Identifying Materials Using Archimedes' Principle.

[Pete W.]

In this thread, I want to review and demonstrate a method of identifying materials (mostly metals) using Archimedes' Principle.

I'm not targeting all metals (especially NOT gold, for which Archimedes thought up this whole idea to start with!) but mainly distinguishing between aluminium alloy and the zinc-based die-casting alloys (aka 'Zamak').

In my own situation, this facility helps me fend off any scrap metal dealer who tries to avoid paying me the aluminium price for my old computer hard drive chassis by claiming that they are zinc-based.  I suggest that the method will also be useful to Mad Modders who do their own casting.

What Archimedes realised was that an object immersed in a liquid experiences a reduction in weight equal to the weight of the liquid it displaces, i.e. a quantity of the liquid that has the same volume as the immersed object.  Using this principle enables us to determine the specific gravity of the object - knowing the specific gravity, we can consult any of the various tables of physical properies of materials and hence identify the metal.

Specific gravity is defined as the ratio of the weight of a given volume of the material to the weight of the same volume of water.  If you're a physicist, you'd want that to be pure water at Standard Temperature & Pressure (aka 'STP') but I'm a Mad Modder and so I'm using ordinary tap water!

I plan to explain three methods of getting to the specific gravity of our 'mystery object', the first one requires a weighing machine that suspends the object rather than having it in a scale pan.  The fisherman's spring balance would do, I've used a modern digital gizmo (see photos, below).

(My scales read in grams or kilograms but pounds will work too, you just have to use the same units for all measurements.  If your scales read in pounds and ounces, you'll have to express the ounces as decimal parts of a pound.  So, for example, 2 lbs 4 oz = 2.25 lbs.)

The second & third methods substitute a ruler and a counterweight for the weighing machine.  All three methods require a bucket of water and some string.  All three methods require the 'mystery object' to be of only one material, so if it's mostly e.g. aluminium, you need to remove any nuts, bolts, brackets, pins etc., etc., that are made of steel or other material different from that comprising the bulk of the object.  But then if your mystery object is destined for either the crucible or the scrap metal dealer, you'd have to have done that anyway!

The first method is very simple, first weigh the object.  Note its weight - call it 'D'.  Then lower the object into the water, ensuring that it doesn't touch the sides or bottom of the bucket.  Also, make sure the object isn't trapping any air.  Again, note the weight - call it 'W'.

Now, from Archimedes' Principle, the weight of the 'same volume of water' is the difference between D and W, i.e. in mathematical terms (D-W).  So the specific gravity of our 'mystery object' is given by D/(D-W).  In my physics classes, years ago, we used the symbol ρ to denote specific gravity.

OK so far?  Here's an example:

Object #1, dry weight:



then, object #1, immersed weight:



Sorry about the fuzzy focus.  So, we have D=510 grams and W=315 grams, so (D-W)=195 grams. So the specific gravity, ρ=315/195=2.62  From my reference tables, aluminium alloys have specific gravity values around 2.6 so object #1 is aluminium alloy.

Now here's another example:

Object #2, dry weight:



then object #2, immersed weight:



So, we have D=275 grams and W=240 grams, so (D-W)=35 grams.  So the specific gravity, ρ=275/35=7.8  From my reference tables, zinc-based alloys have specific gravity values around 7 to 8 so object #2 is zinc-based alloy.

Notice that the denser the material of the 'mystery object', the smaller is the relative change in weight.

I wrote earlier that I'm using ordinary tap water.  If all you have is sea water, you could use that - you'd just have to multiply the (D-W) term by the specific gravity of sea water, i.e. 1.02, but that's probably splitting hairs!

In my next post, I'll explain methods two and three.  I'll post that as soon as I've drawn and scanned some diagrams to illustrate the methods.

Methods #2 & #3 are going to involve some algebra but if you don't 'do' maths just trust me and go straight to the conclusion!           

[Pete W.]

OK, now let's deal with the case where we don't have any scales.

I don't want to confuse anyone but I'm changing the numbering system slightly, this method is method 2a).

What we need is a rigid beam and some string.  Suspend the beam using one piece of string from approximately its mid-point - support it in any way that's convenient.  Then suspend the 'mystery object' on one side of the beam and a second object, used as a counter-weight, on the other side of the beam.  The string supporting the 'mystery object' needs to be long enough for the object to be immersed in the water in the bucket during the second phase of the operation; for now, leave both the 'mystery object' and the counter-weight swing clear in the air.  Adjust the positions of the pieces of supporting string so as to bring the system into balance.

When the 'mystery object is immersed in the water, it will lose weight so its suspension will have to be moved further out from the main suspension point to restore balance - take this into consideration in positioning the 'mystery object' and the counter-weight.

