2
$\begingroup$

I am taking an introductory class in seismology, but have some difficulties understanding the logic behind the formula used to calculate the time it takes for a refracted wave to return to the surface from which it was sent out originally.

So, say we have a seismic wave sent out from a horizontal layer. We also assume that the interfaces below are horizontal, and parallel to the top layer. If the wave hits the first interface at the so-called critical angle, it will be reflected back to the surface with the same angle at which it hit the interface. Now, the formula given in the book states that in order to calculate travel time, we use:

$$t = \frac{2 h_1 cos(i_c)}{v_1}$$

where $h_1$ is the vertical height of the layer, $i_c$ is the critical angle, and $v_1$ is the velocity of the wave as it travels through the layer.

As mentioned, I don't quite see the logic of this formula. I know that time = distance/velocity. Surely the shortest possible distance to travel back and forth between two horizontal layers, separated with distance $h_1$ is simply $2 h_1$. So by that logic, the shortest possible time a wave could possibly use, would be:

$$t = \frac{2 h_1}{v_1}$$

But in the given formula we multiply the numerator with $cos(i_c)$, which would yield an even shorter value for distance. And I just can not see how this makes sense at all. Now, if we were to use the cosine-term in the denominator, then the formula would make perfect sense to me. But why isn't it so? Any help here would be greatly appreciated!

Improve this question
$\endgroup$
2
  • $\begingroup$ I think, and I remark that I think (I need to review my optics notes) that you can derive that formula via Snell's law and imposing minimum condition, so that the sine in Snell gives you a cosine in the derivative $\endgroup$
    – J L
    Commented Jan 24, 2013 at 0:37
  • $\begingroup$ Thanks. I would really, really be grateful if you could provide me with the derivation. $\endgroup$
    – user12277
    Commented Jan 24, 2013 at 8:19

1 Answer 1

Reset to default
1
$\begingroup$

I think the formula you show is not the travel time for a refracted ray. It is T0, the intercept of the time-distance curve with the x-axis for the refracted ray.

See this figure, and associated formula, and the difference should be apparent. Perhaps your text is in error or unclear.

Keep in mind that multiple ray paths can reach the receiver location - including a direct ray, reflected ray, and refracted ray. Which ray is the first arrival depends upon the underlying structure and distance from the source. The refracted ray is the first arrival at distances greater than XCROSS1.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Ah! Thanks a lot. Yes, you are absolutely right. I mixed up the t-intercept and the travel time. Now it all makes sense. $\endgroup$
    – user12277
    Commented Jan 27, 2013 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.