# Suppressing multiples by NMO differences

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.11

A primary and a multiple each arrive at 0.600 s at ${\displaystyle x=0}$; their stacking velocities are 1800 and 1500 m/s, respectively. Calculate the residual NMO (after NMO correction for the primary velocity) for offsets of 300 ${\displaystyle n}$, where ${\displaystyle n=1,2,...}$. What is the shortest offset that will give good multiple suppression for a wavelet with a 50-ms dominant period?

### Solution

The distance to the primary reflector is ${\displaystyle (0.600\times 1800)/2=540\ {\rm {m}}}$ and to the reflector responsible for the multiple, assuming it is simply a double bounce, is ${\displaystyle (0.600\times 1500)/4=225\ {\rm {m}}}$. NMO is given by equation (4.1c), ${\displaystyle \Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}}$. We obtain the following values for the moveouts:

 Offset 300 m 600 m 900 m Primary NMO 0.023 s 0.093 s 0.208 s Multiple NMO 0.033 s 0.133 s 0.300 s NMO Difference 0.010 s 0.040 s 0.092 s

Multiple suppression should be maximum when the NMO difference approximates half the wavelet period so that some of the traces are out-of-phase, which is achieved at offset ${\displaystyle x}$ where

{\displaystyle {\begin{aligned}x^{2}/1.200\times 1500^{2}-x^{2}/1.200\times \;1800^{2}\;=0.050,\;x=660\ {\rm {m}}.\end{aligned}}}
The text passage contains a mathematical problem involving the calculation of residual NMO (Normal Moveout) for a primary and a multiple reflection, and it also provides the solution to this problem, including the final answer for the shortest offset that will give good multiple suppression.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Suppressing multiples by NMO differences

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.11

A primary and a multiple each arrive at 0.600 s at ${\displaystyle x=0}$; their stacking velocities are 1800 and 1500 m/s, respectively. Calculate the residual NMO (after NMO correction for the primary velocity) for offsets of 300 ${\displaystyle n}$, where ${\displaystyle n=1,2,...}$. What is the shortest offset that will give good multiple suppression for a wavelet with a 50-ms dominant period?

### Solution

The distance to the primary reflector is ${\displaystyle (0.600\times 1800)/2=540\ {\rm {m}}}$ and to the reflector responsible for the multiple, assuming it is simply a double bounce, is ${\displaystyle (0.600\times 1500)/4=225\ {\rm {m}}}$. NMO is given by equation (4.1c), ${\displaystyle \Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}}$. We obtain the following values for the moveouts:

 Offset 300 m 600 m 900 m Primary NMO 0.023 s 0.093 s 0.208 s Multiple NMO 0.033 s 0.133 s 0.300 s NMO Difference 0.010 s 0.040 s 0.092 s

Multiple suppression should be maximum when the NMO difference approximates half the wavelet period so that some of the traces are out-of-phase, which is achieved at offset ${\displaystyle x}$ where

{\displaystyle {\begin{aligned}x^{2}/1.200\times 1500^{2}-x^{2}/1.200\times \;1800^{2}\;=0.050,\;x=660\ {\rm {m}}.\end{aligned}}}
user
The text passage contains a mathematical problem involving the calculation of residual NMO (Normal Moveout) for a primary and a multiple reflection, and it also provides the solution to this problem, including the final answer for the shortest offset that will give good multiple suppression.

TYPE: SINGLE_QUESTION_WITH_ANSWER