### Earth vs. Billiard Ball, round 2

I ran into the following quote on Wikipedia (http://en.wikipedia.org/wiki/Flat_earth, final paragraph):

The Earth *is* very smooth, which is to say, locally flat. By comparison, if a well-polished billiard ball were enlarged to the size of the Earth, it would have mountains about 50 miles high.

After a bit of research, I can't find a source for the 50-mile figure anywhere, and I think it's probably off by something like an order of magnitude. Here are my own calculations:

First, here's a quote from http://mentock.home.mindspring.com/ifaq.htm :

----------------------------------------------

IFAQ007 Is the Earth really as smooth as a billiard ball?

We contacted an engineer at Brunswick, and he gave us their specifications for their better billiard balls. Each ball is rolled and measured and an average diameter is computed. The average diameter must be no greater than 2.258 inches, and no less than 2.245 inches. Deviation from the average diameter must be no more than .002 inches. So, if you only consider such minor outcrops as Mt. Everest (6 miles high), the Earth, if reduced to the size of a billiard ball, would pass the smoothness test. Unfortunately, because of the Earth's rotation, the Earth's equatorial diameter is 24 miles more than its polar diameter. The Earth is smooth, but it is out of round. No cosmic corner pocket shots for us.

----------------------------------------------

From memory, the Earth is about 8000 miles in diameter. Some Brunswick balls apparently have radial tolerances of .003, or as low as .001, but .002 seems like a good number to go with. Now we can check some ratios --

height of highest mountain : diameter of sphere

----------------------------------------------

billiard ball: .002 inches : ~2.25 inches; ratio = .000889

yet another Internet source (minimum BCA standards from http://www.bca-pool.com/play/tournaments/rules/equip.shtml -- not "better billiard balls", apparently):

worst possible billiard ball: .005 inches : 2.25 inches; ratio = .00222

Earth: 6 miles / 8000 miles = .00075 (Everest), and

7 miles / 8000 miles = .000875 (Marianas Trench), but

24 miles / 8000 miles = .003 for equatorial bulge

26 miles / 8000 miles = .00325 (another Internet source for Earth's deviation from round)

Earth-sized billiard ball --

multiply by factor of 3555 miles per inch-of-billiard-ball:

7.11 miles / 8000 miles (Brunswick "better billiard balls")

17.775 miles / 8000 miles (worst possible BCA-standard billiard ball)

But because the Brunswick engineer is actually talking about deviations from the *average diameter*, the above assumes that both Everest and the Marianas Trench have land at about sea level on the other side of the Earth. Haven't checked that yet, but that's probably close enough for this calculation (certainly there are no other Everests or Marianas Trenches over there...)

Doesn't really matter anyway. In any case Earth does _not_ pass the billiard ball test, due to the equatorial bulge, by a factor of three or so. Also, even with the above simplistic analysis, the mountains on a "better billiard ball" the size of the Earth would be just slightly higher than Everest -- assuming the 7-mile-high lump was all on one side of the sphere. A worst-possible standard billiard ball might just possibly have 18-mile-high mountains -- but not mountains "about 50 miles high".

More importantly, that's the _maximum tolerable_ deviation we're talking about; the *average* "better billiard ball" probably doesn't have such high deviations in the diameter. Half of that might be a good bet (but I personally don't go around measuring deviant billiard balls, so obviously I don't have exact figures). In other words, the odds seem to be good that the average "better billiard ball", fresh from the factory, is just as smooth as, or smoother than, the Earth -- in terms of the distance between its ideal surface and its actual surface, relative to its diameter.

And MOST importantly, the billiard ball has been polished, so there aren't necessarily any local spikes or dents in it that look anything like Everest or the Marianas Trench! Any "mountains" on an Earth-sized billiard ball are likely to be smooth bulges without any well-defined top -- there would be many places where a marble would roll in one direction or another, but there might not be any vertical cliffs bigger than a few feet high -- very different from Earth.

In point of fact, the above billiard ball test is entirely a test of roundness, not of smoothness! And it may not even be a very good test of roundness -- you can make significantly non-circular shapes that have the same diameter everywhere, and it seems possible that this could be extended to three dimensions (?). Let's have a quick look at the smoothness of a few sample things:

CAMI 100-grit sandpaper has an average grit particle size of .0055 inches, which is way too big; let's find something smaller. The CAMI scale appears to go down to 1000-grit or so on the table I found,

http://www.woodturners.org/tech_tips/misc-pages/abrasive_grading.pdf ,

with an average grit size of .00042 inches. So if you stuck 1000-grit sandpaper on a billiard ball, the sand grains would be about twice as high as Everest (not counting the thickness of the paper, which would add a few more Everests to that).

So I suspect that an accurate billiard-ball scale copy of the Earth (minus the oceans, of course) might still feel very slightly sandpapery. Certainly there are things out there that "feel smoother" -- probably including nearly all billiard balls, unless they've been treated badly.

As an upper limit, the Hubble Space Telescope was at one point the "smoothest optical mirror ever polished", with a surface tolerance of one millionth of an inch. IANALLP (I am not a licensed lens-polisher), but assuming I understand what "surface tolerance" means... if a billiard-ball-sized Earth were machined to *that* tolerance, I think the tallest mountains would be less than ten feet high:

Billiard ball machined to Hubble Space Telescope tolerances:

.000001 inches : 2.25 inches; ratio = .00000045

Multiplying by 3555, we get a maximum deviation on an Earth-sized Hubbilliard-ball of .00158 miles, or 8.34 feet.

Conclusion:

The Earth is definitely not quite as round as a billiard ball, and it probably isn't nearly as smooth as a billiard ball, either -- but I must admit that I don't have absolutely conclusive data available...

## 1 Comments:

Um... wow.

Post a Comment

<< Home