Can you display a 1 meter sphere of pure gold on a flat surface without it deforming?

Factual Questions
21

According to the World Gold Council, all the gold ever mined (as of the end of 2006) totals about 158,000 tonnes (1 tonne = 1,000kg = ~2,200lbs). So, at 10 tonnes the 1m sphere would be about 1/15,800 (or .0632 %) of all gold ever mined.

As for the volume of that gold, we can reverse hajario’s calculations.

Density of gold = 19300 kg/m[sup]3[/sup] = 19.3 tonnes/m[sup]3[/sup]

Volume of 158,000 tonnes, in cubic meters, is:

158,000/19.3 = 8187 m[sup]3[/sup]

This would fill a pool 50m x 20m x 8.187m.

According to Wikipedia, a standard Olympic pool is 50m x 25m x 2m (min depth). This totals 2500m[sup]3[/sup]. So, all the gold ever mined would actually fill three standard Olympic pools, with quite a bit of gold left over.

(Please let me know if i’ve fucked up any of the calculations, or gotten any of my decimals in the wrong place).

22

It amazes me that it’s such a small volume. It seems like centuries of mining the stuff would have netted a bit more than that.

23

You mean, on a thin cushion of water - like those indoor ball fountain things? - I think the initial deformation would be too great to let the system start properly - the ball would settle into the socket and the water would find the easiest route outwards, which probably wouldn’t be an even, thin layer underneath the ball.

Or do you mean to actually levitate the ball on jets? - if so, I think you’d just end up eroding it under their pressure - as in a commercial water jet cutting machine.

You could float the ball on mercury - except for the inconvenient fact that gold readily forms an amalgam with it - so your gold would gradually disappear into the mercury.

24

Gold’s diamagnetic - just stick an unfeasibly humungous magnet under it and it will levitate. :smiley:

25

astro, just offhand, did you get this notion of a gold sphere from a certain comic book?
Dunno how to do the stress modulus/tensile strength thing, but I can’t see how something that soft, with that much force on it (gravity x a LOT of mass) would stay a sphere. It wouldn’t turn into a puddle or anything, but the OP mentioned a perfect sphere.
Of course, you could have a dimple on the flat surface, and float your gold on a liquid. But then you’d need some type of seal over the liquid, to keep it from being squirted out by the aforementioned force. This seal would need to be anchored to the sphere, so I guess you gotta drill holes in it…

26

How bad would it distort?

Are we talking sunny side up egg here? Or “Wait… from this angle it looks… well… a little… well… maybe…”

27

I’ve done some back-of-envelope calcs, and deformation is not a problem.

There’s two material issues. The first is that gold has a low yield point, i.e. it takes a relatively low stress to squidge it so it stays squidged. The second is that it has a low elastic stiffness, so it takes a relatively low stress to deform it like a rubber ball.

Now, when you place a perfect sphere on a perfectly flat plane, the contact area is zero, so the contact stress is infinity. That’s always the case, even for a hard steel ball-bearing on a glass sheet. So in the real world, the flat plate will fractionally deform into a cup, and the sphere will fractionally flatten and they conform to each other, creating a finite contact area.

With a ball-bearing on a glass sheet, these distortions are infinitessimal and elastic - lift the ball away and it pings back, as does the miniscule deformation in the glass. A plasticine ball on a balsawood plank OTOH may leave a permanently deformed ball and a dented plank.

There’s not much that can be done about the elastic stiffness of pure gold, but the yield point can be pushed up by work-hardening. Gold famously doesn’t work harden very much - that’s why it can be beaten so thin and drawn out into wire so easily. But if your gold ball factory squidges the ball into an ellipse, and then back into a ball, a few thousand times, changing the ellipse axis each time, the ball will work-harden.

Now, some mechanical data and calcs: from this pdf paper, section 3.1: http://composite.kaist.ac.kr/public/NPE91.pdf

Yield strength of annealed pure gold is 112 MPa
Yield strength of heavily work-hardened pure gold is 325 MPa

The ball exerts a force of 100,000 N, near enough. For soft annealed gold, using a conservative yield strength figure of 100 MPa, this means a cross-sectional area of 1/1000 square metres, or 10 square centimetres, will support the weight of the sphere. Or, a 4cm diameter, circular flat on the sphere will be enough to prevent any permanent deformation if you sit it on its flat. The 30cm deep cup is WAY overkill, a very shallow cup 4cm in diameter and 0.5mm deep is enough. And this is with annealed gold, no work hardening required!

I doubt the elastic distortion is enough to worry about, but if you insist on a perfect sphere, start off very slightly ellipsoid, park it on its point in the cup, and it will elastically deform into a sphere under its own weight. I really doubt you could tell by eye before and after.

28

Gold sphere? I shall alert the Beagle Boys!

29

You don’t have to float it. The sphere sits on top of a tube of about 1/2 the sphere’s diameter, sealing it. The edges of the tube are shaped to conform to the sphere.

You pump water up the tube at sufficient pressure to lift the sphere. The tiny gap allows the water to dribble through and out, and you send it round and back through the pump.

for your ball, the water pressure required is only 73 psi, for a tube 50cm in diameter. Your gold sphere will only protrude 7cm into the tube - most of it will be visible.

There’s a 1m diameter, granite ball fountain of this type in Rockingham, Australia. Gold is 3-4 times denser than granite (my WAG!) but the principle is the same.

For other rolling ball fountains, look here. They link to video:
http://www.atlanticfountains.com/ball_fountains.htm
I think they’re a wonderful demonstration of pressurised fluid bearings, the same principle that keeps all our cars running. I also think you should definitely do it with your gold ball, just to show off a tiny bit more!

Calcs: - tube cross section =pi x r x r = 3.14x25x25 =1963 cm2.
Ball weight=10,000kgf, pressure required = 5.1 kg / cm2
Converted to psi for familiarity. 1kg/cm2 = 14.2 psi.

30

This thread shows why Stranger On A Train and matt are two of my favorite Dopers.
RR

31

Yeah, mine either, but the Hope diamond resting on the top of it has dented it slightly. Boy, am I ever glad there’s not much crime in northern Japan!

32

Don’t piss off Magica.

33

Those of you planning on walking away with one of these should keep in mind that ten thousand kilograms is the equivalent of eleven tons.

34

Fantastic posts, matt. Thanks.

35

Who said anything about walking? It is a (as close as possible) gold sphere. Once we pop it up out of that mounting dimple we just roll it away.

Wall in the way? It’s a 10,000 kg gold ball, just get it rolling at a good pace and the drywall will crumble like paper.

So yeah, stopping it would be a small challenge.

36

What if it was just a hollow sphere, like a giant sold gold basketball, ala gold leaf moulded onto a collapsable sphere of something else? You still get your sphere but at a fraction of the weight and cost.

37

Yeah, until you command a vector change and forget to strap your ball down. Then you will have a whole new set of problems.

38

Roll it away, huh? You ever try to steal a locomotive engine by pushing it down the track?

39

Last time I build my secret hideaway on the top of a mountain… ::grumbles::

40

And I would’ve gotten away with it, too, if it hadn’t been for those meddling kids and that dog…