Dip, cross-dip, and angle of approach

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Problem 4.2a

Show that for a bed dipping in the direction of the -axis (see Figure 4.2a), equation (4.1a) becomes


(4.2a)

Background

If the normal from the origin to a plane has direction cosines and length , the equation of the plane is (see Sheriff and Geldart, 1995, problem 15.9b).

Solution

In Figure 4.2a, the traveltime at equals (assuming constant velocity). We use the cosine law to express IR in terms of IS and SR. This gives

Figure 4.2a.  Raypaths for dipping reflector.

where is the dip. This equation is also used when is not constant by replacing by the average velocity , but because usually increases with depth, this underestimates .

Problem 4.2b

For two receivers spaced a distance away from the source in opposite directions, show that to the first approximation the dip is given by


(4.2b)

where is the difference in traveltimes at the two receivers.

Solution

We rewrite equation (4.2a) replacing with and get for the traveltime at


(4.2c)

Expanding and taking the first approximation gives

Figure 4.2b.  Dip moveout on zero-offset sections.

Next we take a receiver at a distance to the left of the source to get the traveltime . Subtracting the expressions for from that for gives the result

where moveout.

Problem 4.2c

Figure 4.2b represents a vertical section in the direction of dip of a bed CD, and being sources on the surface. Show that for receivers located at the sources, equation (4.2b) can be used provided is the difference between the two-way traveltimes at the two sources.

Solution

In Figure 4.2b the 2-way traveltime at is . The wavefront arrives at after a further time interval of . Because the traveltimes are 2-way, the distance , so

Problem 4.2d

In Figure 4.2c, CA is a wavefront approaching two receivers and on the surface. Derive the following expression for the angle of approach :


(4.2d)

where is the difference in traveltimes at and , and is the near-surface velocity.

Figure 4.2c.  Angle of approach of a wave.

Solution

If the wavefront arrives at at time , it will arrive at at time where Therefore the angle of approach is given by


(4.2e)

A variation of this equation is


(4.2f)

that is, the distance between geophones on the surface divided by the difference in their arrival times is the apparent velocity .

Problem 4.2e

Discuss the relationship between equation (4.2b) as applied in parts (b) and (c) and equation (4.2e).

Solution

In equation (4.2b) in part (b) is a time difference between two rays generated by the same source and is therefore a time interval due to the dip of the bed and the fact that the two receivers are a distance apart.

When equation (4.2b) is used in part (c), is a two-way time difference between waves generated by two different sources a distance apart, each wave returning directly to the source where it was generated. In Figure 4.2c, is a plane wave that travels from to in time . Equation (4.2d) would be identical with equation (4.2b) if the source were at the midpoint of AB, in equation (4.2d) taking the place of in equation (4.2b).

A further difference between the equations is related to the velocity . In equation (4.2b) is assumed to be constant; if it is not, an average velocity for the section between the surface and the reflector is used. In equation (4.2e), is the average velocity over the short interval ; this distance is usually small enough that velocity variations can be neglected.

Problem 4.2f

Show that the quantity can be considered as a vector or component of a vector, according as corresponds to the total dip or a component of dip.

Figure 4.2d.  3D view of reflection path.

Solution

Referring to Figure 4.2d, we take the -axis as positive vertically downward and consider a plane reflector which intersects the three axes at , , and , the strike being at the angle to the -axis. The vertical depth to the plane is , which is a function of and . We write (see Sheriff and Geldart, 1995, Section 15.1.2c)


(4.2g)

It is shown in Sheriff and Geldart, 1995, problem 15.6a that is perpendicular to the strike (because contours on the reflector are parallel to the strike). It is therefore in the direction of dip, that is, OA in Figure 4.2e is the horizontal projection of the total dip moveout. If we take the -axis in this direction, then

where is a unit vector along the -axis and is given by equation (4.2b) with replacing . Thus we can regard the dip as a vector whose magnitude is given by equation (4.2b), being measured in the direction of dip.

We now take the -axis and a unit vector along at an angle to in Figure 4.2e. The component of dip [apparent dip—see part (g)] in this direction is the projection

Figure 4.2e.  Dip and strike from nonperpendicular measurements.

(or component) of in this direction (see Sheriff and Geldart, 1995, problem 15.6c), so the quantity given by equation (4.2b) when applied to measurements along this axis is the component of the total dip in this direction. Since is multiplied by the scalar quantity to get dip, we can consider as a vector or component of a vector depending on whether is measured using a spread in the direction of dip or at an angle to this.

Problem 4.2g

When a profile is not in the direction of dip, the value given by equation (4.2b) is less than the true dip and is called the apparent dip. Show how to calculate the true dip and the direction of strike from values of the apparent dip measured along the - and -axes.

Solution

The direction cosines of the normal to the reflector (see Figure 4.2d) are the cosines of the angles , , and , being the angle of dip and the angle of strike relative to the -axis. The coordinates of the image point are . For a point on the -axis with coordinates , we have

(note that ). Expanding this result as in part (b) and subtracting the values for two geophones located a distance on either side of the source, we arrive at the result

Repeating the above procedure for a spread along the -axis we get


(4.2h)

Thus, is times the hypoteneuse of a right-angled triangle whose sides are the two apparent dips.

To get the angle of strike , we use the equation of the plane in Figure 4.2d, that is, (see Sheriff and Geldart, 1995, problem 15.9b), and set . This gives the equation of the line NM, namely , from which we get , , so or


(4.2i)

Problem 4.2h

If the two spreads in part (g) are at an angle to each other, show how to calculate the total dip and the strike direction.

Solution

We take and in Figure 4.2e equal to the components of dip moveout along the - and -axes. Then,


(4.2j)

We measure the value of the left-hand side of equation (4.2j) and the apparent dip moveout along the -axis, which gives us , so we can calculate , given both and . We can now find the total dip and the angle of strike as in part ().

In Figure 4.2e, and are the projections of the total dip onto the - and -axes. The perpendiculars to at and at must thus pass through , the terminus of the total-dip vector. Hence we can solve the problem graphically as shown in Figure 4.2e to give the magnitude and direction of the true dip moveout.

Problem 4.2i

Verify that the result in part (h) can be obtained graphically by combining the apparent dips as in Figure 4.2e.

Solution

From part (f) we see that in Figure 4.2e is the projection of the total dip moveout onto the -axis and that is the projection onto the -axis. Therefore the lines and must be perpendiculars to and . Thus, if we are given and , drawing perpendiculars at and locates point , so is the total dip moveout and the strike is normal to .

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