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I hope this is the right place to ask this question.

Suppose I found a small irregular shaped rock, and I wish to find the surface area of the rock experimentally. Unlike for volume, where I can simply use Archimedes principle, I cannot think of a way to find the surface area. I would prefer an accuracy to at least one hundredth of the stone size.

How can I find the surface area experimentally?

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24 Answers 24

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I would ignore answers that say the surface area is ill-defined. In any realistic situation you have a lower limit for how fine a resolution is meaningful. This is like a pedant who says that hydrogen has an ill-defined volume because the electron wavefunction has no hard cutoff. Technically true, but practically not meaningful.

My recommendation is an optical profilometer, which can measure the surface area quite well (for length scales above 400nm). This method uses a coherent laser beam and interferometry to map the topography of the material's surface. Once you have the topography you can integrate it to get the surface area.

Advantages of this method include: non-contact, non-destructive, variable surface area resolution to suit your needs, very fast (seconds to minutes), doesn't require any consumables besides electricity.

Disadvantages include: you have to flip over your rock to get all sides and stitch them together to get the total topography, the instruments are too expensive for casual hobbyists (many thousands of dollars), no atomic resolution (but Scanning tunneling microscopy is better for that).

The optics for these instruments look like below From omniscan

And it gives a topographic map like below. enter image description here
(source: psu.edu)

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    $\begingroup$ If the rock has little tunnels inside of it this method would need to somehow be able to scan inside the tunnels. $\endgroup$
    – Thomas
    Commented Nov 11, 2019 at 6:08
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    $\begingroup$ "doesn't require anything besides electricity": um. It requires very expensive and sophisticated equipment. Electricity is the least of your concerns if you want to use this (very good) suggestion. $\endgroup$
    – terdon
    Commented Nov 11, 2019 at 10:19
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    $\begingroup$ @terdon, on the scale of standard laboratory analysis equipment, these optical profilometers are neither expensive nor sophisticated (compared to e.g., STM, AFM, SEM, FTIR, XRD). I just meant to say that it doesn't require consumables for analysis, you just need a wall outlet. I've edited accordingly. Though I totally agree that it this is too expensive for non-commercial hobbyists. $\endgroup$
    – KF Gauss
    Commented Nov 11, 2019 at 15:04
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    $\begingroup$ @KFGauss , "Once you have the topography you can integrate it to get the surface area." - NOT TRUE. I am not sure how you propose to integrate, but if it is via a 3d mesh, like some other people here, then, depending on the mesh, it can produce an arbitrarily large result, infinity in the limit, even if the shape is perfectly smooth and convex. Even if you keep adding points to the mesh, and shrinking the triangles, it DOES NOT lead to the convergence of the result to the true area!!! $\endgroup$
    – Kphysics
    Commented Nov 12, 2019 at 10:08
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    $\begingroup$ @Kostas, clearly you didn't read the first paragraph of my answer. Treating real life materials as fractals is incorrect and overly pedantic. $\endgroup$
    – KF Gauss
    Commented Nov 12, 2019 at 13:14
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The problem is that as you increase the measurement precision, so will increase the result you get. The result of a meaningful experiment should converge with the increase of the precision, this does not.

This is a 3D analogy of the coastline paradox: the surface of the rock is fractal-like, just like the coasts of the countries:

enter image description here

The result will say more from your measurement precision, than from a real surface area.

Obviously if the stone is not very fractal-like (like it is a sphere, or a skulpture), then the result will converge, but not this is the general case in practical stones.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris ♦
    Commented Nov 10, 2019 at 23:41
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    $\begingroup$ This $\endgroup$
    – Felipe Gutierrez
    Commented Nov 12, 2019 at 17:44
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The way I would do it is to first dip the rock in thinned fingernail polish. Let that dry, and then dip the rock into hot liquid wax. Let the wax cool. Peel the wax off the rock and measure the thickness of the wax layer. Melt the peeled-off wax and measure its volume. Divide the volume by the thickness, and you have the area.

