The British Caving Association

Any views expressed are not necessarily those of the BCA
It is currently Sat 16 Dec 2017 23:22

All times are UTC [ DST ]




Post new topic Reply to topic  [ 12 posts ] 
Author Message
PostPosted: Sun 08 Dec 2013 21:58 
Offline

Joined: Sun 08 Dec 2013 20:56
Posts: 5
It looks like it's a bit quiet round here but I thought I'd ask a question anyway...

I have a site with a few small sinks which for a variety of reasons are unlikely to be dug so I was wondering what, if any, home-brew geophysical techniques might be practical? I think the pie-in-the-sky/holy-grail would be to use the water as some sort of antenna and be able to trace it's location by detecting the signal emitted from it, but I assume that this is not practical.

Seismic seems like one relatively easy option but I'm not sure of it's usefulness for locating caves. Anybody got any idea what would be involved in building a resistivity type device - i.e. voltages etc.?

Any thoughts?


Top
 Profile  
 
PostPosted: Thu 12 Dec 2013 13:53 
Offline
User avatar

Joined: Thu 16 Mar 2006 23:45
Posts: 460
TheBitterEnd wrote:
It looks like it's a bit quiet round here but I thought I'd ask a question anyway...

I have a site with a few small sinks which for a variety of reasons are unlikely to be dug so I was wondering what, if any, home-brew geophysical techniques might be practical? I think the pie-in-the-sky/holy-grail would be to use the water as some sort of antenna and be able to trace it's location by detecting the signal emitted from it, but I assume that this is not practical.

Unfortunately, that is not only "not practical" but defies physics to some extent - unless of course you believe in dowsing. For large bodies of water, there may be some geophysical effect but for a small sink, as you describe, Im afraid your scheme is not plausible.
Quote:
Seismic seems like one relatively easy option but I'm not sure of it's usefulness for locating caves.

Im not sure I would describe seismic as "easy" but it depends on what equipment you have access to, I guess. My knowledge of seismic techniques is almost zero but I have heard that the problem is that the technique does not readily distinguish between one large air space and multiple small air spaces. So you could not tell the difference between navigable cave and, say, a large area of gravel or fractured rock through which the water is (possibly) percolating.
Quote:
Anybody got any idea what would be involved in building a resistivity type device - i.e. voltages etc.?

Many people have investigated resistivity techniques but, again, the interpretation is very difficult. You really need a tomographic array, and tomographic processing techniques and, even then, success is limited. The way tomographic algorithms tend to work is that they assume an answer and then see if the assumption was valid - in other words, they can only find what you are expecting to find. There have been quite a few claims where people say that ERT (electrical resistance tomography) has "found caves" but I remain a little skeptical. The technique certainly wont find detailed small passages - it suffers from the same problem as a seismic tomography technique. All it can do is tell you if there is likely to be a hole underneath you.

As for voltages and currents - there is a very large range of ground conductivity to contend with. perhaps you should be prepared to use the maximum "safe" voltage, say 25V, and to use a very narrow bandwidth for the measurement. After all, the voltage doesnt really matter - what matters is the signal/noise ratio. You might find a few useful references in the CREG journals http://bcra.org.uk/cregj , but most of these are not yet online, apart from a contents list.


Top
 Profile  
 
PostPosted: Thu 12 Dec 2013 14:40 
Offline

Joined: Sun 08 Dec 2013 20:56
Posts: 5
Thanks for taking the time to respond. In digging around this subject I came across this

http://scholarcommons.usf.edu/cgi/viewcontent.cgi?article=1020&context=ijs

which seems to confirm that even when you know there is a cave it still at best hard to interpret.

But my interest is piqued and the big difference I see over traditional geophys is that I know where the water sinks and water is more conductive than rock. I might keep it on the project pile...


Top
 Profile  
 
PostPosted: Thu 12 Dec 2013 16:38 
Offline
User avatar

Joined: Thu 16 Mar 2006 23:45
Posts: 460
TheBitterEnd wrote:
Thanks for taking the time to respond. In digging around this subject I came across this

http://scholarcommons.usf.edu/cgi/viewcontent.cgi?article=1020&context=ijs

which seems to confirm that even when you know there is a cave it still at best hard to interpret.

