## Is the Earth Like a Billiard Ball Or Not?

This is the second time I have come across this particular anecdote in a few months, so I thought I would weigh in on it as well.  Last night I saw on Reddit that “The Earth is relatively smoother than a billiard ball.”  This was a reference to a Wikipedia article about the shape of the Earth, quoted here in case someone gets around to fixing it:

“Local topography deviates from this idealized spheroid, though on a global scale, these deviations are very small: Earth has a tolerance of about one part in about 584, or 0.17%, from the reference spheroid, which is less than the 0.22% tolerance allowed in billiard balls.”

A few months ago, I also came across a Discover Magazine blog post from 2008 titled “Ten things you don’t know about the Earth.”  The first two “things” on the list dealt with this same issue of the shape of the Earth compared to that of a billiard ball.  The article correctly recognized the need to distinguish between smoothness and roundness, but in my opinion still came up with a wrong answer on both counts.

Before getting started, why do we care about any of this?  The motivation for this anecdote begins with the fact that the Earth is not a perfect sphere, viewed on either a large or small scale.  On a large scale (think roundness), because the Earth is spinning, its shape is best approximated not by a sphere but by an ellipsoid– specifically, an oblate spheroid, with a larger radius at the equator than at the poles.  On a smaller scale (think smoothness), the Earth is more obviously seen to not be a perfect sphere, nor is it a perfect ellipsoid, since it has all sorts of ridges, grooves, etc. corresponding to mountains, rivers, ocean trenches, etc.

The question is, how “non-spherical” is it?  The comparison with a billiard ball would be interesting if it were true, because to us a billiard ball seems to be a nearly perfect sphere, at least to the naked eye.  The source of this comparison– and, I think, the source of much of the confusion– is what the World Pool-Billiard Association has to say about a regulation billiard ball:

“All balls must be composed of cast phenolic resin plastic and measure 2-1/4 (+/-.005) inches [5.715 cm (+/- .127 mm)] in diameter.”

The key observation, I think, is that this description has nothing whatever to say about the smoothness of a billiard ball.  It does not mean, as the Discover article states, that a billiard ball “must have no pits or bumps more than 0.005 inches [sic] in height.”  Even such a small pit, bump, or groove would be easily noticeable on a billiard ball; manufacturing capabilities and requirements for smoothness are on the order of microns, much less than 0.005 inch.  (I emailed the WPA asking for clarification on this requirement, but have so far received no response.  I wonder if they get a lot of questions about this.)

The Wikipedia entry makes the same mistake, comparing the 0.22% (0.005/2.25) “tolerance” of a billiard ball to the 0.17% figure corresponding to the largest deviation of the Earth’s surface from the reference ellipsoid.  The latter almost certainly refers to the Mariana Trench, 10,911 m below sea level.  (Actually, this figure should be 0.0855%, not 0.17%, since the referenced billiard ball tolerance is relative to its diameter, not its radius.)

In any case, before we can comment on the smoothness of the Earth compared with a billiard ball, I think we require more information on either WPA rules or manufacturing standards.  [Edit: Thanks to commenter Mark Folsom for providing the following clarification of just how smooth a billiard ball is:

“125 microinches rms is a really rough surface–much more so than any billiard ball I have seen. In my estimation, a new billiard ball has a surface finish no worse than 32 microinches…”

Comments on the Bad Astronomy post give similar estimates.  And “Dr. Dave” provides some actual measurements with photos and plots showing deviations of approximately 20 microinches.  In comparison, at the scale of a billiard ball, the Mariana Trench is a groove almost 2000 microinches deep. So it seems the Earth is nowhere near as smooth as a billiard ball.]

So let us move on to roundness.  The following is quoted from the Discover blog:

“If you measure between the north and south poles, the Earth’s diameter is 12,713.6 km. If you measure across the Equator it’s 12,756.2 km, a difference of about 42.6 kilometers. Uh-oh! That’s more than our tolerance for a billiard ball. So the Earth is smooth enough, but not round enough, to qualify as a billiard ball.”

First, a minor nit: neither of the quoted diameters is correct to the given number of significant digits.  But that will not affect our calculations here.  What the article seems to miss is that the stated tolerance of a billiard ball diameter is plus or minus 0.005 inch.  That is, the diameter may be as small as 2.245 inches, or as large as 2.255 inches.  Enlarging this 0.01 inch difference to the scale of the Earth, the allowable difference in diameters is about 56.6 km, more than the actual difference of 42.8 km.  So the Earth is indeed as round as a regulation billiard ball.

