# Chapter 12 — Directional Drilling Calculations

## Learning Objectives

By the end of this chapter, students should be able to:

1. Construct a wellbore survey table from a planned well description, identifying the key depth stations and their inclinations and azimuths.
2. Use the radius of curvature method to calculate ΔTVD and ΔHD through vertical, build/drop, and tangent sections.
3. Convert horizontal departure to northing and easting components using the well azimuth.
4. Build a complete wellbore position table (MD, INC, AZ, TVD, HD, N, E) from a well plan.
5. Given a wellbore survey, reconstruct the well path and identify its profile type.
6. Explain the minimum curvature method and apply it to a survey interval to compute position increments.
7. Describe what torque and drag are, explain why they occur in deviated wells, and identify the well design factors that increase them.

---

## 12.1 The Calculation Problem

Chapter 11 introduced the geometric vocabulary for directional wells and the concept of the wellbore survey. This chapter works through the actual numerical procedure for translating survey measurements into 3D wellbore position.

The core problem is this: the survey tool measures three quantities at each station — measured depth (MD), inclination (INC), and azimuth (AZ). None of these directly gives position. From these three measurements, we want to calculate the position of the well in three dimensions: true vertical depth (TVD, the z-coordinate), northing (N, the y-coordinate), and easting (E, the x-coordinate).

The procedure is additive: starting from a known starting position (typically the wellhead, at position 0, 0, 0), each survey interval contributes increments ΔTVD, ΔN, and ΔE. These increments are summed to give the cumulative position at each station.

```{admonition} Key Term
:class: tip
The **radius of curvature method** is the standard technique for 2D wellbore survey calculation taught in the sophomore course. It treats each curved section as a circular arc and uses the radius and included angle to compute ΔTVD and ΔHD exactly for a planned well. The **minimum curvature method** is the industry-standard 3D technique used to compute position from actual drilled survey data and is introduced in Section 12.6.
```

---

## 12.2 The Radius of Curvature Method (2D)

The radius of curvature method applies to a 2D well — one that changes inclination but not azimuth (or whose azimuth change is handled separately). It handles the three section types that make up any standard 2D well profile:

1. **Vertical section**: inclination = 0°, azimuth irrelevant.
2. **Build (or drop) section**: inclination changes at a constant DLS.
3. **Tangent section**: inclination is constant, azimuth is constant.

### 12.2.1 Vertical Section

For a vertical section of measured length $L$ (from MD$_1$ to MD$_2 = $ MD$_1 + L$):

$$\Delta TVD = L$$
$$\Delta HD = 0$$

There is no horizontal displacement in a perfectly vertical section.

### 12.2.2 Build (or Drop) Section

A build section changes inclination from $\alpha_1$ to $\alpha_2$ at a constant dogleg severity DLS. The radius of curvature of the arc is:

$$R = \frac{5729.6}{\text{DLS} \; [^\circ/100\text{ft}]} \quad \text{(ft)}$$

The measured length of drill string in the build section is:

$$\text{MD}_\text{build} = \frac{|\alpha_2 - \alpha_1|}{\text{DLS}} \times 100 \quad \text{(ft)}$$

The vertical and horizontal increments through the build are:

$$\boxed{\Delta TVD_\text{build} = R(\sin \alpha_2 - \sin \alpha_1)}$$

$$\boxed{\Delta HD_\text{build} = R(\cos \alpha_1 - \cos \alpha_2)}$$

For a drop (inclination decreasing), $\alpha_2 < \alpha_1$. The formula as written gives negative ΔTVD, which is physically wrong — the well still goes deeper during a drop. The correct form for a drop is derived by recognizing that a drop is simply a build traversed in reverse:

$$\Delta TVD_\text{drop} = R(\sin \alpha_1 - \sin \alpha_2)$$
$$\Delta HD_\text{drop} = R(\cos \alpha_2 - \cos \alpha_1)$$

Both ΔTVD and ΔHD are positive for a drop in the 0–90° inclination range.

```{admonition} Warning: Degree Mode
:class: warning
Always use degree mode when evaluating sin and cos in these formulas. The angles $\alpha_1$ and $\alpha_2$ are wellbore inclinations in degrees. Entering degrees into a calculator set to radians will give completely wrong answers.
```

### 12.2.3 Tangent Section

A tangent section holds constant inclination $\alpha$ for a measured length $L$:

$$\boxed{\Delta TVD_\text{tangent} = L \cos \alpha}$$

$$\boxed{\Delta HD_\text{tangent} = L \sin \alpha}$$

### 12.2.4 Converting HD to Northing and Easting

The horizontal departure ΔHD lies in the horizontal plane and points in the direction given by the well azimuth AZ. Azimuth is measured clockwise from north. Therefore:

$$\Delta N = \Delta HD \cdot \cos(\text{AZ})$$

$$\Delta E = \Delta HD \cdot \sin(\text{AZ})$$

---

## 12.3 Worked Example: 2D Build-and-Hold Well

```{figure} ../figures/ch12/fig_survey_example_problem.png
:name: fig-survey-example-problem
:width: 85%
The planned well for this worked example. A build-and-hold well with kickoff at 5,000 ft, DLS = 5°/100 ft, tangent at 45°, pointing southeast (azimuth 135°).
```

