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I can understand the choice of material, silicon 28, but why is it a sphere rather than (say) a cube? Article here

I would have thought that a sphere would have been the hardest shape to machine accurately.

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  • $\begingroup$ shape doesn't matter after all? i mean whether or not the sphere has a perfect shape $\endgroup$ – RE60K Mar 23 '15 at 13:55
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    $\begingroup$ To the person voting to close as "opinion based" - clearly the people who chose this shape had some idea of what they were doing; the OP is asking what their reasoning was. That is not a matter of opinion - the reasoning used is what it is. $\endgroup$ – Floris Mar 23 '15 at 14:07
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    $\begingroup$ Spheres are by far the most common shape machined with extremely close tolerances: for ball bearings. I can go onto McMaster's website and buy some steel balls with a tolerance of better than 1 part in 10000. Commodity-grade ball bearings generally have tolerances from around 10 to 1 micrometers (hair is about 10x larger). $\endgroup$ – Nick T Mar 23 '15 at 15:59
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    $\begingroup$ I'd expect the corners and edges of a cube to be more susceptible to erosion (evaporation?) than the surface of a sphere. $\endgroup$ – David Richerby Mar 23 '15 at 16:58
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    $\begingroup$ @Michael on the subject of rolling it, 1 kg is about the right weight for candlepins. $\endgroup$ – Random832 Mar 25 '15 at 14:22
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If you know the diameter of the sphere, you know everything you need to know about the dimensions. It all comes down to one single value.

Any other shape requires multiple dimensions and thus multiple values. Further, measuring a cube or another shape for accuracy is harder than measuring a sphere.

Making very accurate spheres is not as difficult as you might think - it's no different than making optical glass or mirrors using grinding techniques, and, in fact, they are measured much the same way with lasers for very high accuracy.

This video goes into a little more detail as to why they are doing this, how they achieved it, and how the sphere is made.

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    $\begingroup$ Upvoting for the awesome video! $\endgroup$ – Floris Mar 23 '15 at 16:15
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    $\begingroup$ Not sure about the 'only need 1 measurement' thing. Assuming it is a perfect sphere, you only need to know diameter. Assuming a perfect cube, you only need to know 1 side length. The problem is if the sphere is oblate or the cube is actually rectangular, not square then you need to know more than 1 $\endgroup$ – Scott Mar 23 '15 at 20:54
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    $\begingroup$ @Scott if you watch the process video, I think you'll find that making it from one value is (relatively) easy with a sphere. The process, by design, produces a perfect sphere (within a degree of error). However, there's no process that produces a perfect cube. You have to grind six faces separately. The error will necessarily be much larger simply due to the fact that each face is cut and found independently of the other faces. $\endgroup$ – Adam Davis Mar 23 '15 at 22:24
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    $\begingroup$ Then you have to measure it. That's just as complicated as making it, and requires more surface area to measure. Rotating machinery is easier to build with a high tolerance than translating machinery. This all doesn't mean that it's impossible, but spheres are easier on a number of levels compared to other solids. $\endgroup$ – Adam Davis Mar 23 '15 at 22:29
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    $\begingroup$ Also - they measured from 60,000 directions to confirm it is "flat everywhere". So while you end up specifying the object with a single parameter that doesn't mean a single measurement suffices. $\endgroup$ – Floris Mar 24 '15 at 1:06
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There is a nice article in the New Scientist that describes how these spheres were made.

Sphere can be made very precisely (and their shape measured accurately) simply because of their symmetry - and therefore their volume can be determined most accurately. The video clip in the above article shows this in detail.

Of course a sphere also has the lowest surface area (less contamination) but I don't think that comes into it.

UPDATE

I wonder how much the sphere distorts under its own weight... if you use the Hertz equation (see for example here)

$$F(x) = \frac{2E\sqrt{R}}{3(1-\sigma^2)}\cdot x^{3/2}$$

and we set $F = mg = 10 N$, then for the sphere in question, with $E= 150 GPa$ (sources vary, this is for estimating only), $\sigma=0.3$ and $R = 50 mm$ (again, rounding 96.75/2) we get x = 870 nm. That would be the distance by which the surface is indented when the sphere rests on a flat surface; the contact area would be extremely small at $\pi R d = 0.14 mm ^2$.

There will also be a much smaller distortion of the diameter of the sphere. Using simple dimensional arguments, the stress of the weight will be proportional to the cross sectional area of the sphere, which would make the distortion along the diameter on the order of

$$\frac{\Delta x}{x} \propto \frac{mg}{\pi R^2 E} \approx 10^{-8}$$

If that is correct, it implies a distortion of less than a nm for the sphere. Which is a reassuringly small number. I suspect that, on the scale of this experiment, it does not need to be taken into account (once somebody has done the calculation more accurately...)

