1. Petroleum Engineering 406
Lesson 20
Directional Drilling
(continued)

2. Lesson 17 - Directional Drilling cont’d
Tool-Face Angle
Ouija Board
Dogleg Severity
Reverse Torque of Mud Motor
Examples

3. Homework:
READ:
Applied Drilling Engineering”, Chapter 8
(to page 375)

4. Tool Face (g)
Solution
Fig : Graphical Ouija Analysis.

5. Initial Inclination = 16o
Over one drilled interval (bit run)
GIVEN:
a = 16o
De = 12o aN = 12o
Tool Face (g)
Solution
New Inclination - 12o
De = 12o
g = ? o b = ? o
Initial Inclination = 16o
Fig : Graphical Ouija Analysis.

6. Fig. 8.33
Basis of chart construction is a trigonometric relationship illustrated by two intersecting planes b a aN
b = dogleg angle De

7. Problem 1
Determine the new direction (eN) for a whipstock set at 705 m with a tool-face setting of 450 degrees right of high side for a course length of 10 m.
The inclination is 70 and the direction is N15W. The curve of the whipstock will cause a total angle change of 30/30 m.

8. Problem 1 a = 7o (inclination) e = 345o (azimuth)
g = 45o (tool face angle)
L = 10 m (course length)
d = 3o/ 30 m (dogleg severity)
eN = ? o
g = 45o

9. Solution to Problem 1, part 1
I. Use Equation 8.43 to calculate
The dogleg severity,

10. Solution to Problem 1, part 2
2. Use Equation 8.42 to calculate the direction change.
New direction = = = N9.7W

11. Problem 2
Determine where to set the tool face angle, for a jetting bit to go from a direction of 100 to 300 and from an inclination of 30 to 50. Also calculate the dogleg severity, assuming that the trajectory change takes 60 ft.
a = 3
e = 10
Find

12. Solution to Problem 2, part 1
1. Find b using Equation 8.53

13. Solution to Problem 2, part 2
2. Now calculate  from equation 8.48.

14. Solution to Problem 2, part 3
3. The dogleg severity,
 = 4.01o / 100 ft
Alternate solution: Use Ouija Board

15. Fig. 8.31: Solution to Example 8.6.

16. Problem 3
Determine the dogleg severity following a jetting run where the inclination was changed from 4.3o to 7.1o and the direction from N89E to S80E over a drilled interval of 85 feet.
1. Solve by calculation.
2. Solve using Ragland diagram
L = 85 ft
Da = = De = = 11

17. Solution to Problem 3, part 1
1. From Equation 8.55
b = 3.01o

18. Solution to Problem 3, part 1
1. From Equation 8.43
the dogleg severity,

19. Solution to Problem 3, part 2
2. Construct line of length  (4.3o)
Measure angle  (11o )
Construct line of length N (7.1o)
Measure length 
(Measure angle )
4.3
11o b
Ragland Diagram
7.1

20. Some Equations to Calculate b

21. Overall Angle Change and Dogleg Severity
Equation 8.51 derived by Lubinski is used to construct Figure 8.32,
a nomograph for determining
the total angle change  and
the dogleg severity, .

22. Fig. 8.32: Chart for determining dogleg severity

23. (a+aN)/2 = 5.7o
aN - a = 2.8o
b = 3o
De = 11o
d = 3.5o/100 ft

24. (a+aN)/2
= 5.7o
De = 11o

25. aN - a = 2.8o
b = 3o

26. b = 3o
d = 3.5o/100 ft

27. (a+aN)/2 = 5.7o
aN - a = 2.8o
b = 3o
De = 11o
d = 3.5o/100 ft

28. Problem 4 - Torque and Twist
1. Calculate the total angle change of 3,650 ft. of 4 1/2 inches (3.826 ” ID) Grade E #/ft drill pipe and 300 ft. of 7” drill collars (2 13/16” ID) for a bit-generated torque of 1,000 ft-lbf. Assume that the motor has the same properties as the 7” drill collars. Shear modules of steel, G = 11.5*106 psi.
2. What would be the total angle change if ,300 ft. of drill pipe were used?

29. Solution to Problem 4
From Equation 8.56,

30. Solution to Problem 4, cont.
radians

31. Solution to Problem 4, cont.
If Length of drillpipe = 7,300 ft.,
M =  * ]
= 4.77 radians *
~ 3/4 revolution!
137.2

32. Example 8.10 Design a kickoff for the wellbore in Fig. 8.35.
e = S48W = 228o eN = N53W = 307o
a = 2o L = 150 ft aN = 6o
De = 79o Find b, g and d
From Ouija Board, b = 5.8o, g = 97o

33. Fig. 8.36: Solution for Example 8.10.
New Direction
Where to Set the Tool Face
b = 5.8o
g = 97o
High Side
High Side
Present Direction
Fig. 8.36: Solution for Example 8.10.

34. Dogleg Severity
From Equation 8.43
the dogleg severity,

35. Fig. 8.36: Solution for Example 8.10.
With jetting bit: 325o
345o
qM = 20o
307o
Fig. 8.36: Solution for Example 8.10.
228o

36. Fig. 8.36: Solution for Example 8.10.
Tool Face Setting
Where to Set the Tool Face
Compensating for Reverse Torque of the Motor
New Direction
Present Direction
High Side
High Side
Fig. 8.36: Solution for Example 8.10.