Friday, 25 September 2015

Directional Surveying Calculations (Minimum Curvature Method)

Surveying is an inseparable part of directional drilling.
Surveys are recorded at regular intervals while drilling.

Reasons for Taking Surveys
  1. To allow accurate determination of well coordinates at a series of measured depths and determine the current location.
  2. To plot the well path over the measured depth.
  3. To measure the inclination and azimuth at the bottom of the hole and hence determine where the well is heading.
  4. To determine the orientation of tool face of deflection tools or steerable systems.
  5. To locate dog legs and allow calculation of dogleg severity values.



Accurate Knowledge of the Course of a Borehole is Necessary:
  1. To hit geological target.
  2. To avoid collision with other near by wells.
  3. To define the target of a relief well in the event of a blowout.
  4. To provide a better definition of geological and reservoir data to allow for optimization of production.
  5. To fulfill the requirements of local legislation if any.
For a directional driller, to successfully drill a well to the specified targets, all that's required is inclination and azimuth.
Now a days, there  are many types of advanced tools used and those along with the directional survey, provides the required geophysical characteristics of the well.

There are
  1. MWD (Measurement While Drilling) tools and 
  2. LWD (Logging While Drilling) tools. 
  • As the name suggests, MWD tools mainly measures the values of inclination and azimuth while drilling whereas, LWD tools in addition to it measures geophysical characteristics of the formations encountered while drilling.
  • LWD tools are more advance and sophisticated. Use of these tools eliminate the need of separate wireline logging; thereby saving rig time.
  • Although hiring of LWD tools and engineers are costly as compared to MWD's.


Survey Calculations

Directional survey in terms of 'Inclination' & 'Azimuth' of a wellbore at certain 'Measured Depth' is taken. This information is then used to calculate the actual position of the wellbore relative to the surface location. When assuming an idealized well path between the two survey stations, many mathematical models can be used such as;
  1. Tangential method
  2. Balanced Tangential method
  3. Average Angle method
  4. Radius of Curvature method
  5. Minimum Curvature method

 Among above mathematical models, 
Tangential method is the least accurate, it assumes a straight line well path taking into consideration the inclination and azimuth at upper survey station and lower station is not accounted.

Balanced Tangential Method takes into account the upper and lower survey station and approximates well path by two  equal straight line segments. The upper line segment is defined by inclination and azimuth at upper survey station and the respective values at lower survey station.

Average Angle Method assumes one straight line defined by averaging inclination and azimuth at both survey stations, intersects both upper and lower survey stations.

Radius of Curvature Method assumes that well path is not a straight line but a circular arc tangential to inclination and azimuth at each survey station.

Minimum Curvature Method is the most accurate, it further adds a Ratio Factor to smoothen the spherical arc formed by using radius of curvature method. This is practically used and accepted calculation method among all and deserve to be discussed in detail.

Survey station is the measured depth at which survey is taken.
Course length (CL) is the difference between two survey stations.


Minimum Curvature Method

The inclination and azimuth at each survey station define two vectors namely inclination vector (lying in the vertical plane) and azimuthal vector (lying in the horizontal plane); and both are tangential to the wellbore trajectory. The only other piece of information available from a survey is the course length (the difference in survey measured depths) between the two stations. Minimum Curvature Method most accurately creates idealized well path between the upper and lower stations. The accuracy of the final coordinates generated by it approximates the actual trajectory of the borehole.
Let's explain the formula's used with an example. 

 Consider inclination and azimuth at these two survey stations.

Azimuth of target is 316°.
Determine next set of values?

