Directional survey calculations

Quick refresher

A reading gives three numbers: the measured depth $L$ , the inclination $\theta$ and the azimuth $\varphi$ . The coordinates are the result of a calculation from point to point, starting from a point with a known position, and the shape of the well between the readings has to be assumed.
The tangential method replaces an interval with a segment along the lower reading. Its error grows in proportion to the spacing, and on a build section it reaches metres. The errors of the balanced tangential method and of the average angle method are of the order of the square of the dogleg (the angle by which the well bends over the interval), so they are only centimetres.
The minimum curvature method is the industry standard. The well between the readings is an arc of a circle, and the calculation comes down to the balanced tangential method multiplied by the factor $RF = \tan(\beta/2)/(\beta/2)$ .
The dogleg $\beta$ is the total angle by which the well bends between two readings. It takes into account the change of both the inclination and the azimuth. The dogleg severity is this angle per 30 m (10 m, 100 ft) of the well.
The departure and the departure azimuth are the position of a point in the plan view (the view from above) given in polar coordinates, that is, as a distance and a direction. The vertical section is the projection of the departure on the vertical plane of the target, and the offset from this plane is the sideways error of the steering.
All azimuths are brought to one north. The true azimuth equals the magnetic azimuth plus the declination (the angle between true north and the north of the compass), and the grid azimuth equals the true azimuth minus the grid convergence (the angle between true north and the north of the coordinate grid).

Key relationships

\Delta H = \tfrac{\Delta L}{2}\,(\cos\theta_1 + \cos\theta_2)\,RF

minimum curvature method; $\Delta N$ and $\Delta E$ are the same, with $\sin\theta\cos\varphi$ and $\sin\theta\sin\varphi$

\cos\beta = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2\cos(\varphi_2 - \varphi_1)

dogleg over an interval

RF = \dfrac{\tan(\beta/2)}{\beta/2}, \qquad i = \dfrac{\beta\,l_0}{\Delta L}

the factor of the minimum curvature method ( $\beta$ in radians) and the dogleg severity ( $\beta$ in degrees)

A = \sqrt{N^2 + E^2}, \qquad A_{\text{VS}} = A\cos(\varphi_A - \varphi_{\text{plan}})

departure and vertical section

\varphi_{\text{g}} = \varphi_{\text{m}} + \delta - \gamma

grid azimuth from the magnetic azimuth: the declination $\delta$ and the grid convergence $\gamma$ , east positive

Common pitfalls

Averaging the azimuths 350° and 10° as plain numbers. The result is 180° and not 0°.
Taking seriously the azimuth of an almost vertical reading. At an inclination of a fraction of a degree it means almost nothing.
Mixing magnetic, true and grid azimuths in one calculation.
Putting angles in degrees into the factor $RF$ and into the radius of curvature formulas where radians are needed.
Calculating small doglegs with the arccosine of a rounded cosine. Losing one or two digits gives an error of tenths of a degree.

After this lesson you can

list what a directional survey reading gives and what must be added to it to get the coordinates of the well
bring the azimuths of the readings to one north, allowing for the magnetic declination and the grid convergence
calculate the increments of the coordinates over an interval between two readings by the tangential, balanced tangential, average angle, radius of curvature and minimum curvature methods, and explain why the results differ
calculate the dogleg and the dogleg severity over an interval
find the departure and the departure azimuth, the vertical section and the offset from the planned plane, and compare the actual well with the planned one
estimate how the choice of the method and the survey spacing affect the accuracy of the path

What a survey reading gives

A survey station is a place in the well where an instrument has measured the direction of the well: the inclination $\theta$ , which is the deviation of the axis of the well from the vertical, and the azimuth $\varphi$ , the direction of the horizontal projection of the axis, measured from north clockwise (Directional wells: applications and profile). The third number is the measured depth $L$ . The driller finds it from the length of the pipe run into the hole, less the distance from the bit to the sensor.

A reading does not give the position of a point. The coordinates are found by calculation, step by step, starting from a tie-in point, a place with a known position. This can be the rotary table or the wellhead, and when drilling goes on, it is the last calculated point of the well or a point found by a gyroscopic instrument (an instrument with a gyroscope, which finds true north without a compass). For each next point we calculate the increments of the true vertical depth $\Delta H$ and of the displacements to the north $\Delta N$ and to the east $\Delta E$ over an interval of length $\Delta L$ , and add them to the previous coordinates $H$ , $N$ , $E$ . In the lesson on the well profile the displacements of the target from the wellhead are written $\Delta N$ and $\Delta E$ . Here the coordinates of the points of the well relative to the wellhead are $N$ and $E$ , and the sign $\Delta$ means the increment over the interval between two readings.

Direction of the well at a survey station

\mathbf{t} = (\sin\theta\cos\varphi;\ \sin\theta\sin\varphi;\ \cos\theta)

where:

$\mathbf{t}$ unit vector along the axis of the well, pointing down the well: its components to the north, to the east and downwards, —
$\theta$ inclination, degrees
$\varphi$ azimuth from the chosen north, degrees

Units: dimensionless components; angles in degrees

Holds when

The $N$ axis points north, the $E$ axis east and the $H$ axis vertically down. The azimuths of all the readings are measured from the same north.
At $\theta = 0$ the first two components are zero, and the azimuth does not affect the result: a vertical well has no azimuth.

A segment of length $l$ going in this direction gives the increments $l\sin\theta\cos\varphi$ to the north, $l\sin\theta\sin\varphi$ to the east and $l\cos\theta$ vertically. All the calculation methods differ only in which segments or arcs they put in place of the unmeasured well between two survey stations.

SchematicWhat the methods assume about the well between two readings (a sketch in one plane; the angles are greatly exaggerated). The thin arc is the well, going with a constant curvature, and the small circle is its true end. The tangential method leads the whole interval along the direction of the lower reading. The balanced tangential method goes half along the upper and half along the lower. The average angle method uses one segment along the averaged direction. The radius of curvature and minimum curvature methods use an arc, and in a flat case they agree.

How often to survey is set by the drilling programme. On sections where the angle builds or the well turns, readings are usually taken at every stand (two or three joints of pipe screwed together), about every 25–30 m. On straight tangent sections (straight inclined parts of the well) they can be taken less often, if the tolerance of the target and the distance to neighbouring wells allow it. The less frequent the readings and the sharper the bend, the more the choice of the method matters.

