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