Thanks for that. The paper runs mentions all the common techniques and there is scope for investigating them further. But interpretation is the big problem!

Quote:
But my interest is piqued and the big difference I see over traditional geophys is that I know where the water sinks and water is more conductive than rock. I might keep it on the project pile...

Ah! But the problem is that water is not more conductive than rock. Not unless it is sea water. Some types of dry rock are actually more conductive than (pure) water; and many rocks will be saturated with water anyway. The assumption that free water is somehow a distinctive conductive path is not necessarily true at all - it depends on both the rock and the water. (The main distinction between water and rock is actually not its conductivity, but its high electrical permittivity. But, again, it depends how saturated the rock is). The idea that you could (say) stick two electrodes in an underground stream, upstream and downstream, and track the flow by looking for the magnetic field produced by the current flowing through the water is, unfortunately, flawed because the current would dissipate into the surrounding medium in a very short distance - unless the rock is dry, impervious, and has an inherent conductivity that is several orders of magnitude lower than the water. In general none of that is true, unfortunately. And even it is was... if you calculate the magnetic field resulting from a practical value of applied voltage you will undoubtedly find that it is too small to be detectable. However - at least the model is straightforward enough to be analysed; and a careful analysis might give some useful indications.

If you have a large underground lake or river then the possibilities for detection may be slightly better than than trying to detect the water flowing into a small sink-hole. Or you could put your faith in dowsing. The supporters of dowsing claim that it works really well. Whilst it is difficult to understand how it might work in terms of basic physics, that fact in itself doesnt mean that it doesnt work :-)


Top
 Profile  
 
PostPosted: Thu 12 Dec 2013 17:33 
Offline
User avatar

Joined: Thu 16 Mar 2006 23:45
Posts: 460
TheBitterEnd wrote:
It looks like it's a bit quiet round here but I thought I'd ask a question anyway...

you might want to consider asking it on the Speleonics List as well. That's also "quiet" but, being a traditional Mailing List, your post will be emailed to everyone. See http://lists.altadena.net/mailman/listinfo/speleonics


Top
 Profile  
 
PostPosted: Thu 12 Dec 2013 19:50 
Offline

Joined: Sun 08 Dec 2013 20:56
Posts: 5
I suppose I am a bit shy of mailing lists really, forums seem less pushy but thanks for the details, I hadn't heard of it.

I take all your points but my hunch is that dense limestone is, in its self, not all that hydraulically permeable, unlike say sandstone or chalk. Figures I've seen suggest 2x10^6 ohm-m for surface water and >10^8 ohm-m for dense limestone so a comparison of a "standard" surface resistivity survey with one where one electrode is in the water may be interesting - then again it may not!

...well there you go, I had nice ohm symbols in the above and got a SQL error when posting the reply...


Top
 Profile  
 
PostPosted: Fri 13 Dec 2013 14:18 
Offline
User avatar

Joined: Thu 16 Mar 2006 23:45
Posts: 460
TheBitterEnd wrote:
my hunch is that dense limestone is, in its self, not all that hydraulically permeable, unlike say sandstone or chalk. Figures I've seen suggest 2x10^6 ohm-m for surface water and >10^8 ohm-m for dense limestone

I need to convert those to conductivity to understand them. So... Surface water at 0.5micro-S/m and dense limestone at less than 0.01 micro-S/m. Hmmm! Those are nowhere close to the figures we (in the cave electronics world) are used to dealing with! Of course, nobody has actually done a comprehensive and systematic investigation of rock conductivity in relation to through-rock penetration of radio signals, but everyone has their familiar ball-park range of conductivity. One thing I have noticed is that, in the UK, the rock does seem particularly "wet". Communication in dry limestone in North America seems much easier than through saturated rock in the UK. "We" usually give a range for limestone of 0.1mS/m to 0.1S/m - i.e. a range of 1000:1 for limestone in the UK, and that is many orders of magnitude greater than your figure! I am not a geologist but I would guess that even though, as you say, "dense limestone is, in its self, not all that hydraulically permeable" the permeability would be greatly affected by microfractures and limestone does joint rather well.