Bringing the azimuths to one north

Magnetic instruments measure the azimuth from magnetic north (the direction of a compass needle), and gyroscopic ones from true (geographic) north. The coordinates of the wellhead, of the target and of neighbouring wells are given in a rectangular grid, which has its own grid north. An azimuth measured from grid north is a grid azimuth. In the surveying of the CIS countries (the Commonwealth of Independent States, which includes Kazakhstan) such an angle is called a directional angle (Directional wells: applications and profile). Before the coordinates are calculated, all the azimuths are brought to one north, usually to grid north, in which the target and the neighbouring wells are given.

Converting a magnetic azimuth into a true and a grid azimuth

\varphi_{\text{t}} = \varphi_{\text{m}} + \delta, \qquad \varphi_{\text{g}} = \varphi_{\text{t}} - \gamma

where:

$\varphi_{\text{m}}$ magnetic azimuth, from the direction to magnetic north, degrees
$\varphi_{\text{t}}$ true azimuth, from the geographic meridian, degrees
$\varphi_{\text{g}}$ grid azimuth, from the vertical line of the coordinate grid, degrees
$\delta$ magnetic declination, the angle from true north to magnetic north; east is plus, west is minus, degrees
$\gamma$ grid convergence, the angle from true north to grid north; plus if grid north is east of true north, degrees

Units: degrees; the result is brought to the range 0–360°

Holds when

The declination is taken from a geomagnetic model for the place and the date of the survey. It changes from year to year, and in some places noticeably.
The grid convergence depends on the position of the point in the zone of the coordinate grid. In the northern hemisphere, east of the central meridian of the zone it is positive, and west of it negative. As a rough rule, $\gamma \approx \Delta\lambda\sin B$ , where $\Delta\lambda$ is the difference of the longitudes of the point and of the central meridian, and $B$ is the latitude.
The signs follow the definitions above. In programs and documents the conventions may be different, so the sign convention is always checked.

The correction is the same for all the readings of a section, so an error in it is not random: it turns the whole path around the wellhead. An error of 1° at a departure of 1000 m is 17 m to the side.

Worked exampleFrom a magnetic azimuth to a grid azimuth

Given (assumed data)

Magnetic azimuth of the last reading: 51.5°
Magnetic declination at the date of the survey: +6.0° (east)
Grid convergence at the wellhead: +2.3° (the wellhead is east of the central meridian of the zone)
Departure of the target in the plan: 1100 m

Solution

The true azimuth: $51.5 + 6.0 = 57.5^\circ$ .
The grid azimuth: $57.5 - 2.3 = 55.2^\circ$ . This is the one that stands in the table of readings below.
If the magnetic azimuth were taken as the grid azimuth, the error would be $55.2 - 51.5 = 3.7^\circ$ . At the departure of the target this is $1100\cdot\sin 3.7^\circ = 71$ m to the side.
If only the grid convergence were forgotten, the error of 2.3° would give $1100\cdot\sin 2.3^\circ = 44$ m.

Answer: The grid azimuth is 55.2°. The corrections that were left out would take the bottom tens of metres away from the target.

What it means. The corrections are small compared with the azimuth itself, but they act on the whole path at once, so they matter more than many subtleties of the calculation. The values of the declination and of the grid convergence that were used are written into the well documents together with the readings. The data in this example are assumed for illustration.

Straight segment methods: tangential, balanced and average angle

The simplest way is to replace the whole interval with one straight segment along the direction measured at its lower point. This is the tangential method, or the method of tangents. It appeared in the first decades of directional drilling, when readings were taken rarely and the calculation was done by hand, and today it is not used for real wells. On a build section it always takes the greatest inclination of the interval, so it overestimates the departure and underestimates the true vertical depth, and when the angle drops, it is the other way round.

The tangential method

\Delta N = \Delta L\sin\theta_2\cos\varphi_2, \qquad \Delta E = \Delta L\sin\theta_2\sin\varphi_2, \qquad \Delta H = \Delta L\cos\theta_2

where:

$\Delta N,\ \Delta E,\ \Delta H$ increments of the displacement to the north, to the east and of the true vertical depth over the interval, m
$\Delta L$ length of the interval along the hole between the readings, $L_2 - L_1$ , m
$\theta_2,\ \varphi_2$ inclination and azimuth at the lower point of the interval, degrees

Units: lengths in metres (or in feet, the same for all)

Holds when

The well over the interval is a straight segment with the direction of the lower reading. The upper reading is not used.

The error over an interval is proportional to the length of the interval and to the change of the angle over it. On a build section it builds up in one direction.

If both readings are used, the accuracy rises sharply. The balanced tangential method takes the upper half of the interval along the direction of the upper reading, and the lower half along the direction of the lower one.

The balanced tangential method

\begin{aligned} \Delta N &= \frac{\Delta L}{2}\,(\sin\theta_1\cos\varphi_1 + \sin\theta_2\cos\varphi_2),\\ \Delta E &= \frac{\Delta L}{2}\,(\sin\theta_1\sin\varphi_1 + \sin\theta_2\sin\varphi_2),\\ \Delta H &= \frac{\Delta L}{2}\,(\cos\theta_1 + \cos\theta_2) \end{aligned}

where:

$\theta_1,\ \varphi_1$ inclination and azimuth at the upper point of the interval, degrees
$\theta_2,\ \varphi_2$ the same at the lower point, degrees
$\Delta L$ length of the interval along the hole, m

Units: lengths in metres

Holds when

The well over the interval is two straight segments of equal length $\Delta L/2$ with the directions of the upper and of the lower readings.

The two links form a broken line. Its end lies on the chord of the arc, but it does not reach the end of the arc: the links of length $\Delta L/2$ are shorter than the tangent segments that run to the point where the tangents meet (see the figure in the section on the arc methods). So the method slightly underestimates the increments, and the more so the greater the bend over the interval.

The average angle method takes one segment of length $\Delta L$ along the direction with the average inclination and the average azimuth. It is easy to calculate even on a pocket calculator, so it was used on rigs for a long time.

The average angle method

\begin{aligned} \Delta N &= \Delta L\sin\bar\theta\cos\bar\varphi, \qquad \Delta E = \Delta L\sin\bar\theta\sin\bar\varphi, \qquad \Delta H = \Delta L\cos\bar\theta,\\ \bar\theta &= \frac{\theta_1 + \theta_2}{2}, \qquad \bar\varphi = \varphi_1 + \frac{\Delta\varphi}{2} \end{aligned}

where:

$\bar\theta,\ \bar\varphi$ average inclination and azimuth of the interval, degrees
$\Delta\varphi$ change of azimuth over the interval, brought to the range from −180 to +180°, along the shorter arc, degrees

Units: lengths in metres; angles in degrees

Holds when

The well over the interval is a straight segment of length $\Delta L$ with the averaged direction.
The azimuths are averaged along the shorter arc: for 350° and 10° the change is +20° and the average azimuth is 0°, not 180°.
If one of the readings is almost vertical, its azimuth is random, and it cannot be averaged (an example is in the section on comparing the methods).