Sol'n:

 Upper Survey Station (MD1) is at 1914.75m
I1 = 13.6°; A1 = 315.2°; TVD1 = 1827.53m; N/S1 = 311.70m;
E/W1 = -299.27m; VS1 = 432.11m; CD1 = 432.11m; CA1 = 316.17°

 Lower Survey Station (MD2) is at 1940.30m
I1 = 10.7°; A1 = 314°
Course Length (CL) = Δ MD = MD2 – MD1
                                = 1940.30 – 1914.75 = 25.55m
Dog Leg = cos–1 [{sinI1 × sinI2 × cos(A2 – A1)} + {cosI1 × cosI2}]
= cos–1[{sin(13.6) × sin(10.7) × cos(314 – 315.2)} + {cos(13.6) × cos(10.7)}]
= 2.91
DLS = (DL × 30)/CL, when calculated per 30m.
DLS = (DL × 100)/CL, when calculated per 100ft.
DLS = (2.91 × 30)/25.55 = 3.42
Ratio Factor (R.F) is simply a smoothing factor used in the following calculations. It has no other significance.
RF = Tan(DL/2) × (180/π) × (2/DL)
      = Tan(2.91/2) × 180/π × (2/2.91)
      = 1
“0” Dogleg Exception
When the inclination and the direction do not change between two survey stations, the dogleg and dogleg severity are equal to 0. When the dogleg is equal to 0, the formula for ratio factor (R.F.) is undefined. In this case, simply assign the ratio factor the value of 1.0.
Change in N/S coordinate
Δ N/S = [(sinI1 × cosA1) + (sinI2 × cosA2)] [R.F. × (ΔMD/2)]
= [(sin(13.6) × cos(315.2)) + (sin(10.7) × cos(314))] [1 × (25.55/2)]
= 3.78
Total N/S (or) (N/S)2 = (N/S)1 + ΔN/S  
                                   = 311.7 + 3.78
                                   = 315.48
Change in E/W coordinate
Δ E/W = [(sinI1 × sinA1) + (sinI2 × sinA2)] [R.F. × (ΔMD/2)]
= [(sin(13.6) × sin(315.2)) + (sin(10.7) × sin(315))] [1 × (25.55/2)]
= –3.79
Total E/W (or) (E/W)2 = (E/W)1 + ΔE/W 
                                     = –299.27 + –8.04

                                      = –303.06
Change in TVD
Δ TVD = [cosI1 + cosI2] [R.F. × (Δ MD/2)]
             = [cos(13.6) + cos(10.7)] [1 × (25.55/2)]
             = 24.97
Total TVD (or) TVD2 = TVD1 + ΔTVD
                                    = 1827.53 + 24.97
                                    = 1852.50
Closure Distance
CD = [(N/S)2Total + (E/W)2Total]1/2
      =  [(315.48)2 + (–303.06)2]1/2
      =  437.46
Closure Azimuth
CA = Tan–1[(E/W) Total / (N/S) Total]
       = Tan–1[–303.06 / 315.48]
       = –43.84°
       = 360°– 43.84°
       = 316.16°

 Note:
If the given target azimuth lies in b/w 
0° to 90°, then CA = Tan–1[(E/W) Total / (N/S) Total]
90° to 180°, then CA = 180° –  Tan–1[(E/W) Total / (N/S) Total]
180° to 270°, then CA = 180° + Tan–1[(E/W) Total / (N/S) Total]
270° to 360°, then CA = 360° – Tan–1[(E/W) Total / (N/S) Total]
Directional Difference (DD) is the angle between target azimuth and closure azimuth.
DD = Azimuthtarget – CA
       = 316° – 316.6°
           = –0.16°

VS = CD × cos(DD)
      = 437.46 × cos(–0.16°)
      = 437.46 

 Hence,



 Note:
The calculation for dogleg and dogleg severity, closure and vertical section do not change when different survey methods are used.
To summarize:

*Course Length (CL) = Δ MD = MD2 – MD1

*Dog Leg = cos–1 [{sinI1 × sinI2 × cos(A2 – A1)} + {cosI1 × cosI2}]
  DLS = (DL × 30)/CL, when calculated per 30m.
  DLS = (DL × 100)/CL, when calculated per 100ft.