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    $\begingroup$ is assumes equal thickness. I appreciate the finger nail polish so that wax is not absorbedm but I think it will not peal off. What about the same logic with caramel? Let it solidify, then break it off and weight it. $\endgroup$
    – anna v
    Commented Nov 9, 2019 at 15:13
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    $\begingroup$ Probably the right way is to weigh yourself, then eat the caramel, then weigh yourself again. But it would be necessary to know the density of the caramel. $\endgroup$
    – S. McGrew
    Commented Nov 9, 2019 at 17:33
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    $\begingroup$ This would give a lower bound on the surface area, in the same way that you could get a lower bound on the length of England's coastline by using a one-page map of England. $\endgroup$
    – user4552
    Commented Nov 9, 2019 at 23:26
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    $\begingroup$ That is true. The "actual" surface is (approximately) fractal, and on the order of the surface area of a crystal grain times the number of grains in the rock. However, I'm guessing that the OP is looking for the surface area that the "caramel method" measures. $\endgroup$
    – S. McGrew
    Commented Nov 9, 2019 at 23:45
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    $\begingroup$ -1 from me. This is going to be really imprecise because the wax will get dragged to very inequal thickness by gravity. $\endgroup$
    – leftaroundabout
    Commented Nov 11, 2019 at 4:04
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I would prefer an accuracy to at least one hundredth of the stone size.

  1. Weigh the stone.
  2. Dip the stone in thin paint; let excess drip off.
  3. Weigh the stone.
  4. Repeat steps 1-3 with a square, 1 cm2 object.

Divide the weight of the stone's paint by the weight of the square's paint to get the surface area of the stone.

This assumes you've defined the "surface area" of a real object, and "the stone size" means a circle of diameter equal to the largest diameter of the stone (or some other reasonable interpretation).

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    $\begingroup$ Liquids would fill cracks where it would not drip out off because of surface tension. $\endgroup$
    – user137289
    Commented Nov 10, 2019 at 8:48
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    $\begingroup$ @Pieter: Those cracks could reasonably be considered to contribute to surface area. $\endgroup$
    – dotancohen
    Commented Nov 10, 2019 at 22:55
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    $\begingroup$ @dotancohen Yes, but the width of liquid in a crack could be much larger than the thickness on a free surface. $\endgroup$
    – user137289
    Commented Nov 10, 2019 at 23:19
  • $\begingroup$ The tradeoff here is that the thinner the paint, the less mass is left behind. This is somewhat equivalent to a different length ruler. You could try a dilute solution of polystyrene in acetone. The rock should then be heated (ideally in a partial vacuum) to help remove the acetone. The thinner the solution, the less effect you will have from surface tension. $\endgroup$
    – cmm
    Commented Nov 11, 2019 at 13:34
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    $\begingroup$ I wonder if one could use a sequence of measurements with increasingly thick "paints" or higher solute load to get a number of measurements at different length scale, and then use that sequence to determine the fractal dimension of the rock surface. $\endgroup$
    – cmm
    Commented Nov 11, 2019 at 13:37
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  1. Fully wrap stone really tight in aluminium foil. (Of course it will crinkle; press the crinkles tightly down.)
  2. Soot the whole thing with a candle, just enough so it's completely black.
  3. Carefully unwrap foil.
  4. Photograph flattened foil together with a reference scale square. Make sure there's a light background (e.g. white ceiling) opposite the foil, so it'll appear bright on the photo in the un-sooted area.
  5. Measure area of soot, using image processing software. This can be done by first using a perspective-correction tool, noting the size of the reference square, then cropping the relevant area and showing a histogram of the brightness values.

Instead of soot, you could also use spray paint, but it would likely sip into the crinkles more. Or you could wrap in paper instead of alu and use a pencil, but that would smear and be harder to see on the photo.

I don't think this method will get 1/100 accuracy, but it gives at least a decent estimate and doesn't require special equipment.

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The task isn't well-defined. Do you include cracks? If yes, you'll see finer and finer cracks adding to the surface, and in the end you'll be at atomic level and find it hard to even define what's part of the rock and what isn't. If you don't include cracks: What's your rule for distinguishing a mere unevenness from a crack?

