of the collars as a beam on two supports, the bit and the stabiliser.

The pendulum force at the bit

F = \tfrac12\,w\,L_{\text{p}}\sin\theta, \qquad w = m\,g\,k_{\text{b}}

where:

$F$ side force with which the bit presses on the low side of the hole, N
$w$ weight of a metre of collars in the mud, N/m
$m$ mass of a metre of collars, kg/m
$g$ acceleration of gravity, 9.81, m/s²
$k_{\text{b}}$ buoyancy factor $1 - \rho_m/\rho_{\text{steel}}$ : $\rho_m$ is the density of the mud, $\rho_{\text{steel}}$ that of steel (7.85 g/cm³), —
$L_{\text{p}}$ length of the pendulum: from the bit to the first stabiliser, or to the point where the collars touch the low side, m
$\theta$ inclination, degrees

Units: SI

Holds when

The free section of the collars is a beam on two supports under a uniform cross-hole load $w\sin\theta$ ; each support carries half of this load.
The weight on the bit is not taken into account: by bending the collars it reduces the pendulum effect.
The collars do not touch the low side between the bit and the stabiliser. If they do, the length of the pendulum is up to the point of contact, and raising the stabiliser higher no longer increases the force.

The force grows with the inclination. At an angle of a few degrees it is small, and the pendulum straightens the hole slowly. At a large angle the drop is more vigorous. A long pendulum is stronger, but only as long as the collars have not lain down on the low side.

Worked exampleThe pendulum force at 25° and 5°

Given (assumed data)

Drill collars: outside diameter 203 mm, bore 80 mm; steel 7850 kg/m³, modulus of elasticity $2.07\cdot10^{11}$ Pa
Mud: 1.20 g/cm³
First stabiliser: 15 m above the bit
Bit diameter: 295.3 mm

Solution

The mass of a metre of collars: $m = \pi/4\cdot(0.203^2 - 0.080^2)\cdot 7850 = 214.6$ kg/m; $k_{\text{b}} = 1 - 1.20/7.85 = 0.8471$ ; $w = 214.6\cdot 9.81\cdot 0.8471 = 1783$ N/m.
At an angle of 25°: $F = 0.5\cdot 1783\cdot 15\cdot\sin 25^\circ = 5.65$ kN. At an angle of 5°: $F = 1.17$ kN.
A check that the collars do not lie on the wall: the sag in the middle of a beam on two supports is $5\,w\sin\theta\,L_{\text{p}}^4/(384\,EI)$ , with the moment of inertia of the cross-section $I = \pi(D^4 - d^4)/64 = 8.135\cdot 10^{-5}$ m⁴. At an angle of 25° it is 29.5 mm, which is less than the clearance between the collars and the wall, $(295.3 - 203)/2 = 46.15$ mm.

Answer: The pendulum presses the bit to the low side with a force of about 5.65 kN at 25° and only 1.17 kN at 5°.

What it means. This is a few per cent of the weight on the bit, but the force is directed across the hole, and the bit, cutting into the low side, gradually drops the angle. At an angle of 45° the sag of the same section would reach 49.4 mm, more than the clearance. The collars would lie on the wall, and the length of the pendulum would become less than 15 m. The data in this example are assumed for illustration.

Assemblies without a motor are cheap and reliable, but their rate is hard to predict. It depends on the weight on the bit and the rotary speed, on the hardness and the bedding of the rocks, and on the wear of the stabilisers. They hardly control the azimuth. Today they are used mostly to drill vertical holes and tangent sections, while complex steering is done with deflecting motors and rotary steerable systems.

Rotary steerable systems

A rotary steerable system (RSS) deflects the bit without stopping the rotation of the string. Its control unit stands right above the bit. Electronics with sensors of inclination, azimuth and the position of the unit hold the set direction by themselves and carry out commands from the surface. By the way they deflect the bit, the systems fall into two groups.

