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Forgive the lack of physics math in this question; I'm not well versed in physics. I have a wonder and this seems to be the place to ask it.

Assume you have a pipe, the inside of which is perfectly round. And you have a ball, which is perfectly spherical. You decide for the fun of it to drop the ball into the top of the pipe and see what happens.

Now if the diameter of the ball is smaller than the inside diameter of the pipe and we assume plain-old earth gravity, the ball will roll down the pipe. ( Let's say for givens that the pipe is in a straight line, pointing straight down, and the bottom is open to the air. )

And obviously if the ball is larger than the pipe diameter it won't fit inside.

But what if the diameter of the ball is exactly the same as the inside pipe diameter? What then? Does the ball fit?

I'm wondering this because of a demonstration on smoothbore muskets. Because the ball is smaller than the barrel of the gun, it bounces around when fired and doesn't leave the barrel true and accurately.

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    $\begingroup$ You may be interested to read about the wringing of gauge blocks, eg physics.stackexchange.com/q/702456/123208 $\endgroup$
    – PM 2Ring
    Commented Sep 23 at 10:46
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    $\begingroup$ BTW if the pipe is vertical, the ball won't roll down it. It will either not fit, or it will fall without rolling (due to the symmetry). Rolling implies a slope $\endgroup$
    – Chris H
    Commented Sep 23 at 15:35
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    $\begingroup$ Is part of the confusion here the implicit assumption that "rolling" is the way that spheres move? Because "sliding" is also a possibility. Generally we don't think of spheres sliding (in the way cubes tend to) because rolling is more energetically favorable, but under the right condition balls can and do slide. $\endgroup$
    – R.M.
    Commented Sep 23 at 15:45
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    $\begingroup$ It's also interesting to note that if the ball's diameter was as close to the pipe diameter as possible and the ball was close to a perfect sphere as possible (and of course the pipe inside is perfectly circular), the ball after being dropped in would not just fall, but execute simple harmonic motion. As the ball is gently placed into the pipe (zero force), it begins to fall and air that is beneath the ball in the pipe is continually being compressed until the air can be compressed no more, and the ball briefly becomes stationary. (Continue) $\endgroup$
    – joseph h
    Commented Sep 24 at 23:12
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    $\begingroup$ But since the air beneath it is now compressed, it will begin to expand, pushing the ball up to the point where the pressure in the pipe is back to atmospheric pressure. Of course the ball is once again at the top of the pipe and has gravitational potential energy, and so it must start to fall again, and the process repeats. Of course there's friction, so this would be an example of damped harmonic motion. $\endgroup$
    – joseph h
    Commented Sep 24 at 23:12

9 Answers 9

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The answer depends upon what you mean by "fit". Assuming the ball or the pipe is made of some real material, and therefore can be deformed, the ball can be pressed into the pipe. If the ball were exactly the same size as the inner diameter of the pipe, and they were both perfectly round, the force required to press the ball into the hole would be $0$. In reality, neither the ball nor the pipe's inner diameter will be perfectly round, so it will require a small amount of pressure. Even if the ball is slightly larger than the pipe's inner diameter, by say 0.0001 inch or less, and unless the materials are very hard, the ball should be able to be pressed into the pipe without too much force. Such a "fit" is called a "press fit" as opposed to a "slip fit".

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    $\begingroup$ To add to this, in theory you could have a sufficiently small interference between the ball and the shaft that the weight of the ball is enough to provide the required force to press the ball into the pipe. However, the "exact fit" comes from the same universe as the spherical cow and the massless rope. Irregularities in the materials make this impossible, and thermal expansion ensures that even if the materials were perfect, the condition could only exist in a very small temperature range. $\endgroup$
    – anon
    Commented Sep 23 at 0:30
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    $\begingroup$ @MarkMorganLloyd indeed I have not! Didn't know the first thing about where to start! $\endgroup$
    – nuggethead
    Commented Sep 23 at 9:34
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    $\begingroup$ @MarkMorganLloyd - If you don't know the term to google, it might not be so obvious $\endgroup$
    – ScottishTapWater
    Commented Sep 23 at 14:58
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    $\begingroup$ @MarkMorganLloyd someone who knows "fit" as a verb does not necessarily know "fit" as a noun in order to think of searching for it. $\endgroup$
    – llama
    Commented Sep 23 at 19:08
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    $\begingroup$ @MarkMorganLloyd - something "fits" in something else, that's not the same as an engineering "fit" $\endgroup$
    – ScottishTapWater
    Commented Sep 23 at 20:42
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EDIT: I don't have time at the moment to improve this answer, and it contains some claims that are plainly wrong (i.e., using Hydrogen atoms as example was a bad idea as they don't behave like this at all; and the main part of the explanation is not that great, I feel). I wrote it in one sitting and didn't take time to think about it more. I pondered deleting the answer, but since it did garner quite a few upvotes, people seem to see some value in it at least. Check the comments on some problems of it, and maybe take a look at the linked video from an actual physicist.

But what if the diameter of the ball is exactly the same as the inside pipe diameter?

You are hitting a very interesting question here that is a general observation about reality: the concept of physical objects "touching", as we use it in every day life, does not make sense on an atomar physical level, or at the very least is quite difficult to reason about, or even to define what it means.