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    $\begingroup$ This is useful insight, but it does not answer the question directly. $\endgroup$
    – Carter Pape
    Commented Nov 10, 2019 at 8:35
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    $\begingroup$ @CarterPape "The task isn't well-defined." $\endgroup$
    – John Kugelman
    Commented Nov 10, 2019 at 10:43
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    $\begingroup$ From an engineering perspective, the task is well-defined. From a mathematical perspective it is not. As this is a physics website, to which of the professions should we defer? $\endgroup$
    – dotancohen
    Commented Nov 11, 2019 at 7:25
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    $\begingroup$ @dotancohen If it is physically well-defined, are you including the surface of dust particles on the surface? You usually do not want to, but: when is a particle part of the rock? The cutoff will be somewhere on the spectrum of: Gravitational attachment (will fall off when turning the rock), adhesion, water-induced adhesion, adhesion induced by some other substance (grease, resin if it's amber or a concretion that contains amber, etc.), van der Waals bonds. If it got pressed into the surface, maybe some definition like "does not stick out form surrounding grains by more than 50%". Etc. etc. $\endgroup$
    – user132372
    Commented Nov 11, 2019 at 10:54
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    $\begingroup$ @dotancohen I'd argue that it's not well defined from an engineering perspective, because you'd want extra clarification for the purpose of measuring the surface area, which would allow to estimate whether a proposed method will result in a metric that is meaningful for that purpose. $\endgroup$
    – Peteris
    Commented Nov 11, 2019 at 17:41
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Difficult. Adsorb some chemical, heat it up, measure the amount that evaporates?

I would look at the literature, maybe start with a search for "experimental determination of the surface area" in geological contexts.

Edit: a molecular probe should give something close to the maximum value. There is an end to the length scale when dealing with real materials, a rock is not a mathematical fractal. After letting in a suitable kind of molecules and pumping them out, thermally stimulated desorption would measure the absorbing area.

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    $\begingroup$ BET technique is typically used, but that is a molecular technique which will determine the surface area with all the microscopic nooks and crannies. That will be MUCH larger than the gross surface area for a (semi)porous rock surface. $\endgroup$
    – Bill N
    Commented Nov 9, 2019 at 15:33
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    $\begingroup$ I interviewed at a small instrumentation company where this technique is the basis of a major product line. With some materials they use temperature dependent outgassing under vacuum as the probe. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 9, 2019 at 19:28
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    $\begingroup$ @BenCrowell I realize that, but a molecular probe should give something close to the maximum value. There is an end to the length scale when dealing with real materials, a rock is not a mathematical fractal. $\endgroup$
    – user137289
    Commented Nov 10, 2019 at 0:19
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    $\begingroup$ I would suggest you edit your answer to say that. And I'm not really sure that there is a maximum value that can meaningfully be interpreted as an area. Suppose you diffuse helium into sandstone. Aren't you then really measuring the vacant volume of the sandstone, not its surface area? $\endgroup$
    – user4552
    Commented Nov 10, 2019 at 1:42
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    $\begingroup$ @Ben These measurements are generally made for a reason. The problem isn’t to define the epistemology of the question. It is to match the meanings of the measurement to that needed for application. When the application is catalyzing gas phase reactions as the reactants pass though a porous plug the gas-absorption or out-gassing measurements serve well. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 10, 2019 at 6:07
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For non-convex bodies of arbitrary shape, as many people already pointed out, there is no reasonable answer in general. For convex bodies the answer is mathematically and physically well defined. The method is based on integral geometry, if I remember right, the formula is due to Steiner or Crofton. Nevertheless it is a practical method, and stable. The formula gives the surface area in terms of the (average area) of the projection of the body along all directions $\vec{n}$: $$ S = \frac{1}{\pi} \int d\Omega_{\vec n} ~ S(\vec n) = 4 \times \left< S(\vec n) \right>$$ So all you need to do is place a lamp high above, hold the stone in many random directions, calculate the average area of the shadow, and multiply by 4. For 1% accuracy, ten thousand (10,000) random projections will be enough.