Push-the-bit: extendable pads press against the wall of the hole and push the bit to the opposite side. This is a side force. In some designs the pads sit on a sleeve that hardly rotates, and a shaft rotates inside it. In others the pads rotate with the housing and are pushed out by the pressure of the mud each time they pass the required side of the hole.
Point-the-bit: in a non-rotating housing an eccentric mechanism bends the shaft that carries the bit, and the axis of the bit tilts. This is the same tilt as in a deflecting motor, but with a rotating string.

There are also hybrid systems, and when the bit needs a higher rotational speed, a downhole motor is put above the RSS. The system receives commands from the surface. The flow rate of the mud is changed in an agreed sequence, and the RSS recognises it as a code (a downlink). The transmission takes minutes. In reply the system reports through the MWD that the command has been received.

Continuous rotation gives the main advantages. The cuttings do not settle, the risk of stuck pipe is lower, the hole is smoother, the friction is smaller, and the weight reaches the bit even in long horizontal holes. This is why extended-reach wells cannot do without an RSS. The rate of penetration is higher than in sliding, and the path can be corrected little by little all the time, without stopping the drilling.

What the driller sees: MWD and LWD

The hole cannot be steered blind. The position of the hole and of the toolface is reported to the driller by measurement while drilling (MWD). Sensors in a non-magnetic drill collar above the motor or the RSS measure the directions of gravity and of the magnetic field of the Earth, and from them the inclination, the azimuth and the toolface are calculated. The data go to the surface as pressure pulses in the flow of mud (mud-pulse telemetry), a few bits per second, so only the most necessary data are sent.

A survey gives the inclination and the azimuth at the depth of the sensor. It is taken at a connection (when a new stand is added to the string), when the string is stationary, usually at every stand. The coordinates of the hole are calculated from the surveys (Directional survey calculations).
The toolface is updated every few seconds during sliding. The driller uses it to hold the deflection tool in the required position.
Logging while drilling (LWD) gives the gamma ray log, the resistivity, the density and the porosity, often with azimuthal images of the hole. They show where the hole is in the formation, and the path is steered by the geology and not only by the geometry. This is called geosteering (Open-hole logging: why, how and when).

The sensors usually stand 10–20 m above the bit, and in a long assembly with LWD tools even farther. So a survey describes a part of the hole that was drilled some time ago. The engineer estimates where the bit is now by a forecast from the last surveys and from what the assembly was doing. This is why the sensors are being brought closer to the bit: modern systems have near-bit sensors of inclination and gamma radiation. Non-magnetic collars are needed because the steel of the string becomes magnetised and distorts the readings of the magnetometers. How the instruments are built and what errors they have is the topic of the lessons Survey instruments: magnetic, gyroscopic and inertial and Measurement while drilling.

Terms and symbols

BHA: The bottom hole assembly: the drill collars, stabilisers, subs, the downhole motor or the RSS, and the MWD and LWD sensors.
Deflection tool: The general name of a tool that deflects the bit. In the narrow sense, a deflecting motor: a downhole motor with a bent sub or a bent housing.
Whipstock: A steel wedge that pushes a mill or a bit to the side. It is needed first of all to start a sidetrack out of a cased well.
Bent sub, bent housing: The source of the tilt of the bit axis relative to the string. The bend is above the motor, or in its housing closer to the bit.
PDM: Positive displacement motor. The speed depends on the flow rate, and the torque on the pressure drop.
Turbodrill: A multistage turbine. It has a high rotational speed, and its torque falls as the speed grows.
Sliding and rotating: The two modes of a steerable assembly. When the string stands still, the hole curves. When the string rotates, the hole goes almost straight.
Toolface $\omega$: The direction of the bend, measured clockwise from the high side of the hole (in an almost vertical hole, from north).
Reaction torque: The torque with which the stator of the motor twists the string counter-clockwise.
Stabiliser: An element of the BHA with blades almost as wide as the hole. It is a point of support of the assembly.
RSS: Rotary steerable system. It deflects the bit with pads or a tilt of the shaft while the string rotates.
Geosteering: Steering the hole by the LWD data about the formation.