A segment of length $\Delta L$ is longer than the chord of an arc of the same length, so the method slightly overestimates the increments, by about half as much as the balanced tangential method underestimates them.

Arc methods: radius of curvature and minimum curvature

A real well between readings does not break but bends smoothly, so it is more accurate to describe it by an arc. The radius of curvature method assumes that the inclination changes uniformly along the length of the well, and that in the plan view the well follows an arc of a circle along which the azimuth changes uniformly. Integrating such relations gives the formulas below (the derivation is in the advanced section).

The radius of curvature method

\begin{aligned} \Delta H &= \Delta L\,\frac{\sin\theta_2 - \sin\theta_1}{\theta_2 - \theta_1}, \qquad \Delta A = \Delta L\,\frac{\cos\theta_1 - \cos\theta_2}{\theta_2 - \theta_1},\\ \Delta N &= \Delta A\,\frac{\sin\varphi_2 - \sin\varphi_1}{\varphi_2 - \varphi_1}, \qquad \Delta E = \Delta A\,\frac{\cos\varphi_1 - \cos\varphi_2}{\varphi_2 - \varphi_1} \end{aligned}

where:

$\Delta A$ length of the horizontal projection of the well over the interval, m
$\theta_2 - \theta_1,\ \varphi_2 - \varphi_1$ changes of the inclination and of the azimuth; in the denominators they are in radians, and the azimuth is taken along the shorter arc, rad

Units: lengths in metres; the differences of angles in the denominators in radians

Holds when

The inclination changes linearly along the length of the well, and the azimuth linearly along the length of the horizontal projection.
If $\theta_1 = \theta_2$ , then $\Delta H = \Delta L\cos\theta$ and $\Delta A = \Delta L\sin\theta$ . If $\varphi_1 = \varphi_2$ , then $\Delta N = \Delta A\cos\varphi$ and $\Delta E = \Delta A\sin\varphi$ . These are the limits of the formulas for a zero denominator.
In space such a curve is not an arc of a circle but a helix (a spiral). For an almost vertical reading the azimuth is random, and the method “trustfully” uses it.

For a section with a constant azimuth and a constant build-up rate the method is exact. With a small survey spacing its results hardly differ from those of the minimum curvature method.

The minimum curvature method draws through two survey stations an arc of a circle that touches both measured directions. It is the smoothest curve that is compatible with the readings. The arc lies in the plane of the two direction vectors, and the angle between these vectors is the dogleg $\beta$ , the total angle by which the well bends over the interval.

The dogleg over an interval

\cos\beta = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2\cos(\varphi_2 - \varphi_1)

where:

$\beta$ the angle between the directions of the well at the upper and the lower points of the interval, degrees
$\theta_1,\ \theta_2$ inclinations at the survey stations, degrees
$\varphi_1,\ \varphi_2$ azimuths at the survey stations, degrees

Units: degrees

Holds when

The formula is the scalar product of two unit direction vectors.
For small angles it is more accurate to use the equivalent formula with the sines of half-angles (the advanced section): the cosine is close to one, and significant digits are lost in rounding.

This is the angle through which the well has turned in all, by the inclination and by the azimuth together. It is not equal to the sum of the changes of the angles. A change of azimuth at a small inclination hardly bends the well, and at a large inclination it bends it strongly.

The minimum curvature method

\begin{aligned} \Delta N &= \frac{\Delta L}{2}\,(\sin\theta_1\cos\varphi_1 + \sin\theta_2\cos\varphi_2)\,RF,\\ \Delta E &= \frac{\Delta L}{2}\,(\sin\theta_1\sin\varphi_1 + \sin\theta_2\sin\varphi_2)\,RF,\\ \Delta H &= \frac{\Delta L}{2}\,(\cos\theta_1 + \cos\theta_2)\,RF, \qquad RF = \frac{2}{\beta}\tan\frac{\beta}{2} \end{aligned}

where:

$RF$ ratio factor of the method; $\beta$ in it is in radians; for $\beta = 0$ it equals 1, —
$\beta$ dogleg over the interval, rad
$\theta,\ \varphi,\ \Delta L$ as in the balanced tangential method, degrees, m

Units: lengths in metres; $\beta$ in the factor in radians

Holds when

The well over the interval is an arc of a circle that touches the directions of both readings.
The factor is greater than one and grows with the bend: for $\beta = 3^\circ$ it is 1.00023, and for $10^\circ$ it is 1.0025.

It is the balanced tangential method multiplied by a correction. The two tangent segments, each of length $R\tan(\beta/2)$ (the figure below), lead exactly from the start of the arc to its end, and $RF$ is the ratio of this length to half of the interval. The minimum curvature method is the industry standard: it is used by trajectory calculation software and in the exchange of data between companies.

SchematicThe geometry of the minimum curvature method in the plane of the arc (a sketch; the dogleg is exaggerated). An arc of radius

R

with the centre O joins the survey stations A and B. The tangents to it at these points meet at the point C, and the segments AC and CB equal

R\tan(\beta/2)

. The broken line of the balanced tangential method (dashed) goes along the same directions, but by

\Delta L/2

each, and does not reach B. The factor

RF

stretches its links to AC and CB.

Dogleg severity

The dogleg alone says little without the length of the interval: 3° per 30 m and 3° per 300 m are very different wells. So it is related to the length of the well, and this gives the dogleg severity (DLS). In a flat profile with a constant azimuth it is the same as the build-up or drop-off rate (Directional wells: applications and profile), and in the general case it also takes the turn of the azimuth into account.

Dogleg severity over an interval

i = \frac{\beta\,l_0}{\Delta L}

where:

$i$ dogleg severity: degrees per the reference length $l_0$ , °/30 m
$\beta$ dogleg over the interval, degrees
$\Delta L$ length of the interval along the hole, m
$l_0$ reference length: 30 m, 10 m or 100 ft (30.48 m), m

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

Holds when

The bend is taken as uniform over the interval: this is an average rate. Local bends inside the interval cannot be seen from two readings.

The dogleg severity is limited by the drilling programme and by the requirements of the equipment. In a sharply curved hole, rotating drill pipe is bent back and forth and fails sooner from fatigue, the casing and the tool joints wear faster, key seats (grooves worn in the wall of the hole by the pipe) form, and it is harder to run a casing string and pumping equipment. So it is calculated for every reading.

