PETE 661
Drilling Engineering
Lesson 13
Directional Drilling
Slide 1
Directional Drilling
I II III
KOP
When is it used?
EOC
Type I Wells
Type II Wells Build Build-Hold Continuous
and Hold and Drop Build
Type III Wells
Directional Well Planning & Design
Survey Calculation Methods
Slide 2
Read ADE Ch.8 (Reference)
HW #7
Cementing
due 10-25-02
Slide 3
Inclination Angle
, I
Direction Angle
, A
Slide 4
Slide 5
Max.
Horiz.
Depart.
?
Slide 6
Slide 7
Slide 8
Slide 9
Slide 10
Slide 11
Type I Type II Type III
KOP
EOC
Build Build-Hold Continuous
and Hold and Drop Build
Slide 12
In the BUILD
Section
r x = r (1 - cos I)
y = r sin I
I
y L L = r rad
r
L = r Ideg
180
I
18,000
r
x * BUR
Slide 13
Slide 14
Fig. 8.11
r1 x 3 and r1 r2 x 4
Slide 15
r1 x 3 and r1 r2 x 4 3D Wells
Slide 16
N18E S23E
Azimuth
Angle
N55W
S20W
Slide 17
Slide 18
Example 1: Design of Directional Well
Design a directional well with the following
restrictions:
• Total horizontal departure = 4,500 ft
• True vertical depth (TVD) = 12,500 ft
• Depth to kickoff point (KOP) = 2,500 ft
• Rate of build of hole angle = 1.5 deg/100 ft
• Type I well (build and hold)
Slide 19
Example 1: Design of Directional Well
(i) Determine the maximum hole angle
required.
(ii) What is the total measured depth (MD)?
(MD = well depth measured
along the wellbore,
not the vertical depth)
Slide 20
(i) Maximum
Inclination
Angle
18,000
r1
.
15
r2 0
D 4 D1
12,500 2,500
10,000 ft
Slide 21
(i) Maximum Inclination Angle
D D x 2 ( D D ) 2 2( r r ) x
max 2 tan 1 4 1 4 4 1 1 2 4
2(r1 r2 ) x 4
10,000 4,500 2
10,000 2
2(3,820)4,500
2 tan
-1
2(3,820) 4,500
max 26.3
Slide 22
(ii) Measured Depth of Well
x Build r1 (1 cos )
3,820(1 - cos 26.3 )
395 ft
x Hold 4,500 395
4,105 ft
L Hold sin 4,105
L Hold 9,265 ft
Slide 23
(ii) Measured Depth of Well
MD D1 r1 rad L Hold
26.3
2,500 3,820 9,265
180
MD 13,518 ft
Slide 24
* The actual well path hardly ever coincides with
the planned trajectory
* Important: Hit target within specified radius
Slide 25
What is known?
I1 , I2 , A1 , A2 ,
L=MD1-2
Calculate
= dogleg angle
DLS =L
Slide 26
Slide 27
(20)
Slide 28
Wellbore Surveying Methods
Average Angle
Balanced Tangential
Minimum Curvature
Radius of Curvature
Tangential
Other Topics
Kicking off from Vertical
Controlling Hole Angle
Slide 29
I, A, MD
Slide 30
Example - Wellbore Survey Calculations
The table below gives data from a directional survey.
Survey Point Measured Depth Inclination Azimuth
along the wellbore Angle Angle
ft I, deg A, deg
A 3,000 0 20
B 3,200 6 6
C 3,600 14 20
D 4,000 24 80
Based on known coordinates for point C we’ll calculate
the coordinates of point D using the above information.
Slide 31
Example - Wellbore Survey Calculations
Point C has coordinates:
x = 1,000 (ft) positive towards the east
y = 1,000 (ft) positive towards the north
z = 3,500 (ft) TVD, positive downwards
C N (y) C
N
Z z
D D
y
E (x) x
Slide 32
Example - Wellbore Survey Calculations
I. Calculate the x, y, and z coordinates
of points D using:
(i) The Average Angle method
(ii) The Balanced Tangential method
(iii) The Minimum Curvature method
(iv) The Radius of Curvature method
(v) The Tangential method
Slide 33
The Average Angle Method
Find the coordinates of point D using
the Average Angle Method
At point C, X = 1,000 ft
Y = 1,000 ft
Z = 3,500 ft
Measured depth from C to D, MD 400 ft
I C 14
A C 20
I D 24 A D 80
Slide 34
The Average Angle Method
Measured depth from C to D, MD 400 ft
I C 14
A C 20
I D 24
A D 80
C
N (y)
C
Z D N
z
E (x) y D
x
Slide 35
The Average Angle Method
Slide 36
The Average Angle Method
This method utilizes the average of I1 and I2 as an
inclination, the average of A1 and A2 as a direction, and
assumes all of the survey interval (MD) to be tangent
to the average angle.