**Given:** A build-and-hold well has the following plan:

- Kickoff point (KOP): 5,000 ft MD (well is vertical from surface to KOP)
- DLS: 5°/100 ft
- Target inclination: 45°
- Tangent length: 5,000 ft (after end of build)
- Azimuth: 135° (southeast) throughout the deviated section

**Required:** Construct the complete wellbore position table: MD, INC, AZ, TVD, HD, N, E at each key station.

---

### Step 1: Construct the Survey Table (MD, INC, AZ)

| Station | MD (ft) | INC (°) | AZ (°) |
|---------|---------|---------|--------|
| A — Surface | 0 | 0 | 0 |
| B — KOP | 5,000 | 0 | 0 |
| C — End of Build (EOB) | ? | 45 | 135 |
| D — TD | ? | 45 | 135 |

**Finding MD at End of Build (Station C):**

$$\text{MD}_\text{build} = \frac{45^\circ - 0^\circ}{5^\circ/100\text{ft}} \times 100 = 900 \text{ ft}$$

$$\text{MD}_C = 5000 + 900 = 5900 \text{ ft}$$

**Finding MD at TD (Station D):**

$$\text{MD}_D = 5900 + 5000 = 10{,}900 \text{ ft}$$

Updated survey table:

| Station | MD (ft) | INC (°) | AZ (°) |
|---------|---------|---------|--------|
| A | 0 | 0 | 0 |
| B — KOP | 5,000 | 0 | 0 |
| C — EOB | 5,900 | 45 | 135 |
| D — TD | 10,900 | 45 | 135 |

---

### Step 2: Calculate Position Increments for Each Section

**Section A→B (Vertical, 5,000 ft):**

$$\Delta TVD = 5000 \text{ ft}, \quad \Delta HD = 0$$

**Section B→C (Build, 0° to 45°, DLS = 5°/100 ft):**

$$R = \frac{5729.6}{5} = 1145.9 \text{ ft}$$

$$\Delta TVD = 1145.9(\sin 45° - \sin 0°) = 1145.9 \times 0.7071 = 810.4 \text{ ft}$$

$$\Delta HD = 1145.9(\cos 0° - \cos 45°) = 1145.9(1 - 0.7071) = 335.6 \text{ ft}$$

**Section C→D (Tangent, 45°, 5,000 ft):**

$$\Delta TVD = 5000 \cos 45° = 5000 \times 0.7071 = 3535.5 \text{ ft}$$

$$\Delta HD = 5000 \sin 45° = 5000 \times 0.7071 = 3535.5 \text{ ft}$$

---

### Step 3: Sum Increments to Get Cumulative Position

```{figure} ../figures/ch12/fig_survey_calc_table.png
:name: fig-survey-calc-table
:width: 90%
Calculation table showing cumulative TVD, HD, northing, and easting at each survey station.
```

Starting from Station A at (TVD, HD, N, E) = (0, 0, 0, 0):

| Station | MD (ft) | INC (°) | AZ (°) | TVD (ft) | HD (ft) | N (ft) | E (ft) |
|---------|---------|---------|--------|----------|---------|--------|--------|
| A | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| B — KOP | 5,000 | 0 | 0 | 5,000 | 0 | 0 | 0 |
| C — EOB | 5,900 | 45 | 135 | 5,810.4 | 335.6 | −237.4 | 237.4 |
| D — TD | 10,900 | 45 | 135 | 9,345.9 | 3,871.1 | −2,737.4 | 2,737.4 |

**Converting HD to N and E at each station** using AZ = 135°:

At Station C:
$$\Delta N = 335.6 \cos 135° = 335.6 \times (-0.7071) = -237.4 \text{ ft}$$
$$\Delta E = 335.6 \sin 135° = 335.6 \times 0.7071 = +237.4 \text{ ft}$$

At Station D (total HD = 3,871.1 ft):
$$N = 3871.1 \cos 135° = -2737.4 \text{ ft}$$
$$E = 3871.1 \sin 135° = +2737.4 \text{ ft}$$

The negative northing confirms the well is traveling south of the wellhead; the positive easting confirms it is traveling east. Together, southeast — consistent with AZ = 135°.

```{figure} ../figures/ch12/fig_survey_calc_results.png
:name: fig-survey-calc-results
:width: 90%
Final wellbore position table with TVD, HD, northing, and easting at each station. The side view and map view can be drawn from these results.
```

---

## 12.4 Worked Example: Reading a Survey and Drawing the Well

**Given:** The following wellbore survey is provided for an existing well:

| MD (ft) | INC (°) | AZ (°) |
|---------|---------|--------|
| 0 | 0 | 0 |
| 3,000 | 0 | 0 |
| 3,400 | 40 | 270 |
| 6,000 | 40 | 270 |
| 6,400 | 0 | 90 |
| 9,000 | 0 | 90 |

**Required:** (a) Describe the well path. (b) Calculate TVD, HD, N, and E at each station. (c) Sketch the side view and map view.