*RF = Tan(DL/2) × (180/π) × (2/DL)

*Δ N/S = [(sinI1 × cosA1) + (sinI2 × cosA2)] [R.F. × (ΔMD/2)]
  Total N/S (or) (N/S)2 = (N/S)1 + ΔN/S  

*Δ E/W = [(sinI1 × sinA1) + (sinI2 × sinA2)] [R.F. × (ΔMD/2)]
  Total E/W (or) (E/W)2 = (E/W)1 + ΔE/W 

*Δ TVD = [cosI1 + cosI2] [R.F. × (Δ MD/2)]
  Total TVD (or) TVD2 = TVD1 + ΔTVD

*CD = [(N/S)2Total + (E/W)2Total]1/2

*CA = Tan–1[(E/W) Total / (N/S) Total]
  If the given target azimuth lies in b/w 
   0° to 90°, then CA = Tan–1[(E/W) Total / (N/S) Total]
   90° to 180°, then CA = 180° – Tan–1[(E/W) Total / (N/S) Total]
   180° to 270°, then CA = 180° + Tan–1[(E/W) Total / (N/S) Total]
   270° to 360°, then CA = 360° – Tan–1[(E/W) Total / (N/S) Total]

*DD = Azimuthtarget – CA
  VS = CD × cos(DD)
These days Directional Drillers need not perform manual calculations instead, many high-end software are used for well planing such as COMPASS, WELL PATH, WIN SERVE, etc.
But, remembering these formulas is useful in the due course.




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Tuesday, 22 September 2015

UTM COORDINATE SYSTEM (UNIVERSAL TRANSVERSE MERCATOR)

What is a map projection? And why do we need it?

Map projection is a geographical transformation of earth's curved surface on to a flat map surface.

The earth's surface, as we all know is neither a perfect sphere nor a perfect ellipsoid. Although, globes provide the most accurate picture of the earth and basic geometric properties such as distance, shape,direction and area are all preserved; they have certain disadvantages too such as:
  • In globe simultaneous view of whole of the earth is not possible, at most only half hemisphere is visible.
  • They're bulky and difficult in handling or storage and a small scale globe is of practically no use.
  • Instruments and techniques that are suited for measuring distance, direction and area on spherical surfaces are relatively complicated. 
  • Globes construction is laborious and costly relative to maps.
  • Globes don 't offer the variety and details as offered by maps.


The universal transverse Mercator (UTM) is a map projection that is based completely on the cylindrical transverse Mercator, with a secant cylinder.





Why do we use UTM system in directional drilling industry, when there're other systems too?

Advantages of UTM coordinate system:

  1. Cylindrical projection preserves Area, i.e. exact representation of area is possible.
  2. UTM is a planar coordinate system which provides a constant distance relationship anywhere on the map. In angular coordinate systems like latitude and longitude, the distance covered by a degree of longitude differs as you move towards the poles.
  3. All UTM coordinates are measured in metric units.
  4. No negative numbers.
  5. Simple Cartesian coordinate mathematics can be used to find distances, thereby eliminating complex mathematical calculations. 
The UTM coordinate system is universally used plane coordinate system (except for polar regions). 
What is a UTM coordinate system? 

In UTM grid, 


(1). Area of earth between 84°N to 80°S is divided into 60 north-south columns of 6° wide longitude called Zones. So, UTM covers most of the world. 
{Rest part of the world 84°N to 90°N and 80°S to 90°S is covered by UPS system (Universal Polar Stereographic).}

(2). Zones are numbered eastwards from 1 to 60; starting from 180th meridian. Zones in northern and southern hemisphere are labelled N and S respectively.
o Zone 1 is from 180°W to 174°W,
o Zone 2 is from 174°W to 168°W, and so on.




                                                              

UTM zones are flattened using the Transverse Mercator Projection, and a rectangular grid network of straight horizontal and vertical lines is superimposed on each zone.