Your figure for surface water is also surprising. Ive just done a quick check on the Web and http://en.wikipedia.org/wiki/Conductivity_(electrolytic) says "High quality deionized water has a conductivity of about 5.5 micro-S/m, typical drinking water in the range of 5-50 mS/m, while sea water about 5 S/m". So, if that is correct, your figure for "surface water" is out by perhaps at least 1000 times. (You do have to be careful. There is a lot of bad data out there. One website I found just now quoted seawater as 5mS/m which is out by 1000 times!).

But, supposing your figures are a result of a data conversion error (e.g. cm for m), that may not matter. The important thing is the ratio. You are saying that you expect the ratio of the conductivity of water to limestone to be 50:1. The figures "we" tend to assume would give the ratio as anything from... well, lets say 30:1 for dry rock to 1:30 for wet rock, so the rock property is critical!

I think the conclusion is that we need some enthusiastic experimenter to actually go and make some measurements.

Quote:
...well there you go, I had nice ohm symbols in the above and got a SQL error when posting the reply...

This bulletin-board software parses your typed text for a restricted range of pseudo-html tags (known as BBcode http://en.wikipedia.org/wiki/BBCode). If it was proper HTML you would be able to embed an "HTML entity" like Ω for a capital Greek omega, but I cant see any way of entering that data. Some BBs allow you to embed HTML but I think that's been switched off here, as it introduces a security loophole.


Top
 Profile  
 
PostPosted: Fri 13 Dec 2013 15:13 
Offline

Joined: Sun 08 Dec 2013 20:56
Posts: 5
I've just been and pressed my multimeter probes on a piece of limestone from the garden. The rock was bioclastic great scar limestone and was about 50mm thick and 150x75mm across. One side was "dry" and the other side was "wet" - i.e. it's drizzling and one surface was actually damp but the other side looked dry since the rock was sheltered and on a dry paving slab. I realize this is hopelessly unscientific but it's a start. Roughly the results were anywhere between 0.8M Ohm and 17M Ohm, for either wet or dry sides or between to two sides, the modal reading was about 3M Ohm. There was no correlation between Probe separation resistance.

So, off the back of that little test I'm assuming that there are a variety of factors such as probe contact that come into play. The little reading I've done on this suggests that AC may be better than DC and that the measurements are likely to be frequency dependent. A quick Google did not find anything on standard methods for measuring resistivity of rock.

I feel a bridge and oscillator coming on...


Top
 Profile  
 
PostPosted: Fri 13 Dec 2013 19:14 
Offline
User avatar

Joined: Thu 16 Mar 2006 23:45
Posts: 460
Unfortunately, your readings may not bear any relationship to the intrinsic conductivity (or resistivity if you prefer), because - as you suggest - the probe is a key factor. It is essential to use different sets of probes for injecting the current and measuring the voltage, otherwise you cannot isolate the bulk conductivity figure from effects due to contact resistance of the probe (Clearly the actual resistance you measure depends on the size of the probe. The smaller the probe the larger the resistance). The standard method is a four-terminal array known as a Schlumberger array. However, you need to adapt that for measuring rock samples because the current flow in a small lump of rock is different to that in a large outcrop of rock. The techniques are covered well in geophysics text books. One method involves cutting a measured cylinder of rock and using a particular design of electrodes - the end electrodes of the array, traditionally, made use of a mercury bath, to ensure uniform contract across the sample (but you dont need to go to that extreme. You could use some scrunched up aluminium foil); the middle electrodes are annular. You could use any rough "slab" or "brick" of rock, provided you place the electrodes correctly - a 4-terminal array with the two end electrodes applying a uniform current across the sample and the two central electrodes (e.g. annular rings of copper wire) placed correctly "around the middle". You may get a very different answer to a simple two-terminal ohm-meter reading with pointed probes!

Remember - you are not interested primarily in the "resistance" of the rock sample - you want a method of inferring its bulk resistivity from whatever measurements you can make. Resistance and resistivity are not the same thing - they do not have the same units! Or, to put it another way.... if you measured 1 megohm, how does that relate to the resistivity in "ohm metres"? Where are you going to get your "metres" from? And what do those "metres" actually represent? (Answer: its not a single physical "length"; its a combination of dimensions, hence the need for a carefully-shaped sample of rock).

I would suggest you look at a geophysics book that deals with electrical techniques. Be very cautious of believing anything on the web. I have seen many web pages that have thoroughly confused "resistance" with "resistivity" and which have got very muddled trying to explain how to measure it.


