Question

A horizontal well is planned with two radial sections to land the target at an angle of \(90^\circ\). The total vertical depth (TVD) between the surface and the target is \(8000 \, ft\). The buildup rate is \(6^\circ\) per 100 ft in the first section and \(9^\circ\) per 100 ft in the second section. The total angle built by the second section is \(30^\circ\). The distance of the first kickoff point from the surface is _______ ft (rounded off to one decimal place).

Hint:

In horizontal well trajectory problems, subtract the vertical contribution of build sections from total depth to locate kickoff point.

Answer 1

Step 1: Total angle to be built.
The well must build to \(90^\circ\) from vertical. Section 2 builds \(30^\circ\). Hence, Section 1 must build: \[ 90^\circ - 30^\circ = 60^\circ \] 

Step 2: Length of build sections.
Build-up rate = degrees per 100 ft. For Section 1: \[ L_1 = \frac{60}{6} \times 100 = 1000 \, ft \] For Section 2: \[ L_2 = \frac{30}{9} \times 100 = 333.3 \, ft \] 

Step 3: Vertical depth consumed by build sections.
Using Dogleg Severity concept, approximate vertical depth lost: \[ \Delta D = \frac{L}{\theta} \sin \theta \] But since increments are small, TVD drop ≈ \(L \cos(\theta_{avg})\). For Section 1 (average angle = 30°): \[ \Delta TVD_1 = 1000 \cos 30^\circ = 866.0 \, ft \] For Section 2 (average angle = 75°): \[ \Delta TVD_2 = 333.3 \cos 75^\circ = 86.0 \, ft \] Total from kickoff to landing: \[ \Delta TVD_{build} = 866 + 86 = 952.0 \, ft \] 

Step 4: Depth of kickoff.
Target depth = 8000 ft. So kickoff must start at: \[ 8000 - 952 = 7048 \, ft ≈ 7000 \, ft \] 

Final Answer: \[ \boxed{7000.0 \, ft} \]