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    $\begingroup$ This is interesting and I +1'd it, but a rock is highly unlikely to be convex. I wonder how the error in applying this method to non-convex bodies varies relative to some reasonable measure of how non-convex they are (perhaps something bounded-variation-like?). $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Nov 11, 2019 at 13:03
  • $\begingroup$ @R.. For some non-convex shapes, you might be able to slice the object with a flat cut, then measure and add up the area of the resulting convex pieces then substract twice the area of the cuts. If the object is "fractal" so that no finite slicing will produce convex pieces, then I suggest to just stop there, as this body does not have a well defined area at all. Not mathematically, and not physically. $\endgroup$
    – Kphysics
    Commented Nov 12, 2019 at 9:45
  • $\begingroup$ Formula is due to Cauchy $\endgroup$
    – Bananach
    Commented Apr 25, 2020 at 22:35
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I'm vouching for vapor deposition or application of coating which can be measured by weight.

Depends on your rock size. Open cell pumice and porous limestone bored by rain and animals will be difficult to measure. Limestone can be micro-porous and can have hundreds of square meters of surface area. Consider this micrograph of chalk.

Using a substance that adheres strongly and evenly to the surface of rocks regardless of their pH and chemical affinities, vapor-treat or dip the rock in the coating substance, use an effective way of removing the excess, and weigh the rock/substance afterwards. Perhaps you can get a degree of precision if there is a substance which can be applied in a perfectly even coat for all the different samples.

For the first attempt, I'd use water-vapour. Weigh the rock dry, subject it to a few moments in a high-humidity environment, and weigh it again afterwards.

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    $\begingroup$ This is just adding details to the general method that S. McGrew mentioned in the very first answer to this question, and should probably be a comment on that answer. $\endgroup$
    – dotancohen
    Commented Nov 11, 2019 at 7:27
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Take a lot of pictures from different angles that enable you to create a 3D mesh using photogrammetry (I recommend Meshroom). You could also use a LIDAR to capture the point-cloud then use Meshroom to mesh it (apparently there are some cheap ones from less than 2000 dollars). Calculate the area of the mesh (I recommend Rhinoceros3d). There are many open-source tools that can help you with the process. EDIT: someone already gave a similar answer, so I added some software recommendations (I know they are usually out of place for stackexchange but if OP really wants to sove the problem rather than hypothetically post a cool question the recommendations could be handy). If you are going the photogrammetry way, remember that if the surface is specular you will need to coat it in a diffuse paint.

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  • $\begingroup$ Might want to coat the rock somehow before picturing it, to give it favorable optical properties. $\endgroup$
    – Nat
    Commented Nov 12, 2019 at 2:18
  • $\begingroup$ @Nat thanks for that I am gonna update the answer I forgot about that. $\endgroup$
    – Felipe Gutierrez
    Commented Nov 12, 2019 at 2:24
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    $\begingroup$ -1 This method, depending on the 3d mesh can produce an arbitrarily large result, infinity in the limit, even if the shape of the stone is perfectly smooth and convex. Adding points to the mesh, and shrinking the triangles DOES NOT lead to the convergence of the result to the true area!!! $\endgroup$
    – Kphysics
    Commented Nov 12, 2019 at 10:13
  • $\begingroup$ I didn't see this answer when I wrote a similar one, but this predates mine by 2 days, which I suspect is why mine was downvoted. Deleted mine, upvoted this, as generating a point cloud is clearly the most-correct answer for any object that lacks occlusive overhangs. I mean, there's literally an app for this! I'd note that this fails for occlusive overhangs (fossilized skulls, pumice, etc), though. $\endgroup$
    – Dewi Morgan
    Commented Nov 14, 2019 at 22:23
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Similar to the gas adsorption/BET and @McGrew wax technique. You would need a sensitive scale with centigram precision.