---

**Step 1: Identify sections:**

- MD 0→3,000: Vertical (INC = 0)
- MD 3,000→3,400: Build from 0° to 40° over 400 ft
- MD 3,400→6,000: Tangent at 40°, AZ 270° (due west)
- MD 6,000→6,400: Drop from 40° to 0° over 400 ft
- MD 6,400→9,000: Vertical again

This is an **S-type** well.

---

**Step 2: Compute increments:**

DLS for build and drop: $\frac{40°}{400 \text{ ft}} \times 100 = 10°/100 \text{ ft}$

$$R = \frac{5729.6}{10} = 572.96 \text{ ft}$$

**Section 1 (Vertical, 3,000 ft):**
$\Delta TVD = 3000$, $\Delta HD = 0$

**Section 2 (Build, 0°→40°):**
$$\Delta TVD = 572.96(\sin 40° - \sin 0°) = 572.96 \times 0.6428 = 368.2 \text{ ft}$$
$$\Delta HD = 572.96(\cos 0° - \cos 40°) = 572.96(1 - 0.7660) = 134.1 \text{ ft}$$
(Pointing west: AZ = 270°, so $\Delta N = 134.1 \cos 270° = 0$, $\Delta E = 134.1 \sin 270° = -134.1$ ft)

**Section 3 (Tangent, 40°, 2,600 ft, AZ 270°):**
$$\Delta TVD = 2600 \cos 40° = 2600 \times 0.7660 = 1991.6 \text{ ft}$$
$$\Delta HD = 2600 \sin 40° = 2600 \times 0.6428 = 1671.3 \text{ ft}$$
($\Delta N = 1671.3 \cos 270° = 0$, $\Delta E = 1671.3 \sin 270° = -1671.3$ ft)

**Section 4 (Drop, 40°→0°):**
Using the drop form of the formula ($\alpha_1 = 40°$, $\alpha_2 = 0°$):

$$\Delta TVD_\text{drop} = 572.96(\sin 40° - \sin 0°) = 572.96 \times 0.6428 = 368.2 \text{ ft}$$
$$\Delta HD_\text{drop} = 572.96(\cos 0° - \cos 40°) = 572.96 \times 0.2340 = 134.1 \text{ ft}$$

```{admonition} Key Concept: Build vs. Drop Formula
:class: tip
Both build and drop sections are circular arcs. For a build from $\alpha_1$ to $\alpha_2$ ($\alpha_2 > \alpha_1$):
$$\Delta TVD = R(\sin \alpha_2 - \sin \alpha_1) \quad \Delta HD = R(\cos \alpha_1 - \cos \alpha_2)$$

For a drop from $\alpha_1$ to $\alpha_2$ ($\alpha_2 < \alpha_1$, both in 0°–90°):
$$\Delta TVD = R(\sin \alpha_1 - \sin \alpha_2) \quad \Delta HD = R(\cos \alpha_2 - \cos \alpha_1)$$

In both cases, $\Delta TVD > 0$ (the well goes deeper) and $\Delta HD > 0$ (the well moves further from the surface location).
```

**Section 5 (Vertical, 2,600 ft):**
$\Delta TVD = 2600$, $\Delta HD = 0$

---

**Step 3: Cumulative position table:**

| Station | MD (ft) | INC (°) | TVD (ft) | HD (ft) | N (ft) | E (ft) |
|---------|---------|---------|----------|---------|--------|--------|
| Surface | 0 | 0 | 0 | 0 | 0 | 0 |
| KOP | 3,000 | 0 | 3,000 | 0 | 0 | 0 |
| EOB | 3,400 | 40 | 3,368.2 | 134.1 | 0 | −134.1 |
| EOT | 6,000 | 40 | 5,359.8 | 1,805.4 | 0 | −1,805.4 |
| EOD | 6,400 | 0 | 5,728.0 | 1,939.5 | 0 | −1,939.5 |
| TD | 9,000 | 0 | 8,328.0 | 1,939.5 | 0 | −1,939.5 |

The well moves due west (N = 0, E negative throughout) and arrives at a horizontal distance of 1,939.5 ft west of the wellhead, at a TVD of 8,328 ft.

In map view, the well traces a straight line due west. In side view, it shows the characteristic S-shape: vertical → build → tangent → drop → vertical.

---

## 12.5 Practice Problems

**Problem 12.1** — A well is planned with the following description: vertical to 6,000 ft, build at 4°/100 ft to 50°, then hold 50° for 8,000 ft measured depth. The azimuth throughout the deviated section is 45° (northeast).