From: API Bulletin D20. Dec. 31, 1985 Slide 37
The Average Angle Method
I C I D 14 24
I AVG 19
2 2
AC AD 20 80
AAVG 50
2 2
North MD sin I AVG cos AAVG
400 sin19 cos 50 83.71 ft
Slide 38
The Average Angle Method
East MD sin I AVG sin AAVE
400 sin19 sin 50 99.76 ft
Vert 400 cos I AVG
400 cos19 378.21 ft
Slide 39
The Average Angle Method
At Point D,
X = 1,000 + 99.76 = 1,099.76 ft
Y = 1,000 + 83.71 = 1,083.71 ft
Z = 3,500 + 378.21 = 3,878.21 ft
Slide 40
The Balanced Tangential Method
This method treats half the measured distance (MD/2) as
being tangent to I1 and A1 and the remainder of the
measured distance (MD/2) as being tangent to I2 and A2.
From: API Bulletin D20. Dec. 31, 1985 Slide 41
The Balanced Tangential Method
MD
North (sin I C cos A C sin I D cos A D )
2
400
(sin 14 cos 20 sin 24 cos 80 )
2
59.59 ft
Slide 42
The Balanced Tangential Method
MD
East (sin I C sin A C sin I D sin A D )
2
400
(sin 14 sin 20 sin 24 sin 80 )
2
96.66ft
Slide 43
The Balanced Tangential Method
MD
Vert (cos I D cos I C )
2
400
(cos 24 cos14 ) 376.77ft
2
Slide 44
The Balanced Tangential Method
At Point D,
X = 1,000 + 96.66 = 1,096.66 ft
Y = 1,000 + 59.59 = 1,059.59 ft
Z = 3,500 + 376.77 = 3,876.77 ft
Slide 45
Minimum Curvature Method
Slide 46
Minimum Curvature Method
This method smooths the two straight-line segments of the
Balanced Tangential Method using the Ratio Factor RF.
RF = (2/DL) * tan(DL/2) (DL= and must be in radians)
Slide 47
Minimum Curvature Method
The dogleg angle, , is given by:
Cos cos(I D I C ) sin I C sin I D (1 cos(A D A C ))
cos(24 - 14) - sin14 sin 24 (1 cos(80 20))
0.935609
20.67 0.36082 radians
Slide 48
Minimum Curvature Method
2
The Ratio Factor, RF tan
Z
2
2 20.67
RF * tan 1.01099
0.3608 2
MD
North (sin I C cos A C sin I D cos I D )RF
2
59.59 *1.01099 60.25 ft
Slide 49
Minimum Curvature Method
MD
East (sin I C sin A C sin I D sin A D )RF
2
96.66 *1.01099 97.72 ft
MD
Vert (cos I C cos I D )RF
2
376.77 *1.01099 380.91 ft
Slide 50
Minimum Curvature Method
At Point D,
X = 1,000 + 97.72 = 1,097.72 ft
Y = 1,000 + 60.25 = 1,060.25 ft
Z = 3,500 + 380.91 =3,888.91 ft
Slide 51
The Radius of Curvature Method
2
MD(cos I C cos I D )(sin A D sin A C ) 180
North
(I D I C )(A D A C )
2
400(cos14 cos 24 )(sin 80 sin 20 ) 180
(24 14)(80 20)
79.83 ft
Slide 52
The Radius of Curvature Method
MD (cos cosIDIDcos )D 180
22
MD cos ICI Ccos )(cos
AAC AA
cos 180
East
East C D
(I IDIIC)( A
D C AC )
A D A
D C
2
400(cos14 cos 24 )(cos 20 cos 80 ) 180
(24 14)(80 20)
95.14 ft
Slide 53
The Radius of Curvature Method
MD(sin I D sin I C ) 180
Vert
ID IC
400(sin24 sin 14 ) 180
377.73 ft
24 14
Slide 54
The Radius of Curvature Method
At Point D,
X = 1,000 + 95.14 = 1,095.14 ft
Y = 1,000 + 79.83 = 1,079.83 ft
Z = 3,500 + 377.73 = 3,877.73 ft
Slide 55
The Tangential Method
Measured depth from C to D, MD 400 ft
I C 14
A C 20
I D 24
A D 80
North MD sin I D cos AD
400 sin 24 cos 80 28.25 ft
Slide 56
The Tangential Method
East MD sin I D sin AD
400 sin24 sin 80 160.22 ft
Vert 400 cos I D
400 cos 24 365.42 ft
Slide 57
The Tangential Method
At Point D,
X 1,000 160.22 1,160.22 ft
Y 1,000 28.25 1,028.25 ft
Z 3,500 365.42 3,865.42 ft
Slide 58
Summary of Results (to the nearest ft)
X Y Z
Average Angle 1,100 1,084 3,878
Balanced Tangential 1,097 1,060 3,877
Minimum Curvature 1,098 1,060 3,881
Radius of Curvature 1,095 1,080 3,878
Tangential Method 1,160 1,028 3,865
Slide 59
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Building
Hole Angle
Slide 62
Holding
Hole Angle
Slide 63
Slide 64
CLOSURE
(HORIZONTAL) DEPARTURE
LEAD ANGLE
Slide 65
Slide 66
Tool Face Angle
Slide 67