Having said all this, I think this entire analysis abuses the spirit of the law, so to speak.  The WPA probably does not intend to allow such ellipsoidal billiard balls onto pool tables around the world, but rather to allow some variability in the size of nearly-spherical balls.  That is, the intent of the regulation is more likely that a ball should be spherical with a fixed diameter, but that diameter may be 2.245 inches for one ball, and 2.255 inches for another ball.

In summary:

1. Is the Earth as smooth as a billiard ball?  Answer: I’m not sure.  The question may be rephrased by comparing again with the Mariana Trench: can one detect a 49-micron groove in a billiard ball, and if so, would it be acceptable to play with?  [Edit: As mentioned in the edit above, I think the answer is now a definite No.]
2. Is the Earth as round as a billiard ball?  Answer: technically, yes… but you probably wouldn’t want to play with it.
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### 24 Responses to Is the Earth Like a Billiard Ball Or Not?

1. Chris says:

Nice attention to detail! I had also seen that post on reddit, but had blindly accepted it as true, probably because I had heard it elsewhere.

2. Thanks– I think this caught my eye in part because, like you, I had heard it before. But I imagine it also reminded me of a similar analysis from several years ago involving my home state of Kansas (someone on Reddit brought this up as well). The final assessment is reasonably accurate: it is “damn flat.”

3. Villard says:

Has anyone actually measured the diameter differences on a billiard ball? I guess the quality manufactures make them as round as possible and with better smothness and tolerances than the extreme tolerances allow.
The question should be : Is earth smoother than a billiard ball with the worst tolerances?

• Good question. I don’t have a pool table, and haven’t tried to make any measurements myself. I emailed the WPA– a few months ago now– and asked for clarification of their requirements, but have not received a response.

4. An. says:

Hey,
You assumption that the nominal diameter variation is the same as the sphericity tolerance is almost certain wrong.
You can guess the sphericity tolerance based on this nominal diameter tolerance given by the WPA, take a look at ball bearing grades, on the metric grade
http://en.wikipedia.org/wiki/Ball_%28bearing%29
the nominal diameter tolerance is exactly the same as the grade 1000, so my guess is that the sphericity should be about the same too(5 times lower), so the earth is not as round as a billard ball.

• Hmmm. I am not sure where I assumed that the nominal diameter variation “is the same as” the sphericity tolerance. My point in the post was simply (1) to note, as you also point out, the difference between these two measures, and the resulting ambiguity in the WPA’s stated requirements; and (2) to carry the sphericity interpretation– assumed elsewhere, not by me– to its (corrected) logical conclusion.

I find it interesting that the ball bearing specifications that you reference are independent of the absolute diameter of the bearing. That is, a G1000 bearing must have a sphericity within 0.001 inch, even if it is the size of a basketball, or billiard ball, or a BB. And an admittedly quick web search found no manufacturers selling ball bearings anywhere near 2+ inches in diameter. So I’m not so sure that the WPA had ball bearings in mind when they wrote their spec.

So it seems to me the question remains as it did in the original post: what does the WPA mean by its “+/- 0.005 inch” requirement? If they really meant “G1000,” I would think they would have indicated as such.

5. An. says:

There are ball bearing bigger than 2″, I have one here that is 3″.
My point is that this tolerance, although not clearly said, can only be the nominal diameter tolerance just by the way it’s written
2-1/4 (+/-.005) inches [5.715 cm (+/- .127 mm)] in diameter.
a sphericity tolerance would not be written like this. I don’t think they would have ball bearing grade in mind while setting the stardard, as this is a billiard ball, but my point is that an equipment working with this tolerance for the nominal diameter would probably have have the same sphericity diameter tolerance as for the g1000 grade.
I think they would even have invent a machine that could do that(work with such a relatively bad nominal diameter tolerance and use it too for the sphericity tolerance)

• We seem to be in violent agreement. I agree, as also indicated in the last paragraph of the post, that nominal diameter tolerance (or something like it) is probably the more likely intended meaning of the WPA requirement. But what, *if any*, is the corresponding sphericity requirement? If it is in fact G1000, then you are also correct that the Earth is not as *spherical* as a billiard ball. But I think we need more information, preferably from an actual billiard ball manufacturer, before simply assuming that billiard ball manufacturing (where resin is the typical material being manipulated) borrows all equipment, specifications, etc., from (steel) ball bearing manufacturing.