Common mistakes and misconceptions

Wrong

“Since the assembly has a motor with a bend, the hole will curve.”

In fact

Only in sliding. When the string rotates, the bend goes round a circle, and the hole goes almost straight. The average rate is set by the share of sliding.

Wrong

“The standpipe pressure says nothing about the work of a downhole motor.”

In fact

In a PDM the pressure drop is proportional to the torque: the rise of the pressure when the bit is put on the bottom is the main indicator of the load and a sign of an approaching stall. Only in a turbodrill does the pressure drop hardly depend on the load, and its regime cannot be seen from the gauge.

Wrong

“A toolface of 90° means that the hole will go east.”

In fact

The toolface is measured from the high side of the hole and not from north: 90° is a turn to the right of the current azimuth, whatever it is. Only the magnetic toolface in an almost vertical hole is measured from north.

Wrong

“To drop the angle, put a stabiliser right at the bit.”

In fact

A near-bit stabiliser works as the fulcrum of a lever and builds the angle. To drop the angle, the first stabiliser is put high above the bit, leaving a free-hanging section of collars, a pendulum.

Wrong

“A rotary steerable system is always better.”

In fact

It is better where a large departure, a smooth hole and speed matter. In simple wells a deflecting motor is cheaper, and in very hot wells the electronics may not survive.

In depth: advanced detail

Derivation of the three-point formula

Let the bit B, the lower stabiliser S and the upper stabiliser U lie on a circle of radius $R$ . The chords B–S and S–U rest on arcs with the central angles $\alpha_1$ and $\alpha_2$ . The exterior angle of the inscribed triangle at the vertex S, which is the angle between the extension of the chord U–S and the chord S–B, equals half the sum of these arcs: $\delta = (\alpha_1 + \alpha_2)/2$ . This is a direct consequence of the inscribed angle theorem: the interior angle at S rests on the remaining arc $360^\circ - \alpha_1 - \alpha_2$ and equals half of it.

A chord of length $L_k$ subtends an arc $\alpha_k = 2\arcsin(L_k/2R)$ , and for small angles $\alpha_k \approx L_k/R$ in radians. Then $\delta \approx (L_1 + L_2)/(2R)$ , and the curvature is $1/R = 2\delta/L$ radians per metre. To get degrees per the reference length $l_0$ , we multiply the curvature by $l_0$ and by $180/\pi$ . If $\delta$ is also expressed in degrees, the factors $180/\pi$ and $\pi/180$ cancel, and what is left is $i = 2\delta\,l_0/L$ .

For the example of the lesson, the exact solution, a circle drawn through the three points without simplifications, gives a radius of 206.27 m instead of 206.26 m. The difference is less than a centimetre on a radius of two hundred metres. The errors of the three-point estimate come not from the geometry but from what it leaves out: the clearances, the bending of the assembly, and the properties of the rock and of the bit.

The exact calculation from the toolface

The approximate formulas of the section on sliding are convenient, but they are wrong for a small inclination and a large dogleg. The exact calculation treats the directions of the hole as unit vectors. Let $\mathbf{t}_1$ be the direction of the hole at the start of the interval, $\mathbf{h}$ the unit vector to the high side, and $\mathbf{r}$ the unit vector to the right. If the hole has bent through the angle $\beta$ along an arc of a circle at the toolface $\omega$ , the new direction is $\mathbf{t}_2 = \mathbf{t}_1\cos\beta + (\mathbf{h}\cos\omega + \mathbf{r}\sin\omega)\sin\beta$ . The formulas below come from the vertical and horizontal components of this vector.

