Geophysics question - How do you compute the depth of a layer of bedrock, and how far should the geo

farmerman should be able to answer your question. I'll try to find him, and let him know about your question.

Im not going to give an answer but your on the righttrack with the exception that you should go back to Snells Law and rearrange some of the postulate wrt to it. You apparently already have the initial data for first and second refractors and you can therefore calculate a salient point where the
velocities cross and that is the depth to the refractor. Remember to consider your correction factors for thinner layers effect on velocities of propogation.

you've already got your v1/v2 numbers so you 'could" solve this by just plotting your given times against the velocities and then (based upon your known depth of the overburden) arrange your geophones graphically (e have programs that do that, maybe all hes trying to do is make you think about the rules and how you can bend them while learning them)

Theres so much analyses that are done by computer graphic methods and crunching long algebraic functions that , by understanding what goes into them (sorta thinking for the program) you will be given some extra insights for when you do some field work and are given a budget.

Snells Law is - N1Sin(angle) = N2Sin(angle)

N is the refractive 'indice', which i assume is the same as the refraction coeffeicant? Because i see no mention of indices, only coefficients in my slides.

And the refractive index is worked out by (Z2-Z1)/(Z2+Z1)
where Z=pv p=Density v=velocity
Correct?
Which gives me a refractive index of 0.26

But i have no information about what angle the wave is refracting at/what angle its entering at, so what now?

Snells Law is - N1Sin(angle) = N2Sin(angle)

N is the refractive 'indice', which i assume is the same as the refraction coeffeicant? Because i see no mention of indices, only coefficients in my slides.

And the refractive index is worked out by (Z2-Z1)/(Z2+Z1)
where Z=pv p=Density v=velocity
Correct?
Which gives me a refractive index of 0.26
Nevermind, i dont have density values so that number was gotten using velocities only and so must be wrong.
I dont even no what the first layer is so it is impossible to know its density. Maybe i am supposed to know the density of Sandstone off hand, but for the first layer all i have is velocity and an approx thickness. Can i work out density using thickness and velocity??

And then i have no information about what angle the wave is refracting at/what angle its entering at, despite knowing the law of refraction is Sin(angle)/v1 = Sin (angle)/v2 if i dont know what angle its entering out, what am i supposed to do?

I know absolutely nothing about this topic, but found this on the net.

First, is this to be a reflective survey, or a refractive survey.

If it is a reflective survey, one only needs the overburden velocity, v1, and not v2, which typically is a larger value to help reflect seismic wave energy upward. The equation for this would then be (without advantage of a diagram):

V^2*t^2 = x^2 + 4*h^2

where V = seimic speed of wave in medium, in this case v1,
t = travel time of wave from surface entry to surface exit,
h = height or thickness of layer.

I assume that the height of the layer is 4 miles (or 4 km?), not 4 meters, although not clear in the description.

If h, x = 4km, then t = 15.012 seconds.

The geophone spacing should be about a half wavelength apart in the direction of the source so that ground waves are nulled.

If this is a refraction survey, the second layer velocity is important, as one wants to know the critical angle, or where the incident wave angle with respect to the surface normal will not absorb into the second layer, but rather glide along it while continually sending refracted energy back to the surface.

The critical angle is computed as:

sin(i1) = sin(i2)*v1/v2 = 1*1.4/2.4 = .5833, with i2 = 90 deg.
i1 = i(critical) = 35.68 deg.

With this information, one assumes that the refracted wave arrival time at the geophones is the up and down travel times of a simple reflected wave to the layer at 35.68 degrees, using V^2*t^2 = x^2 + 4*h^2, plus the travel along the top of the second layer at its speed of 2.4 km/s. If the geophones are far enough away from the shot source, the refracted wave will arrive before the ground wave, unlike the reflection survey, since the second layer's seismic velocity is faster than the first layer.

Refraction surveys are usually done with shallow layers, since the refracted wave dampens with distance through the upper layer and along the lower layers.

Walter Goedecke
Boulder, Colo.

Its a refractive survey, i believe. And m would be for meteres.

I dont understand how you did this bit
sin(i1) = sin(i2)*v1/v2 = 1*1.4/2.4 = .5833,
I follow upto here, but then this,
with i2 = 90 deg.i1 = i(critical) = 35.68 deg.
Why is i2/i1 90? And where does 35.68 come from?

Ive done a seach for 'critical angle' and it gave me the formula Sin(crit angle)=V1/V2, So i would have stopped after .5833 and said 58degrees was the critical angle!

The rest though is absolutely fantastic, thank you so much

To invert for the incident angle that causes the second layer ray to glide along the interface:

sin(i1)/sin(i2) = v1/v2, by Snell's law
sin(i1) = v1/v2, since sin(i2) = sin(90 deg) = 1.0 at i1 = i(critical)
i1 = arcsin(v1/v2) = arcsin(.5833) = 35.683 deg.

Walter