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PETE 411

Well Drilling

Lesson 35

Wellbore Surveying Methods

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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 Azimuth along the wellbore Angle Angle

ft I, deg A, deg

A 3,000 0 20 B 3,200 6 6 C 3,600 14 20 D 4,000 24 80

Based on known coordinates for point C we’ll calculate the 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

yx

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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 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

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The Average Angle Method

80 A 24I

20 A 14I

ft 400MD D, to C from depth Measured

DD

CC

z

E (x)

N (y)

C

D

zN

D

C

yx

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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 (MD) to be tangent to the average angle.

From: API Bulletin D20. Dec. 31, 1985

2

III 21AVG

AVGAVG AsinIsinMDEast

AVGIcosMDVert

2

AAA 21

AVG

AVGAVG AcosIsinMDNorth

13

192

2414

2

III DCAVG

The Average Angle Method

502

8020

2

AAA DC

AVG

AVEAVG AsinIsinMDEast

50sinsin19400x

ft76.99x

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The Average Angle Method

AVGIcos400Vert cos19400z

AVGAVG AcosIsinMDNorth

ft 71.83y

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 (MD/2) as being tangent to I1 and A1 and the remainder of the measured distance (MD/2) as being tangent to I2 and A2.

From: API Bulletin D20. Dec. 31, 1985

2211 AsinIsinAsinIsin2

MDEast

2211 AcosIsinAcosIsin2

MDNorth

12 IcosIcos2

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

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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= and must be in radians)2tan

2RF

RFAcosIsinAcosIsin2

MDNorth 2211

RFAsinIsinAsinIsin2

MDEast 2211

RFIcosIcos2

MDVert 21

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Minimum Curvature Method

)AAcos(1IsinIsinIIcoscos CDDCCD

)2080cos(124sin14sin1424cos o00ooo

cos = 0.9356

= 20.67o = 0.3608 radians

The Dogleg Angle, , is given by:

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Minimum Curvature Method

The Ratio Factor,

2tan

2RF

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

ft 14.59 x

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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

ft 79.83 y

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The Radius of Curvature Method

180

II

)IsinI(sinMDVert

CD

CD

ft 73.773 z

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

ft 400MD D, to C from depth Measured

80 A 24I

20 A 14I

DD

CC

80sinsin24400

DD AsinIsinMDEast

ft 22.160x

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The Tangential Method

DIcosMDVert 24cos400

ft 42.365z

DD AcosIsinMDNorth

ft 25.28y

oo 80cos24sin400

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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

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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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Tool Face Angle