In other unit systems (3)

Degrees per 30 m

i = \frac{30\,\beta}{\Delta L\,[\text{m}]}

$\beta$ in degrees, $\Delta L$ in metres.

30 is the reference length in metres.

Degrees per 10 m

i = \frac{10\,\beta}{\Delta L\,[\text{m}]}

$\beta$ in degrees, $\Delta L$ in metres.

three times smaller than per 30 m.

Degrees per 100 ft (oilfield units)

i = \frac{100\,\beta}{\Delta L\,[\text{ft}]} = \frac{30.48\,\beta}{\Delta L\,[\text{m}]}

$\beta$ in degrees, $\Delta L$ in feet or metres.

100 ft $= 30.48$ m.

Worked exampleThree intervals: a build with a turn, a pure turn and a kick-off

Given (assumed data)

Interval 690–720 m: $\theta$ : 19.6° → 22.5°; $\varphi$ : 57.6° → 56.3°
An interval of 30 m on a tangent section: $\theta = 30^\circ$ with no change; $\varphi$ : 50° → 56°
Interval 480–510 m at the kick-off point: $\theta$ : 0.3° → 1.6°; $\varphi$ : 215° → 63°

Solution

A build with a turn: $\beta = 2.94^\circ$ over 30 m, that is $i = 2.94$ °/30 m $= 0.98$ °/10 m $= 2.98$ °/100 ft. The inclination grew by 2.9°, the azimuth decreased by 1.3°, and the total turn of the well is a little greater than the change of the inclination.
A pure turn at an inclination of 30°: $\beta = 3.00^\circ$ , $i = 3.00$ °/30 m. The azimuth changed by 6°, but the well bent by only 3.00°: the change of azimuth enters the dogleg with a factor close to $\sin\theta = 0.5$ .
The kick-off: the inclination changed by only 1.3°, but $\beta = 1.87^\circ$ . The almost vertical well changed the direction of its lean, from south-west to north-east, and this turn added to the dogleg.

Answer: The dogleg severities are 2.94, 3.00 and 1.87°/30 m. In the second case the inclination did not change at all, and the well is bent as much as with a build of 3° per 30 m.

What it means. The dogleg severity cannot be judged by the inclination alone or by the azimuth alone: the dogleg decides. In the table of readings it is recalculated for every interval and compared with the value allowed by the drilling programme. The data in this example are assumed for illustration.

One interval by five methods

Let us compare the methods on one interval of the well from the lesson on the profile. It was drilled with a kick-off (the point where the well leaves the vertical) at 500 m and a build of about 3° per 30 m. To make the differences visible, let us take a long interval, from the reading at 660 m straight to the reading at 750 m, as if the two intermediate readings had not been taken. The result from all the readings at 30 m is known and will serve as a reference.

Worked exampleInterval 660–750 m by the straight segment methods

Given (assumed data)

Upper reading: $L_1 = 660$ m, $\theta_1 = 16.4^\circ$ , $\varphi_1 = 59.0^\circ$
Lower reading: $L_2 = 750$ m, $\theta_2 = 25.4^\circ$ , $\varphi_2 = 55.2^\circ$
Length of the interval: $\Delta L = 90$ m

Solution

The direction at the upper point: $\sin\theta_1\cos\varphi_1 = 0.14542$ , $\sin\theta_1\sin\varphi_1 = 0.24201$ , $\cos\theta_1 = 0.95931$ . At the lower point: 0.24480, 0.35222 and 0.90334.
Tangential: $\Delta N = 90\cdot 0.24480 = 22.03$ m, $\Delta E = 90\cdot 0.35222 = 31.70$ m, $\Delta H = 90\cdot 0.90334 = 81.30$ m.
Balanced tangential: $\Delta N = 45\cdot(0.14542 + 0.24480) = 17.56$ m, $\Delta E = 45\cdot(0.24201 + 0.35222) = 26.74$ m, $\Delta H = 45\cdot(0.95931 + 0.90334) = 83.82$ m.
Average angle: $\bar\theta = 20.9^\circ$ , $\bar\varphi = 57.1^\circ$ ; $\Delta N = 90\sin 20.9^\circ\cos 57.1^\circ = 17.44$ m, $\Delta E = 26.96$ m, $\Delta H = 90\cos 20.9^\circ = 84.08$ m.

Answer: The tangential method gave a departure over the interval of 38.60 m, the balanced tangential method 31.99 m, and the average angle method 32.11 m.

What it means. The tangential method drew the whole interval at an inclination of 25.4°, which the well reached only at the end. This is where the extra metres in the plan view and the shortfall in the vertical come from. The other two methods differ from each other by tens of centimetres. The arc methods will show which of them is closer to the truth. The data in this example are assumed for illustration.

Worked exampleThe same interval by the arc methods

Given (assumed data)

Readings: the same: 16.4° and 59.0° at 660 m, 25.4° and 55.2° at 750 m
Result of the balanced tangential method: $\Delta N = 17.56$ m, $\Delta E = 26.74$ m, $\Delta H = 83.82$ m

Solution

The dogleg: $\cos\beta = \cos 16.4^\circ\cos 25.4^\circ + \sin 16.4^\circ\sin 25.4^\circ\cos(-3.8^\circ) = 0.98742$ ; $\beta = 9.10^\circ = 0.15877$ rad; the dogleg severity is 3.03°/30 m.
The factor: $RF = \tan(0.15877/2)/(0.15877/2) = 1.00211$ .
Minimum curvature: $\Delta N = 17.56\cdot 1.00211 = 17.60$ m, $\Delta E = 26.74\cdot 1.00211 = 26.80$ m, $\Delta H = 83.82\cdot 1.00211 = 84.00$ m.
Radius of curvature: the differences of the angles in radians are $\theta_2 - \theta_1 = 0.15708$ and $\varphi_2 - \varphi_1 = -0.06632$ . $\Delta H = 90\cdot(\sin 25.4^\circ - \sin 16.4^\circ)/0.15708 = 83.99$ m; $\Delta A = 90\cdot(\cos 16.4^\circ - \cos 25.4^\circ)/0.15708 = 32.07$ m.
Next, by the azimuth: $\Delta N = 32.07\cdot(\sin 55.2^\circ - \sin 59.0^\circ)/(-0.06632) = 17.42$ m, $\Delta E = 32.07\cdot(\cos 59.0^\circ - \cos 55.2^\circ)/(-0.06632) = 26.92$ m.
The reference is the calculation from all the readings at 30 m by the minimum curvature method: $\Delta N = 17.62$ m, $\Delta E = 26.97$ m, $\Delta H = 83.94$ m.

