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Math Science

Perfect Silicon Sphere to Redefine the Kilogram 453

MrCreosote writes "The Age reports optical specialists at CSIRO are helping create a new standard for the kilogram, based on a precise number of atoms in a perfect sphere of silicon. This will replace the International Prototype, a lump of metal alloy in a vault in Paris."
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Perfect Silicon Sphere to Redefine the Kilogram

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  • First of all (Score:2, Insightful)

    by alx5000 ( 896642 ) <alx5000&alx5000,net> on Friday June 15, 2007 @07:37AM (#19517105) Homepage
    ... no sphere made of atoms will ever be a perfect one.
    Second, if that rusty lump in Paris defines what a kilogram is, in no way is this sphere gonna change that.
  • by Aladrin ( 926209 ) on Friday June 15, 2007 @07:38AM (#19517115)
    No, it's impossible. What they -really- mean is that it'll be perfect as far as we are able to measure it. And it has absolutely nothing to do with what is really important here: They are counting the atoms of silicon in a kilogram and will use that measurement as the basis for the kilogram, instead of some lump of metal in a vault.

    The kilogram will not change, only a proposed scientific definition of it.

    The sphere doesn't mean -anything- except that it'll weight exactly a kilogram and be amazingly round.

    There's either a lot of media spin, or someone's attempt to get his work recognized and used. From what I can see, there's not a single soul that has dedicated to USING this new scientific definition, other than those directly involved with the project.
  • by oliverthered ( 187439 ) <oliverthered@nOSPAm.hotmail.com> on Friday June 15, 2007 @07:43AM (#19517161) Journal
    A kilogram was equal to 1000 millilitres of water and that 1000 millilitres of water would fit into a space 10cm cubed.

    If they've already defined the metre using constants, isn't something like this the best way of defining a kilogram.
  • by setagllib ( 753300 ) on Friday June 15, 2007 @07:44AM (#19517173)
    The point of having a physical object is that it can be used as a root for calibrating devices. From there you can calibrate more devices on each other. The further you get, the less likely you are to be precise, but the chances are pretty good that little deviations up and down will cancel out overall. But it's absolutely important to have an exact starting point, and a physical object is the only way to do that.

    It's a lot easier to measure a large object than a small one and multiply it, since a small error will also multiply out. What I don't get is how they intend to build an exact number of atoms into the sphere. You would need some other exact measurement, like number of electrons for calculating precise electrolysis procedures.
  • by JanneM ( 7445 ) on Friday June 15, 2007 @07:48AM (#19517193) Homepage
    This method works along essentially the same principles as the Watt Balance. In it, a superconductor of a known mass is placed within a superconducting coil.

    If you have a lump of anything of a known mass, why bother with the rest?

  • by jimstapleton ( 999106 ) on Friday June 15, 2007 @08:05AM (#19517319) Journal
    maybe because you are using the lump of known mass to measure something else.

    Duck Measurer: "I put a duck on one side of the scale, and use weights (lumps of known mass) on the other side to determine the mass of the duck."
    Some Guy: "Umm, but you already know the mass of the weights, why are you bothering?"
  • by Delirium Tremens ( 214596 ) on Friday June 15, 2007 @08:18AM (#19517423) Journal
    You are confusing mass and weight.
    Mass = how much matter there is in an object.
    Weight = how much pull does a particular gravity (like Earth's g) has on that quantity of matter.

    That's why you could be floating (weightless) in a space ship without having lost any of your fingers or other parts of your body (mass) ;-)
  • Re:Huh? (Score:5, Insightful)

    by nine-times ( 778537 ) <nine.times@gmail.com> on Friday June 15, 2007 @08:27AM (#19517495) Homepage

    Ok, people are giving snarky answers here, but I'll try to give you a more straight answer.

    The only way we have to keep a standard unit is to have an object with that unit and call that the standard. Let's say you were building some sort of a scale that would measure weight in kilograms; you'd have to calibrate it first. This means that you'd have to find an existing weight that was one kilogram, put it on the scale, and mark that this weight is a kilogram. But then how do you find a 1 kilogram weight? You have to measure it on some scale that's already calibrated correctly. This chain continues, and has to end somewhere.

    So the two questions I anticipate are:

    1. why not keep an already calibrated scale?
    2. why do we need a particular weight stored somewhere, instead of continually basing the measurement on kilograms measured on other scales?

    To answer the first question, a scale would be harder to maintain accurately. It could break, and calibrations don't hold forever. You'd have to re-calibrate it every so often, and how do you do that without an object known to be exactly 1kg?

    The answer to the second question (which I imagine might have been your question all along) is a little more complicated. Let's imagine that we have no exact 1kg object stored anywhere that we use as the standard. So one guy in a lab is using an iron ball as his 1kg weight, calibrating scales with it, and selling scales to others. The iron ball slowly rusts over time, and the weight of the ball changes a little. Someone takes one of the scales calibrated with the rusty balls and does the same thing, but this time with his own hunk of iron, but the environmental conditions in this guy's lab aren't as controlled, and he tends to get water condensation on his iron ball, meaning it rusts faster and each calibration varies depending on how much water has collected.