  1. Make a mono-layer of sand of a known area (for example a square meter). Measure the mass of that sand. This is your calibration/conversion ratio.
  2. Measure the mass of the rock.
  3. Wet the rock and coat it with a mono-layer of sand. Re-measure the mass and calculate the mass of the sand attached.
  4. Use your calibration from #1 to find the area.
  5. Repeat 3 or 4 times to determine an average and uncertainty.
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    $\begingroup$ Wouldn't that measure the mass of the sand + water? I would think that'd be non-negligible. What about a variant in which after you coat the rock, you wash that sand off into a clean tray whose mass you know, then let that water evaporate, and then re-measure the tray and subtract to find the mass of the sand? $\endgroup$
    – yshavit
    Commented Nov 10, 2019 at 0:34
  • $\begingroup$ The result here is going to depend on how fine-grained the sand is. In other words, this answer is yet another example of the fact that the quantity being discussed in not well defined, unless you specify some other parameter that sets a scale. $\endgroup$
    – user4552
    Commented Nov 10, 2019 at 1:44
  • $\begingroup$ You could try charging the rock electrostatically and coating it with a monolayer of polystyrene beads. This would probably be impractically fiddly, but it would remove the water mass problem. $\endgroup$
    – N. Virgo
    Commented Nov 11, 2019 at 8:17
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enter image description hereYou can also measure the location of points on the stone for some fixed point say the (0,0,0) point in close proximation.

Map the points using Octave (free and open-source) or Matlab Mathematical software. Form 3d triangle meshes with that point. Compute the area of the triangles. Add them. And that's it. The Surface Area.

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    $\begingroup$ As a practical matter of mensuration, this is do-able. The real purpose of the measurement, though, will determine the scale of the necessary model (paint coverage, you'd want point spacing comparable to paint layer thickness). $\endgroup$
    – Whit3rd
    Commented Nov 11, 2019 at 3:54
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    $\begingroup$ The first word of the question is "how". How does one measure the location of points on the stone? $\endgroup$
    – dotancohen
    Commented Nov 11, 2019 at 7:29
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    $\begingroup$ @dotancohen: You could certainly construct a device like a caliper, but with an arm assembly with 3 or more (probably more) degrees of freedom, that gives you cartesian coordinates of the tip relative to the base as output. I don't know if such a thing is readily available, and it would be some work to create one, but it's definitely doable. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Nov 11, 2019 at 12:49
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    $\begingroup$ An alternative would be to setup jigs along each of three walls whereby you can move a laser distance finder to each point in the two axes of the wall you want to test and get the third coordinate as the distance measured. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Nov 11, 2019 at 12:52
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    $\begingroup$ @R: Again, how? Notice the word "experimentally" in the title question. To what degree of precision (never mind the accuracy) do you think that you could design (never mind build) such a contraption? In theory I'm sure that "build a machine" is an answer, but in practicality how does one design and build such a machine (never mind the cost)? $\endgroup$
    – dotancohen
    Commented Nov 11, 2019 at 15:16
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If you have access to a planimeter, then you might try the method used in this research paper on the strength of cements used on teeth.

In order to compare the strength of the cement, the authors needed to separate the effect due to the cement from effect due to the varying surface areas of the real teeth used in the tests.

For each tooth used, the authors laid aluminum foil over the teeth and used a burnishing tool to make the foil follow the contour of the surface of each tooth. Overlapping areas were then cut away and the foil was removed from the the tooth then pressed flat. A tracing of the outline of each piece of foil was made, and the area measured using a planimeter.

I happen to have bought a planimeter of the exact same model as used in the referenced paper, and actually found that paper while searching the internet for information about the planimeter I had just bought at a fleamarket.

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You can put the rock in an MRI scanner and get a 3D profile of it (and therefore, the volume and surface area). If it doesn't have spins that are useful for NMR, you can dunk the rock in something that does (ie. water or mineral oil), and then image that, and the void will give you the 3D profile of the rock (which you can then use to calculate the area).

The main problem of using NMR is that if the magnetic susceptibility of your rock is very different than that of vacuum... you will get image artifacts. But there are tricks around this though.

As an example: here is a lithium dendrite inside a battery imaged using MRI.

enter image description here

Alternatively, you can use Xray images of your rock from many different angles and reconstruct the 3D profile of the rock using the inverse 3D Radon Transform. Having the 3D profile, you can easily calculate the area.