Answer: The arc methods give

\Delta H

of 84.00 and 83.99 m. From all the readings it is 83.94 m. The tangential method was wrong by 2.6 m in the vertical and by 6.4 m in the plan view.

What it means. For a dogleg of 9° over the interval, the correction $RF$ changes the result of the balanced tangential method by only 0.2%. The arc methods and the straight segment methods (except the tangential) differ by centimetres to decimetres, and from the calculation with frequent readings by tenths of a metre. No method knows how the well went between rare readings. The error of the tangential method is metres, and on every build interval it is added in the same direction. The data in this example are assumed for illustration.

Interval 660–750 m: the increments by the five methods

Method: Tangential
$\Delta H$ , m: 81.30
$\Delta N$ , m: 22.03
$\Delta E$ , m: 31.70
Departure, m: 38.60

Method: Balanced tangential
$\Delta H$ , m: 83.82
$\Delta N$ , m: 17.56
$\Delta E$ , m: 26.74
Departure, m: 31.99

Method: Average angle
$\Delta H$ , m: 84.08
$\Delta N$ , m: 17.44
$\Delta E$ , m: 26.96
Departure, m: 32.11

Method: Radius of curvature
$\Delta H$ , m: 83.99
$\Delta N$ , m: 17.42
$\Delta E$ , m: 26.92
Departure, m: 32.07

Method: Minimum curvature
$\Delta H$ , m: 84.00
$\Delta N$ , m: 17.60
$\Delta E$ , m: 26.80
Departure, m: 32.06

Method: From all the readings at 30 m
$\Delta H$ , m: 83.94
$\Delta N$ , m: 17.62
$\Delta E$ , m: 26.97
Departure, m: 32.21

Method	$\Delta H$ , m	$\Delta N$ , m	$\Delta E$ , m	Departure, m
Tangential	81.30	22.03	31.70	38.60
Balanced tangential	83.82	17.56	26.74	31.99
Average angle	84.08	17.44	26.96	32.11
Radius of curvature	83.99	17.42	26.92	32.07
Minimum curvature	84.00	17.60	26.80	32.06
From all the readings at 30 m	83.94	17.62	26.97	32.21

The departure is the horizontal displacement over the interval. The last row is the minimum curvature method from the readings at 660, 690, 720 and 750 m.

Why the methods differ in just this way. The broken line of the balanced tangential method goes along the directions of the tangents, but its links are shorter than they should be, and it does not reach the end of the arc. This is where the smaller increments come from. The segment of the average angle method points almost along the chord of the arc, but its length equals the arc itself, and the chord is shorter than the arc. This is where the small excess comes from. In space there is also the distribution by azimuth. The average angle and the radius of curvature methods take the average azimuth of the interval. The balanced tangential and minimum curvature methods give more weight to the direction of the point where the well is inclined more. So they share the increment between north and east a little differently.

Departure, departure azimuth and the projection on the plane of the target

The coordinates of a point of the well are conveniently given in two more ways. The departure $A$ and the departure azimuth $\varphi_A$ (called the closure distance and direction) are the distance from the wellhead to the projection of the point on the horizontal plane and the direction to it. These are the same quantities as for the target in the lesson on the profile (departure). And for the vertical profile of the well, the departure is projected on the vertical plane through the wellhead at the planned azimuth, usually towards the target. This projection is the vertical section. The distance of the point from this plane shows how far the well has gone sideways.

Departure, departure azimuth and the projections on the plane of the target

\begin{aligned} A &= \sqrt{N^2 + E^2}, \qquad \tan\varphi_A = \frac{E}{N},\\ A_{\text{VS}} &= A\cos(\varphi_A - \varphi_{\text{plan}}) = N\cos\varphi_{\text{plan}} + E\sin\varphi_{\text{plan}},\\ A_{\text{lat}} &= A\sin(\varphi_A - \varphi_{\text{plan}}) = E\cos\varphi_{\text{plan}} - N\sin\varphi_{\text{plan}} \end{aligned}

where:

$N,\ E$ coordinates of the point: the displacements to the north and to the east from the wellhead, m
$A,\ \varphi_A$ departure and departure azimuth, m, degrees
$\varphi_{\text{plan}}$ planned azimuth, the azimuth of the vertical plane of the profile, degrees
$A_{\text{VS}}$ vertical section, the displacement along the planned azimuth, m
$A_{\text{lat}}$ offset from the planned plane: plus is to the right, looking along the planned azimuth, m

Units: metres; angles in degrees

Holds when

The quadrant of the departure azimuth is chosen by the signs of $N$ and $E$ , as for the azimuth of the target.
All azimuths are from the one north in which the coordinates are given.

The vertical profile is drawn in the coordinates of “vertical section against true vertical depth”, and the plan view in “east against north”. While the well is close to the planned plane, the vertical section is almost equal to the departure. The sideways offset shows whether the azimuth needs correcting.

Readings of the well of the example (the azimuths are grid azimuths)

$L$ , m: 480
$\theta$ , °: 0.3
$\varphi$ , °: 215.0
$i$ , °/30 m: —

$L$ , m: 510
$\theta$ , °: 1.6
$\varphi$ , °: 63.0
$i$ , °/30 m: 1.87

$L$ , m: 540
$\theta$ , °: 4.3
$\varphi$ , °: 59.0
$i$ , °/30 m: 2.71

$L$ , m: 570
$\theta$ , °: 7.2
$\varphi$ , °: 58.0
$i$ , °/30 m: 2.90

$L$ , m: 600
$\theta$ , °: 10.3
$\varphi$ , °: 58.5
$i$ , °/30 m: 3.10

$L$ , m: 630
$\theta$ , °: 13.5
$\varphi$ , °: 59.4
$i$ , °/30 m: 3.21

$L$ , m: 660
$\theta$ , °: 16.4
$\varphi$ , °: 59.0
$i$ , °/30 m: 2.90

$L$ , m: 690
$\theta$ , °: 19.6
$\varphi$ , °: 57.6
$i$ , °/30 m: 3.23

$L$ , m: 720
$\theta$ , °: 22.5
$\varphi$ , °: 56.3
$i$ , °/30 m: 2.94

$L$ , m: 750
$\theta$ , °: 25.4
$\varphi$ , °: 55.2
$i$ , °/30 m: 2.93

$L$ , m	$\theta$ , °	$\varphi$ , °	$i$ , °/30 m
480	0.3	215.0	—
510	1.6	63.0	1.87
540	4.3	59.0	2.71
570	7.2	58.0	2.90
600	10.3	58.5	3.10
630	13.5	59.4	3.21
660	16.4	59.0	2.90
690	19.6	57.6	3.23
720	22.5	56.3	2.94
750	25.4	55.2	2.93

The tie-in point is at 480 m: by a gyroscopic reading the bottom is here practically under the wellhead, and its coordinates are taken as zero. An azimuth of 215° at an inclination of 0.3° means almost nothing.

