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5.2: Seismic Refraction (Single Layer)

  • Page ID
    3542
  • One of the most important uses of seismic waves is for seismic surveys. Seismic surveys can be used to determine shallow (~100s of meters) crustal structure. The surveys shed light on the layers of rock, including each ones thickness and seismic velocity (and from seismic velocity we can estimate rock type). We can also determine structures such as faults and basins, history of deposition, and look for oil and gas. Primarily, we use P-waves for seismic surveys as they are faster and easier to see on seismograms. There are two types of seismic surveys

    1. Refraction ie 'critical refraction' which is useful for determining velocities of layers
    2. Reflection, which is useful for determining layers and structure

    In seismic surveys we can use geometry and travel time of waves to determine structure and velocity. Today we are going to focus on refraction surveys. First, we are going to quickly review what we learned in the last section: slowness/Snell's Law and Reflection and Transmission 

    5.2 Lecture 2 Review Snell's Law.png

    Figure \(\PageIndex{1}\): Snell's Law

    \[\frac{\sin \theta_i}{v_1}=\frac{\sin \theta_r}{v_2}\]

    \(v_2>v_1\) is the most common interface in the earth, and \(\theta_r>\theta_i\) is due to \(v_2>v_1\).

    Now that we've reviewed, let's first look at two sample ray paths.

    The direct wave travels horizontally along the surface. The Rs represent receiving stations. What is the arrival time at each station? 

    5.2 DIrect Ray.png

    Figure \(\PageIndex{2}\): Direct Wave

    \[distance=velocity\cdot time\]

    \[x=v_1\cdot t\]

    Solve for t.

    \[t=\frac{x}{v_1}\]

    The above equation is the direct ray travel/arrival time. The equation is a line whose slope is \(\frac{1}{v_1}\) and intercept=0.

    5.2 Plot Time.png

    Figure \(\PageIndex{3}\): Plot of Direct Ray Travel/Arrival Time

    Using the vertical incidence of a reflected ray, we can derive an equation to calculate the time the ray took to reach the receiver:

    \[d=v\cdot t\]

    \[2h=v_1\cdot t\]

    \[t=\frac{2h}{v_1}\]

    5.2 Vertical Incidence Reflected Ray.png

    Figure \(\PageIndex{4}\): Vertical Incidence of a Reflected Ray

    Now, let's look at what happens as \(\theta_i\) increases.

    5.2 Draw Big So Have Space.png

    Figure \(\PageIndex{5}\): Changes as \(\theta_i\) Increases

    We can see from the figure that after the hammer blow/weight drop/small explosion (the seismic wave source), the waves go out in all directions. If there is a large angle reflection of a wave, all energy is reflected and none is refracted. This is called total reflection. When \(\theta_r\) increases until 90o, the wave is critically refracted. The critically refracted ray is crucial to surveys. 

    5.2 Critically Refracted Ray.png

    Figure \(\PageIndex{6}\): Critically Refracted Ray

    The critically refracted ray generates a 'head wave'. Receivers (R1, R2, R3) on the surface detect head wave at t1, t2, t3, which correspond with increasing distance from the source. We can use geometry and arrival times to determine v1, v2, and h. One of the cons of momentum and fitting boundary conditions is that it sends energy back into the top layer. 

    Now we're going to look at refraction geometry 

    5.2 Refraction Geometry.png

    Figure \(\PageIndex{7}\): Refraction Geometry

    We know that 

    \[\frac{\sin\theta_{ic}}{v_1}=\frac{sin90^o}{v_2}\]

    \[\sin\theta_{ic}=\frac{v_1}{v_2}\]

    To get refraction, we must have v2>v1. Looking at the above figure, how do we figure out what are the distances s1, s2, and s3? What are velocities v1, v2, and v3? To get the travel time for each segment,

    \[t=t_1+t_2+t_3\]

    \[t=\frac{s_1}{v_1}+\frac{s_2}{v_2}+\frac{s_3}{v_3}\]

    We know that s1=s3 so:

    \[t=\frac{2s_1}{v_1}+s_3\]

    Now we need to know 'what is s1'?

    5.2 s1 Triangle.png

    Figure \(\PageIndex{8}\): Refraction Geometry Variables

    From the triangle we can derive that:

    \[s_1=\frac{h}{\cos\theta_{ic}}\]

    \[x_1=h\cdot tan\theta_{ic}\]

    \[s_2=x-2x_1\]

    Substituting:

    \[s_2=x-2h\cdot tan\theta_{ic}\]

    \[t=\frac{2h}{v_1\cos\theta_{ic}} +\frac{(x-2h\cdot tan\theta_{ic})}{v_2}\]

    We can use 

    \[tan\theta_{ic}=\frac{\sin\theta_{ic}}{\cos\theta_{ic}}\]

    We know that \(\sin\theta_{ic}=\frac{v_1}{v_2}\) from \(\frac{\sin\theta_{ic}}{v_1}=\frac{sin(90)}{v_2}\) and that \(sin^2\theta_{ic}+cos^2\theta_{ic}=1\)

    \[t=\frac{x}{v_2}+\frac{2h(v^2_2-v^2_1)^{\frac{1}{2}}}{v_1v_2}\]

    Or we can write it as

    \[t=\frac{x}{v_1}+\frac{2h}{v_1}(1-\frac{v^2_1}{v^2_2})^{\frac{1}{2}}\]

    So what kind of equation is this exactly? Its a line with slope \(m=\frac{1}{v_2}\) and \(t(o)=\frac{2h(v^2_2-v^2_1)^{\frac{1}{2}}}{v_1v_2}\)

    5.2 x crit.png

    Figure \(\PageIndex{9}\): Xcritical

    From the above figure we can also derive that 

    \[x_{crit}=2x_1\]

    \[x_{crit}=2h\cdot tan\theta_{ic}\]

    Another important value that can be graphed is the crossover distance, or xco

    5.2 Xcros.png

    Figure \(\PageIndex{10}\): Xcrossover

    The xco is when the tdir=thead(rfr)

    \[x_{co}=2h(\frac{v_2+v_1}{v_2-v_1})^{\frac{1}{2}}\]

    To summarize what we learned in this section:

    How do you determine subsurface structure?

    • Identify direct wave arrivals
      • Measure slope→m1
      • Calculate v1(layer 1)→\(v_1=\frac{1}{m_1}\)
    • Identify headwave arrivals
      • Measure slope, m, intercept, to
      • Calculate v2(layer 2)→\(v_2=\frac{1}{m_2}\)
    • Use to, v1, and v2 to calculate h
      • \(t_o=\frac{2h(v^2_2-v^2_1)^{\frac{1}{2}}}{v_1v_2}\)
      • \(h=\frac{t_ov_1v_2}{2(v^2_2-v^2_1)^{\frac{1}{2}}}\)