# Streamer feathering due to cross-currents

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 7 221 - 252 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 7.10a

A 96-channel streamer with 25-m groups has the hydrophones spaced uniformly throughout its length. The lead-in and compliant sections together are 200 m in length and the tail section and buoy connection are 150 m long. Assume a source 100 m behind the ship, ship’s speed of 5.8 knots, and a 1.9 knot current perpendicular to the direction of travel.

What are the perpendicular and inline components of the distance to the farthest active group center with respect to the traverse direction?

Figure 7.10a  Marine streamer.

### Background

Marine seismic work usually employs hydrophones to receive the seismic signals. These consist of piezoelectric crystals which generate a voltage difference between opposite faces when subjected to a pressure. The hydrophones are enclosed in a neutrally buoyant tube called a streamer (Figure 7.10a) which is towed by the ship. A number of hydrophones, spread over as much as 25 m, are connected together to form each group. Lead-in and compliant sections are inserted between the ship and the near end of the streamer; these increase the distance between the ship and the cable and reduce the effects of sudden changes which might jerk or break the streamer. A location sensor and radar target are usually placed on a tail buoy so that the end of the streamer can be located and the amount of feathering (sideways drift of the streamer) determined.

Common-midpoint (CMP) recording is discussed in problem 5.12.

### Solution

We assume that the source is in the line between the ship and the streamer. We take the origin at the source and the ${\displaystyle x}$-axis along the line of traverse. With 200 m lead-in and compliant sections and active streamer length of ${\displaystyle 96\times 25=2400}$ m, the farthest group center is ${\displaystyle \left(200+2400-12\right)=2590}$ m from the ship (assuming distances are to the group centers). The last group has a velocity of 5.8 knots along the ${\displaystyle x}$-axis and 1.9 knots along the ${\displaystyle y}$-axis, hence the streamer will be at the angle ${\displaystyle \tan ^{-1}\left(1.9/5.8\right)=18^{\circ }}$ (see Figure 7.10b). The coordinates of this group center with respect to the source are ${\displaystyle x=2490\cos 18^{\circ }=2370}$ m, ${\displaystyle y=2490\sin 18^{\circ }=770}$ m.

## Problem 7.10b

If the average velocity to a reflector 2.00 km below the source is 3.00 km/s and if the reflector dips ${\displaystyle 20^{\circ }}$ perpendicular to the traverse direction, (i) by how much will the arrival time be changed for the far trace, and (ii) if this should be attributed to a change in velocity rather than cross-dip, what velocity would it imply?

Figure 7.10b  Drift of streamer in crosscurrent.

### Solution

(i) We shall use a derivation similar to that in problem 4.2g except that the problem is now three dimensional, so we take the receiver coordinates as ${\displaystyle \left(x,y,0\right)}$. Because the dip is along the ${\displaystyle y}$-axis, the coordinates of the image point ${\displaystyle I}$ are ${\displaystyle (0,2h\sin \xi ,2h\cos \xi )}$, ${\displaystyle h}$ being the slant depth (2.00 km). We now have

 {\displaystyle {\begin{aligned}(Vt)^{2}=x^{2}+(y\pm 2h\sin \xi )^{2}+(2h\cos \xi )^{2}\\=x^{2}+y^{2}+4h^{2}\pm 4hy\sin \xi ,\end{aligned}}} (7.10a)

the plus sign for the last term being used when the dip and feathering are in the same direction.

When the streamer is tracking behind the boat, ${\displaystyle y=0}$ and equation (7.10a) gives

 {\displaystyle {\begin{aligned}t=\left(1/3.00\right)(2.49^{2}+4^{2})^{1/2}=22.1^{1/2}/3.00=1.571\ \mathrm {s} .\end{aligned}}} (7.10b)

When the feathering is in the downdip direction, ${\displaystyle x=2.37}$ km, ${\displaystyle y=0.77}$ km, and

 {\displaystyle {\begin{aligned}t=\left(1/3.00\right)(2.37^{2}+0.77^{2}+4^{2}+2\times 4\times 0.77\times \sin 20^{\circ })^{1/2}\\=24.32^{1/2}/3.00=1.644\ \mathrm {s} .\end{aligned}}} (7.10c)

When the streamer is drifting updip, the sign of the ${\displaystyle \sin \xi }$-term is reversed, so

 {\displaystyle {\begin{aligned}t=\left(1/3.00\right)(2.37^{2}+0.77^{2}+4^{2}-2\times 4\times 0.77\times \sin 20^{\circ })^{1/2}\\=20.103^{1/2}/3.00=1.495\ \mathrm {s} .\end{aligned}}} (7.10d)

(ii) At the source we observe the traveltime

 {\displaystyle {\begin{aligned}t_{0}=2h/V=4.00/3.00=1.333\ \mathrm {s} .\end{aligned}}} (7.10e)

If we attribute the difference in traveltimes because of dip to a velocity change rather than to dip, we can determine the velocity as follows:

{\displaystyle {\begin{aligned}(Vt)^{2}=x^{2}+y^{2}+4h^{2}=x^{2}+y^{2}+(Vt_{0})^{2},\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{so}}\qquad V^{2}\left(t^{2}-t_{0}^{2}\right)=x^{2}+y^{2}=\left(2.37^{2}+0.77^{2}\right)=6.21\ \mathrm {km} ^{2},\end{aligned}}}

{\displaystyle {\begin{aligned}{\text{and}}\qquad V=2.49/(t^{2}-1.33^{2})^{1/2}\ \mathrm {km/s} .\end{aligned}}}

When the streamer is feathering in the down-dip direction,

{\displaystyle {\begin{aligned}V=2.49/(1.64^{2}-1.33^{2})^{1/2}=2.49/0.96=2.57\ \mathrm {km/s} .\end{aligned}}}

For updip feathering,

{\displaystyle {\begin{aligned}V=2.49/(1.45^{2}-1.33^{2})^{1/2}=2.49/0.694=3.56\ \mathrm {km/s} .\end{aligned}}}

Compared with the actual value of 3.00 km/s, these results are in error by 14% and 19% for the downdip and updip cases, respectively.