Coordinates of the survey stations (minimum curvature method)

$L$ , m: 480
$H$ , m: 480.00
$N$ , m: 0.00
$E$ , m: 0.00
$A_{\text{VS}}$ , m: 0.00

$L$ , m: 510
$H$ , m: 510.00
$N$ , m: 0.13
$E$ , m: 0.33
$A_{\text{VS}}$ , m: 0.34

$L$ , m: 540
$H$ , m: 539.95
$N$ , m: 0.90
$E$ , m: 1.67
$A_{\text{VS}}$ , m: 1.87

$L$ , m: 570
$H$ , m: 569.80
$N$ , m: 2.47
$E$ , m: 4.22
$A_{\text{VS}}$ , m: 4.86

$L$ , m: 600
$H$ , m: 599.45
$N$ , m: 4.87
$E$ , m: 8.11
$A_{\text{VS}}$ , m: 9.41

$L$ , m: 630
$H$ , m: 628.80
$N$ , m: 8.05
$E$ , m: 13.41
$A_{\text{VS}}$ , m: 15.56

$L$ , m: 660
$H$ , m: 657.78
$N$ , m: 12.02
$E$ , m: 20.05
$A_{\text{VS}}$ , m: 23.25

$L$ , m: 690
$H$ , m: 686.31
$N$ , m: 16.90
$E$ , m: 27.94
$A_{\text{VS}}$ , m: 32.49

$L$ , m: 720
$H$ , m: 714.30
$N$ , m: 22.78
$E$ , m: 36.96
$A_{\text{VS}}$ , m: 43.24

$L$ , m: 750
$H$ , m: 741.72
$N$ , m: 29.64
$E$ , m: 47.02
$A_{\text{VS}}$ , m: 55.40

$L$ , m	$H$ , m	$N$ , m	$E$ , m	$A_{\text{VS}}$ , m
480	480.00	0.00	0.00	0.00
510	510.00	0.13	0.33	0.34
540	539.95	0.90	1.67	1.87
570	569.80	2.47	4.22	4.86
600	599.45	4.87	8.11	9.41
630	628.80	8.05	13.41	15.56
660	657.78	12.02	20.05	23.25
690	686.31	16.90	27.94	32.49
720	714.30	22.78	36.96	43.24
750	741.72	29.64	47.02	55.40

The planned azimuth is 53.1°, the direction to the target from the lesson on the profile.

Worked exampleWhere the well is at 750 m relative to the plan

Given (assumed data)

Last reading, minimum curvature method: $H = 741.72$ m, $N = 29.64$ m, $E = 47.02$ m
Planned azimuth: $\varphi_{\text{plan}} = 53.13^\circ$
The plan at 750 m along the hole: inclination 25.0°, true vertical depth 742.14 m, departure 53.68 m

Solution

The departure: $A = \sqrt{29.64^2 + 47.02^2} = 55.58$ m. The departure azimuth is $\varphi_A = \arctan(47.02/29.64) = 57.78^\circ$ . Both displacements are positive, so no correction of the quadrant is needed.
The planned azimuth was set by the displacements of the target, 660 m to the north and 880 m to the east, so $\cos\varphi_{\text{plan}} = 660/1100 = 0.6$ and $\sin\varphi_{\text{plan}} = 880/1100 = 0.8$ . The vertical section is $A_{\text{VS}} = 29.64\cdot 0.6 + 47.02\cdot 0.8 = 55.40$ m.
The offset from the planned plane: $A_{\text{lat}} = 47.02\cdot 0.6 - 29.64\cdot 0.8 = 4.50$ m, to the right. The same is given by $A\sin(\varphi_A - \varphi_{\text{plan}})$ .
Comparison with the plan: along the planned plane the well is ahead by $55.40 - 53.68 = 1.72$ m, in the vertical it is higher by $742.14 - 741.72 = 0.42$ m, and sideways it has gone off by 4.50 m.

Answer: At 750 m the well is 4.50 m to the right of the planned plane and 1.72 m ahead of the plan in departure.

What it means. The build goes almost as planned, and a small lead is not a problem. But the sideways drift of 4.50 m is already noticeable: the well went at an azimuth of about 58° from the start instead of 53.1°. While the target is more than a kilometre away, it is corrected by a small turn to the left, without changing the build-up rate (Deflection tools and downhole motors). The data in this example are assumed for illustration.

CalculatedThe plan view of the well of the example (calculated by the minimum curvature method, with equal scales on the axes): the survey stations from 480 to 750 m and the planned line to the target at an azimuth of 53.1°. The well goes to the right of the planned line. At 750 m the offset is about 4.5 m.

How the method and the survey spacing affect the result

Over one interval the difference between the methods is small, but the calculation goes from point to point, and the errors add up. Let us calculate the whole table of readings of the well of the example with each of the five methods.

The position of the point at 750 m by the five methods

Method: Tangential
$H$ , m: 740.21
$N$ , m: 33.37
$E$ , m: 52.34
From the MC result, m: 6.67

Method: Balanced tangential
$H$ , m: 741.66
$N$ , m: 29.63
$E$ , m: 47.01
From the MC result, m: 0.06

Method: Average angle
$H$ , m: 741.74
$N$ , m: 29.10
$E$ , m: 47.05
From the MC result, m: 0.54

Method: Radius of curvature
$H$ , m: 741.72
$N$ , m: 29.20
$E$ , m: 46.96
From the MC result, m: 0.45

Method: Minimum curvature
$H$ , m: 741.72
$N$ , m: 29.64
$E$ , m: 47.02
From the MC result, m: —

Method	$H$ , m	$N$ , m	$E$ , m	From the MC result, m
Tangential	740.21	33.37	52.34	6.67
Balanced tangential	741.66	29.63	47.01	0.06
Average angle	741.74	29.10	47.05	0.54
Radius of curvature	741.72	29.20	46.96	0.45
Minimum curvature	741.72	29.64	47.02	—

MC is the minimum curvature method. The last column is the distance in space between the point calculated by the given method and the point by the minimum curvature method.

