TVD=

2900

0ft

R1=5730 ft

R1

X3=5420 ft

X2

Water depth 4921 ft

D 1=

2500

00 ft

D 2 =

2588

2 ft

ϴ=8.85°

Ϯ=4.4°

=68.2 ft

Kick off point

Ω=13.25°

X =

882

FTO

C

B

D

A

D/

Appendix

Total Vertical Depth (TVD), D3 = 29000 feet

Kick off depth, D1 = 25000 feet

Horizontal departure from target X3 = 5420 feet

Rate of build, q = 1°/100ft

Radius of curvature, R1 = 180π× 1q

R1 = 180π × 100 ft1° = 5730ft

Since X3 ˂ R1, ϴ = Ω – Ϯ

Tan Ϯ = R1 – X3/D3-D1

= 5730-5420/29000-2500

Tan Ϯ = 0.0775

Tan-1 Ϯ = 4.4°

Sin Ω = R1 / D1

= 5730/25000

Sin Ω =0.2292

Sin-1 0.2292 = 13.25°

Maximum inclination angle ϴ = Ω – Ϯ

ϴ = 13.25° - 4.4° = 8.85°

Horizontal departure to end of build X2 = R1 (1 – cos ϴ)

X2 = 5730 (1 – cos 8.85°) = 68.2 ft

LCB = R1/Tan Ω

LCB = 5730/ Tan 13.25° = 24344 ft

LDC = π180 × R1 × ϴ = ϴ/q

LDC = 8.85° × 100 ft/ 1° = 885 ft

Measured depth to end of build, Dm = D1 + LDC + LCB

Dm = 25000 + 885 + 24334 = 50219 ft

Solving for D2 from triangle OCD/

Sin 8.85°= x/R1

x = R1 Sin 8.85° = 5730 × 0.154

x = 882 ft

D2 = x + D1 = 882 + 25000 = 25882 ft

O

C

D/

R1=5730x

8.85°