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    $\begingroup$ If he bathes the rock in a lead/radioactive based solvent which is very visible in MRI/Xray that would be cool, although expensive because MRI's are about 300-500 per hour. $\endgroup$
    – bandybabboon
    Commented Nov 11, 2019 at 7:36
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because the rock is in irregular shapes (mostly), it is hard to use some normal surface measuring method for the regular 3-D objects. Of course one can use closed integrals to go calculation, but that is boring. it will be easier if we can change the 3-D object surface into 2-D.

I will recommend if you have a bucket of sticky fluid, you can dip the rock into it and let it dry. then use some papers to fit into it and you can get the result. however, this is not accurate.

I will more recommend you to scan the rocket into 3-D models to make the computer do the job using accurate algorisms.

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  • $\begingroup$ This is just adding details to the general method that S. McGrew mentioned in the very first answer to this question, and should probably be a comment on that answer. $\endgroup$
    – dotancohen
    Commented Nov 11, 2019 at 7:27
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In most cases like this, the best option is to scan the object into a dense cloud somehow and measure the approximate surface area using the tools provided. Although I am certain that a variety of exotic methods exist for generating dense clouds, your best options would be to either use a LIDAR unit of some kind or to use a camera and a photogrammetry program. Depending on how detailed you want your estimate to be, you could use anything from a dedicated 3D scanning setup to a few tens of photos taken on your phone and one of many free photogrammetry programs.

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I would use rice or sand, With the volume measured you can measure the area by pouring the sand or rice into a flat tray and making sure that you have a one-grain thickness carried over the tray, You will then be able to not only see the physical representation but measure it as well. I have done this many times myself when figuring out the outer surface areas of my parts.

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    $\begingroup$ Can you try to explain better? I don't follow what you mean by this. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Commented Nov 11, 2019 at 13:01
  • $\begingroup$ You aren't going to get the surface that way: Pouring the sand will destroy that information immediately. $\endgroup$
    – user132372
    Commented Nov 12, 2019 at 10:08
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Here's a more solution-oriented answer which takes the clarification into account:

  • Spray the stone with some conductive paint.
  • Electroplate it.
  • Measure the amount of metal being deposited on the stone.

This is essentially equivalent to the wax method, except electroplating is not affected by gravity.

I am a bit hazy about how to best measure the amount of metal; please feel free to suggest improvements or directly edit this answer.
The most direct approach I can think of would be to measure the loss of metal on the counterelectrode.

Do you want to restore the stone to its pre-measurement state?
You'd probably select a metal and a paint that are easy to remove.
Again, somebody with more practical knowledge of electroplating may be able to help with some advice what materials to use.

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  • $\begingroup$ Downvoter(s?), please add a comment so I know what can be improved about this answer. $\endgroup$
    – user132372
    Commented Nov 15, 2019 at 8:47
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Why not try electrophoretic deposition? You would know the average thickness based on the statistics from engineering specs/data for whatever deposited material. You would also be able to calculate the added volume from Archimedes' principle. You also know the mass density so you could then calculate the surface area of the deposited film/material.

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Dip the rock into motor oil. Take it out and and let it drip dry for about a hour. Then place the rock into a container filled with water. For the next few days occasionally move the rock around in the water with the objective of removing the oil from the rock. The area of oil slick on the water’s surface will match the surface area of the rock. If needed you can maneuver the slick into a geometric shaped that can easily be measured.

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  • $\begingroup$ No reason to believe the area of the oil slick equals the area of the rock - though it might be proportional. To do this experiment you would take a piece of the same rock but of known dimensions, and repeat with that. This gives a scaling term that you would most certainly need. $\endgroup$
    – Floris
    Commented Mar 27, 2021 at 14:27
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I think it can work this way also .........use a fast-drying paint bottle to sprinkle the rock...1. measure the rock volume (Archimede law)...2....measure the 'shortest' distance between two selected points on oposite sides of the rock by a caliper 3. spray the rock uniformly(or just use a paintbrush)....4...measure the new distance between same two selected points of the rock...5. ...measure the new rock volume... ---step 4 gives you indirectly how much tickness the paint layer has because you must subtract the new distance from the old one and divide with number 2.... ---step 5 gives you how much volume the paint layer has which you get from subtracting the old rock volume from the new rock volume... Now procede to step 6. which is an arithmetic job of dividing volume of the paint layer by its tickness to get the area of the paint that is approximately the area of the rock....

