<p>This is conceptually relatively simple if you look at where the reflection (and refraction) happen, and in fact you have articulated the key concept in your question:</p>

<blockquote>
  <p>The reflected wave ...Over time it becomes more horizontal and therefore more like a direct wave which takes a shorter and quicker path across the surface. </p>
</blockquote>

<p>This is absolutely correct. So if you are close to the seismic source (distance a, where a is small), the direct wave is quickest, as the reflected wave has to first travel down to the boundary before being reflected back. It travels a little over 2b, where b is the depth of the boundary layer. That "little over" depends on the angle, which is related to the distance a.</p>

<p>As an example, look at the image below, from <a href="https://slideplayer.com/slide/5354822/" rel="nofollow noreferrer">this presentation on reflected, refracted and diffracted waves</a>. The 2nd of the reflected waves is not much larger than the first, however the direct wave has travelled 4 times as far.</p>

<p><a href="https://i.sstatic.net/lBAEG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lBAEG.png" alt="enter image description here"></a></p>

<p>If a is much larger than b, then that 2b portion becomes less and less significant, and in fact you can say the distance travelled by the reflected wave tends towards a.</p>

<p>So the graph does not show that the velocity of the wave is slowing down, more that the distance relationship with the angle of reflection tends to an asymptote.</p>

CC BY-SA 4.0
2019-05-01T20:47:06.470Z
On the plot though, the reflected wave curves upward to approach the direct wave.  It's slope increases which on this diagram means it's slowing down,  not speeding up.  How could it have to slow down to converge with the direct wave? It doesn't make physical sense. Unless this is just a side effect of the reflection equation being nonlinear with respect to distance
MattGeo
CC BY-SA 4.0
2019-05-01T20:52:50.260Z
No, the graph indicates the equation relationship, not a slowing down.
Rory Alsop
CC BY-SA 4.0
2019-05-01T16:45:39.210Z
2019-05-03T12:13:39.343Z
2019-05-03T12:13:39.343Z
Rory Alsop
<p>You might be viewing the problem in wrong light. Given this graph we want to find the velocity V1 (a parameter that defines the shape of hyperbola). V1 has effect on shape of hyperbola as whole. </p>

<p>Instantaneous velocity does not make sense here. I would term it as "apparent velocity", as you are using single point data rather than the whole hyperbola to judge velocity of the medium.</p>

<p>Hyperbolic equation is <span class="math-container">$t^2= (x/v1)^2 + (2h/v1)^2$</span></p>

<p>You can use <span class="math-container">$X^2-T^2$</span> plot and find the slope and therefore <span class="math-container">$v1$</span> using linear regression. </p>

CC BY-SA 4.0
2019-06-14T22:20:09.220Z
2019-06-14T22:20:09.220Z
shul
<p>I do not understand how the reflected wave in a seismic survey has a steepening slope (slows down)  Please bear with me as I lay out my thoughts because I feel like I can grasp all of this except this one thing and I am struggling to reconcile it. </p>

<p>If a wave source produces waves at a point, as in a seismic geophysical survey there are 3 types of waves which we consider. There is the direct wave, the reflected wave, and the refracted wave. The reflected wave is never a first arrival. The refracted wave is first lagging behind the direct wave before reaching it at the crossover point and overtaking it. This makes sense because the refracted wave has a higher speed but it does not register its propagation at the first geophone until a certain time has elapsed since the initial pulse at the wave source; and the direct wave has been traveling from the source all the while. </p>

<p>On an arrival time plot where time is the vertical y-axis and where distance is the horizontal x-axis we see that the refracted wave has an unphysical beginning at the y intercept and the direct wave begins normally from the origin. Where the intersect the refracted wave overtakes the direct wave. On this type of diagram the smaller the slope, the faster the wave. The slope of the refracted wave is 1/v2 and the slope of the direct wave is simply 1/v1.   </p>

<p><a href="https://i.sstatic.net/Iefet.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Iefet.jpg" alt="enter image description here"></a></p>

<p>What I do not understand is why the reflected wave has a hyperbolic or curved trajectory on this plot. The slope steepens as it asymptotically approaches the direct wave. It makes sense that over long distances that the direct wave and the reflected wave should merge to almost being indistinguishable, but how is it possible that the reflected wave plot as a steepening slope i.e. is slowing down?  The reflected wave travels less  surface distance per unit time the smaller its angle of incidence ( meaning closer to the source) because it is more vertical than horizontal. Over time it becomes more horizontal and therefore more like a direct wave which takes a shorter and quicker path across the surface.  What am I missing? Why does the reflection plot curve as it does?</p>

