PETE 411 Well Drilling Lesson 35 Wellbore Surveying Methods 1 Wellbore Surveying Methods Average Angle Balanced Tangential Minimum Curvature Radius of Curvature Tangential Other Topics Kicking off from Vertical Controlling Hole Angle 2 Read: Applied Drilling Engineering, Ch.8 (~ first 20 pages) Projects: Due Monday, December 9, 5 p.m. ( See comments on previous years’ design projects ) 3 Homework Problem #18 Balanced Cement Plug Due Friday, December 6 4 I, A, DMD 5 Example - Wellbore Survey Calculations The table below gives data from a directional survey. Survey Point A B C D Measured Depth along the wellbore ft 3,000 Inclination Angle I, deg 0 3,200 3,600 4,000 Azimuth Angle A, deg 20 6 14 24 6 20 80 Based on known coordinates for point C we’ll calculate the coordinates of point D using the above information. 6 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 Dz C N (y) N Dz D E (x) D Dy Dx 7 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 8 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, DMD 400 ft IC 14 A C 20 ID 24 A D 80 9 The Average Angle Method Measured depth from C to D, DMD 400 ft IC 14 A C 20 ID 24 A D 80 C N (y) Dz C D N Dz Dy D E (x) Dx 10 The Average Angle Method 11 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 the entire survey interval (DMD) to be tangent to the average angle. IAVG A AVG I1 I2 2 A1 A 2 2 DEast DMD sin IAVG sin A AVG DNorth DMD sin IAVG cos A AVG DVert DMD cos IAVG From: API Bulletin D20. Dec. 31, 1985 12 The Average Angle Method IAVG A AVG IC ID 14 24 19 2 2 A C AD 20 80 50 2 2 DEast DMD sin IAVG sin A AVE Dx 400 sin19 sin 50 Dx 99.76 ft 13 The Average Angle Method DNorth DMD sin IAVG cos A AVG Dy 400 sin19 cos 50 Dy 83.71 ft DVert 400 cos IAVG Dz 400 cos19 Dz 378.21 ft 14 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 15 The Balanced Tangential Method This method treats half the measured distance (DMD/2) as being tangent to I1 and A1 and the remainder of the measured distance (DMD/2) as being tangent to I2 and A2. DEast DMD sin I1 sin A1 sin I2 sin A 2 2 DNorth DMD sin I1 cos A1 sin I2 cos A 2 2 DMD cos I2 cos I1 DVert 2 From: API Bulletin D20. Dec. 31, 1985 16 The Balanced Tangential Method DMD sin IC sin AC sin ID sin AD DEast 2 400 sin 14 o sin 20o sin 24 o sin 80o 2 Dx 96.66 ft 17 The Balanced Tangential Method DMD sin IC cos AC sin ID cos AD DNorth 2 400 sin 14 o cos 20o sin 24 o cos 80o 2 Dy 59.59 ft 18 The Balanced Tangential Method DMD DVert 2 cos ID cos IC 400 cos 24 o cos 14 o 2 Dz 376.77 ft 19 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 20 Minimum Curvature Method b 21 Minimum Curvature Method This method smooths the two straight-line segments of the Balanced Tangential Method using the Ratio Factor RF. 2 b RF tan b 2 (DL= b and must be in radians) DMD sin I1 sin A1 sin I2 sin A 2 RF DEast 2 DMD sin I1 cos A1 sin I2 cos A 2 RF DNorth 2 DMD cos I1 cos I2 RF DVert 2 22 Minimum Curvature Method The Dogleg Angle, b, is given by: cos b cos ID IC sin IC sin ID 1 cos(AD AC ) cos 24o 14o sin 14o sin 240 1 cos(800 20o ) cos b = 0.9356 b = 20.67 = 0.3608 radians o 23 Minimum Curvature Method The Ratio Factor, 2 b RF tan b 2 20.67o 2 RF tan 0.3608 2 RF 1.0110 24 Minimum Curvature Method DMD sin IC sin AC sin ID sin AD RF DEast 2 400 sin 14 o sin 20o sin 24 o sin 80o 1.0110 2 96.66 * 1.011 97.72 ft Dx 97.72 ft 25 Minimum Curvature Method DMD sin IC cos AC sin ID cos AD RF DNorth 2 400 sin 14 o cos 20o sin 24 o cos 80o 1.0110 2 59.59 * 1.011 60.25 ft Dy 60.25 ft 26 Minimum Curvature Method DMD DVert 2 cos ID cos IC RF 400 cos 24 o cos 14 o 1.0110 2 376.77 * 1.0110 380.91 ft Dz 380.91 ft 27 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 = 3,880.91 ft 28 The Radius of Curvature Method DMD cos IC cos ID cos A C cos AD 180 DEast ID IC AD AC 2 400 cos 14 cos 24 cos 20 cos 80 180 24 14 80 20 o o o o 2 Dx 95.14 ft 29 The Radius of Curvature Method DMD (cos IC cos ID ) (sin AD sin A C ) 180 DNorth (ID IC ) ( AD A C ) 400(cos14 cos 24 )(sin 80 sin 20 (24 14)(80 20) 2 ) 180 2 Dy 79.83 ft 30 The Radius of Curvature Method DMD (sin ID sin IC ) 180 DVert ID IC 400 (sin 24o sin 14o ) 180 24 14 Dz 377.73 ft 31 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 32 The Tangential Method Measured depth from C to D, DMD 400 ft IC 14 A C 20 ID 24 A D 80 DEast DMD sin ID sin AD 400 sin24 sin 80 Dx 160.22 ft 33 The Tangential Method DNorth DMD sin ID cos AD 400 sin 24o cos 80o Dy 28.25 ft DVert DMD cos ID 400 cos 24 Dz 365.42 ft 34 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 35 Summary of Results (to the nearest ft) x Average Angle y z 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 1,028 3,865 1,160 36 37 38 Building Hole Angle 39 Holding Hole Angle 40 41 CLOSURE (HORIZONTAL) DEPARTURE LEAD ANGLE 42 b 43 Tool Face Angle 44