Now, with the system in the air, measure the distance of the 'mystery object' from the main suspension point of the beam.  Note the value.  (If you wish, you can also measure the corresponding distance to the counter-weight but it doesn't get used!)

Next, either raise the water-filled bucket or lower the beam's main suspension so as to immerse the 'mystery object' in the water.  The same precautions regarding contact with the sides or bottom of the bucket and regarding trapped air apply here as in Method #1. 

Immersing the 'mystery object' disturbs the balance of the system and you will need to re-adjust - in this method, you move ONLY the position of the suspension of the 'mystery object', leave the position of the counter-weight unchanged.

When you're satisfied with the new balance conditions, measure the distance of the 'mystery object' suspension from the main beam suspension point again, note the value.

To follow the next bit, you need to understand 'moments'.  It's not difficult, it's all to do with leverage.  Consider a see-saw with one side twice as long as the other - then it'll need two people sitting on the short side to balance one person sitting on the long side.  At balance, both sides have the same 'moment' about the pivot.  The 'moment' is equal to the force multiplied by the distance at which it acts.

Now here comes the algebra:

The volume of the 'mystery object' is represented by the symbol 'V', the specific gravity by the symbol 'ρ' and the acceleration due to gravity by the symbol 'g'.  The measured lengths are represented by the symbols on the diagrams that follow.



(I hope that my algebra is readable.)  Remember, in algebra, a point means 'multiplied by'.

In my next post, I'll proceed to Method 2b) in which the 'mystery object' remains in the same place but it's the counter-weight suspension that's moved to restore balance.  That means a similar but slightly different bit of algebra! 

[Pete W.]

I hope you're all still with me!  Here comes method 2b)

The procedure is basically the same as in method 2a) except that to restore balance, it's the counter-weight suspension that's moved and the 'mystery object' suspension is left unchanged.

This needs a slightly different set of algebra, with symbology consistent with that used in method 2a), as follows:



I hope that what I've written is clear - if not I'll try to answer questions.

Right now, I've just been called for the evening meal!!           

[awemawson]

I'm definitely sticking with my Analloy Analyser:

http://madmodder.net/index.php/topic,9244.msg102310.html#msg102310


I seem to remember that aluminium and zinc based alloys can be distinguished by their reaction to acids quite easily

BTW do you have huge quantities of disc chassis, or is this a theoretical analysis?

[Pete W.]

Hi there, Andrew,

Thank you for your post.  I did intend to refer to your marvellous gadget in my introduction - it must be a useful asset but it's one not many of us have.

Regarding your question - why do you want to know?

(There's an anecdote lurking in the undergrowth of that topic but I'm not sure I should recount it on-list!)          

But here's a clue:  Some time ago, my lovely but shy assistant sold quite a few neodymium-iron-boron magnets on eBay - the buyers were motorcyclists whose tank-bags wouldn't stay on at speed. 
One buyer came back with the report that with 'our' magnets, his tank-bag would stay on at 100+ mph, I always wondered how he knew!!          

[awemawson]

Scrap value of ones and twos is peanuts. A big heap makes analysis worthwhile hence the question. If your back garden is piled 10 foot high in scrap disk chassis I'll lend you the analyser to sort through them, but I very much doubt that it is!

When I moved 7 years ago I had accumulated a big pile of alloys that I had cast to ingot for convenience. (Mainly the chassis from several Dataproducts line printers) - rather than strap up yet another pallet to move I sold it all (for a very good price) rationalising that I could thus afford to buy ingots of the specific alloy that I wanted when I set up the furnace again.

[SwarfnStuff]

Ha, thanks for the physics revision. Back down the deep dark recesses among the cobwebs among my grey matter. You were right, I skipped to the answer (final line) of the algebra. Amazing just what stuff we learned actually does come in handy later in life - just have to recall it though.
John B

[vtsteam]

I recently read an article in a woodland management magazine written by a guy who was trying to calculate what tha actual volume of wood was in a cord, minus the air. He went to great lengths trying different methods to calculate it based on measurements, estimations of volumes and densities of diffeent wood species, etc. He still wasn't satisfied with his own answer.

I was tempted to write in and suggest he take a sample of say 8 cubic feet stacked, and then immerse the pieces individually in a graduated tub, to get the exact volume of each irregular piece add them together and divide by 8 to get the wood volume per cubic foot of stack. but figured it was too much trouble to write to a magazine. So here it is, courtesy of Pete, anyway! 

[DavidA]

I'm probably being a bit dim here.

How does this apply to materials that have a SG greater than 1 ?

I.e.  materials that sink.