The tangential method is off by 6.7 m: in each of the nine build intervals it took the greatest inclination of the interval.
The balanced tangential method lags behind minimum curvature by a few centimetres in the vertical. This is the correction $RF$ , accumulated over nine intervals.
The average angle and radius of curvature methods differ from minimum curvature by 0.54 and 0.45 m, and almost all of this built up in the first interval, at the kick-off point (the example below).

Worked exampleWhy the average azimuth lets us down at the kick-off point

Given (assumed data)

Reading at 480 m: $\theta = 0.3^\circ$ , $\varphi = 215^\circ$ : the well is barely inclined to the south-west
Reading at 510 m: $\theta = 1.6^\circ$ , $\varphi = 63^\circ$ : the well is inclined to the north-east

Solution

The average angle is $\bar\theta = 0.95^\circ$ . The change of azimuth along the shorter arc is $63 - 215 = -152^\circ$ , and the average azimuth is $215 - 152/2 = 139^\circ$ . This is south-east, where the well was not inclined at either of the two points.
The average angle method: $\Delta N = -0.38$ m, $\Delta E = 0.33$ m.
The minimum curvature method: $\Delta N = 0.13$ m, $\Delta E = 0.33$ m. The contribution of the upper reading is multiplied by $\sin 0.3^\circ \approx 0.0052$ , and its azimuth hardly matters.
The difference over this interval is 0.50 m. Of the half-metre at 750 m, almost all built up here.

Answer: The average azimuth 139° of the almost vertical interval takes the increment to the south-east. The error is 0.50 m already in the first interval.

What it means. The azimuth of an almost vertical reading is noise: the sensor sees only a barely noticeable lean of the well. The methods that average the azimuth or take it as changing uniformly (average angle, radius of curvature) take this noise seriously. The methods in which the contribution of a reading is proportional to the sine of the inclination (balanced tangential, minimum curvature) hardly notice it. An error of half a metre is not large, but it is carried over to the whole well below. The data in this example are assumed for illustration.

How accurate the methods are at different survey spacings can be seen on a path for which the answer is known in advance: a build section from vertical to 30° with a constant rate of 3° per 30 m and a constant azimuth. Such a well is an arc of a circle, and the arc methods are exact here. The graph shows the errors of the others.

CalculatedThe calculated error of the position of the end of a build section (300 m of the hole, from 0 to 30° at 3° per 30 m, with a constant azimuth) against the survey spacing. The error scale is logarithmic. The error of the tangential method grows in proportion to the spacing and is metres. The errors of the balanced tangential and the average angle methods grow in proportion to the square of the spacing and are centimetres. The radius of curvature and minimum curvature methods are exact for such a well.

Terms and symbols

Directional survey, survey station: A measurement of the inclination and the azimuth at a known depth along the hole. The place of the measurement is a survey station.
Tie-in point: A point with known coordinates from which the calculation starts.
Tangential method: The interval is a segment along the direction of the lower reading. It is systematically wrong on curved sections.
Balanced tangential method: The two halves of the interval go along the directions of the upper and the lower readings.
Average angle method: One segment along the average inclination and azimuth.
Radius of curvature method: The inclination changes uniformly along the hole, and the azimuth along the horizontal projection.
Minimum curvature method: An arc of a circle that touches both readings. It is the industry standard.
Dogleg $\beta$: The angle between the directions of the well at two survey stations.
Dogleg severity $i$: The dogleg per 30 m (10 m, 100 ft) of the well.
Departure and departure azimuth: The distance from the wellhead to the projection of a point in the plan view, and the direction to it.
Vertical section $A_{\text{VS}}$: The projection of the departure on the vertical plane of the profile.
Magnetic declination $\delta$ , grid convergence $\gamma$: The corrections that bring a magnetic azimuth to a true one, and a true one to a grid one.

Common mistakes and misconceptions

Wrong

“All the methods give almost the same result, so we can use any.”

In fact

The methods that use both readings are indeed close when the readings are frequent. The tangential method is not. On a build section its error builds up in one direction and reaches metres.

Wrong

“The dogleg severity is the change of the inclination per 30 m.”

In fact

Only when the azimuth does not change. In general we calculate the angle in space between the directions. A turn of the azimuth at a large inclination bends the well no less than a build.

Wrong

“The average azimuth is the half-sum of the azimuths.”

In fact

Only if there is no crossing of north between them and the readings are not almost vertical. The change of azimuth is taken along the shorter arc, and the azimuth of an almost vertical reading is not included in the average.

Wrong

“The corrections for the declination and the grid convergence are small, a couple of degrees.”

In fact

A couple of degrees at a departure of a kilometre is tens of metres to the side, for the whole path at once. The corrections must be applied and recorded together with the readings.

Wrong

“If the method is exact, the path is exact.”

In fact

The method decides only how to join the readings. Between rare readings the well could have gone differently, and the readings themselves contain instrument errors, which are estimated separately.

In depth: advanced detail

Where the factor RF comes from, and how to calculate small angles

Let $\mathbf{t}_1$ and $\mathbf{t}_2$ be the unit direction vectors at the points A and B, and let the well between them be an arc of a circle of radius $R$ and length $\Delta L$ . The central angle of the arc equals the angle between the tangents, that is $\beta$ , so $R = \Delta L/\beta$ ( $\beta$ in radians). The tangents at the ends of the arc meet at the point C, and by symmetry $AC = CB = R\tan(\beta/2)$ .

The chord AB equals the sum of the two tangent segments: $\overrightarrow{AB} = R\tan(\beta/2)\,(\mathbf{t}_1 + \mathbf{t}_2)$ . If we put in $R = \Delta L/\beta$ , we get a chord that equals $\tfrac{\Delta L}{2}(\mathbf{t}_1 + \mathbf{t}_2)$ multiplied by $\tan(\beta/2)/(\beta/2)$ . This is the balanced tangential method with the factor $RF$ . The series expansion $RF \approx 1 + \beta^2/12$ shows that the correction is of the second order: for a dogleg of 3° over an interval it is less than 0.03%.

The dogleg through half-angles

\sin^2\frac{\beta}{2} = \sin^2\frac{\theta_2 - \theta_1}{2} + \sin\theta_1\sin\theta_2\,\sin^2\frac{\varphi_2 - \varphi_1}{2}

where:

$\beta$ dogleg, degrees or radians
$\theta_1,\ \theta_2,\ \varphi_1,\ \varphi_2$ inclinations and azimuths at the survey stations, degrees

Units: any units of angle, the same for all the quantities

Holds when

It follows from the formula with cosines by the substitution $1 - \cos x = 2\sin^2(x/2)$ ; it is equivalent to it for any angles.