The new inclination and the change of azimuth after a bend at a given toolface

\begin{aligned} \cos\theta_2 &= \cos\theta_1\cos\beta - \sin\theta_1\sin\beta\cos\omega,\\ \tan\Delta\varphi &= \frac{\tan\beta\,\sin\omega}{\sin\theta_1 + \tan\beta\,\cos\theta_1\cos\omega} \end{aligned}

where:

$\theta_1,\ \theta_2$ inclination at the start and at the end of the interval, degrees
$\beta$ dogleg over the interval, degrees
$\omega$ toolface at the start of the interval, from the high side clockwise, degrees
$\Delta\varphi$ change of the azimuth, degrees

Units: angles in degrees; the quadrant of $\Delta\varphi$ is chosen by the signs of the numerator and the denominator

Holds when

The hole over the interval is an arc of a circle in one plane (the toolface stays the same relative to this plane).
The inverse problem, finding $\beta$ and $\omega$ from two directions, is solved in the same way: $\beta$ is the angle between the vectors $\mathbf{t}_1$ and $\mathbf{t}_2$ , and $\omega$ is the direction of the projection of $\mathbf{t}_2$ on the plane of the vectors $\mathbf{h}$ and $\mathbf{r}$ .

For $\omega = 0$ the first formula gives $\theta_2 = \theta_1 + \beta$ , a pure build, and the second gives $\Delta\varphi = 0$ . For $\omega = \pm 90^\circ$ the inclination still grows a little. The plane of the bend passes through the axis of the hole and the horizontal direction to the side, and by turning in it the hole becomes slightly flatter. So, to turn without building the angle, the toolface is set a little beyond $\pm 90^\circ$ , towards the low side.

Worked exampleWhere the approximate formulas are good, and where they are not

Given (assumed data)

Case 1: inclination 20°, dogleg 1.8° at a toolface of −60° (to the left and up)
Case 2: inclination 3°, dogleg 1.8° at a toolface of 90° (to the right)
Case 3: the correction of the sliding example: from 16.4° and 59.0° to 19.4° and 57.5°

Solution

Case 1, by the approximate formulas: $\Delta\theta = 1.8\cos(-60^\circ) = 0.90^\circ$ , $\Delta\varphi = -4.46^\circ$ . By the exact formulas: $\Delta\theta = 0.96^\circ$ , $\Delta\varphi = -4.36^\circ$ . The difference is a few hundredths to tenths of a degree.
Case 2, by the approximate formulas: $\Delta\theta = 0$ , $\Delta\varphi = 1.8/\sin 3^\circ = 34.4^\circ$ . By the exact formulas: the inclination has grown to $3.50^\circ$ , and the azimuth has turned by $31.0^\circ$ .
Case 3, the exact solution of the inverse problem: $\beta = 3.035^\circ$ , $\omega = -9.5^\circ$ , against the approximate $3.035^\circ$ and $-8.7^\circ$ .

Answer: At an inclination of 20° the approximate formulas are accurate to tenths of a degree. For an almost vertical hole they are wrong by degrees: at an inclination of 3° a “pure turn” also builds the inclination by 0.5°.

What it means. In case 3 the difference in the toolface is less than a degree, while the reaction torque turns the toolface at the bottom by tens of degrees. So the approximate formulas are enough for orientation. The exact ones are needed by trajectory calculation programs and for a kick-off from an almost vertical hole. The data in this example are assumed for illustration.

Connections to other topics

Directional wells: applications and profile — the profile that has to be followed: the kick-off point, the build-up rate, the tangent section.
Directional survey calculations — how the position of the hole and the dogleg severity that the assembly gave are found from the surveys.
How a well is drilled: the process and the people involved — the place of deflection tools in the construction of a well, and the roles of the service companies.
Casing design — the buoyancy factor, and the loads on a string run into a curved hole.
Drilling fluids: functions and properties — mud for inclined holes: cuttings removal, lubricity, compatibility with the elastomer of a PDM.
Open-hole logging: why, how and when — logging while drilling, and the delivery of tools into inclined and horizontal holes.