6. Pingback: Billard Ball | Colorado Cue Club

7. Mark Folsom says:

When I was studying surface finishes in engineering school, there was a roughness reference with sample areas of different specified rms roughness values. 125 microinches rms is a really rough surface–much more so than any billiard ball I have seen. In my estimation, a new billiard ball has a surface finish no worse than 32 microinches

• This is the additional information we are looking for. In comparison, at the scale of a billiard ball, the Mariana Trench is a groove almost 2000 microinches deep. So it seems the Earth is nowhere near as smooth as a billiard ball.

8. D Ryan Anderson says:

I saw this on Reddit a while ago and remembered it while reading about the Gravity Probe B satellite. It has some very spherical spheres in the main instrument. The Wikipedia article even compares them to the Earth. Worth a read, if you have already gone this deep down the spherical rabbit hole.

• You’re right, this is another useful comparison. (For others, the article referred to is here. To put all three objects’ “smoothness” on the same scale: at billiard-ball size, the spacecraft gyroscopes’ deviations would be less than half a micro-inch, compared with 20-30 micro-inches for an actual billiard ball, and almost 2000 micro-inches for the Earth.)

• Mark Folsom says:

OMG–someone saying something sensible!

9. Steve Garcia says:

You are using the wrong terms and concepts.

There is no such term as smoothness in machining, which is what is used to make billiard balls. Instead of smoothness you should be talking about “surface roughness,” more commonly called, simply, roughness. Do NOT confuse with tolerances.

Also, you are correct in talking about the tolerance “regardless of feature size”. There are SIZE tolerances, and then there are variation tolerances.

The WPA is insufficiently precise in their specification terminology, in giving simply 2-1/4″ +/- .005″. There should ALSO be a ROUNDNESS tolerance, or in this case, a SPHERICAL tolerance.

So right now you – and anyone else who delves into this – is handicapped by their imprecise specification.

Mark Folsom is almost certainly in error with his 32 microinch value for surface finish. First of all, there is a surface finish BEFORE painting/coating. Then there is the allowable surface finish AFTER coating. THAT, IMHO, is a very low surface finish, something on the order of a 2 or a 4 microinch surface finish.

I have 35+ years in machine design engineering and have spelled out more microfinishes, size tolerances, and roundness tolerances than you can shake a stick at. Rule #1: ALWAYS use the roughest finish that will get the job done reliably time after time or cycle after cycle. For a billiard ball, those would all be held very tightly. That +/-.005″ would NEVER be used for a spherical tolerance by anybody but an idiot. For a SIZE tolerance for pool balls? Probably not a big deal.

Rule #2: Tolerances are REALLY given for the INSPECTOR, not the machinist. The machinist today will use some form of CNC, and that will give a +/-.0002″ on everything it does. IN older days, the machinist would have chosen the machining process based on the tolerance for that individual feature, spelled out on the drawing he is working from. That last is no longer true at all, though, not since CNC came into universal use. So, with CNC the spherical tolerance NOW will be +/-.0002″, regardless of what a drawing says. So will the size be within +/-.0002″ – no matter what the specified tolerances are.

For older billiard balls this may not be true, though they may meet that brain-dead specification of 2-1/4″ +/-.005″.

As to roughness (surface roughness), the microinches is not a MAXIMUM, but an average of several measurements. After all, the full terminology is “32 microinches RMS” which means “root mean square.” Look up statistics for that one and learn something. What it means is that there can be – WILL BE – some extreme variations larger than the 32 microinches. Most of the time the variations will be about half peaks and half valleys – just like in geography, except there are few flat “plains.” Even with one spike or deep valley being measured, the specified roughness can be attained and be acceptable – not only to the machinist, but to the parts inspector and, most of all, to the functioning of the part.

In my professional opinion, the WPA meant to say a ball can have a nominal SIZE of 2.245″/2.255″. They do not (in THAT document) spell out anything else. That would be covered in whatever technical drawings they provide for machinist vendors, but it should ALSO have been given in that document, IMHO. It’s sloppy on their part, them being a rules organization.

In the absence of their professionalism, basically your are in the dark about what the actual spherical tolerances are, or the surface roughness, either.