For a small angle $\cos\beta$ is close to one, and all the information about the angle is in the last digits of the cosine. The arccosine of a rounded number is wrong by more the smaller the angle is. Half-angles do not have this trouble, so programs calculate with this formula.

Worked exampleA small dogleg: the arccosine against half-angles

Given (assumed data)

Readings at 30 m on a tangent section: $\theta$ : 32.0° → 32.5°; $\varphi$ : 53.0° → 54.5°

Solution

By half-angles: $\sin^2(\beta/2) = \sin^2 0.25^\circ + \sin 32.0^\circ\sin 32.5^\circ\sin^2 0.75^\circ = 6.782\cdot 10^{-5}$ , so $\beta = 0.944^\circ$ .
By the cosine: $\cos\beta = 0.9998644$ . If we round it to four digits (0.9999), the arccosine gives 0.81°, and to five (0.99986), 0.96°.

Answer: The dogleg is 0.944°. Rounding the cosine to four digits would have understated it by 14%.

What it means. The smaller the angle, the more digits of the cosine are needed so as not to lose it. In a hand calculation this is easy to miss, and an understated dogleg severity is a missed bend of the well. The data in this example are assumed for illustration.

Derivation of the radius of curvature formulas

If the inclination changes linearly along the length of the well, $\theta(l) = \theta_1 + (\theta_2 - \theta_1)\,l/\Delta L$ , then $\dd l = \Delta L\,\dd\theta/(\theta_2 - \theta_1)$ , and the increment of the true vertical depth is the integral $\int\cos\theta\,\dd l = \Delta L\,(\sin\theta_2 - \sin\theta_1)/(\theta_2 - \theta_1)$ . The length of the horizontal projection, the integral $\int\sin\theta\,\dd l$ , gives $\Delta A$ with a difference of cosines.

Then, in the plan view, the azimuth changes linearly along the length of the horizontal projection. The projection is an arc of a circle, and the integrals $\int\cos\varphi\,\dd a$ and $\int\sin\varphi\,\dd a$ along it give $\Delta N$ and $\Delta E$ . Both curves are arcs of circles: the one in the vertical view (unrolled along the well) and the one in the plan view. This gives the method its name. In space such a line looks like a turn of a spiral. It coincides with the arc of a circle of the minimum curvature method only if the azimuth is constant (a flat build) or if the inclination is constant at 90° (a horizontal turn).

Connections to other topics

Directional wells: applications and profile — the planned profile, the zero datums, the departure and the azimuth of the target, the three norths.
Deflection tools and downhole motors — what is used to correct the path when the readings show a departure from the plan.
Open-hole logging: why, how and when — converting the log depths along the hole into true vertical depths from the survey data.
Pore pressure and overburden pressure — why pressures are calculated from the true vertical depth that the path calculation gives.
Casing design — bending stresses in a string run into a curved hole.

Check yourself

1
Which method systematically overestimates the departure and underestimates the true vertical depth on a build section?
1. A.The balanced tangential method
2. B.The average angle method
3. C.The tangential method
4. D.The minimum curvature method
Show the answer
Correct answer: C. The tangential method
The tangential method leads the whole interval at the inclination of the lower reading, which is the greatest on a build section. So every interval comes out flatter than it really is: extra displacement in the plan view and a shortfall in the vertical, and all this builds up from interval to interval.
2
Readings: 1200 m, $\theta = 30^\circ$ , $\varphi = 120^\circ$ ; 1230 m, $\theta = 32^\circ$ , $\varphi = 124^\circ$ . Find the increments by the average angle method and the dogleg severity.
Show the answer
The average angles are $\bar\theta = 31^\circ$ and $\bar\varphi = 122^\circ$ . $\Delta H = 30\cos 31^\circ \approx 25.72$ m. The horizontal projection is $30\sin 31^\circ \approx 15.45$ m, so $\Delta N \approx 15.45\cos 122^\circ \approx -8.19$ m and $\Delta E \approx 15.45\sin 122^\circ \approx 13.10$ m. The dogleg: $\cos\beta = \cos 30^\circ\cos 32^\circ + \sin 30^\circ\sin 32^\circ\cos 4^\circ \approx 0.99875$ , $\beta \approx 2.87^\circ$ . The dogleg severity is about 2.87° per 30 m.
3
What average azimuth should we take for readings with the azimuths 350° and 20°?
1. A.15°
2. B.185°
3. C.355°
4. D.5°
Show the answer
Correct answer: D. 5°
The change of azimuth along the shorter arc is $20 - 350 = -330^\circ$ , plus 360°, that is +30°. The average azimuth is $350 + 15 = 365^\circ$ , that is 5°. The half-sum, 185°, would turn the interval to the south.
4
Why can the minimum curvature method be called the balanced tangential method with a correction, and when is this correction close to one?
Show the answer
If the well is an arc of a circle, the tangents to it at the ends of the interval meet, and the tangent segments equal $R\tan(\beta/2)$ . The chord of the arc is the sum of these segments, that is, the same two directions as in the balanced tangential method, but with the lengths $R\tan(\beta/2)$ instead of $\Delta L/2$ . The ratio of these lengths is $RF = \tan(\beta/2)/(\beta/2) \approx 1 + \beta^2/12$ . For a small bend over the interval (frequent readings, a small dogleg severity) it is almost equal to one.
5
The magnetic azimuth of a reading is 210.0°, the magnetic declination is +7.0° (east), and the grid convergence is −1.5°. Find the grid azimuth.
Show the answer
The true azimuth is $210.0 + 7.0 = 217.0^\circ$ . The grid azimuth is $217.0 - (-1.5) = 218.5^\circ$ . A negative convergence means that grid north is west of true north (the point is west of the central meridian of the zone), so the grid azimuth is greater than the true azimuth.
6
The bottom is at the point $N = -300$ m, $E = -400$ m, and the planned azimuth is 200°. Find the departure, the departure azimuth, the vertical section and the offset from the planned plane.
Show the answer
The departure is $A = \sqrt{300^2 + 400^2} = 500$ m. The arctangent is $\arctan(-400/-300) = 53.13^\circ$ , but $N < 0$ , so we add 180°: $\varphi_A = 233.13^\circ$ , south-west. The difference from the planned azimuth is 33.13°: $A_{\text{VS}} = 500\cos 33.13^\circ \approx 418.7$ m and $A_{\text{lat}} = 500\sin 33.13^\circ \approx 273.3$ m, to the right of the planned plane.

Updated: 5 October 2026