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Depends on the tools you have at your disposal; I'll describe an expensive and inexpensive approach:

  • Expensive: scan the rock, use software to process it & compute area. Medical imaging involves topologies lot harder to measure than a rock, but it's done.
  • Inexpensive: wrap a balloon, or a more stretchable and flexible fabric fully around the rock, cut it at the wrapping tip; unwrapped fabric's area a lot easier to measure/calculate.

The underlying idea's the same: we map 1D slices of the rock onto a 2D surface to model its 3D shape, then obtain a surface area estimate. With the 'expensive' option, this mapping is very granular and precise - with latter, it's as good as your balloon and your wrapping procedure (how it covers bumps, ridges, whether there are empty gaps, etc) - but won't compete with a scan.

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  • $\begingroup$ Both these methods have been mentioned in previous answers. $\endgroup$
    – dotancohen
    Commented Nov 11, 2019 at 7:30
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    $\begingroup$ How would stretchy material help? When you remove it the area will change, so you won't get a good measurement. $\endgroup$
    – N. Virgo
    Commented Nov 11, 2019 at 8:19
  • $\begingroup$ @Nathaniel "cut it at the wrapping tip" - so whatever remains, measure its surface area (i.e. by unwrapping it) $\endgroup$
    – OverLordGoldDragon
    Commented Nov 11, 2019 at 16:18
  • $\begingroup$ If the material is stretchable then it’s area when wrapping the surface is not necessarily the same as it’s area when relaxed and laid out. That issue comes up after you clip the excess. And it’s not the only issue: this finds a convex hill, not the actual surface. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 11, 2019 at 22:52
  • $\begingroup$ @dmckee Indeed, hence the "as good as your balloon and your wrapping procedure"; it's inexpensive, so it can only do so well. With the right fabric (of which I wouldn't know), however, all these 'artifacts' can be diminished, maybe to even within 1% accuracy $\endgroup$
    – OverLordGoldDragon
    Commented Nov 11, 2019 at 22:56
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Peterh is correct that the task is ill-defined, while the various suggestions offer ways to measure it that define what is being measured and then calculate that measurement to various degrees of accuracy. However I would say that all of the offered definitions of measurement are highly arbitrary; what you really want is a method that is, in some sense, a natural meaning for surface area.

I suggest to you that a natural definition of surface area is the area through which heat is lost since this represents a real and well defined physical property of the object.

The rate of heat loss of a body is proportional to this surface area; therefore in order to calculate the surface area of your rock, what you need to do is raise its temperature to a known value and then calculate how long it takes for the rock to lose temperature. From this measurement you can calculate how fast the rock is losing heat energy. To convert that into an actual surface area you will need to understand the thermal properties of the rock, and so will either need a sample of similar rock or need to sacrifice a proportion of the rock for testing.

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3
  • $\begingroup$ To do this, you would need a very accurate estimate of the heat transfer coefficient; which depends on the geometry itself. I'm not sure how accurate you would expect this to be, especially since this wouldn't necessarily work for conduction or radiation either. $\endgroup$
    – JMac
    Commented Nov 12, 2019 at 14:47
  • $\begingroup$ Heat loss will be lower per unit area for a highly concave surface such as pumice, so this wouldn't work so well. $\endgroup$
    – Dewi Morgan
    Commented Nov 12, 2019 at 15:33
  • $\begingroup$ The rate of heat loss is proportional to the effective surface area. Concave sections of the surface have less effective area than convex sections. $\endgroup$
    – Mark
    Commented Nov 16, 2019 at 1:53
-6
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Use Nuclear Magnetic Resonance imaging to calculate the position of every atom in the rock. Then count how many atoms border an empty space that is connected to the space outside the rock.

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