Tech Drilling WellSurveyMeth

8/11/2019 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



3

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.




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




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



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




80 A 24I

20 A 14I


DD

CC

z

E (x)

N (y)

C

D

z

N

D

C

y

x




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



192

2414

2

III DCAVG


502

8020

2

AA

ADC

AVG

AVEAVG AsinIsinMDEast D

50sinsin19400x

ft76.99x




AVGIcos400Vert

cos19400z

AVGAVG AcosIsinMDNorth D

ft71.83y

50cossin19400y

ft21.378z




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



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




DDCC AsinIsinAsinIsin2

MDEast

oooo 80sin24sin20sin14sin2

400

ft66.96x




DDCC AcosIsinAcosIsin2

MDNorth

oooo 80cos24sin20cos14sin2

400

ft59.59y




CD IcosIcos2

MDVert

oo 14cos24cos2

400

ft77.376z




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



Minimum Curvature Method

b



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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



23


)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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The Ratio Factor,

2

tan2

RF b

b

2

67.20tan

3608.0

2RF

o

0110.1RF



25


RFAsinIsinAsinIsin2

MDEast DDCC

0110.180sin24sin20sin14sin2

400 oooo

ft72.97x

ft72.97011.1*66.96



26


RFAcosIsinAcosIsin2

MDNorth DDCC

ft25.60y

ft25.60011.1*59.59

0110.180cos24sin20cos14sin2

400 oooo



27


RFIcosIcos2

MDVert CD

0110.114cos24cos2

400 oo

ft91.380z

ft91.3800110.1*77.376



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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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2

CDCD

CDDC 180

)AA()II(

)AsinA(sin)IcosI(cosMDNorth

2180

)2080)(1424(

)20sin80)(sin24cos400(cos14

ft79.83y



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180

II

)IsinI(sinMDVert

CD

CD

ft73.773z

180

1424

)14sin24(sin400 oo



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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


80 A 24I

20 A 14I

DD

CC

80sinsin24400

DD AsinIsinMDEast D

ft22.160x



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DIcosMDVert D

24cos400

ft42.365z

DD AcosIsinMDNorth D

ft25.28y

oo 80cos24sin400



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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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42

CLOSURE

LEAD ANGLE

(HORIZONTAL) DEPARTURE



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b



Tool Face Angle

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