(3). Longitude lines run towards true north whereas, grid lines don't, instead they run towards grid north and are parallel to the central meridian of the zone.
Hence, grid north are latter corrected to true north.
Latitude lines are parallel to equator.

Each zone has a central meridian.
o Zone 1 central meridian is 177°W,
o Zone 2 central meridian is 171°W, and so on.


(4). UTM system is a coordinate system and therefore, it will have a x-axis, y-axis and an origin.
For a zone,
X-axis is the equator;
Y-axis is the central meridian of the zone; and
an origin is taken as there intersection point.
As (x,y) are used to denote a point in the coordinate system, Easting and Northing are used in UTM coordinate system.

Easting are used to determine east-west position w.r.t central meridian which is assigned an arbitrary value of 500,000m. All the points lying east will have easting value > 500,000m and points lying east will have < 500,000m w.r.t central meridian.

{Longitude lines are farthest apart at equator, hence max. width occur at equator with a max value of 834,000m and min. width near poles with a min. value of 166,000m. As a result, there'll be no negative easting.}

A point lying 10m east of central meridian has an easting of 500000 + 10 = 500010mE. The easting of a point 350m west of central meridian would be 500000 – 350 = 499650mE. The east-west distance between two points is obtained by the difference of their easting values. The distance between the above points is 500008 – 499650 = 360m.
(If eastings are < 500,000m it means point lies on the west of central meridian and for eastings > 500,000m point lies on east of central meridian.)

Northing are used to determine north-south position w.r.t equator which is assigned a value of 0m north for northern hemisphere and an arbitrary value of 10,000,000m south for referencing northing coordinates in southern hemisphere. The northing of a point south of the equator is equal to 10,000,000m minus its distance from the equator. In both northern and southern hemispheres, northing values increase from south to north.

{In the southern hemisphere the northing values range from 10,000,000m at the equator to approximately 1,100,000m at the 80°S. In the northern hemisphere the northing values stretch from 0m at the equator to around 9,350,000m at the 84°N. As a result, there'll be no negative northing.}

A point south of equator with a northing of 7,560,897mN is 10,000,000 – 7,560,897 = 2,439,103m south of the equator. A point located 54m south of the equator has a northing of 99,99,946mN, while a point 34m north of the equator has a northing of 00,00,034mN. The north-south distance between two points north of equator with northings of 4,657,134mN and 4,212,189mN is 4,657,134 – 4,212,189 = 444,945m.
Easting and northing are often referred as a False Easting and False Northing.


How to locate places on earth with UTM coordinates?


 To explain this let's consider following examples.

 Example 1:
The latitude and longitude of The Taj Mahal in Agra, Uttar Pradesh, India is 78.0419°S and 27.1750°N.
The Taj Mahal as per UTM coordinate system lies in,
UTM Zone 44R.
Easting & Northing coordinates are is 206,890.80mE, 3,009,276.00mN


We can see from map agra lies in b/w 84°E to 90°E longitudes. 87°E is the central meridian of this zone
From northing value we know the point lies 3,009,276m north to the equator in the zone 44R. 
From easting value we know the point lies at 500,000 - 206,890.80 = 293,109.20m west from the central meridian.
Hence, the Taj Mahal lies at the intersection of these two points.



Example 2:
The latitude and longitude of The Sydney Opera House in Australia is 33.8587°S and 151.2140°N.
The Sydney Opera House as per UTM coordinate system lies in,
UTM Zone 56H.
Easting & Northing coordinates are is 334,895.26mE, 6,252,359.77mN

We can see from map sydney opera house lies in b/w 150°E to 156°E longitudes. 153°E is the central meridian of this zone
From northing value we know the point lies 10,000,000 - 6,252,359.77 = 3,747,640.23m  south to the equator in the zone 44R. 
From easting value we know the point lies at 500,000 - 334,895.26 = 165104.74m west from the central meridian.
Hence, the Sydney Opera House lies at the intersection of these two points.



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