(a) Construct the MD–INC–AZ survey table for key stations.
(b) Calculate ΔTVD and ΔHD for each section.
(c) Build the complete position table (TVD, HD, N, E) at each station.

```{dropdown} Solution
**(a) Survey table:**

Build arc length: $(50° - 0°) / (4°/100 \text{ ft}) \times 100 = 1250 \text{ ft}$

| Station | MD (ft) | INC (°) | AZ (°) |
|---------|---------|---------|--------|
| Surface | 0 | 0 | 0 |
| KOP | 6,000 | 0 | 0 |
| EOB | 7,250 | 50 | 45 |
| TD | 15,250 | 50 | 45 |

**(b) Section increments:**

$R = 5729.6 / 4 = 1432.4 \text{ ft}$

Vertical (A→KOP): $\Delta TVD = 6000$, $\Delta HD = 0$

Build (KOP→EOB):
$$\Delta TVD = 1432.4(\sin 50° - \sin 0°) = 1432.4 \times 0.7660 = 1097.2 \text{ ft}$$
$$\Delta HD = 1432.4(\cos 0° - \cos 50°) = 1432.4(1 - 0.6428) = 511.4 \text{ ft}$$

Tangent (EOB→TD): $L = 8000$ ft
$$\Delta TVD = 8000 \cos 50° = 8000 \times 0.6428 = 5142.3 \text{ ft}$$
$$\Delta HD = 8000 \sin 50° = 8000 \times 0.7660 = 6128.4 \text{ ft}$$

**(c) Position table (AZ = 45°: $\cos 45° = \sin 45° = 0.7071$):**

| Station | TVD (ft) | HD (ft) | N (ft) | E (ft) |
|---------|----------|---------|--------|--------|
| Surface | 0 | 0 | 0 | 0 |
| KOP | 6,000 | 0 | 0 | 0 |
| EOB | 7,097.2 | 511.4 | 361.5 | 361.5 |
| TD | 12,239.5 | 6,639.8 | 4,694.0 | 4,694.0 |
```

**Problem 12.2** — A wellbore survey is given:

| MD (ft) | INC (°) | AZ (°) |
|---------|---------|--------|
| 0 | 0 | 0 |
| 4,500 | 0 | 0 |
| 5,100 | 60 | 180 |
| 9,100 | 60 | 180 |

(a) Identify the profile type and azimuth direction.
(b) Compute the DLS of the build section.
(c) Compute TVD, HD, N, and E at each station.

```{dropdown} Solution
**(a) Profile type:** Build-and-hold, pointing due south (AZ = 180°).

**(b) DLS:**
$$\text{DLS} = \frac{60°}{600 \text{ ft}} \times 100 = 10°/100 \text{ ft}$$

**(c) Calculations:**

$R = 5729.6 / 10 = 572.96 \text{ ft}$

Vertical: $\Delta TVD = 4500$, $\Delta HD = 0$

Build (0°→60°):
$$\Delta TVD = 572.96(\sin 60° - \sin 0°) = 572.96 \times 0.8660 = 496.2 \text{ ft}$$
$$\Delta HD = 572.96(\cos 0° - \cos 60°) = 572.96(1 - 0.5) = 286.5 \text{ ft}$$

Tangent (60°, 4,000 ft):
$$\Delta TVD = 4000 \cos 60° = 2000 \text{ ft}$$
$$\Delta HD = 4000 \sin 60° = 3464.1 \text{ ft}$$

AZ = 180°: $\cos 180° = -1$, $\sin 180° = 0$

| Station | TVD (ft) | HD (ft) | N (ft) | E (ft) |
|---------|----------|---------|--------|--------|
| Surface | 0 | 0 | 0 | 0 |
| KOP | 4,500 | 0 | 0 | 0 |
| EOB | 4,996.2 | 286.5 | −286.5 | 0 |
| TD | 6,996.2 | 3,750.6 | −3,750.6 | 0 |

The well is directly south of the wellhead (N negative, E = 0). The total HD at TD is 3,750.6 ft.
```

**Problem 12.3** — A directional driller is steering a well that is currently at the following position: MD = 7,200 ft, TVD = 6,800 ft, INC = 30°, AZ = 225°, N = −850 ft, E = −850 ft. The plan calls for the well to continue at 30° / 225° for another 1,500 ft of MD.