    Now, imagine it keeps on like this for 75 years, with different guys selling scales, getting their original measure from someone else, and then using their less-than-perfect means to continue calibrating and making scales. After 75 years, there are some drastically different "kilograms" floating around I buy a scale, measure out 1 kilogram, take it to a different scale and get 1.5 kilograms, while another says .75 kilograms. In this case, who's kilogram is "correct"? When the issue was raised, people would say, "Oh, if only we had a standard "kilogram" to compare them to!"

    And so we have someone keep a physical reference object under very controlled conditions and of materials that will prevent corrosion or other corruption to the material.

  • by Oersoep ( 938754 ) on Friday June 15, 2007 @08:34AM (#19517559)
    Yeah, but to know the known mass you'll need the silicon sfere cause that's the most precise known mass.
  • by Anonymous Coward on Friday June 15, 2007 @08:39AM (#19517603)
    The problem isn't with the various definitions of STP, but rather that the P element in each of them (pressure) has mass as a component term, introducing circularity into the definition. (Pressure is in Pascals, where 1 Pa = 1 N / m^2; the Newton is 1 kg * m / s^2; there is no practical alternative force component).

    A silicon crystal sphere will have the same radius over a much wider range of atmospheric pressures (and ambient temerpatures), and will be more chemically stable than the platinum-iridium ingot used to avoid the pressure-kilogram circularity.

  • by Anonymous Coward on Friday June 15, 2007 @09:15AM (#19517905)
    He's not. But everything's method is trying to define the kilogram (a unit of mass) by weighing it, and that's the problem.
  • by Software ( 179033 ) on Friday June 15, 2007 @10:03AM (#19518437) Journal
    At the risk of being pedantic, people in spaceships orbiting earth are not weightless; the Earth's gravity has an effect on them (namely, keeping them in orbit). Their sensation of weight [wikipedia.org] will be zero, because they are essentially in free fall, but they still have weight.

    To be fair, though, you didn't specify that the person was in orbit. Maybe you were thinking of in interstellar space, where a person would be weightless.

  • Repercussions (Score:2, Insightful)

    by Tabernaque86 ( 1046808 ) on Friday June 15, 2007 @10:29AM (#19518803)
    If they redefine the kilogram, what happens to coefficients that rely on kilograms? For example, for the equation for Universal Gravitational Energy [G=(coefficient)(m1xm2)/d^2)], is the coefficient going to change?
  • by jimstapleton ( 999106 ) on Friday June 15, 2007 @10:31AM (#19518833) Journal
    they are using flux to measure mass.

    Actually, the discussion involves using EM Flux as a way to calibrate mass measurements.

    Or, using the duck analogy further - then finding (or making) a duck that precisely matches the needed measurements...

    Regardless, the goal here is to get a reliable way to reproduce accurate mass meausre, without having a chunk of known mass available at or available to the reproduction site.
  • by thegnu ( 557446 ) <thegnu.gmail@com> on Friday June 15, 2007 @10:33AM (#19518873) Journal

    Maybe you were thinking of in interstellar space, where a person would be weightless.

    Not to be pedantic here, either, but maybe you'll just be in an imperceptibly slow freefall in an imperceptibly large orbit.
  • by gardyloo ( 512791 ) on Friday June 15, 2007 @11:46AM (#19519881)
    Haha. Love the Douglas Adams reference. Such a lead-in, too!

        Seriously, though, photons of anywhere near visible frequencies won't displace the atoms; light bouncing is almost always a purely electronic transition thing. And if this thing is ultrapure silicon, atoms are NOT going to want to displace. No worries there.
  • by AlecC ( 512609 ) <aleccawley@gmail.com> on Friday June 15, 2007 @11:56AM (#19520037)
    And you are you going to determine if you have x atoms? It will take a long time to count them out. That is the point of this project: to create an object, the sphere, whose number of atoms is known as precisely as possible. If it is a perfect sphere, its volume is precisely known. And if the crystal spacing of silicon is known (which it is, very well), then you have a very precisely known number of silicon atoms. This means that you have a macro-scale object, suitable for use in a balance or equivalent, whose mass is known in terms of unique natural substance rather than an arbitrary lump.
  • by painandgreed ( 692585 ) on Friday June 15, 2007 @12:12PM (#19520321)

    WHAT WILL WE GOT THEN: 1kg = M atoms of silicon. This definition will never change, and if the silicon spheres rust or break or change weight by any circunstances, we make new ones with M atoms and we have a forever-constant definition of a kilogram. This is GOOD.

    Yes, but which isotope of Si? Three occur in nature in various quanities. Did they use just a single isotope? If not, what are the various percentages?

  • by mrvan ( 973822 ) on Friday June 15, 2007 @04:02PM (#19523773)
    Probably shouldn't be feeding the trolls here, but GP is probably Dutch from his nick, and in Dutch the same word is spelled with 'sf'.

    If being unAmerican is the same as being mentally retarded it seems we're right back in the days of the terrorism^H^H^H communism scare

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