## Problem 7.10c

Assume that the amount of sideways drift of the streamer is ascertained by radar sighting on the tail buoy with an accuracy of only ${\displaystyle \pm 3^{\circ }}$; (i) how much change will this produce in locating the far group? (ii) How much change in arrival time will be associated with this uncertainty?

### Solution

i) The radar sighting gives the angle of feathering ${\displaystyle \theta }$. The center of the last group is 2590 from the ship, so taking an origin at the ship, the ${\displaystyle x'}$ and ${\displaystyle y'}$ coordinates of the group are ${\displaystyle x'=2590\cos \theta }$, ${\displaystyle y'=2590\sin \theta }$; ${\displaystyle \theta =18^{\circ }}$. The errors in ${\displaystyle x'}$ and ${\displaystyle y'}$ due to ${\displaystyle \Delta \theta =3^{\theta }=0.052}$ radians are

{\displaystyle {\begin{aligned}\Delta x'=\Delta x=-2590\sin \theta \Delta \theta =-\left(2590\times \sin 18^{\circ }\right)\times 0.052=-41\ \mathrm {m} ,\\\Delta y'=\Delta y=2590\cos \theta \Delta \theta =\pm 128\ \mathrm {m} .\end{aligned}}}

ii) When ${\displaystyle \Delta \theta }$ is positive, i.e., ${\displaystyle \theta +\Delta \theta =21^{\circ }}$, ${\displaystyle \Delta x=-42}$ m, ${\displaystyle \Delta y=+128}$ m. When ${\displaystyle \Delta \theta }$ is negative, the sign of ${\displaystyle \Delta y}$ is reversed. We use equation (7.10a) to find the traveltimes, that is,

{\displaystyle {\begin{aligned}(Vt)^{2}=x^{2}+y^{2}+4h^{2}\pm 4hy\sin \xi ,\end{aligned}}}

where ${\displaystyle x=2.49}$ km, ${\displaystyle y=0.77}$ km, ${\displaystyle h=2.00}$ km, ${\displaystyle V=3.00}$ km/s, ${\displaystyle \xi =21^{\circ }}$, and we use the plus sign for the ${\displaystyle \xi }$-term when the cable is drifting downdip, the minus sign when the drift is updip. We have four cases to consider: the drift is down- or updip, and ${\displaystyle \Delta \theta }$ is ${\displaystyle +3^{\circ }}$ or ${\displaystyle -3^{\circ }}$.

Drift downdip

{\displaystyle {\begin{aligned}\Delta \theta =+3^{\circ }:t=0.333[(2.49-0.04)^{2}+(0.77+0.13)^{2}\\\qquad \qquad +16+8\times \left(0.77+0.13\right)\sin 21^{\circ })]^{1/2}=1.676\ \mathrm {s} .\\\Delta \theta =-3^{\circ },t=0.333[2.45^{2}+(0.77-0.13)^{2}+16\\\qquad \qquad +8\times 0.64\sin 15^{\circ }]^{1/2}=1.639\ \mathrm {s} .\end{aligned}}}

Drift updip

{\displaystyle {\begin{aligned}\Delta \theta =+3^{\circ },t=0.333(2.45^{2}+0.64^{2}+16-8\times 0.64\sin 20^{\circ })^{1/2}=1.515\ \mathrm {s} ,\\\Delta \theta =-3^{\circ },t=0.333(2.45^{2}+0.90^{2}+16-8\times 0.90\sin 20^{\circ })^{1/2}=1.504\ \mathrm {s} .\end{aligned}}}

## Problem 7.10d

The midpoint traces that are to be combined in a CMP stack will be distributed over what distance?

### Solution

We determine the shifts from the line of traverse of groups #96 and #1, subtract these shifts, then divide by 2 to get the shifts of the midpoint. Thus, the shift relative to the ship of group center #96 is

{\displaystyle {\begin{aligned}\Delta x=2590\left(\cos 18^{\circ }-1\right)=-129\ \mathrm {m} ,\qquad \Delta y=2590\sin 18^{\circ }=800\ \mathrm {m} .\end{aligned}}}

For group #1,

{\displaystyle {\begin{aligned}\Delta x=212\left(\cos 18^{\circ }-1\right)=-10\ \mathrm {m} ,\qquad \Delta y=212\sin 18^{\circ }=66\ \mathrm {m} .\end{aligned}}}

The midpoints are shifted by half of these amounts, so the spread of the midpoints is: ${\displaystyle \Delta x=\left(-128+10\right)/2=-59}$ m, ${\displaystyle \Delta y=\left(800-66\right)/2=367}$ m.