Effect of velocity change on VSP traveltime

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Problem 13.5a

A source is offset 3.00 km west of a vertical well in which a geophone is suspended. There is a vertical north-south fault 0.80 km west of the well with velocities $ V=4.00 $ and 3.00 km/s west and east of the fault, respectively. A horizontal reflector is present west of the fault at a depth of 2.00 km. Find the reflection traveltime for geophone depths of 0.60 and 1.20 km.

Solution

This problem must be solved by trial-and-error methods, either graphical or numerical. We have chosen the latter, because it is easier, quicker, and more accurate than graphical methods.

Figure 13.5a shows the geometry. The horizontal reflector at a depth of 2 km has an image point 4.00 km below the source. A typical ray is shown going from the image point to a geophone located in the well at a depth $ h $.

Figure 13.5a.  Geometry of the problem.

The angle of incidence $ \theta _{1} $ and of refraction $ \theta _{2} $ where the ray passes through the fault are related by Snell’s law: that is, $ \sin \theta _{2}=(3.00/4.00)\sin \theta _{1}=0.75\sin \theta _{1} $. We have a further constraint in that the vertical distance between the image and the geophone is $ (4.00-h) $. These relations fix $ h $:

  1. $ \sin \theta _{2}=0.75\sin \theta _{1} $;
  2. $ 2.20\tan \theta _{1}+0.80\tan \theta _{2}=(4.00-h) $.

We denote the left-hand side of (b) by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): L . For the geophone at depth 0.60 km, the bending of the ray will have a small effect because of the shallowness and proximity to the fault, so we take as our first approximation the straight path,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} \theta _{1} \approx \tan^{-1} (4.00-0.60)/3.00\approx 49^{\circ}. \end{align}

To allow for some bending, we try Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{1} =51^{\circ} and get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{2} =35.7^{\circ} . Substituting theses values in relation (b) we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} L=2.20\tan 51^{\circ} +0.80\tan 35.7^{\circ} =3.29\ {\rm km}. \end{align}

This should be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): (4.00-0.60) =3.40 , so we must increase Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{1} by a small amount; taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{1} =52^{\circ} gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{2} =36.2^{\circ} and $ L=3.40\ {\rm {m}} $. We now find the traveltime:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} {\rm time} =(2.20/4.00\cos \theta _{1} )+(0.800/3.00\cos \theta _{2} )=1.224\ {\rm s}. \end{align}

For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): h=1.20\ {\rm km} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): L=2.80\ {\rm km} . We know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{1} will be less than for the shallower geophone, so we take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{1} =45^{\circ} and get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{2} =32.0^{\circ} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): L=2.70 . Taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{1} =47.0^{\circ} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{2} =33.3^{\circ} , $ L=2.88 $. The change of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): 2^{\circ} increased Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): L by 0.18 m; we need an increase of 0.10, so we take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{1} =45^{\circ} +2^{\circ} \times (0.10/0.18)=46.1^{\circ} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \theta _{2} =32.7^{\circ} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): L=2.80\ {\rm km} . The traveltime is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): \begin{align} {\rm time} =(2.20/4.00\cos 46.1^{\circ} )+(0.800/3.00\cos 32.7^{\circ} )=1.110\ {\rm s}. \end{align}

Problem 13.5b

What is the deepest geophone location for which the reflection can be recorded?

Solution

The deepest geophone position is that for which the ray passes through the intersection of the fault and reflector. For this point,

$ {\begin{aligned}\theta _{1}=\tan ^{-1}(2.00/2.20)=42.3^{\circ },\quad \theta _{2}=30.3^{\circ },\\L=2.20\tan 42.3^{\circ }+0.80\tan 30.3^{\circ }=2.47\ {\rm {km}},\\(400-h)=2.47,\quad h=1.53\ {\rm {km}}.\end{aligned}} $

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Effect of velocity change on VSP traveltime
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