If the object has a volume of,  say,  100 cc and it is made of steel then it will sink and displace water similar to an object of the same volume made of brass.

This kind of problem usually goes back to Archimedes and the kings crown.

But that was a straight forward matter of weight and displaced water volume of a known material,  gold, and what  ever the crown was made out of.

Dave.

[vtsteam]

David, whatever the object weighs, divded by its submerged displaced volume gives the mean density of the object.

For objects with density greater than water, you can just let it sink on its own. For objects less dense than water, you have to hold them under.

Displacement in naval architecture is a little different. It is the weight of a volume of water displaced by a floating object. The displaced volume is affected by the density of the water (salt vs fresh for instance), because the relative density of the object and the water determine how deeply it sets. Displacement does not vary between fresh water and salt, though displaced volume does, because displacement in this field is expressed as weight.

When objects are completely submerged, the conventional displacement is always equal to the volume of the object, no  matter what the relative densities are. So it's very handy for determining the volume of irregular objects.

[DavidA]

VT,

Thanks for that.  But that is confirming what I said.

Maybe it is the way that Pete phrased it that is causing my confusion.

...What Archimedes realised was that an object immersed in a liquid experiences a reduction in weight equal to the weight of the liquid it displaces, i.e. a quantity of the liquid that has the same volume as the immersed object.

This seems to imply that two object with equal volume but made of different metals will displace different weights of water even though  the displacement will be the same.

A cubic foot of lead weigh much more than a cubic foot of aluminium,  but they both displace the same cubic foot of water.

Dave.

[Pete W.]

Hi there, David,

VT,

Thanks for that.  But that is confirming what I said.

Maybe it is the way that Pete phrased it that is causing my confusion.

...What Archimedes realised was that an object immersed in a liquid experiences a reduction in weight equal to the weight of the liquid it displaces, i.e. a quantity of the liquid that has the same volume as the immersed object.

This seems to imply that two object with equal volume but made of different metals will displace different weights of water even though  the displacement will be the same.

A cubic foot of lead weigh much more than a cubic foot of aluminium,  but they both displace the same cubic foot of water.

Dave.

With all respect, I challenge your implication.  The sentence of mine that you quoted is entirely consistent with your own last sentence.

Archimedes Principle is explained well on Wikipedia, try http://en.wikipedia.org/wiki/Archimedes%27_principle

If you have a spring balance, try it out for yourself. 

[mklotz]

If it's practical to cut a small sample from the subject material it can be machined into a convenient shape (eg, cylinder, cube) and the volume calculated.  Once weighed this sample's density can be computed and found in online tables.

Once the density is known, the volume of the subject material can be found from its weight.

[vtsteam]

The confusion comes I believe, because of the different uses of the word displacement. In naval architecture it is equal to the weight of a floating object. Thus two objects of the same shape and size (and overall volume) may have different displacements. They would have correspondingly different weights out of water, as well. And different effective densities.

But in other uses the word displacement is a volume measure, and is always the same for the same size and shape object. It's equal to the volume of water that an object displaces when submerged. Two objects of different weights and effective densities would always have the same volume and the same displacement.

In fact, this kind of displacement has nothing to do with water, or any fluid. You can measure an object's displacement in alcohol, mercury, or sand, and it will not vary. And the object itself be made of lead or balsa, and it would still have the same displacement.

But, you certainly can determine an object's density using this fact IF you also weigh it, and then divide the weight by the displacement.

[vtsteam]

To add to the fun, here, I can think of a way of determining the weight and density of a material of irregular shape with a calibrated tub, some fresh water, and a balloon, without a scale. Can you?

[DavidA]

Pete,

I am not implying that what you wrote is wrong. I am just having a bit of trouble getting my head around this particular description.  Please do not feel offended.
l
I actually do have a personal interest in this subject as one of my last tasks at my workplace was to make a device that allowed a sample to be suspended in liquid on a very sensitive digital balance.
I never saw how they were using it. but they still have it so it must have worked.

Dave.

[DavidA]

Marv,

Even easier,  just measure the displacement (volume) of the whole thing and then dry weigh it.

Once you have the grams per cubic Centimetre (or whatever) refer to the tables.

Job done.

Dave.

[mklotz]

Even easier,  just measure the displacement (volume) of the whole thing and then dry weigh it.

Measuring the displacement accurately is not very easy.  Calculating the volume of a (simple) shape, accurately measured, can be done more precisely.

[DavidA]

Agreed,

As I do have a balance that weighs to 0,1 gram and can get hold of some Sharpy test pieces (very precise metal blocks 55 X 10 X 10 Millimetre in different metals I am going to give this a try.

Dave.

