Petroleum Engineering

Directional
Drilling
 Projected Trajectory Projected Trajectory
with Left Turn to Hit
Targets

Target 1
Target 2
Target 3

Directional well used to intersect multiple

targets
 N18E S23E
A = 157o

Directional
quadrants and
compass
measurements
N55W
S20W
A = 305o
 Projected Well Path
Lead Angle
Surface
Location
for Well
No. 2

Target at a
Lake TVD
9,659
Plan View
 Horizontal
N View

Vertical
View
We may plan a 2-D well, but we always
get a 3D well (not all in one plane)
 MD,     


MD
 = dogleg
angle   

A curve representing a wellbore between

survey stations A1 and A2
 Directional Drilling
1.Drill the vertical (upper) section of
the hole.
2.Select the proper tools for kicking off to
a non-vertical direction
3.Build angle gradually
 Directional Tools
(i) Whipstock
(ii) Jet Bits
(iii) Downhole motor and bent sub
 Whipstocks

Standard retreivable Circulating Permanent Casing

 Setting a Whipstock
 Small bit used to start
 Apply weight to:
– set chisel point &
– shear pin

 Drill 12’-20’
 Remove whipstock
 Enlarge hole
 Jetting Bit
 Fast and Small Jets

economical
 For soft
formation
 One large - two
small nozzles
 Orient large
nozzle
 Spud
periodically
 Jetting
 Wash out pocket
 Return to normal
drilling
 Survey
 Repeat for more
angle if needed
 Mud Motors
Drillpipe
Non-magnetic
Drill Collar
Bent Sub
Mud Motor
Rotating
Sub
 Increasing Inclination
 Limber assembly
 Near bit stabilizer
 Weight on bit forces
DC to bend to low
side of hole.
 Bit face kicks up
 Hold Inclination
Packed hole
assembly
Stiff assembly
Control bit weight
 Decrease Inclination
 Pendulum effect
 Gravity pulls bit
downward
 No near bit
stabilizer
 Packed Hole Assemblies

Drill
String String String NB
pipe
Stabilizer
Stabilizer Stabilizer Stab
Monel
HW DP Steel DC Steel DC DC
Vertical Calculation Horizontal Calculation
3D View Dog Leg Angle
 Deflecting Wellbore Trajectory
0

270 90

180
 Bottom Hole Location
Direction : N 53
E
D
T iVs D
t a :n c e1: 0 ,20, 05 05 0 f t

E  2,550 sin 53 
 2,037 ft
N  2,550 cos
53 
 1 , 5 3 5 ft
Closure  2,550  E2 N2

E
Closure Direction  t a n - 1   
N
53 o
 Survey Calculation Methods

1. Tangential Method
= Backward Station Method
= Terminal Angle Method

Assumption: Hole will maintain

constant inclination and azimuth
angles between survey points
A Known : Location of
A Distance
AB
Angles I , I
IA A B

Angles A A ,
IB AB
Calculation : VAB 
AB cosIB
B
H
Poor accuracy!!
AB

IB
 AB
Average Angle Method
= Angle Averaging Method

Assumption: Borehole is parallel to the

simple average drift and bearing angles
between any two stations.

Known: Location of A, Distance AB,

Angles I A , IB , A A ,
AB
A (i) Simple enough for field use
(ii) Much more accurate than
IA “Tangential” Method

IB  I A I 
I avg


B 

IAVG 2 

B  AA  A 
A avg  B 

  2 
IAVG
A Average Angle Method
Vertical Plane:
IA
 I A I 
IB I avg


B 
2 
IAVG V AB  avg
ABcosI
B  AB
H AB
sinI av g
IAVG
N Average Angle Method
Horizontal Plane:

 ABsin Iavg
AB H AB

B E  ABsinI a v g sin A a v g
AAVG N  AB sinI a v g cos
N
AA Aa v g
E Z  AB cosI a v g
A
E
Change in position towards the east:
x  E  L  I A I B sin  A A  A B ..(1)
 2   2 
sin  in position towards the north:
Change
y  N  L  I A I B cos  A A  A B  ..(2)
 2   2 
sin 
Changeindepth:
 I A I B  ..(3 )
Z  L c o s 2 
Where L is the measured distance
between the two stations A & B.
Example
The coordinates of a point in a wellbore
are:
x = 1000 ft (easting)
y = 2000 ft (northing)
z = 3000 ft (depth)

At this point (station) a wellbore survey shows

that the inclination is 15 degrees from vertical,
and the direction is 45 degrees east of north.
The measured distance between this station and
the next is 300 ft….
Example
The coordinates of point 1 are:
x1 = 1000 ft (easting)
o
y1 = 2000 ft (northing) I1 = 15
o
z1 = 3000 ft (depth) A1 = 45
L12 = 300 ft

o o
At point 2, I2 = 25 and A2 = 65

Find x ,y and z
Solution
 I1  I 2   15  
   
I avg  2  25 2  20
   45  
A avg  1
  
2

  2  55 65 2 
A  A
H12 = L12 sin Iavg = 300 sin 20 = 103 ft
E = H12 sin Aavg = 103 sin 55 = 84 ft
N = H12 cos Aavg = 103 cos 55 = 59 ft
Z = L12 cos Iavg = 300 cos 20 = 282 ft
Solution - cont’d

E = 84 ft
N = 59 ft
Z = 282 ft

x2 = x1 + E = 1,000 + 84 ft = 1,084 ft
y2 = y1 + N = 2,000 + 59 ft = 2,059 ft
z2 = z1 + Z = 3,000 + 282 ft = 3,282 ft