Exercise 2

1237.5 ft

V1

A

O

R1

α V2

ft 10230

β

B α

Φ

2 3 1

V4

C

R2

V3

F

V5

α

D E

X

ft 7590

μ

Y = 7590 – 2 * R

Y

W

Z = 2* R = 5729 ft

W = ( 1861 ^ 2+ 5729 ^ 2)^0.5 W= 6023.68 ft μ( = cos-1(1861/6023.68 α º = 72 =

Z

ft = 1861 ft 5729 – 7590 =

Φ

R = 5729.5/BU = 5729/2 = 2864.5 ft W

X = 7590 – 2 * R = 7590 – 5729 = 1861 ft

H

H = (6023.68 ^ 2 – 1861 ^ 2)^0.5 ft 5729 = Φ( = tan -1 ( 1861/ 5729 º 18 = (MD= KOP + Tangent Section + The Two Curves+(v5-v4 Length of the Curve = (α/BU)*100 = (72/2)*100 = 3600 ft Tangent Section = Y = 1861 ft v5-v4)=(10230-1237.5-5729) = 3263.5 ft) Hence, MD= 1237.5+1861+2*3600 + 3263.5 = 13562 ft

X

Intermediate Casing

ft 1237.5 ft 2640

J

)J = R* cos (90-α cos(18) = 2724.3 ft *2864.5 = KOP + J = 1237.5+2724.3 = 3961.8 ft P = cos-1(1402.5/2864.5) = 60.68º

R

P

Length of the new curve = (60.68/2)*100 ft 3034.23 = MD2=1237.5+3034.23 ft = 4271.74 ft X1 = (R^2-1402.5^2 )^0.5 = 2497.67 ft

x1

ft 1402.5

Summary

Max. Inclination reached =α = 72 º Total casing length from the top to the bottom = MD = 13562 ft Length of intermediate casing = MD2= 4271.74 ft Horizontal departure at the casing shoe = x1 = 2497.67 ft