[RussellT]

As no one else has responded to your challenge Steve.  I think I can do it.  I think it would be easier to use a second tub which would fit inside the first rather than a balloon.

Russell

[Pete W.]

Hi there, all,

Thank you all for your contributions.          

I feel I need to reiterate my opening remarks as to the scope of this process/method.

First of all, while Marv's input is true, machining a precise sample of the material is more something for forensics than for diferentiating between two similar-looking materials prior to charging the foundry crucible or going to the scrap metal merchant.

Secondly, I used the word 'displaced' (adjective), not 'displacement' (noun).  It's usually stated that Archimedes was inspired by seeing the water overflowing from his bath but the customary expression of his principle refers to 'displaced liquid', not 'overflowing liquid'.

I do agree that one could assess the specific gravity of the 'mystery object' by catching the overflowing liquid and measuring either its weight or its volume but, as Marv says, that is not without its difficulties.  Ideally, you'd want to have a controlled size and position of overflow pipe or spout to simplify catching the overflowing liquid and the bath/bucket would need to be precisely pre-filled to the overflow level.  But David A and, I think, Steve, have offered that process in conjunction with weighing the object dry.  I submit that my method #1, comprising two weighings with the same instrument (one dry, the other wet) is much quicker, simpler and more in keeping with the overall aim.

My methods #2a & #2b were included to show how the result could be achieved without a quantitative weighing device.  The use of a constant counter-weight plus a ruler or tape measure is an effective substitute, though I do agree that managing the things dangling from a stick would take a bit of practice.

As regards Steve's balloon or Russell's second tub, I'm retiring to the side-lines and changing to spectator-mode!                   

[vtsteam]

My thought on using the balloon if you didn't have a scale to weigh something would be to check the water level with nothing in the tub, then check with the mystery irregular object in the tub (I assumed it was a metal denser than water) and then attach the balloon to the object and fill the baloon with air untill the combo reached neutral buoyancy, and then check the water level again.

Divide the difference between the first and last measurements by the difference between the first two measurements. That will give you the specific gravity of the object. That's the density.

Multiply the specific gravity times the difference between the first and second measurements (the volume of the object) and the density of water (in whatever units you favor), and you have the weight

I can see Russell's method working, too. In this case you basically have a cylinder shaped tub, acting as a boat with what are esentially plimsoll marks, and you place the object in it, read the difference and get the weight.

But you haven't measured density of the irregular object. I guess you could first just use one tub for that as I did above.

[DavidA]

Pete,

.. machining a precise sample of the material is more something for forensics than for diferentiating between two similar-looking materials prior to charging the foundry crucible or going to the scrap metal merchant...

That is exactly what I was engaged in at work.  I was 'the sampler'. My job was to produce precise samples from workpieces and make them suitable for mechanical, chemical and spectrograph analysis.

Dave.

[RussellT]

I thought about filling the balloon to reach neutral buoyancy and then decided it would be easier just to fully inflate the balloon so the object floated attached to the balloon and the right volume of water would be displaced.

My second tub method does the same thing but without having to tie things together.  Fill the first tub and put the second tub in.  Then add the mystery object and note the volume displaced (1 litre = 1 kg).  That gives you the weight.  Fill the first tub and put the mystery object in.  Note the volume displaced.  That gives you the volume.  Density, SG etc can be calculated from there.

I have to say though that if you have a spring balance in the right weight range then Pete's method is quick and easy.

Now - on to making spring balances.  If you could make a sping balance where you could adjust the scale (for example by having a scale on a piece of elastic or a spiral scale on a disc or drum) then you could weigh your mystery object, set the scale to read 100 and then dunk it in water and have a direct readout of the material.

Russell

[vtsteam]

I thought about filling the balloon to reach neutral buoyancy and then decided it would be easier just to fully inflate the balloon so the object floated attached to the balloon and the right volume of water would be displaced.

Good idea!

Homemade ordinary balances are fun, too. I once made one of paper to weigh paper airplanes with fractional gram accuracy. I calibrated it initially with coins. I made gram weights out of cardstock squares. The cost was a little paper, some fishing line a thumbtack and a paper clip.

You can do the same thing on a larger scale (pardon the pun) with water to calibrate initially.

edit: Oh yeah Russell, that just reminded me -- fifteen years ago I moved my houseboat up north on a trailer. I figured I needed about 500 pounds tongue weight a few minutes before taking off on the 1500 mile trip with a rental truck. How to weigh? Then it ocurred to me to just put a fulcrum under a 2 by 4 on edge, measure the distance to the tongue and figure out where I could press down with my 175 lb. human weight on the lever opposite to just lift and indicate 500 lbs tongue weight. I shifted the houseboat on the trailer slightly to adjust -- worked perfectly!