Directional Drilling Calculation Example for J-Profile Well

This article demonstrates how to design the well trajectory in J-shape from the surface location to the required target depth (TD).

Information Given

  • The surface location coordinate of Well-A is 6,543,065.00N 416,695.00E and the target is located at 6,542,213.00N 415,456.00E and the UTM zone is 31N.
  • Kick off depth = 4,200’MD/4,200’TVD
  • Planned build up rate = 2 degree/100 ft
  • Well profile = J-profile (build and hold)

The surface location coordinate of Well-A is 6,543,065.00N 416,695.00E and the target is located at 6,542,213.00N 415,456.00E and the UTM zone is 31N. Therefore, the surface and the target for Well-A can be illustrated is Figure 1.

Figure 1- Well-A surface location in reference to subsurface location

Distance from Surface Location to Target

ΔN = 6,542,213.00 – 6,543,065.00 = – 852 m

ΔE = 415,456.00 – 416,695.00 = -1,239 m

Map Distance = 1503.671 m

Figure 2 – Distance from surface to the target illustration

True Distance

This distance must be applied the scale factor in order to get the true distance.

True Distance (m) = Map Distance ÷ scale factor

True Distance (m) = 1503.671 ÷ 0.999685 = 1504.144 m

Convert from meter to feet

True Distance (ft) = True Distance (m) x 3.281

True Distance (ft) = 1504.144 x 3.281

True Distance (ft) = 4934.86 ft

The true distance is the horizontal displacement referencing from surface location to the target.

Target Azimuth

The target in reference to the surface location is located in 3rd quadrant; therefore the target azimuth is calculated by the following relationship;

Target AZI = 180 + tan-1(ΔE÷ΔN)

Target AZI = 180 + tan-1((-1,239) ÷ (-852))

Target AZI = 235.486 degree

Directional Plan and Calculation

In order to design well trajectory for this well, the illustration (Figure 3) is made to determine all required parameters.

Figure 3 – Well-A Section View Plan

Calculations

Radius build (ft) = (180 ÷ ϖ) × (100 ÷2)

Radius build (ft) = 2,864.79 ft

a = TVDtarget – TVDkop

a = 9,500- 4,200 = 5,300 ft

b = Horizontal Displacement to Target – Radius Build

b = 4,934.86 – 2,864.79= 2070.07 ft

c = (a^2 + b^2)^1/2

c = (5,300^2 + 2,070.07^2)^1/2 = 5,689.92 ft

d= (c^2 – e^2)^1/2

d= (5,689.92^2 – 2,864.79^2)^1/2 = 4916.11 ft

e = Radius build

e = 2,864.79 ft (Radius build)

f = g ÷ cos(θ1)

f = 750÷ cos(51.565) = 1,206.52 ft

g = Planned Depth TVD– Target Depth TVD

g = 10,250 – 9,500 = 750 ft

h = g × tan(θ1)

h = 750 × tan(51.565) = 945.09 ft

θ1 = 180- ?1- α1

θ1 = 180 – 59.769 – 68.665 = 51.565 degree

θ2 = 90- θ1

θ2 = 90- 51.565 = 38.435 degree

Distance build (ft) = Change in Inclination ÷ Build Rate

Inclination increases from 0 degree to 51.57 degree at the end of build

Distance build = 51.57 degree ÷ (2 degree/100ft) = 2578.27 ft

Measured Depth at End of Build = Kick of Depth + Distance Build

Measured Depth at End of Build = 4,200 + 2,578.27 = 6,778.27 ft MD

TVD at End of Build = Kick of Depth TVD + Radius of Build x sin (θ1)

TVD at End of Build = 4,200 + 2864.79x sin (51.57) = 6,444.04 ft TVD

Horizontal departure at EOB = Radius Build × (1- cos(θ1))

Horizontal departure at EOB = 2864.79x × (1- cos(51.57)) = 1083.98 ft

Horizontal Departure at TD = Horizontal Departure at Target + h

Horizontal Departure at TD = 4,934.86 + 945.05 = 5879.91 ft

Figure 4 illustrates all calculated parameters for Well-A

Figure 4 – Well-A Section View with All Calculated Figures

North-South and East-West Calculation

Since there is no change in Azimuth direction, North-South and East-West can be calculated by the following equations

North-South = Horizontal Departure x cos (Wellbore Azimuth)

East-West = Horizontal Departure x sin (Wellbore Azimuth)

North-South and East-West at Kick Off Point (KOP)

At KOP, there is not horizontal departure so North-South and East-West direction are 0.

North-South and East-West at End of Build (EOB)

North-South = Horizontal Departure at EOB x cos (Wellbore Azimuth)

North-South = 1083.98x cos (235.49) = -614.20 ft

East-West = Horizontal Departure at EOB x sin (Wellbore Azimuth)

East-West = 1083.98 x sin (235.49) = -893.18 ft

North-South and East-West at Target

North-South = Horizontal Departure at Target x cos (Wellbore Azimuth)

North-South = 4934.86 x cos (235.49) = -2,796.16 ft

East-West = Horizontal Departure at Target x sin (Wellbore Azimuth)

East-West = 4934.86 x sin (235.49) = -4,066.24 ft

North-South and East-West at Total Depth (TD)

North-South = Horizontal Departure at TD x cos (Wellbore Azimuth)

North-South = 5879.95x cos (235.49) = -3,331.66ft

East-West = Horizontal Departure at TD x sin (Wellbore Azimuth)

East-West = 5879.95x sin (235.49) = -4,844.98ft

Figure 5 shows the plan view of Well-A and Table 1 demonstrates all critical points.

Figure 5 – Plan View of Well-A

Table 1 – Summary for Well-A

Calculation Summary

Well-A is planned for 2-D well design with a J-shape trajectory plan. The well will be drilled vertically to 4,100’ MD/4,100’ TVD. Then, the kick off starts at 4,200’ MD/4,200’ TVD (100 ft below 16” shoe) in a firm shale zone and the designed build up rate is 2 degree/100 ft. End of build depth is 6,778’MD/6,444’TVD. The target depth is 11,694’MD/9,500’ TVD KRB and the target depth of the well is 12,901’MD/10,250’ TVD KRB.

Reference – https://www.linkedin.com/pulse/well-trajectory-j-profile-example-directional-suwat-pongtepupathum

Tagged , , , . Bookmark the permalink.

About DrillingFormulas.Com

Working in the oil field and loving to share knowledge.

4 Responses to Directional Drilling Calculation Example for J-Profile Well

  1. Mike says:

    Hello,
    Thanks for this. Please, what would the azimuth be if the target reference is in the 1st, 2nd or 4th quadrant?

  2. Abdulla says:

    Hello, I hope you guys fix the picture cuz all of them are not showing

Leave a Reply

Your email address will not be published.

This site uses Akismet to reduce spam. Learn how your comment data is processed.