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  1. Standard member sonhouse
    Fast and Curious
    18 Nov '11 00:41
    http://www.physorg.com/news/2011-11-world-lightest-material.html
  2. 18 Nov '11 10:00
    It it theoretically possible to design a lighter than air material ie one that doesn't make use of vacuums or light gasses?
  3. 18 Nov '11 12:39 / 1 edit
    Well air has a density of about 1.275mg/cm3 whereas this is being reported as having a density of 0.9mg/cm3 presumably due to having vacuums inside it.
    Not sure how you make a materiel less dense than a gas without having vacuum spaces inside it, otherwise it gets air inside it which contributes to it's density.
  4. Standard member sonhouse
    Fast and Curious
    19 Nov '11 07:47
    Originally posted by googlefudge
    Well air has a density of about 1.275mg/cm3 whereas this is being reported as having a density of 0.9mg/cm3 presumably due to having vacuums inside it.
    Not sure how you make a materiel less dense than a gas without having vacuum spaces inside it, otherwise it gets air inside it which contributes to it's density.
    If it really had a 0.9 density it would float but there was no mention of that. It may not be strong enough to have vacuum inside, the 14.7 PSI of air would squash it flat.
  5. 19 Nov '11 22:20
    Originally posted by sonhouse
    If it really had a 0.9 density it would float but there was no mention of that. It may not be strong enough to have vacuum inside, the 14.7 PSI of air would squash it flat.
    It's possible (maybe, the 'articles' I have seen on it are very non specific.) that the metal tubes
    it's made from have vacuum inside, they are nano-scale structures and very strong.

    Or it might be that the stated weight doesn't include the weight of air inside the structure.
  6. Standard member sonhouse
    Fast and Curious
    28 Nov '11 01:56
    Originally posted by googlefudge
    It's possible (maybe, the 'articles' I have seen on it are very non specific.) that the metal tubes
    it's made from have vacuum inside, they are nano-scale structures and very strong.

    Or it might be that the stated weight doesn't include the weight of air inside the structure.
    Nano is not small enough to exclude air. Picosized devices are.
  7. 28 Nov '11 09:23
    Originally posted by sonhouse
    Nano is not small enough to exclude air. Picosized devices are.
    Your going to have to expand on what you mean because their are macro sized devices that 'exclude air' so I really don't know what you mean.
  8. Subscriber WoodPush
    Pusher of wood
    02 Dec '11 21:49
    They aren't reporting density of total material including the air it contains, they are reporting density of the material itself excluding air. The article is actually pretty explicit that the material contains 99.99% air (not vacuum).

    Compare to the density of styrofoam, which at .089 mg/cc, also doesn't measure the air inside. We all know styrofoam doesn't float
  9. Subscriber WoodPush
    Pusher of wood
    02 Dec '11 23:04 / 1 edit
    Oops, figure on styrofoam was wrong it's more like 9 mg/cc, but anyway... you get my point.

    Another way to look at it is that the figure presented is the density of the material in a vacuum. Reporting anything else wouldn't really make sense -- they'd have to report the density of the measured air, since it would dominate the measurement.
  10. Standard member sonhouse
    Fast and Curious
    05 Dec '11 02:07
    Originally posted by WoodPush
    Oops, figure on styrofoam was wrong it's more like 9 mg/cc, but anyway... you get my point.

    Another way to look at it is that the figure presented is the density of the material in a vacuum. Reporting anything else wouldn't really make sense -- they'd have to report the density of the measured air, since it would dominate the measurement.
    Would air contribute to the weight if the pores were not sealed? I don't think it would in that case. I think it would weigh the same in a vacuum.
  11. Subscriber WoodPush
    Pusher of wood
    05 Dec '11 17:29 / 3 edits
    Originally posted by sonhouse
    Would air contribute to the weight if the pores were not sealed? I don't think it would in that case. I think it would weigh the same in a vacuum.
    You might be confusing weight and density, which aren't the same (the figures in the article and most of this thread were density).

    But to answer your question:

    Air doesn't contribute to weight measured by a scale, in general, because the scale is already calibrated with air pressure considered. So unless you change the density of the air (and with open pores, you aren't), there won't be a change in weight measured.

    But if you move the scale into a vacuum, it would have to be recalibrated, because it would no longer have atmospheric pressure applied.

    I suppose it might be difficult to say how much the scale would register the difference because it depends on the shape of the scale and how the air pressure was originally distributed on the scale's parts.

    So, to sum up: no the material itself would never weigh different in a vacuum - weight, or force of gravity, isn't affected by air pressure. But the scale might have a different reading due to the change in air pressure.
  12. 05 Dec '11 17:47
    Originally posted by WoodPush
    So, to sum up: no the material itself would never weigh different in a vacuum - weight, or force of gravity, isn't affected by air pressure. But the scale might have a different reading due to the change in air pressure.
    That is clearly incorrect. A simple example of a helium filled balloon which would have a weight in a vacuum, but has negative weight in air.
  13. Subscriber WoodPush
    Pusher of wood
    05 Dec '11 17:51
    Originally posted by twhitehead
    That is clearly incorrect. A simple example of a helium filled balloon which would have a weight in a vacuum, but has negative weight in air.
    No, it's not incorrect. Weight is defined as the force of an object due to gravity.

    Gravity doesn't push helium balloons away from the earth.

    Your experiment to measure the weight of the helium balloon was done incorrectly.
  14. 05 Dec '11 18:42
    Originally posted by WoodPush
    No, it's not incorrect. Weight is defined as the force of an object due to gravity.

    Gravity doesn't push helium balloons away from the earth.

    Your experiment to measure the weight of the helium balloon was done incorrectly.
    I see. But that means that a scale does not measure weight.
  15. Subscriber WoodPush
    Pusher of wood
    05 Dec '11 19:35
    Originally posted by twhitehead
    I see. But that means that a scale does not measure weight.
    Right, just a matter of using the right instrument and controlling the environment of your experiment so you're measuring what you intend.