(a) What are TVD, N, and E at the end of this tangent section?
(b) If the target is at TVD = 9,000 ft, N = −2,500 ft, E = −2,500 ft, is the well on course?

```{dropdown} Solution
**(a) Tangent section increments:**
$$\Delta TVD = 1500 \cos 30° = 1500 \times 0.8660 = 1299.0 \text{ ft}$$
$$\Delta HD = 1500 \sin 30° = 1500 \times 0.5 = 750 \text{ ft}$$

AZ = 225°: $\cos 225° = -0.7071$, $\sin 225° = -0.7071$
$$\Delta N = 750 \times (-0.7071) = -530.3 \text{ ft}$$
$$\Delta E = 750 \times (-0.7071) = -530.3 \text{ ft}$$

New position: TVD = 6800 + 1299 = **8,099 ft**; N = −850 − 530 = **−1,380 ft**; E = −850 − 530 = **−1,380 ft**

**(b) On course check:**

At the end of the tangent, TVD = 8,099 ft vs. target TVD = 9,000 ft — still 901 ft short vertically.  
N = −1,380 ft vs. target −2,500 ft; E = −1,380 ft vs. target −2,500 ft — still 1,120 ft short horizontally in each direction.

The well is not at target yet — it still needs to drill further. Whether it is *on the planned trajectory* requires knowing the planned well path to this point, which is not given in this problem. What we can say is that the well is 901 ft of TVD and about 1,584 ft of total horizontal distance away from the target.
```

**Problem 12.4** — A horizontal well must be drilled to a target at TVD = 8,000 ft, 6,500 ft north of the wellhead. The DLS for the build is 7°/100 ft. Assume the well builds from 0° to 90°.

(a) What is the TVD gained through the build?
(b) What is the horizontal departure gained through the build?
(c) If the well holds 90° after the build, what is the remaining horizontal departure needed to reach the target?
(d) What is the total tangent length needed?

```{dropdown} Solution
$R = 5729.6 / 7 = 818.5 \text{ ft}$

**(a) TVD through build** (0°→90°):
$$\Delta TVD_\text{build} = R(\sin 90° - \sin 0°) = 818.5 \times 1 = 818.5 \text{ ft}$$

**(b) HD through build** (0°→90°):
$$\Delta HD_\text{build} = R(\cos 0° - \cos 90°) = 818.5(1 - 0) = 818.5 \text{ ft}$$

**(c) Remaining to target:**

If TVD at EOB = 8,000 ft, then the KOP is at TVD = 8,000 − 818.5 = **7,181.5 ft**.

Remaining HD after build: 6,500 − 818.5 = **5,681.5 ft**

**(d) Tangent length** (horizontal section at 90° — $\Delta TVD = 0$, $\Delta HD = L$):
$$L = \Delta HD_\text{needed} = 5681.5 \text{ ft}$$

Total well design:
- Drill vertically to 7,181.5 ft TVD
- Build at 7°/100 ft to 90° (arc length = $90/7 \times 100 = 1285.7$ ft MD, TVD gain 818.5 ft)
- Drill horizontal for 5,681.5 ft to reach target

Total MD ≈ 7,181.5 + 1,285.7 + 5,681.5 = **14,148.7 ft**
```

**Problem 12.5** — A wellbore survey shows the following at MD = 10,000 ft: TVD = 7,500 ft, N = +1,200 ft, E = −800 ft. The surface location is at a true north coordinate of 5,000 ft N, 2,000 ft E on a local coordinate grid.

(a) What are the absolute (grid) coordinates of the downhole point?
(b) What is the total horizontal displacement from wellhead to downhole point?
(c) If the field target box is centered at grid coordinates (6,200 N, 1,200 E) with a radius of 100 ft, is this point inside the target?

```{dropdown} Solution
**(a) Absolute grid coordinates:**

The wellbore survey gives position *relative to the wellhead*. The wellhead is at (5,000 N, 2,000 E).

$$N_\text{abs} = 5000 + 1200 = 6200 \text{ ft N}$$
$$E_\text{abs} = 2000 + (-800) = 1200 \text{ ft E}$$

The downhole point is at **(6,200 N, 1,200 E)** on the grid.

**(b) Total horizontal displacement:**
$$HD = \sqrt{1200^2 + (-800)^2} = \sqrt{1{,}440{,}000 + 640{,}000} = \sqrt{2{,}080{,}000} = 1442 \text{ ft}$$

**(c) Inside target box:**

The target is centered at (6,200 N, 1,200 E) with radius 100 ft. The downhole point is exactly at (6,200 N, 1,200 E) — the center of the target. Distance = **0 ft**. The point is inside the target. ✓
```

---

## 12.6 The Minimum Curvature Method

### 12.6.1 Why a Different Method is Needed

The radius of curvature method is exact for planned well sections, because the engineer specifies a constant DLS for each section. In a real drilled well, however, the survey tool measures inclination and azimuth at discrete stations spaced 100 ft apart (or closer when steering actively). The shape of the wellbore *between* those stations is not known. The calculation method is therefore an interpolation assumption.

The radius of curvature method assumes the well follows a circular arc in a 2D plane between stations — it is exact when azimuth does not change and DLS is truly constant. The **minimum curvature method** (MCM) makes a different, more general assumption: between any two survey stations, the well follows a circular arc in 3D space whose plane is defined by the two survey vectors. This is the industry-standard method for processing actual drilled surveys in three dimensions because it handles simultaneous changes in both inclination and azimuth.

