Profesores:
ING Gustavo Castillo
ING Francisco Rivas Lara
ING José Omar Sánchez
MÉTODOS DE ESTUDIOS DIRECCIONALES
2
I, A, DMD
3
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 calculate the coordinates of point D using the above information.
4
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
Dz
E (x)
N (y)C
DDz
N
D
C
Dy
Dx
5
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
6
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
80 A 24I
20 A 14I
DD
CC
ft 400MD D, to C from depth Measured D
7
The Average Angle Method
80 A 24I
20 A 14I
ft 400MD D, to C from depth Measured
DD
CC
D
Dz
E (x)
N (y)
C
D
Dz
N
D
C
Dy
Dx
8
The Average Angle Method
9
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.
From: API Bulletin D20. Dec. 31, 1985
2
III 21AVG
AVGAVG AsinIsinMDEast DD
AVGIcosMDVert DD
2
AAA 21
AVG
AVGAVG AcosIsinMDNorth DD
10
192
2414
2
III DCAVG
The Average Angle Method
502
8020
2
AAA DC
AVG
AVEAVG AsinIsinMDEast DD
50sinsin19400x D
ft76.99x D
11
The Average Angle Method
AVGIcos400Vert D
cos19400z D
AVGAVG AcosIsinMDNorth DD
ft 71.83y D
50cossin19400y D
ft21.378z D
12
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
13
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
DD
2211 AcosIsinAcosIsin2
MDNorth
DD
12 IcosIcos2
MDVert
DD
14
The Balanced Tangential Method
DDCC AsinIsinAsinIsin2
MDEast
DD
oooo 80sin24sin20sin14sin2
400
ft66.96x D
15
The Balanced Tangential Method
DDCC AcosIsinAcosIsin2
MDNorth
DD
oooo 80cos24sin20cos14sin2
400
ft59.59y D
16
The Balanced Tangential Method
CD IcosIcos2
MDVert
DD
oo 14cos24cos2
400
ft77.376z D
17
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
18
Minimum Curvature Method
b
19
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
RFAcosIsinAcosIsin2
MDNorth 2211
DD
RFAsinIsinAsinIsin2
MDEast 2211
DD
RFIcosIcos2
MDVert 21
DD
20
Minimum Curvature Method
)AAcos(1IsinIsinIIcoscos CDDCCD b
)2080cos(124sin14sin1424cos o00ooo
cos b = 0.9356
b = 20.67o
= 0.3608 radians
The Dogleg Angle, b, is given by:
21
Minimum Curvature Method
The Ratio Factor,
2tan
2RF
b
b
2
67.20tan
3608.0
2RF
o
0110.1RF
22
Minimum Curvature Method
RFAsinIsinAsinIsin2
MDEast DDCC
DD
0110.180sin24sin20sin14sin2
400 oooo
ft72.97x D
ft72.97011.1*66.96
23
Minimum Curvature Method
RFAcosIsinAcosIsin2
MDNorth DDCC
DD
ft25.60y D
ft25.60011.1*59.59
0110.180cos24sin20cos14sin2
400 oooo
24
Minimum Curvature Method
RFIcosIcos2
MDVert CD
DD
0110.114cos24cos2
400 oo
ft91.380z D
ft91.3800110.1*77.376
25
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
26
The Radius of Curvature Method
2
CDCD
DCDC 180
AAII
AcosAcosIcosIcosMDEast
DD
2oooo 180
20801424
80cos20cos24cos14cos400
ft 14.59 x D
27
The Radius of Curvature Method
2
CDCD
CDDC 180
)AA()II(
)AsinA(sin)IcosI(cosMDNorth
DD
2180
)2080)(1424(
)20sin80)(sin24cos400(cos14
ft 79.83 y D
28
The Radius of Curvature Method
DD
180
II
)IsinI(sinMDVert
CD
CD
ft 73.773 z D
180
1424
)14sin24(sin400 oo
29
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
30
The Tangential Method
ft 400MD D, to C from depth Measured D
80 A 24I
20 A 14I
DD
CC
80sinsin24400
DD AsinIsinMDEast DD
ft 22.160x D
31
The Tangential Method
DIcosMDVert DD24cos400
ft 42.365z D
DD AcosIsinMDNorth DD
ft 25.28y D
oo 80cos24sin400
32
The Tangential Method
ft 3,865.42365.423,500z
ft 1,028.2528.251,000 y
ft 1,160.22160.221,000x
D,Point At
33
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