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Transcript of Tech Drilling WellSurveyMeth
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Wellbore Surveying Methods
Average Angle Balanced Tangential
Minimum Curvature
Radius of Curvature
Tangential
Other Topics Kicking off from Vertical
Controlling Hole Angle
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Read:
Applied Drilling Engineering, Ch.8
(~ first 20 pages)
Projects:
Due Monday, December 9, 5 p.m.
( See comments on previous years’ design
projects )
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Homework Problem #18
Balanced Cement Plug
Due Friday, December 6
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I, A, MD
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Example - Wellbore Survey Calculations
The table below gives data from a directional survey.
Survey Point Measured Depth Inclination Azimuthalong the wellbore Angle Angle
ft I, deg A, deg
A 3,000 0 20B 3,200 6 6C 3,600 14 20D 4,000 24 80
Based on known coordinates for point C we’ll calculatethe coordinates of point D using the above information.
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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
z
E (x)
N (y)C
Dz
N
D
C
y
x
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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
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The Average Angle Method
Find the coordinates of point D usingthe Average Angle Method
At point C, x = 1,000 ft
y = 1,000 ft
z = 3,500 ft
80 A 24I
20 A 14I
DD
CC
ft400MDD,toCfromdepthMeasured
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The Average Angle Method
80 A 24I
20 A 14I
ft400MDD,toCfromdepthMeasured
DD
CC
z
E (x)
N (y)
C
D
z
N
D
C
y
x
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The Average Angle Method
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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 betangent to the average angle.
From: API Bulletin D20. Dec. 31, 1985
2III 21
AVG
AVGAVG AsinIsinMDEast D
AVGIcosMDVert D
2
AAA 21
AVG
AVGAVG AcosIsinMDNorth D
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192
2414
2
III DCAVG
The Average Angle Method
502
8020
2
AA
ADC
AVG
AVEAVG AsinIsinMDEast D
50sinsin19400x
ft76.99x
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The Average Angle Method
AVGIcos400Vert
cos19400z
AVGAVG AcosIsinMDNorth D
ft71.83y
50cossin19400y
ft21.378z
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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
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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.
From: API Bulletin D20. Dec. 31, 1985
2211 AsinIsinAsinIsin2
MDEast
2211
AcosIsinAcosIsin2
MDNorth
12 IcosIcos
2
MDVert
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The Balanced Tangential Method
DDCC AsinIsinAsinIsin2
MDEast
oooo 80sin24sin20sin14sin2
400
ft66.96x
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The Balanced Tangential Method
DDCC AcosIsinAcosIsin2
MDNorth
oooo 80cos24sin20cos14sin2
400
ft59.59y
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The Balanced Tangential Method
CD IcosIcos2
MDVert
oo 14cos24cos2
400
ft77.376z
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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
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Minimum Curvature Method
b
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Minimum Curvature Method
This method smooths the two straight-line segments
of the Balanced Tangential Method using the Ratio
Factor RF.
(DL= b and must be in radians) 2tan
2RF
b
b
RFAcosIsinAcosIsin2MDNorth 2211
RFAsinIsinAsinIsin2
MDEast 2211
RFIcosIcos2
MDVert 21
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Minimum Curvature Method
)AAcos(1IsinIsinIIcoscos CDDCCD
)2080cos(124sin14sin1424cos o00ooo
cos b = 0.9356
b = 20.67o = 0.3608 radians
The Dogleg Angle, b, is given by:
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Minimum Curvature Method
The Ratio Factor,
2
tan2
RF b
b
2
67.20tan
3608.0
2RF
o
0110.1RF
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Minimum Curvature Method
RFAsinIsinAsinIsin2
MDEast DDCC
0110.180sin24sin20sin14sin2
400 oooo
ft72.97x
ft72.97011.1*66.96
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Minimum Curvature Method
RFAcosIsinAcosIsin2
MDNorth DDCC
ft25.60y
ft25.60011.1*59.59
0110.180cos24sin20cos14sin2
400 oooo
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Minimum Curvature Method
RFIcosIcos2
MDVert CD
0110.114cos24cos2
400 oo
ft91.380z
ft91.3800110.1*77.376
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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
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The Radius of Curvature Method
2
CDCD
DCDC 180
AAII
AcosAcosIcosIcosMDEast
2oooo 180
20801424
80cos20cos24cos14cos400
ft14.59x
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The Radius of Curvature Method
2
CDCD
CDDC 180
)AA()II(
)AsinA(sin)IcosI(cosMDNorth
2180
)2080)(1424(
)20sin80)(sin24cos400(cos14
ft79.83y
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The Radius of Curvature Method
180
II
)IsinI(sinMDVert
CD
CD
ft73.773z
180
1424
)14sin24(sin400 oo
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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
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The Tangential Method
ft400MDD,toCfromdepthMeasured
80 A 24I
20 A 14I
DD
CC
80sinsin24400
DD AsinIsinMDEast D
ft22.160x
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The Tangential Method
DIcosMDVert D
24cos400
ft42.365z
DD AcosIsinMDNorth D
ft25.28y
oo 80cos24sin400
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The Tangential Method
ft3,865.42365.423,500z
ft1,028.2528.251,000 y
ft1,160.22160.221,000x
D,PointAt
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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
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Building
Hole Angle
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Holding
Hole Angle
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CLOSURE
LEAD ANGLE
(HORIZONTAL) DEPARTURE
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b