## Dimensional Drill Hole Planning

Four types of directional problems may be solved with the following mathematics: (1) planning of a three dimensional well within an oblique plane, (2) calculation of minimum dogleg severity for the intersection of a non-colinear target with the wellbore, (3) planning the locations of two or more non-colinear targets, (4) selection of a plug-back depth (Sidetrack Depth) in a wellbore for the purpose of reducing the dogleg severity to an acceptable value prior to intersecting the target.

Definition of variables in the sketch:

P = oblique plane containing points 2, 3, 4, 5, and 6 a, [3, y, X = corresponding angles d, e, f, h, j, k, m, r, s = corresponding lengths

In the drilling of a directional well the wellbore may wander off the planned traverse to such a distance that a large displacement of the wellbore toward the target may be required to intersect the target. These large displacements lead to dogleg severities of the wellbore in excess of those acceptable in modern drilling practices.

In order to obviate the excessive dogleg severity problem, a wellbore to target dogleg severity calculation is made at each station. If a plugback depth is desired, 3 TARGET

the plugback depth is assigned to that station which has an acceptable wellbore to target dogleg severity.

### MATHEMATICS

The space is defined with the following sketch. Let the point 2 be a directional station in a drill hole as shown in the sketch with coordinates N2, E2, V2, and let point 3 be a point in the target with coordinates N3, E3, V3. Also, let inclination and azimuth be <|> and 0 respectively at point 2.

The radius of curvature r is 18000 TIDLS

The direction cosines of the hole's direction at the kick off point 2 are equal to those of j and are the following.

cos ocj = sin (J) cos 9 cos pj = sin ty sin 9 cos yj = cos (j)

The length and direction cosines of line e are e = a/ (N2 - N3)2 + (E2 - E3)2 + (V2 - V3)2

N3-N2

V3-V2

The angle a is a = acos[cos ocj cos ae + cos ßj cos ße + cos yj cos ye] The lengths e, d, f, and k are e =

sm a tan a d = V^+r2-2ersina The coordinates of point 6 are

The direction cosines of line f are k = Vd2 - r 2

cos af = - cos ctj cos pf = - cos pj cos yf = - cos Yj The coordinates of point 4 are

N4 = f cos af + N6 E4 = f cos pf + E6 V4 = f cos yf + V6

y = p - X if f is positive y = 2% -p - X if f is negative The direction cosines of line r from point 4 to point 2 are

N2-N4 E2-E4 V2-V4

The coordinates of point 5 are h = r cos y m = r sin y

N5 = N4 + h cos ar + m cos aj E5 = E4 + h cos pr + m cos pj V5 = V4 + h cos yr + m cos yj

The direction cosines of the line k are

N3-N5 E3-E5 V3-V5

cos ak = —£— cos Pj^ = —k— cos \ = —k—

The angle of inclination 0 and azimuth 0 into the target is cos Pk

K cos ak 