<p><strong>The $\tau$–$p$ time domain $t$ is a function of offset $d$, so the result is an array of size $n_t \times n_d$, where $n$ is the number of samples.</strong></p>

<p>The inputs are a vector $\tau$ of travel time at zero offset, or $t(0)$, an array $p$ of <a href="http://wiki.seg.org/wiki/Dictionary:Raypath_parameter" rel="nofollow noreferrer">ray parameter</a>, which is in turn a function of velocity $V$ and angle, and a vector $d$ (usually called $x$) of offset.</p>

<p>Here's an example of some visualizations of the $\tau$–$p$ domain from <a href="http://www.ahay.org/RSF/book/milano/taupvel/paper_html/" rel="nofollow noreferrer">a paper</a> by Sergey Fomel; they might help explain what I mean by an 'array' — you can think of them as images:</p>

<p><img src="https://i.sstatic.net/fxLZ1.png" alt="enter image description here"></p>

<p>The resulting travel-time $t(d)$ is an array of size $n_t \times n_d$, where $n$ is the number of samples. For a <em>given</em> offset at a <em>given</em> time sample, the result is a scalar — a single value of $t$.</p>

<p>By the way, <strong>stevej</strong>'s answer is a great explanation of how exactly to calculate the ray parameter term, given limited information. Do follow that link to Yilmaz, and maybe also read <a href="http://wiki.seg.org/wiki/The_radon_transform" rel="nofollow noreferrer">the Radon transform article</a>.</p>

CC BY-SA 3.0
2015-07-14T14:36:15.283Z
2016-10-05T16:04:00.387Z
2016-10-05T16:04:00.387Z
Matt Hall
<p>To be consistent with the references referred to in this post, I will re-write your equation of the $\tau-p$ transform as:
$t = \tau + px$</p>

<p>Where $t$ is two-way time, $x$ is offset, $p$ is ray parameter ($\Delta x / \Delta t$ or $\sin(\theta) / velocity$), $\tau$ is the intercept time when $p=0$ (Yilmaz, 2001). </p>

<p>There seem to be at least two methods for overcoming the finite aperture, or limited offset and ray parameter range, problem when transforming from time-offset ($t-x$) domain to $\tau-p$ domain:</p>

<ol>
<li>Sample ray parameter $p$ such $np = nx$; i.e. number of ray parameter samples equals number of offset samples (Yilmaz, 2001).</li>
<li>Formulate the the computation of the transform as a least-squares estimation problem (Kostov, 1990; Thorson and Claerbout, 1985). </li>
</ol>

<p>Kostov, C., 1990. Multi-channel seismic experiment with a drill-bit source. Phd Thesis, Standford University. <a href="http://sepwww.stanford.edu/data/media/public/oldreports/sep63/63_06.pdf" rel="nofollow">See Appendix C for details on least-squares solution.</a> </p>

<p>Thorson, J. and Claerbout, J., 1985. Velocity-stack and slant-stack stochastic inversion. Geophysics, Vol. 50, N. 12, P. 2727-2741.</p>

<p>Yilmaz, Öz (2001). Seismic data analysis. Society of Exploration Geophysicists. ISBN 1-56080-094-1.  <a href="http://wiki.seg.org/wiki/Slant-stack_parameters" rel="nofollow">See SEG Wiki page on the problem.</a></p>

CC BY-SA 3.0
2015-07-14T14:37:52.470Z
This is a really nice contribution but I think it answers a slightly different question, so I had a go at my interpretation of what the OP wants.
Matt Hall
CC BY-SA 3.0
2015-07-15T01:39:56.457Z
Yeah good call; in your answer the portion in bold answers the OP nicely.
stevej
CC BY-SA 3.0
2015-07-13T03:02:00.013Z
2016-02-20T00:07:25.657Z
2016-02-20T00:07:25.657Z
stevej
<p>In the formula for slant stacking (tau-p transform) a substitution <code>t=tau+d/v</code> is made. </p>

<p>If we consider the substitution as a vector equation then how can vector d/v have the same length as vector t given that in most surveys the number of data points for offset(d) is less than for time (t) .</p>

<geophysics><seismology>
Discrete Slant Stack on seismic data
shrey
# Discrete Slant Stack on seismic data

(Asked by: username_0 on 2015-07-09)

In the formula for slant stacking (tau-p transform) a substitution `t=tau+d/v` is made.

If we consider the substitution as a vector equation then how can vector d/v have the same length as vector t given that in most surveys the number of data points for offset(d) is less than for time (t) .

**Tags:** geophysics, seismology


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## 2 answers

### Answer by username_2 on 2015-07-13. Score: 6 | Accepted answer

To be consistent with the references referred to in this post, I will re-write your equation of the $\tau-p$ transform as:
$t = \tau + px$

Where $t$ is two-way time, $x$ is offset, $p$ is ray parameter ($\Delta x / \Delta t$ or $\sin(\theta) / velocity$), $\tau$ is the intercept time when $p=0$ (Yilmaz, 2001).

There seem to be at least two methods for overcoming the finite aperture, or limited offset and ray parameter range, problem when transforming from time-offset ($t-x$) domain to $\tau-p$ domain:

1. Sample ray parameter $p$ such $np = nx$; i.e. number of ray parameter samples equals number of offset samples (Yilmaz, 2001).
2. Formulate the the computation of the transform as a least-squares estimation problem (Kostov, 1990; Thorson and Claerbout, 1985).

Kostov, C., 1990. Multi-channel seismic experiment with a drill-bit source. Phd Thesis, Standford University. [See Appendix C for details on least-squares solution.](http://sepwww.stanford.edu/data/media/public/oldreports/sep63/63_06.pdf)

Thorson, J. and Claerbout, J., 1985. Velocity-stack and slant-stack stochastic inversion. Geophysics, Vol. 50, N. 12, P. 2727-2741.

Yilmaz, Öz (2001). Seismic data analysis. Society of Exploration Geophysicists. ISBN 1-56080-094-1. [See SEG Wiki page on the problem.](http://wiki.seg.org/wiki/Slant-stack_parameters)

> **Comments:**
> **username_1**: This is a really nice contribution but I think it answers a slightly different question, so I had a go at my interpretation of what the OP wants.  
> **username_2**: Yeah good call; in your answer the portion in bold answers the OP nicely.  

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### Answer by username_1 on 2015-07-14. Score: 7

**The $\tau$–$p$ time domain $t$ is a function of offset $d$, so the result is an array of size $n\_t \times n\_d$, where $n$ is the number of samples.**

The inputs are a vector $\tau$ of travel time at zero offset, or $t(0)$, an array $p$ of [ray parameter](http://wiki.seg.org/wiki/Dictionary:Raypath_parameter), which is in turn a function of velocity $V$ and angle, and a vector $d$ (usually called $x$) of offset.

Here's an example of some visualizations of the $\tau$–$p$ domain from [a paper](http://www.ahay.org/RSF/book/milano/taupvel/paper_html/) by Sergey Fomel; they might help explain what I mean by an 'array' — you can think of them as images:

![enter image description here](https://i.sstatic.net/fxLZ1.png)

The resulting travel-time $t(d)$ is an array of size $n\_t \times n\_d$, where $n$ is the number of samples. For a *given* offset at a *given* time sample, the result is a scalar — a single value of $t$.

By the way, **username_2**'s answer is a great explanation of how exactly to calculate the ray parameter term, given limited information. Do follow that link to Yilmaz, and maybe also read [the Radon transform article](http://wiki.seg.org/wiki/The_radon_transform).


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