Whenever we are investigating multiple layer interfaces in a seismic reflection survey the root-mean-squared velocity is often employed to deduce travel time to a certain n-th layer and also to deduce velocity of any given interval. This is done because it is a way to account for refraction that is also occurring between layers which alters the reflection of a shot made through multiple layers.

What I don't understand is how do we know the travel times in any layer beforehand? How are we distinguishing between different travel times and assigning them to different layers we've distinguished? We need travel times of successive layers in order to compute the root-mean-square, but this is puzzling to me because how could you already know them? Who is to say it is multiple and not just 1? or 3, or 5 or some arbitrary number? What about the arrivals to the geophones tell us this?

I don't get how we know there are extra layers in the first place to which we must assign travel times, and when they do exist, from where do these travel times come? I have included a graphic to help show what I mean. The more I think about this, the more I start to think I don't understand multiple layer seismic reflection in general.

Source: UC Berkeley Dept. Earth & Planetary Science’s Richard Allen’s Applied Geophysics GEO594/GLE594 notes for Lecture 15: Seismic reflection - II

enter image description here

  • $\begingroup$ And that is why oil exploration ,where dozens of different layers of rock are traversed by the sound waves require the biggest computer capacities and significant times to process., $\endgroup$ Jun 18, 2019 at 19:46

1 Answer 1


It is impossible to to convert one recorded time-series into structural information. As you correctly note, the equation $$ v_i = \frac{z_i}{t_i}, \tag{1a}$$ or $$ t_i(z,v) = \frac{z_i}{v_i}, \tag{1b}$$ is difficult to solve when you only a know one time $t_i$. That is, choosing $z_i$ to twice its value would provide twice $v_i$! If you only record time $t$, but try to turn it into two values $(z_i,v_i)$, you have an underdetermined problem.

However, if you have multiple traces, you know from the theory that a flat reflector produces a hyperbolic move-out (I believe it's the following equation, but I'm not 100% certain. Don't quote me on it!): $$ t_i(x,z,v) = t_0 + \frac{x^2}{v_\text{stacking}}. \tag{3}$$ As you know the location $x$, knowledge of two (rather than one) traces would perfectly constrain $t_0$ and $v_\text{stacking}$! Just like the slides you provide in the link, the velocity $v_\text{stacking}$ is usually simply found by trying many values for the move-out correction velocity, and seeing which velocity value flattens the reflector best! From that knowledge, suddenly you can constrain also the depth, etc.

Reflection seismics deals with the situation that is described in the 5 slides preceding the slide you post. We have a lot of traces, and based on the assumption of hyperbolic moveouts we can perfectly obtain the stacking velocities. The slide you post approaches the problem from the other way around: what are the stacking velocities $v_\text{rms}$ you would find if you know the depth and velocity structure already? This is not necessarily the method used to get structural information from your data! I think that is the cause of the confusion..!


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.