For a well drilled in a single vertical plane (no azimuth change), the two methods give nearly identical results. For real wells where azimuth continuously varies, only the minimum curvature method is reliable.

### 12.6.2 The Dogleg Angle

The first step in the minimum curvature method is computing the total angular change — the **dogleg angle** $\beta$ — between two consecutive survey stations. This is the 3D angle between the two survey direction vectors.

Let station 1 have inclination $\alpha_1$ and azimuth $\phi_1$, and station 2 have inclination $\alpha_2$ and azimuth $\phi_2$. The dogleg angle is:

$$\boxed{\cos\beta = \sin\alpha_1 \sin\alpha_2 \cos(\phi_2 - \phi_1) + \cos\alpha_1 \cos\alpha_2}$$

where all angles are in degrees and $\beta$ is in degrees. Note that:
- If azimuth does not change ($\phi_1 = \phi_2$): $\cos\beta = \cos(\alpha_2 - \alpha_1)$, so $\beta = |\alpha_2 - \alpha_1|$. The dogleg angle equals the inclination change.
- If both inclination and azimuth change simultaneously, $\beta$ is always greater than either change alone.

The dogleg severity (DLS) of the interval is:

$$\text{DLS} = \frac{\beta}{\Delta MD} \times 100 \quad [°/100\text{ft}]$$

### 12.6.3 The Ratio Factor

The **ratio factor** $RF$ accounts for the effect of curvature on the average direction vector through the interval. For a straight interval ($\beta = 0$), the average direction is simply the mid-point between the two stations, and $RF = 1$. For a curved interval, the minimum curvature path requires a correction:

$$\boxed{RF = \frac{2}{\beta_\text{rad}} \tan\frac{\beta}{2}}$$

where $\beta_\text{rad}$ is $\beta$ converted to radians ($\beta_\text{rad} = \beta \times \pi / 180$).

When $\beta \to 0$: $RF \to 1$ (straight section, no curvature correction needed).
When $\beta$ is large: $RF > 1$ (curved section, the path is longer than a straight-line average).

### 12.6.4 Position Increment Formulas

The position increments over the interval from station 1 to station 2 are:

$$\boxed{\Delta TVD = \frac{\Delta MD}{2}(\cos\alpha_1 + \cos\alpha_2) \cdot RF}$$

$$\boxed{\Delta N = \frac{\Delta MD}{2}(\sin\alpha_1 \cos\phi_1 + \sin\alpha_2 \cos\phi_2) \cdot RF}$$

$$\boxed{\Delta E = \frac{\Delta MD}{2}(\sin\alpha_1 \sin\phi_1 + \sin\alpha_2 \sin\phi_2) \cdot RF}$$

Note that $\Delta TVD$, $\Delta N$, and $\Delta E$ are computed directly — there is no separate horizontal departure step. Horizontal departure can be recovered afterward as $HD = \sqrt{N^2 + E^2}$.

### 12.6.5 Worked Example: MCM Applied to the Build-and-Hold Well

Apply the minimum curvature method to the B→C interval (build, 0° to 45°) from the well in Section 12.3. This interval has:

- $\Delta MD = 900$ ft
- $\alpha_1 = 0°$, $\phi_1 = 135°$ (at the KOP, azimuth begins)
- $\alpha_2 = 45°$, $\phi_2 = 135°$

**Step 1: Dogleg angle**

$$\cos\beta = \sin(0°)\sin(45°)\cos(0°) + \cos(0°)\cos(45°) = 0 + 1 \times 0.7071 = 0.7071$$

$$\beta = \cos^{-1}(0.7071) = 45°$$

This confirms that for a 2D build with no azimuth change, $\beta$ equals the inclination change — as expected.

**Step 2: Ratio factor**

$$\beta_\text{rad} = 45 \times \frac{\pi}{180} = 0.7854 \text{ rad}$$

$$RF = \frac{2}{0.7854} \tan\frac{45°}{2} = 2.546 \times \tan 22.5° = 2.546 \times 0.4142 = 1.0548$$

**Step 3: Position increments**

$$\Delta TVD = \frac{900}{2}(\cos 0° + \cos 45°) \times 1.0548 = 450 \times 1.7071 \times 1.0548 = 810.2 \text{ ft}$$

$$\Delta N = \frac{900}{2}(\sin 0° \cos 135° + \sin 45° \cos 135°) \times 1.0548$$
$$= 450 \times (0 + 0.7071 \times (-0.7071)) \times 1.0548 = 450 \times (-0.5) \times 1.0548 = -237.3 \text{ ft}$$

$$\Delta E = \frac{900}{2}(\sin 0° \sin 135° + \sin 45° \sin 135°) \times 1.0548$$
$$= 450 \times (0 + 0.7071 \times 0.7071) \times 1.0548 = 450 \times 0.5 \times 1.0548 = +237.3 \text{ ft}$$

The MCM gives ΔTVD = 810.2 ft and ΔHD = $\sqrt{237.3^2 + 237.3^2}$ = 335.5 ft. Compare with the ROC result of ΔTVD = 810.4 ft and ΔHD = 335.6 ft. The difference is less than 0.2 ft — negligible in this case.

