Drilling guides › BHA basics: collars, HWDP, stabilizers and weight on bit

BHA basics: collars, HWDP, stabilizers and weight on bit

By the DrillPro team · Last reviewed 9 November 2026

The bottom hole assembly (BHA) is everything between the bit and the drill pipe. It has three jobs: put weight on the bit without putting the drill pipe into compression, control the direction of the hole, and carry the tools that measure and steer. A well designed BHA drills a smooth, in-gauge hole; a poor one gives doglegs, vibration, fatigue failures and stuck pipe. This guide covers the components and the basic weight calculation that every BHA starts from.

Components

ComponentWhat it does
Bit and bit subCuts the rock. The bit sub connects the bit and often carries a float valve.
Mud motor or rotary steerableTurns the bit independently of the string or steers it, in directional wells.
MWD / LWDMeasures inclination, azimuth, tool face and formation properties, and sends them to surface.
StabilizersCentre the BHA in the hole, control the side force on the bit and so the build, hold or drop tendency.
Drill collarsThick-walled, heavy tubes that supply the weight on bit and the stiffness of the BHA.
Jar and acceleratorDeliver impact to free stuck pipe.
Heavyweight drill pipe (HWDP)Transition between stiff collars and flexible drill pipe; extra weight in deviated wells.
Crossovers, shock sub, reamersJoin connections of different types, damp vibration, open the hole.

Drill collars

Collars are sized by outside and inside diameter. Their weight in air per foot follows from the steel cross-section:

Collar weight (lb/ft) ≈ 2.67 × (OD² − ID²) (OD, ID in inches)

The 2.67 is the weight of steel per foot for each square inch of area (about 3.4 lb/ft per in²) multiplied by π/4. Spiral collars weigh a few percent less than slick collars of the same size, so use the supplier's figure where it is known. Spiral grooves reduce the contact area against the wall, which lowers the risk of differential sticking.

The largest collar that can be safely fished in the hole size is usually chosen for the bottom of the BHA, because a stiff, large collar keeps the hole straighter and reduces the bending at the connection between the collar and the bit.

Heavyweight drill pipe

HWDP has the same OD as drill pipe but a thicker wall and longer tool joints, often with a central wear pad. Placed above the collars, it spreads the change in stiffness so that fatigue does not concentrate in the first joint of drill pipe. In high-angle and horizontal wells, HWDP is often run in compression in the vertical or low-angle part of the well to supply weight, because collars lying on the low side of a horizontal section add drag but almost no weight to the bit. That design needs a buckling check and is beyond this basic guide.

Stabilizers and directional tendency

The position of stabilizers decides how the BHA bends under weight. A near-bit stabilizer with the next one a collar or more above acts as a fulcrum and tends to build angle. Several closely spaced stabilizers make a stiff "packed" assembly that holds angle. No near-bit stabilizer, with the first one well above the bit, lets the collars sag and the bit drop angle, like a pendulum. Undergauge stabilizers change these tendencies, which is why stabilizers are gauged every time they come out of the hole.

Why the weight calculation comes first

Directional behaviour, vibration and hydraulics are all tuned later, but they all depend on a BHA that can deliver the weight on bit the bit needs. If the collar section is too short, the driller either cannot reach the planned WOB or reaches it by pushing drill pipe into compression, which is the start of many twist-offs. If it is far too long, the extra collars add torque, drag, cost and more steel in the hole to fish. The simple calculation below gives the right order of length before the finer design begins.

Buckling and the neutral point

Drill pipe is designed to work in tension. In compression, a long slender pipe buckles, first into a snake shape and then into a helix, which raises the contact forces, the wear and the fatigue at the tool joints. Collars are stiff enough to take some compression.

The neutral point is the point in the string where the axial load changes from compression below to tension above. In a vertical well, it sits at the height of buoyed collar weight equal to the weight on bit. The aim of the basic design is to keep the neutral point inside the collars, with a margin, so the drill pipe and the transition joints stay in tension even when the WOB rises or the driller overshoots.

Buoyancy factor (BF) = 1 − MW (ppg) ÷ 65.5 Neutral point above bit (ft) = WOB (lb) ÷ (collar weight (lb/ft) × BF) Collar length needed (ft) = WOB × DF ÷ (collar weight × BF × cos θ)

DF is a design factor, commonly 1.15 to 1.20, which means only about 85% of the buoyed collar weight is used as WOB. θ is the hole inclination; in a deviated hole only the axial component of the weight, the weight times cos θ, pushes on the bit. 65.5 ppg is the density of steel. In metric, BF = 1 − MW (sg) ÷ 7.85.

Worked example 1: vertical well

Planned WOB 40,000 lb, mud 12.0 ppg, 8 in × 2-13/16 in drill collars, design factor 1.15, vertical hole.

Collar weight = 2.67 × (8² − 2.8125²) ≈ 149.8 lb/ft BF = 1 − 12.0 ÷ 65.5 = 0.8168 Buoyed weight = 149.8 × 0.8168 ≈ 122.3 lb/ft Length needed = 40,000 × 1.15 ÷ 122.3 ≈ 376 ft At 30 ft each = 12.5 collars → run 13 collars (390 ft) Neutral point = 40,000 ÷ 122.3 ≈ 327 ft above the bit

With 390 ft of collars, the neutral point sits about 84% of the way up the collars. The top 63 ft of collars and everything above stay in tension. If the driller later wants 45,000 lb, the neutral point moves to about 368 ft, very close to the top of the collars, so the assembly would need more collars or HWDP.

Worked example 2: deviated hole

WOB 25,000 lb, mud 10.0 ppg, 6.5 in × 2.25 in collars, design factor 1.15, tangent section at 30° inclination.

Collar weight = 2.67 × (6.5² − 2.25²) ≈ 99.3 lb/ft BF = 1 − 10.0 ÷ 65.5 = 0.8473 Axial per foot = 99.3 × 0.8473 × cos 30° = 99.3 × 0.8473 × 0.866 ≈ 72.9 lb/ft Length needed = 25,000 × 1.15 ÷ 72.9 ≈ 395 ft At 30 ft each = 13.2 collars → run 14 collars (420 ft)

In a vertical hole the same job needs only about 342 ft. At 30° the collars give 13% less axial weight, and the side force of the collars on the low side adds drag, so the surface weight indicator overstates the real WOB. At higher angles this method stops working, and a torque and drag model is used instead.

Common mistakes

Good practice

Build and check your BHA in DrillPro

DrillPro does these sums for you on Android, the web and desktop, and exports the result to PDF or Excel.