• Thanks for the detailed information, as well as for providing the specific terminology used in manufacturing (e.g., “roughness”). I think everyone seems to be in agreement that the WPA spec addresses neither roughness nor sphericity (what I colloquially called “smoothness” and “roundness”). I also agree with your speculation that the WPA intended to specify nominal size tolerance (note the last paragraph of the post, as well as the discussion by commenter An.).

So disregarding the spec, we’re left with the question of how an actual billiard ball (either brand new, or “reasonably used”) compares with the Earth in terms of surface roughness. You make two important points:

1. A newly coated billiard ball probably has a surface finish of ~2-4 microinches, as opposed to ~32 microinches as suggested by Mark Folsom. I’m taking both of you at your word, so I’ll let you two fight that out :). But fortunately, it doesn’t matter: either value is still smaller by nearly a couple of orders of magnitude than the scaled roughness of the Earth.

2. Either of the above quoted values are rms, not maxima. (Hint: hopefully “Look up statistics for that one and learn something” is not your typical choice of words in a professional discussion. :)) I admittedly glossed over this… but again, I don’t see that it matters. Both the (absolute value of the) mean and the standard deviation are bounded by the rms value. So in the worst case, even if we assume the larger rms value of 32 microinches (which you disagreed with), and if we assume that both the mean and std. dev. are each also 32 microinches (they can’t be), and if we assume that the deviations are normally distributed (they are not), then even a one-in-a-billion measurement of deviation will still be only ~200 microinches, still significantly “smoother” than the Earth.

• Asher says:

Looks lime I’m late to the party. It sounds like the answer is “That may have been true at onetime, but nowadays we manufacture things very very precisely. It might be similar to a very well used billiards ball.”

10. buttons says:

id say that if you consider the surface of the earth to include the water surfaces, then the rms roughness is likely to drop to a similar level as the billiard ball, statistically anyway. The main issue is the lack of info on sphericity as has been stated.

11. justin roux says:

Very nice article which addresses more detail than the bar-room Earth/Ball quip. I’m certain that Earth’s non-spherical shape would lend it more to lawn bowling than billiards (although it’s not non-spherical enough for that application).

Nontheless, returning to the quip which stated that the Earth, with all its cliffs, mountains and deep oceans, is about as smooth as a billiard ball. Setting the squashed sphere aside, I did a rough calculation based on Everest being about 9km above sea level and the Mariana Trench being about 12km deep and got a “variance of radius to extreme fibre” of around 0.3%. That makes Earth about 33% rougher than a billiard ball but still, therefore, pretty darned smooth if you had it in your pocket. Admittedly, this is back-of-a-cigarette-packet mathematics but, in my experience as an engineer, that’s what practicality breaks down to in the end.

Quite amazing really, when you look at a cliff. Also amazing if you look at a billiard ball and imagine that those invisible scratches are the height of Everest and how flipping tiny we are in comparison.

• justin roux says:

Actually, now I think further (one cup of tea later), 0,3% of a billard ball is around 0.1mm, which is about the diameter of a grain of fine sand – which you would be able to feel on a billiard ball. Not that the Himalayas are shaped like a grain of sand, mind you. This means I need to think about the rate of change of variance in radius in %per degree.

The Earth might be a scratched billiard ball.
I’ve got some of them.
Still quite impressive though.

12. Observant Pleb says:

I noticed a few flaws in this logic:
The tolerance of diameter has nothing to do with sphere-ness of the ball, but the diameter. Even if the tolerance was 10%, it would just be the difference of a ball that could be 2.03 to 2.47 inches in diameter, but could still be a perfect sphere with perfect smoothness.

I thought of a test that’s simple. The math is already done here as well. Let’s take the earth and convert it’s diameter into mm. Then the height of Everest, then divided by the scale. We’ll even go so far as to say the ocean is perfect smoothness and not count below the surface.

Earth diameter, we only need one: 12742km = 12,742,000m = 12,742,000,000mm
Everest height: 8,848m = 8,848,000mm

To get the Earth and Everest to scale, we need to divide Earth by 57.15mm ( 2.25 inches ) to get the scale and then devide Everest to that scale.

Scale divisor ( Earth diameter / Cueball diameter ):
12,742,000,000mm / 57.15mm = 222,957,130.359mm
Everest scale:
8,848,000mm / 222,957,130.359mm = 0.038541402438mm

So let’s do a test we’ve all done already:
Rub your finger over a roll of scotch tape to find the end. Scotch Tape is 1 mil or 0.0254mm, which is less than 1/10th of the roughness that Everest would add to the Earth Cueball.

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