Top
 Profile  
 
PostPosted: Fri 13 Dec 2013 22:12 
Offline
User avatar

Joined: Thu 16 Mar 2006 23:45
Posts: 460
I was thinking further about this... you could make a reasonably good two-terminal measurement if you prepare the electrodes carefully, and use a rock sample with a defined shape. After all, youre only after a ball-park figure - not a laboratory-grade 4-terminal measurement. If you can get a slab of rock that ha a uniform thickness - like a brick or a roof tile shape, for example, you can apply the two electrodes to opposite faces. You need to make a sandwich with layers that are
  • aluminium foil or other metal sheet
  • blotting paper or cloth, soaked in brine
  • slab of rock
  • blotting paper or cloth, soaked in brine
  • aluminium foil or other metal sheet
Press that together firmly and measure the resistance between the two metal sheets. The purpose of the metal is to spread the current over the whole of the face of the rock. The purpose of the cloth is to make good contact with the sample all over the face. The lines of current flow will then be from one face to the other and, you hope, fairly uniform.

You need to measure the area (A), and the thickness (T) of the sample, as well as its resistance R. The resistivity , often given the Greek symbol rho, is then R*A/T. So, for example, if your sample is like a brick and measures 10cm x 20cm x 5cm, it has A = 0.02 m^2, T = 0.05 m, and therefore A/T = 0.4 m. If you measure a resistance of (say) 1000 ohm, the resistivity would therefore be 400 ohm metres, corresponding to a conductivity of 2.5 mS/m.

If your limestone is "dry" you will read a much higher resistance, of course, ... 1 M-ohm would imply 2.5 micro-s/m


Top
 Profile  
 
PostPosted: Sat 14 Dec 2013 13:12 
Offline

Joined: Sun 08 Dec 2013 20:56
Posts: 5
I can certainly do that and its on the project pile. Polishing the ends of a piece of limestone to a reasonably smooth finish is really not that hard so I'm wondering if the brine is really necessary? When I get some time I'll do some tests - I might be more inclined to go with a graphite or aluminium paste of some sort - it's just that brine is horribly corrosive.


Top
 Profile  
 
PostPosted: Sat 14 Dec 2013 14:22 
Offline
User avatar

Joined: Thu 16 Mar 2006 23:45
Posts: 460
TheBitterEnd wrote:
I can certainly do that and its on the project pile. Polishing the ends of a piece of limestone to a reasonably smooth finish is really not that hard so I'm wondering if the brine is really necessary? When I get some time I'll do some tests - I might be more inclined to go with a graphite or aluminium paste of some sort - it's just that brine is horribly corrosive.

The important thing is that you must inject the current over the whole of the face of the rock, otherwise you do not know what the effective area is for your calculation. Im not sure how much contact you would get, even with a polished surface. At a microscopic level, perhaps not a lot? One option, which I have used myself on shale samples I had to investigate, was to use a piece of conductive foam, of the sort that semiconductor chips are transported in. If it is highly compressed it will have a lower resistivity than the rock. But, for you, the same problem would apply - you would not really know the effective contact area. For me, it was not a problem, because I was using a four terminal measurement, for which the contact area is not important. I found these photos from a report I wrote a few years ago. In Fig 2 you can see the two rings of wire I used for the voltage-sensing electrodes. The formula for resistivity is the same as before, i.e. RA/T, but now T is the distance between the two wire rings and R = V/I, where V is the voltage you measure across the two wire rings, and I is the current you feed into the end electrodes. For this set-up, the contact at the ends of the sample does not have to be perfect, because you assume/hope that the current flow "sorts itself out" by the time it gets to the region where you are making the measurement.

Experimental set-up to measure electrical conductivity of a core sample at low frequencies.

Attachment:
File comment: Fig 1: core clamped for 2-terminal measurement using conductive foam pads
temp_adema_D5-1_4v6_image1.jpg
temp_adema_D5-1_4v6_image1.jpg [ 107.74 KiB | Viewed 9152 times ]

Attachment:
File comment: Fig 2: sense electrodes for four-terminal measurement
temp_adema_D5-1_4v6_image2.jpg
temp_adema_D5-1_4v6_image2.jpg [ 100.08 KiB | Viewed 9152 times ]


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 12 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users and 2 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group