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  • $\begingroup$ Given that it's a single crystal, the distortion should be very low, but even further, it's supported by a ring and a much larger surface area than a single point support on a flat surface. $\endgroup$ – Adam Davis Mar 23 '15 at 16:14
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    $\begingroup$ @AdamDavis - the fact that it is a single crystal is captured in the use of a Young's modulus of 150 GPa (I think); and yes, they don't support it on a flat surface so the Hertzian estimate is much larger than the actual distortion. However, the "sagging" under its own weight should not depend on how it is supported and I think that the last part of the calculation (< 1 nm change in diameter) is instructive. I find that it's helpful to put numbers on these things - if only to understand their approximate magnitude. $\endgroup$ – Floris Mar 23 '15 at 16:17
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    $\begingroup$ I think that measurement to under a nm will allow calculation of the number of atoms in the sphere that will correspond to a mass variation well below the variations seen between the prototype kilogram and its comparison control kilograms. So even a number quite a bit bigger than this would allow for a more workable kilogram definition than we already have. Also note that one could get quite sophisticated and measure the sag versus orientation accurately with the sphere in an interferometric test setup. $\endgroup$ – WetSavannaAnimal Mar 25 '15 at 3:28
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I'd like to add a bit more to Adam Davis's concise answer.

Newton's Sphere Production Method

Firstly, note that the lathing process shown in The video that Adam Davis's Answer links to with the sphere rotated and an oscillating abrasive cup sliding over its surface about pseudo-random rotation axes is the standard technique invented by Isaac Newton that is used today to make precise spherical surfaces. The point about the technique is that even if the abrading cup is not quite spherical, the system settles down to a state where both the cup and machined sphere are spherical, as they grind one another until the pressure between them is uniformly distributed across their contact region for all orientations of the pseudo-randomly oriented cup. This surface is of course spherical. One begins, naturally, with an oversize rough sphere and, through practice, tweaks the process so that the steady state is reached precisely when the desired radius is achieved.

Spherical Surfaces are Really Easy to Measure Extremely Well

Now I'd like to talk further about how amazingly accurately one can measure a sphere's radius and its deviation from ideal sphericalhood. The video briefly shows a simple laser ranging verification that would be done as a rough measurement during the sphere's making. To get really serious, one can do waaaaaay better than this, and so the following is my understanding (gleaned from questions at conferences) of how the CSIRO in Australia envisage verification would be done reasonably cheaply and certainly repeatably for the production of national and regional test standard kilograms.

This is done this through absolute interferometry - a completely self-calibrating measurement technique that not only measures the sphere's actual surface, but calibrates the interferometer at the same time. In principle, you can easily understand how interferometry can be self calibrating (this is NOT how it's done): you are counting numbers of wavelengths so "all you need to do" is to count the number of fringes by adjusting the pathlength from zero in an interferometer trained on one point on the sphere, and then shift to another point and do the same. So you can then describe the sphere's surface in terms of co-ordinates grounded on numbers of wavelengths. Not only would this be horribly time consuming, it would be impracticable mechanically, but the idea of characterizing the surface in terms of wavelength grounded co-ordinates is clear.

Here is where the sphere makes it easy - through the amazing Three-Sphere-Test. The paper by its inventor (James Wyant WARNING- Do not read this man's CV - Severe brain envy hazard)):

Katherine Creath and James C. Wyant "Absolute Measurement of Spherical Surfaces" SPIE Vol. 1332 Optical Testing and Metrology ill: Recent Advances in Industrial Opticallnspection (1990)

describes the technique fully, but the principle is as follows.

Fizeau Interferometer

I've sketched here a Fizeau interferometer, which is the commonest test setup used to verify spheres. Here we an incident lightfield $I$ which is focussed to a spherical wave by an extremely well designed antireflexion coated focussing surface $AR$, so that it becomes a spherical wave converging on the tested sphere $TS$ through a spherical reference surface $SR$. One adjusts the interferometer until the tested sphere and the reference surface are nearly concentric, and the reflexions $R_R$ from the reference surface and $R_T$ pass back out through the converging surface $AR$ to become interfering, nominally plane waves that make the interferogram on the interferometer's CCD array. However, the "reference" reflexion $R_T$ is only as spherical as the reference surface $SR$ and converging surface $AR$ will make them, and we don't at the outset know exactly what the forms of these surfaces are. Moreover the probe reflexion $R_T$ is also "contaminated" by imperfections in $AR$ and $SR$.

