Ch 3 - Seismic Exploration - Refraction MethodImportant to our early understanding of Earth structureRefrac also important to oil in 1920s, before reflecStill extensively used for shallow work

I. Review homogeneous subsurfacetime vs distance Fig 3-1 as wavefront reaches each

geophoneslope = 1 V1

thus we calculate velocity easily from a t-d curve (Fig 3-2)(remember to keep track of units!! Like seconds vs milliseconds..

II. Now bring one interface into the pictureFig 3-3 , with the path of the headwave via the critical angle

A. Derive the travel-time equationtotal travel time = d1 + d2 + d3 V1 V2 V1

bunch of math leads to the same result as the direct rayrefracted ray has slope of 1 V2

B. Analyze the arrival times Fig 3-4 and Table 3-1 show the data

look at table 3-1 - at what shot point does the refr wave beat the direct?Can always take the data and compute the lines…, if you need a visual clue.As an exercise, take the data in Table 3-1, make a graph with the 2 line segments

C. Calc thickness

We can calc thickness of layer 1 using the T-t equation, at t = 0:

Equa 3-17: h1 = ti V1xV2

2 (V22 - V1

2)1/2

D. Thickness using crossover distance

alternative equa can be used with crossover distance from travel time graph

Equa 3-21: h1 = xco [ V2 - V1 ]1/2

2 [(V2 + V1)]

E. Critical Distancex crit is the minimum dist required for refracted energy to emerge…any less than this distance, no refractions are possible

Fig 3-5 shows this as essentially a reflection at the critical angleyou need thickness h and velocities v1 and v2 to solve:

x crit = 2 h1 (V2/V1)2 - 1 1/2

now on to the good stuff…

F. Construct a travel-time curve fig 3-7 from a field record Fig 3-61. Pick the first break for each trace (pt when the signal first hits the

geophone 2. enter these data in a table format (table 3-3)

3.construct a time-distance graph (by hand or computer)4. Pick the lines that go through the points5. Calculate the slopes of the lines (1/slope = velocity of that layer)6. Determine intercept time7. Calculate thickness of layer 1

II. 2 interfaces, 3 layersthe model here: 3 diff velocity layers (only 2 diff materials)

Dry sand

Wet sand

Granite

A. Layer thickness Fig 3-9 shows the ray paths through dry sand, wet sand, and then along graniteuse 2-layer procedure to get h1, then use this equa to get h2:

h2= ti2 - 2h1 (V32 - V1

2) 1/2 V3V2 V3V1 2 (V3

2 - V22) 1/2

B. Critical distance equa for multilayer case pretty involved - Equa 3-38

C. Second field seismogram to analyze - Fig 3-13Do same procedure as 2-layer case:

1. Pick first arrivals2. Build a spreadsheet showing first break time vs distance3. Create a travel-time curve from the spreadsheet4. Determine t=0 intercepts, slopes of line segments, velocities

III. Multiple interfaces

Equa 3-39 - travel time equation derived for n layers

time n = x + 2 hi (Vn2 - Vi2)1/2

Vn Vn Vi

IV. Dipping interfacesA. Analyzing problem Fig 3-15

dipping interface produces shorter path to geophones, less time, faster apparent velocity

• Note that the gap between horiz and dipping interface t-t curves INCREASES as signal moves to right..progressively shorter last leg up through V1 causes shorter and shorter travel times to the geophones the further up dip you get

Now move to forward and reverse traverses (source position moved from one end to the other) Fig 3-16

Reciprocity holds with a horizontal interface: Tf = Tr, that is, it takes equal amt of time for signal to go forward to end of line or backward to beginning of line.

•In this case, signal takes 160 ms to get to opposite end of line, 125 m away

but move to a dipping interface Fig 3-17, and t-t curves no longer the same for forward and backward, save the last geophone, where pathf = pathr

so moral of the story is, in the field we always collect data both updip and downdip, so we can see non-horiz interfaces

note all the geophones on the forward traverse receive the signal later than on the reverse traverse, until the last phone…path longer on the forward shoot up until the last phone, when paths are equal

Calculating the travel times for a dipping interfaceV1 values can be found by standard graph, but not V2

V2 will be the avg of V2f and V2r or can determine with ic equation after solving for ic

good rule of thumb - “intercept time is less for the traverse with its energy source at the updip end, where the vertical distance to the interface is the least”