<geophysics><seismology><waves><seismic>
Travel Times for Seismic Wave Types - Why reflected t-x plot curves
MattGeo
# Travel Times for Seismic Wave Types - Why reflected t-x plot curves

(Asked by: username_0 on 2019-04-30)

I do not understand how the reflected wave in a seismic survey has a steepening slope (slows down) Please bear with me as I lay out my thoughts because I feel like I can grasp all of this except this one thing and I am struggling to reconcile it.

If a wave source produces waves at a point, as in a seismic geophysical survey there are 3 types of waves which we consider. There is the direct wave, the reflected wave, and the refracted wave. The reflected wave is never a first arrival. The refracted wave is first lagging behind the direct wave before reaching it at the crossover point and overtaking it. This makes sense because the refracted wave has a higher speed but it does not register its propagation at the first geophone until a certain time has elapsed since the initial pulse at the wave source; and the direct wave has been traveling from the source all the while.

On an arrival time plot where time is the vertical y-axis and where distance is the horizontal x-axis we see that the refracted wave has an unphysical beginning at the y intercept and the direct wave begins normally from the origin. Where the intersect the refracted wave overtakes the direct wave. On this type of diagram the smaller the slope, the faster the wave. The slope of the refracted wave is 1/v2 and the slope of the direct wave is simply 1/v1.

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

What I do not understand is why the reflected wave has a hyperbolic or curved trajectory on this plot. The slope steepens as it asymptotically approaches the direct wave. It makes sense that over long distances that the direct wave and the reflected wave should merge to almost being indistinguishable, but how is it possible that the reflected wave plot as a steepening slope i.e. is slowing down? The reflected wave travels less surface distance per unit time the smaller its angle of incidence ( meaning closer to the source) because it is more vertical than horizontal. Over time it becomes more horizontal and therefore more like a direct wave which takes a shorter and quicker path across the surface. What am I missing? Why does the reflection plot curve as it does?

**Tags:** geophysics, seismology, waves, seismic


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## 2 answers

### Answer by username_1 on 2019-05-01. Score: 3

This is conceptually relatively simple if you look at where the reflection (and refraction) happen, and in fact you have articulated the key concept in your question:

> The reflected wave ...Over time it becomes more horizontal and therefore more like a direct wave which takes a shorter and quicker path across the surface.

This is absolutely correct. So if you are close to the seismic source (distance a, where a is small), the direct wave is quickest, as the reflected wave has to first travel down to the boundary before being reflected back. It travels a little over 2b, where b is the depth of the boundary layer. That "little over" depends on the angle, which is related to the distance a.

As an example, look at the image below, from [this presentation on reflected, refracted and diffracted waves](https://slideplayer.com/slide/5354822/). The 2nd of the reflected waves is not much larger than the first, however the direct wave has travelled 4 times as far.

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

If a is much larger than b, then that 2b portion becomes less and less significant, and in fact you can say the distance travelled by the reflected wave tends towards a.

So the graph does not show that the velocity of the wave is slowing down, more that the distance relationship with the angle of reflection tends to an asymptote.

> **Comments:**
> **username_0**: On the plot though, the reflected wave curves upward to approach the direct wave.  It's slope increases which on this diagram means it's slowing down,  not speeding up.  How could it have to slow down to converge with the direct wave? It doesn't make physical sense. Unless this is just a side effect of the reflection equation being nonlinear with respect to distance  
> **username_1**: No, the graph indicates the equation relationship, not a slowing down.  

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### Answer by username_2 on 2019-06-14. Score: 1

You might be viewing the problem in wrong light. Given this graph we want to find the velocity V1 (a parameter that defines the shape of hyperbola). V1 has effect on shape of hyperbola as whole.

Instantaneous velocity does not make sense here. I would term it as "apparent velocity", as you are using single point data rather than the whole hyperbola to judge velocity of the medium.

Hyperbolic equation is $t^2= (x/v1)^2 + (2h/v1)^2$

You can use $X^2-T^2$ plot and find the slope and therefore $v1$ using linear regression.


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