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On the slide I am reading, it says this:

Any surface related multiple can be construced of primaries:

$\hspace{5cm}$Multiple = Primary$_1$ (green) + Primary $_2$ (red)

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$$T_2 = T_1 + \left( \frac{1}{v} \frac{2h}{\cos (\theta)} \right)$$

where

$$\cos(\theta) = \frac{h}{\sqrt{\frac{x^2}{4} + h^2}}.$$

I think $v$ is supposed to represent the velocity in between the layers of the rocks. My question is where has that $\cos$ bit come from? Why does it equal that?

Also, am I correct in saying that what this equation says is that the time it takes for our second primary to complete is the same as the time it takes the first primary to complete, plus that little bit on the end?

This is all the information on the slide I am reading so sorry there isn't much information.

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  • $\begingroup$ Seismic waves ? $\endgroup$ – John Rennie Nov 26 '13 at 18:44
  • $\begingroup$ @JohnRennie YEapp, surface related multiple attenuation. I'm trying to figure understand why the amplitude changes as you convolve the data with itself, but without much physical maths knowledge, I'm struggling a bit. I just posted another question about converting to frequency domain, if you don't mind having a look? $\endgroup$ – Kaish Nov 26 '13 at 19:16
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In the equation:

$$ T_2 = T_1 + \left( \frac{1}{v} \frac{2h}{\cos (\theta)} \right) $$

the bit in brackets is presumably the time taken for some signal to travel from the surface down the red line, reflect at the first bounday then travel back to the surface again. Let us pick out half the red triangle:

Distance

If the signal velocity is $v$, and the length of the red line is $d$, then the time taken to travel along the red line is just $d/v$, so the total time from the surface and back is:

$$t = \frac{2d}{v} $$

If $\theta$ is the angle shown then basic trigonometry tells us that:

$$ d \cos \theta = h $$

therefore:

$$ d = \frac{h}{\cos\theta} $$

and substituting for $d$ gives:

$$ t = \frac{2h}{v \cos\theta} $$

and this matches the eqution from the slide.

The last equation we get from Pythagoras' theorem that tells us:

$$ d^2 = h^2 + \left(\frac{x}{2}\right)^2 = h^2 + \frac{x^2}{4} $$

so:

$$ d = \sqrt{h^2 + \frac{x^2}{4}} $$

From an earlier equation we know that:

$$ \cos\theta = \frac{h}{d} $$

and substituting for $d$ gives:

$$ \cos\theta = \frac{h}{\sqrt{h^2 + \frac{x^2}{4}}} $$

I'm guessing that this is the measurement of seismic waves. The wave travels down the green line, reflects off the lower boundary, travels back to the surface and then reflects off the surface and follows the red line. When it reaches the surface again the total time is the time taken to travel along the green line ($T_1$ ?) plus the time we've just calculated above.

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