So the three sphere test takes (1) one measurement with spheres as shown, then (2) one rotates either the tested sphere, or the reference lens (wontedly the latter) through 180 degrees to take a second interferogram and (3) lastly, one then focusses the interferometer onto the surface of the tested sphere and takes a third interferogram. This last interferogram is essentially making a near-diffraction limited spot on the surface of the tested sphere, so even rough quality surfaces are excellent over the submicron lateral scales of the spot. Therefore, the third interferogram is influenced by the surfaces $SR$ and $AR$ alone.

It can be fairly straightforwardly shown that one can deduce from these three measurements the shape of the tested sphere's surface and the wavefront distortion imparted by the interferometer, thus not only getting an absolute measurement but also a calibration of the interferometer.

There is an analogous "three flat test". However, it is more time consuming and complicated. The three sphere test captures and wholly characterizes as much of the sphere as the interferometer's numerical aperture will allow (typically about 0.4NA, so we can image a solid angle $\pi\times NA^2$ or about a twentieth of the sphere's surface at a time and then make and link together thirty or so charts of the surface), and the movements between images are easy to automate as this kind of test is best done by a robotic, hands-off procedure.

With instantaneous phase shifting interferometry (where one uses tricks with the light's polarization) one can simultaneously take interferograms with relative phase delays of $\exp\left(\frac{2\,k\,\pi\,i}{N}\right);\,k=0,1,2,\cdots,N-1$ between the reference and probe beams (where $N$ is 3 or 4 in the implementations I have seen), so one can reliably reconstruct the test sphere induced wavefront map to within a tiny fraction of a wavelength. If you use high optical power (to get good SNR notwithstanding quantum noise) and average measurements, well under nanometre accuracy can nowadays be gotten by a test setup costing less than $USD100K.

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  • $\begingroup$ This is a really really good answer! $\endgroup$ – Floris Mar 28 '15 at 14:43
  • $\begingroup$ @Floris Thanks, Floris. If you're interested in this stuff, you might enjoy reading a bit QED Technologies qedmrf.com . In particular, check out their "stitching" technology qedmrf.com/metrology/ssi-technology $\endgroup$ – WetSavannaAnimal Mar 29 '15 at 0:12
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    $\begingroup$ @Floris BTW, I shouldn't be surprised, if the kilogram grounded on a number of $Si$ atoms is adopted, if QED Technologies ends up playing a major role in defining "implementation" standards and technology for different labs to set up their own mass standards. $\endgroup$ – WetSavannaAnimal Mar 29 '15 at 2:39
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Another reason that nobody David Richerby mentioned in a comment, is durability. The corner of a cube would be easy to accidentally break because a small force on it would equate to a large pressure (its area gets smaller as the shape of the cube gets more accurate). In fact I think it would be likely to break during machining, resulting in truncated corners (even if only microscopically).

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    $\begingroup$ @Floris: thanks, didn't see that. Edited answer. $\endgroup$ – Hugh Allen Mar 29 '15 at 0:57
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A sphere might be harder to machine, but easiest to verify the accuracy of, especially when accounting for slight changes due to temperature.

It should be noted that any standard like this must not only have a mass of 1 kg (or whatever) but also have some secondary method of verifying mass, in this case, being able to count (to some accuracy) the number of silicon 28 atoms within the sphere.

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    $\begingroup$ Spheres are the easiest to machine. $\endgroup$ – Nick T Mar 23 '15 at 16:01
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    $\begingroup$ @NickT, really? I'd need a bandsaw and a right angle to machine a cuboid. Spheres might be the easiest to machine to ultra-high accuracy, but I think I said that. $\endgroup$ – Jiminion Mar 23 '15 at 16:09
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    $\begingroup$ @jiminion All I need to make a sphere to ultra-high accuracy is any chunk of homogeneous and isotropic material and an abrasive (not sure if sand will suffice). And (lots of) time. Machine tools are strictly optional. $\endgroup$ – Bernd Jendrissek Mar 24 '15 at 4:04
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    $\begingroup$ @BerndJendrissek More like infinite time. Without precision tooling, you will make mistakes and ruin your work. You'd need tooling just to measure it. (This differs from grinding lenses, which CAN be done accurately without special tooling.) $\endgroup$ – Jiminion Mar 24 '15 at 13:22

protected by Qmechanic Mar 24 '15 at 18:17

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