•Derivation - need to keep the symbols straight…Fig 3-18•We need to solve several parameters, in some type of sequence:

•first, recall that you must measure mu, md, and V1 directly from t-t graphwhere mu is inverse slope of V2 shot with the source updipmd is inverse slope of V2 shot with source downdip

you can calculate ic = sin-1 (V1 md) + sin-1 (V1 mu)(crit angle equa) 2

you can calculate = sin-1 (V1 md) - sin-1 (V1 mu)(slope angle) 2

Determine thickness“we are not quite finished”

thickness h for the updip and downdip portions first we need to get jd and ju Fig 3-19

jd = tid V1

2 cos ic ju = tiu V1

2 cos ic

then hd = jd cos

and hu = jiu cos

Now on to

IV. Multiple dipping interfaces Fig 3-20

travel time equations simply laid out, then Burger suggests getting more familiar with computer programs to solve …..

•Analyzing field seismograms - good exercises•Fig 3-21, forward and reverse profiles

•note reason for 4 lines, not 2….they have to splice one 12-channel line to another…..make sure you see this (handout)

•could be a good computer exercise - Burger picks result in Fig 3-22•sight down those picks..might you do them differently??

•Remember, it’s your call…YOU are the interpreter

•concerning layer 1 - note that only one pt defines…Burger pts out that if you need more definition, put your phones closer together, get more direct wave arrivals to pin down the V of this near-surface layer

•concerning layer 2 - the intercepts are similar in time for forward and reverse - suggesting what about dip of the top if this layer?? Hint…not much

•layer 3 - intercepts suggest signif dip

V. move to the non-ideal subsurfacea variety of topics pertaining to realities of subsurface

A. hidden zone: low velocity layer beneath a high velocity layerconsequences:

•no crit refraction occurs (ray doesn’t move forward)•no refracted energy returns to surface•all energy received at phones is due to direct wave (or air wave)•no evidence of an interface (all we see is the surface layer velocity)

Fig 3-24 shows situation of surf layer, then low V, then high V

First line segment comes from the surface layer, slope 1/V1

second line segment mostly from the third layer, slope 1/V3

there are ways to get around this, but they are difficult..leave it for time being as “inability of refraction method to detect low-velocity layers is one of it’s major shortcomings” (p.99)

B. hidden zone: thin layer Fig 3-25problem: “all head waves from the first interface arrive later than from second interface”..so implication is all the first breaks are coming from head wave from below second interface…if we were able to pick the arrival of head wave immed below the first interface we would see it as the slow segment in Fig 3-25

C. lateral velocity change - refer to Fig 3-27•forward shoot from low to high V looks like a typical 2-layer case

•but reverse shoot looks pretty weird…first segment with the high V, then second segment with low V

•note too that the discontinuity is in the same general area as the crossover pts of the curves

•and note that the inverse slopes represent V1 and V2.

fig 3-28 looks very messy; lateral variation buried under a layer

as in 2-layer case, velocities are accurately represented by inverse slopes, and the discontinuity is near the crossover points of the lines

D. Interface discontinuities (like a step across a fault scarp) fig 3-29

•a different thickness V1 on each side of scarp results in a displaced 1/V2 line on both forward and reverse traverses…line stays parallel to orig line (slope is inverse of V2), but is offset and comes in either:

•earlier in time (reverse case, with a thinner V1 zone and therefore faster time of arrival)•later in time (forward case, with a thicker V1 zone and slower time of arrival)

you can collect the data necessary to find the height of the fault:

z = (ti2 - ti1) V2V1

(V22 - V12)1/2

one last topic in the chapter….

VI. Delay -Time Methoduseful for mapping irregular surfaces

lots of calcs boil down to the ability to place a value hG which is depth to the refractor under each geophone

first find the delay time under each geophone:

TG = tEFG + tERG - tR

2then hG = TG V1V2

(V22 - V12)1/2

skip to last section - VII. Applications using Seismic Refraction•Whately, MA - shallow geology beyond resolution of the geophys tools, but geophys found bedrock, ultimately they discovered a buried river valley that became new aquifer•SE New Hampshire - Geophys helped determine water table geometry, found it mimicked the bedrock elevation more than the surface elevations•Maricopa, AZ - good example of slight diffs from multiple data sets - which data would YOU trust???