Check yourself

1
A steerable motor with a housing bend of 1.5° drills while the string is rotated at 60 rpm. What happens to the path?
1. A.The hole curves towards the bend at the full rate
2. B.The hole goes almost straight: the bend goes round a circle, and its effect averages out
3. C.The motor stops
4. D.The hole turns to the left under the action of the reaction torque
Show the answer
Correct answer: B. The hole goes almost straight: the bend goes round a circle, and its effect averages out
When the string rotates, the bend turns with it and looks in all directions in turn during each revolution. On average the bit goes along the axis of the hole, reaming it a little wider. Only sliding, when the bend is stationary, curves the hole.
2
A motor with a displacement of 12 L/rev works at a flow rate of 24 L/s. When the bit is put on the bottom, the standpipe pressure rises from 15.0 to 18.0 MPa. Estimate the speed and the torque of an ideal motor.
Show the answer
The speed is $n = 60\cdot 24/12 = 120$ rpm. The pressure drop across the motor is the rise of the standpipe pressure, 3.0 MPa. The torque is $M = 3.0\cdot10^6\cdot 0.012/(2\pi) \approx 5730$ N·m, that is about 5.7 kN·m. A real motor will give less because of leaks and friction. How much less is told by its data sheet characteristic.
3
Which assembly without a motor will drop the inclination?
1. A.The first stabiliser 15 m above the bit and free-hanging collars below it
2. B.Three stabilisers within 10 m of the bit on stiff collars
3. C.Any of them: the angle drops under the weight on the bit
4. D.A stabiliser right above the bit and a second one 20 m higher
Show the answer
Correct answer: A. The first stabiliser 15 m above the bit and free-hanging collars below it
The free section of collars works as a pendulum: the cross-hole component of its weight presses the bit to the low side, and the angle drops. The first assembly is a build assembly (the near-bit stabiliser is the fulcrum of a lever), and the second is a stiff one, for holding. The weight on the bit, on the contrary, weakens the pendulum.
4
At what inclination will the same dogleg at a toolface of 90° turn the azimuth more, at 10° or at 30°? By how many times?
Show the answer
At 10°. The turn of the azimuth $\Delta\varphi \approx \beta\sin\omega/\sin\bar\theta$ is inversely proportional to the sine of the inclination: $\sin 30^\circ/\sin 10^\circ = 0.5/0.174 \approx 2.9$ times. This is why the azimuth of an almost vertical hole changes so easily, and why orienting the toolface needs special care.
5
Why are long horizontal holes and extended-reach wells drilled with rotary steerable systems and not with steerable motors?
1. A.Downhole motors cannot be run into a horizontal hole
2. B.An RSS does not need MWD surveys
3. C.In sliding, a stationary string has large friction, the weight does not reach the bit, and cuttings are carried out worse, while an RSS steers the hole while rotating
4. D.An RSS always gives a higher dogleg severity
Show the answer
Correct answer: C. In sliding, a stationary string has large friction, the weight does not reach the bit, and cuttings are carried out worse, while an RSS steers the hole while rotating
The main trouble in a long hole is the friction of the stationary string against the low side in sliding: the weight cannot be passed to the bit, the cuttings settle, and the risk of stuck pipe grows. An RSS changes the direction without stopping the rotation. The greatest dogleg severity of standard RSS is, on the contrary, usually lower than that of a deflecting motor.
6
In sliding, an assembly gives 6° per 30 m, and the plan requires 2° per 30 m. How many metres should be slid on each 30 m stand? How does the answer change if in rotating the assembly itself drops the angle by 0.5° per 30 m?
Show the answer
Without a tendency of its own: $f = 2/6 = 1/3$ , that is 10 m of sliding per stand. With a drop in rotating, $i_{\text{rot}} = -0.5$ : $f = (2 + 0.5)/(6 + 0.5) = 0.385$ , that is about 11.5 m. Part of the build goes to compensating the drop in the rotating sections.

Updated: 19 October 2026