The two methods agree closely for 2D surveys with constant DLS between stations. Differences become significant only when azimuth changes simultaneously with inclination over a long interval.

### 12.6.6 When the Methods Diverge

Consider a survey interval where both inclination and azimuth change:
- $\alpha_1 = 20°$, $\phi_1 = 0°$ (north)
- $\alpha_2 = 35°$, $\phi_2 = 45°$ (northeast)
- $\Delta MD = 300$ ft

The radius of curvature method applied in 2D (treating only the inclination change) would miss the azimuthal component entirely. The minimum curvature method computes:

$$\cos\beta = \sin 20° \sin 35° \cos 45° + \cos 20° \cos 35°$$
$$= 0.3420 \times 0.5736 \times 0.7071 + 0.9397 \times 0.8192 = 0.1387 + 0.7699 = 0.9086$$
$$\beta = \cos^{-1}(0.9086) = 24.6°$$

This dogleg angle (24.6°) is larger than either the inclination change (15°) or the azimuth change (45° weighted by inclination). The ratio factor and 3D position increments correctly account for the combined steering action.

Industry survey processing software always uses the minimum curvature method. When designing a well or checking a survey by hand, the radius of curvature method is appropriate for 2D profiles and for quick estimates; the minimum curvature method is required for rigorous 3D position computation.

---

## 12.7 Introduction to Torque and Drag

### 12.7.1 The Problem

In a perfectly vertical well, the drill string hangs vertically and its weight is supported by the hook at surface. The only force on the drill string (besides gravity) is the hydrostatic buoyancy from the mud. There is no contact between the drill string and the borehole wall, and therefore no friction.

In a deviated well, the situation changes fundamentally. The drill string, following a curved and inclined path, presses against the borehole wall at every bend, every stabilizer contact, and throughout the tangent sections where the string rests on the low side of the hole. These contact points generate a normal force $N$ perpendicular to the string. Friction at each contact point generates forces that oppose the string's motion along the borehole.

This friction manifests in two forms:
- **Drag**: axial friction opposing the string's movement along the borehole — pulling it back when tripping out, holding it back when tripping in. Drag increases the hook load when tripping out and decreases it when tripping in compared to the buoyed string weight.
- **Torque**: rotational friction opposing the string's rotation. As the drill string rotates, friction at each contact point exerts a torque in the opposite direction. This torque must be supplied by the top drive and is measured at surface as surface torque.

### 12.7.2 Where Contact Forces Originate

Contact force arises at two types of locations:

**Doglegs**: At any curvature in the wellbore, the tensioned drill string pushes against the outer wall of the bend. The contact force at a dogleg is proportional to the axial tension in the string and the dogleg severity:

$$N_\text{dogleg} \propto T \cdot \text{DLS}$$

Higher tension (deeper wells, heavier strings) and sharper bends (higher DLS) both increase the contact force and therefore the friction.

**Tangent sections**: In an inclined straight section, the drill string rests on the low side of the borehole under its own buoyed weight per unit length. The normal force per unit length is proportional to the component of the pipe weight perpendicular to the borehole axis:

$$\frac{dN}{ds} = w_b \sin\alpha$$

where $w_b$ is the buoyed weight per unit length and $\alpha$ is the inclination. At 90° (horizontal), the full weight of the pipe presses on the borehole wall. This is why horizontal wells — which have long sections at high inclination — experience the most severe drag and torque.

### 12.7.3 Practical Implications

Torque and drag set real physical limits on how far and how deep a deviated well can be drilled:

**Extended-reach drilling (ERD)**: In wells with very long horizontal sections (sometimes 10,000–20,000 ft or more), the cumulative drag on the drill string can exceed the buckling load of the pipe, causing the string to buckle inside the wellbore. Once buckled, the drill string locks up and cannot be advanced further — the drill-ahead limit has been reached. The ability to land the well at the target distance is constrained as much by torque and drag as by the reservoir depth.

**Casing running**: After drilling the hole, casing must be run to bottom. Casing is heavier, stiffer, and has no bottom-hole assembly to center it. In a tortuous hole (one with a borehole pattern from bent-motor drilling), casing running drag can be severe enough that the casing cannot reach TD under its own weight. Centralizers help by reducing contact force, but their placement must be designed to balance drag against cementing quality.

**Trip monitoring**: On a well with significant drag, the hook load during tripping is not equal to the string weight. When tripping out, the hook load is higher than the string weight (drag adds to the pull); when tripping in, it is lower (drag opposes the motion). The difference between pick-up weight and slack-off weight is called the **friction factor** and is used to calibrate torque and drag models. A suddenly high friction factor (drag increasing unexpectedly) is a warning sign of a developing problem — a ledge, a key seat, or a section of hole that is starting to pack off.

