# 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?

### 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 $x$ -axis along the line of traverse. With 200 m lead-in and compliant sections and active streamer length of $96\times 25=2400$ m, the farthest group center is $\left(200+2400-12\right)=2590$ m from the ship (assuming distances are to the group centers). The last group has a velocity of 5.8 knots along the $x$ -axis and 1.9 knots along the $y$ -axis, hence the streamer will be at the angle $\tan ^{-1}\left(1.9/5.8\right)=18^{\circ }$ (see Figure 7.10b). The coordinates of this group center with respect to the source are $x=2490\cos 18^{\circ }=2370$ m, $y=2490\sin 18^{\circ }=770$ 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 $20^{\circ }$ 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?

### 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 $\left(x,y,0\right)$ . Because the dip is along the $y$ -axis, the coordinates of the image point $I$ are $(0,2h\sin \xi ,2h\cos \xi )$ , $h$ being the slant depth (2.00 km). We now have

 {\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, $y=0$ and equation (7.10a) gives

 {\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, $x=2.37$ km, $y=0.77$ km, and

 {\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 $\sin \xi$ -term is reversed, so

 {\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

 {\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:

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

{\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,

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

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

{\begin{aligned}(Vt)^{2}=x^{2}+y^{2}+4h^{2}\pm 4hy\sin \xi ,\end{aligned}} where $x=2.49$ km, $y=0.77$ km, $h=2.00$ km, $V=3.00$ km/s, $\xi =21^{\circ }$ , and we use the plus sign for the $\xi$ -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 $\Delta \theta$ is $+3^{\circ }$ or $-3^{\circ }$ .

Drift downdip

{\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

{\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

{\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,

{\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: $\Delta x=\left(-128+10\right)/2=-59$ m, $\Delta y=\left(800-66\right)/2=367$ m.