How many traces will be present in a zero-offset section that is generated from the following marine seismic survey? Near offset = 100 m Shot interval = 20 m Receiver interval = 10 m Number of shots = 100 Number of receivers = 1000 (Figure not to scale)
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Total CMP-covered length = spread length + sail-line distance; divide by the CMP bin size (half the receiver interval, since shot interval = 2 x receiver interval here).
A single vessel tows one streamer while steaming along the line, firing a shot every 20 m (the shot interval). The streamer carries 1000 receiver channels spaced 10 m apart, with the nearest channel sitting 100 m behind the source (the near offset). Because a fresh streamer record is captured after every shot, the survey ends up illuminating a stretch of subsurface midpoints that is the SUM of two separate lengths: the physical length of the towed spread, plus the distance the boat itself travels while shooting.
Length of the towed spread, from the source out to the farthest (last) channel: \(L_{spread}= \text{near offset} + (N_{receivers}-1)\times RI = 100 + (1000-1)\times 10 = 100+9990=10090\ \text{m}\)
Distance the vessel sails between the first shot and the last shot: \(L_{sail} = (N_{shots}-1)\times SI = (100-1)\times 20 = 1980\ \text{m}\)
Total subsurface length covered along the profile: \(L_{total}=L_{spread}+L_{sail}=10090+1980=12070\ \text{m}\)
Because the shot interval (20 m) is exactly twice the receiver interval (10 m), this acquisition geometry gives smooth, uniform-fold coverage when the common-midpoint (CMP) bin is taken as half of the receiver interval: \(\text{CMP interval} = \dfrac{RI}{2} = \dfrac{10}{2} = 5\ \text{m}\)
The number of distinct zero-offset (CMP) traces that this profile can generate is the total covered length divided by the CMP bin size: \(N_{traces}=\dfrac{L_{total}}{\text{CMP interval}}=\dfrac{12070}{5}=2414\)
So the zero-offset section produced from this survey will contain \(\boxed{2414}\) traces, option (B).
