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I have an approximately basketball-sized non-hollow piece of aluminum sitting in my house that is of irregular shape. I need to find the volume of it for a very legitimate yet irrelevant reason.

What is the best way I can do this? In fact, what are all the ways I could feasibly do this without going to a lab? (I don't live near any labs)

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    $\begingroup$ It doesnt say about the mass you've only written about the volume of a shape. Sorry about that. $\endgroup$
    – user3574
    Commented May 17, 2011 at 20:57

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Eureka! As Archimedes said, according to legend.

In principle, "TheMachineCharmer's" answer is feasible, but I would recommend recording the change in the volume of water instead (if you need an accurate measurement), because (1) it could be difficult to measure the volume of the spilled water, and (2) it is also a little less accurate to do so. (Some water will be left on the sides of the first container, and inserting the object into the filled container, catching all of the spillover while making sure the water level doesn't drop below the brim, could be difficult.)

If you only need a rough idea, the other way is fine.

If you need a more accurate measurement, try one of these:

  • Get the volume of a container by filling it with water (e.g. from a graduated vessel). Empty the container, then place the object in it. Fill it with water again, measuring how much water you added. Subtract this number from the volume of the container to find the volume of your object. (The order in which you do these doesn't matter, of course.)

  • This one could be a little harder, because you need a large graduated vessel. Fill the empty vessel with roughly enough water to submerge the object. Put the object in and record the change in the water volume.

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    $\begingroup$ Weigh the object. Weigh the container when it's full up to some mark. Then empty the container, put the object in, fill the container to the mark, and weigh it again. The difference between the first two weights and the third is the mass of the displaced water. All you need to complete the calculation is the density of water. $\endgroup$
    – Peter Shor
    Commented May 18, 2011 at 9:58
  • $\begingroup$ @Peter Thank you Peter for the comment. That would be the most accurate provided he has a suitably accurate scales. $\endgroup$
    – Mark C
    Commented May 19, 2011 at 4:30
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Fill a vessel(that can accommodate the irregular shape) with water brim to brim. And put that shape into gently into the vessel.Measure the volume of water spilled.

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Measure the mass. If you know it is solid aluminum, then you know its density, and you can easily calculate its volume.

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  • $\begingroup$ There is aluminum, and then there are aluminum alloys. I wouldn't count on the density being the same as elemental aluminum, so the displacement method should be used. $\endgroup$
    – Omega Centauri
    Commented Jan 2, 2011 at 17:25
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    $\begingroup$ @Omega: That's true, and obviously a direct measurement will (almost) always have less uncertainty. However, it depends on what degree of precision you need, and if you know the alloy of the material in question. If he doesn't need to be more precise than is allowed by variation in the alloy, then whats the difference? and if he knows the alloy, a scale is a lot simpler than a messy bucket of water. $\endgroup$
    – Colin K
    Commented Jan 2, 2011 at 20:46
  • $\begingroup$ Colin, that might the case. No doubt lowering something into a bucket of water and figuring out just how much spilled -and not spilling extra because you disturb the liquid surface and some sloshes out, may make the displacement method difficult in practice. $\endgroup$
    – Omega Centauri
    Commented Jan 2, 2011 at 21:22
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    $\begingroup$ Now, on the other hand if he measures the volume and the mass he may be able to identify the alloy if he didn't already have that information. $\endgroup$
    – Nick Pascucci
    Commented Jan 4, 2011 at 8:34
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Since the object is basketball size, it would displace a significant volume and weight of water when submerged. Weigh a container with some water and take a reading. Then attach a wire to the object, suspend it in the container of water till it's submerged without touching the container, and take a second reading. The difference between the two readings is the water weight that is displaced by the object. Volume (m^3) = weight difference (kg) / 1000 (kg/m^3).

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