### 12.7.4 The Soft-String Model

The simplest torque and drag model, the **soft-string model**, treats the drill string as a cable (flexible, with no bending stiffness). At each point along the string, the model computes the normal force from the local inclination and tension, then applies a Coulomb friction factor $\mu$ to compute the friction force. The model is integrated from the bit to surface, accumulating the drag and torque contributions of each section.

For a tangent section of length $\Delta L$ at inclination $\alpha$:

$$\Delta\text{Drag} = \mu \cdot w_b \sin\alpha \cdot \Delta L$$

$$\Delta\text{Torque} = \mu \cdot w_b \sin\alpha \cdot \Delta L \cdot r$$

where $r$ is the pipe radius. The friction factor $\mu$ is typically 0.15–0.25 for water-based mud in a clean wellbore and higher in oil-based mud or in a rough hole.

The soft-string model is adequate for most planning purposes and is widely used for preliminary ERD screening, casing running design, and trip load prediction. More advanced **stiff-string models** incorporate the bending stiffness of the pipe, which becomes significant at doglegs and in very stiff BHA components, and are used for wells where the soft-string model overpredicts drag because it ignores the stiffening effect.

### 12.7.5 Design Strategies

The drilling engineer uses torque and drag analysis to make design choices that minimize friction:

- **Lower DLS in the build section**: The smoother the curve, the lower the contact force at the doglegs. Rotary steerable systems (which produce smooth, continuously curved wellbores) generate substantially less torque and drag than bent-motor assemblies (which create periodic tortuosity).
- **Mud type**: Oil-based mud typically has a lower friction factor against steel than water-based mud, which is one reason OBM is preferred for long ERD wells even when the formation would otherwise permit WBM.
- **Pipe size and weight**: Lighter drill pipe (lower weight per unit length) reduces the normal force on tangent sections. However, lighter pipe also has less axial strength, which limits how deep the well can be drilled in tension.
- **Lubricants and friction reducers**: WBM systems can incorporate lubricant additives (graphite, glass beads, synthetic lubricants) to reduce the effective friction factor when OBM cannot be used.
- **Short trip intervals**: On high-drag wells, short trips (pulling back 1–2 stands and back to bottom) are used periodically to redistribute cuttings beds and prevent progressive packing that would increase drag over time.

---

## Chapter Summary

This chapter presented the radius of curvature method for 2D wellbore survey calculations. The procedure is straightforward: identify the section type (vertical, build/drop, or tangent), apply the appropriate formula to compute ΔTVD and ΔHD, convert horizontal departures to northing and easting using the azimuth, and sum incrementally to build the position table.

The three formulas to commit to memory are:

| Section type | ΔTVD | ΔHD |
|---|---|---|
| Vertical | $L$ | $0$ |
| Build ($\alpha_1 \to \alpha_2$, $\alpha_2 > \alpha_1$) | $R(\sin\alpha_2 - \sin\alpha_1)$ | $R(\cos\alpha_1 - \cos\alpha_2)$ |
| Drop ($\alpha_1 \to \alpha_2$, $\alpha_1 > \alpha_2$) | $R(\sin\alpha_1 - \sin\alpha_2)$ | $R(\cos\alpha_2 - \cos\alpha_1)$ |
| Tangent (inclination $\alpha$) | $L\cos\alpha$ | $L\sin\alpha$ |

Where $R = 5729.6 / \text{DLS}$ and $L$ is the measured section length. Once position is known, the relationship between HD and N, E is:

$$\Delta N = \Delta HD \cos(\text{AZ}), \quad \Delta E = \Delta HD \sin(\text{AZ})$$

The minimum curvature method is the industry standard for processing real drilled surveys. It computes the 3D dogleg angle $\beta$ from the inclination and azimuth at both endpoints of each interval, applies a ratio factor $RF$ to account for path curvature, and calculates TVD, northing, and easting increments directly without a separate horizontal departure step. For 2D wells (no azimuth change), MCM and the radius of curvature method give essentially identical results; MCM is essential when azimuth changes between survey stations.

Torque and drag arise because the deviated drill string contacts the borehole wall at every bend and rests on the low side of inclined sections. Contact forces create friction that increases the hook load during tripping out, decreases it during tripping in, and generates surface torque during rotation. Torque and drag grow with increasing inclination, longer horizontal sections, higher dogleg severity, and greater string weight. Managing these forces through well path design, mud selection, and real-time monitoring is a central task of the drilling engineer on any deviated well.

Chapter 13 develops both topics quantitatively: the minimum curvature method applied to full 3D survey processing, and the Johancsik soft-string model that turns the qualitative torque and drag picture of Section 12.7 into element-by-element tension and torque predictions.