Atoms are made up of a positively charged core, and a negatively charged cloud of electrons surrounding it (for this particular question it does not matter how the electrons function, exactly). The fundamental interaction of relevance here is that of electromagnetism, we can ignore the other three. Electromagnetism repels particles of equal charge.

Interestingly, if you compare the sizes of the particles involved (both the nucleus as well as the electrons), to the radius of the electron cloud surrounding the core, the latter is larger by orders of magnitude:

  • Nucleus: 1.70 fm (Hydrogen), 11.7 fm (Uranium)
  • Atom: 25 pm (Hydrogen), 170 pm (uranium)

Nevermind the concrete numbers, but observe that the first is femtometers, the second picometers, which is a factor of 1000 already. So the hydrogen atom as a whole is ~14,000 times larger than its core, and, geometrically speaking, nearly all of it is the negatively charged electron cloud.

Also remember that electromagnetism falls off with distance - you know this from everyday magnets; the closer they get, the more significant the forces become, and not in a linear fashion.

To get to the point: imagine two regular, uncharged Hydrogen atoms approaching each other, getting ever closer and closer. Since they are uncharged, when they are relatively far from each other, electromagnetism has no relevant effect, they will approach each other easily. Of course, electromagnetism still is in effect, but with the relatively large distances, the electrons within the one atom will attract the nucleus of the other, and vice versa; but also the electrons in the one atom will repel the electrons from the other. Due to the large distances, these forces cancel out - they will never be mathematically zero, but they will soon be so little that they just have no effect.

Now imagine the atoms getting very very close together. At this point, the electron clouds are, relatively, "closer" to each other than the respective cores. Since the force generated by the electromagnetic interaction is not linear, the effect of the electrons repelling becomes more significant than the electron-nucleus attraction. In the end, the atoms will deflect each other. In an idealistic scenario where they move towards each other "head on", they would slow down and eventually fly apart in the same line (in practice there would always be some sideways component of course, but that's not the point here).

In every day regimes, which we are talking about here, it is utterly impossible that the two atoms do anything we would call "touching" or even "merging"; due to the EM forces they would separate much before they would somehow geometrically reside within the same volume of space in any meaningful manner. This is of course possible, but then we're talking about high-energy experiments at CERN or in fusion reactors, not balls in pipes.

TLDR: individual atoms do not touch or even come close to each other in any significant sense, in everyday regimes.

Now take your macroscopic ball and tube. They are made up of atoms in some kind of 3-dimensional grid structure depending on material. No matter how that exactly works (for example, metals bond differently than other materials), one constant would be that there will always be a (now relatively thin) cloud of electrons on the outside. (Remember, we are always talking about uncharged, everyday materials here.)

The same concept as before applies. If two macroscopic objects "touch", what happens in truth is that their outermost atoms (obv. a great great many of them) get ever closer and closer, until eventually they are close enough that the electromagnetic interaction between the electron clouds of the closest / outermost atoms becomes significant enough to repel. Let's also just now ignore more complex scenarios like very porous, fluid or otherwise different materials (no velcro straps ;) ).

So to get to the final point: it is impossible to have your ball have "exactly the same diameter" as your pipe, as there is no well-defined diameter. There will always be a wishy-washy region where the objects are maybe a little closer than before, or where the forces become too large to continue. If you "press harder", this region might shrink. If the objects start out with significant kinetic energy, the region might shrink. Etc. but physically, the objects will never "touch" at all, unless we get into destructive regimes.

As you can imagine, none of this is up for easy, exact mathematic treatment.

Check out the video "Do Atoms Ever Touch?" from the Sixty Symbols channel for the same explanation in a much more entertaining way.

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    $\begingroup$ Atoms do come close enough to merge in everyday use. What prevents them from getting entangled in metals is the oxide layer that forms in the atmosphere. If you remove the process of oxidation, as in space, things will get stuck together, unless you take special precautions. You don't need CERN levels either, a little bit of heat and ultrasound concentrated on a needle will do the trick. $\endgroup$
    – user71659
    Commented Sep 24 at 19:23
  • $\begingroup$ Two hydrogen atoms will bond when they get close and form H2.... Nobel gases will not, but the reasons for that are the full valence shells and cannot be explained by simple electromagnetism. Simple(ish) electromagnetism leads to the Van der Waals force/bonding which again is not repulsion. $\endgroup$
    – Eph
    Commented Sep 25 at 12:56
  • $\begingroup$ For forces that fall of at $\frac{1}{r^2}$ like electromagnetic charge, and gravity, a spherically symmetric distribution of charge/mass is equivalent to a point charge/mass at the center. So if the electron cloud remained symmetric as atoms approached each other, then the attraction and repulsion would continue to cancel exactly as they approached. However, oscillations in the electron clouds can and do harmonize with each other causing a net attractive force called the Van der Waals force. There is no electro static repulsion the pushes the atoms apart. $\endgroup$
    – Eph
    Commented Sep 25 at 13:50
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    $\begingroup$ All of the talk of atoms is rather irrelevant at the scale of a pipe though. Solid mechanics that treats objects as flexible continuous mediums can describe slip and press fits quite well even with well defined diameters. Contact mechanics is much more relevant to the question than anything to do with atoms. $\endgroup$
    – Eph
    Commented Sep 25 at 13:57
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    $\begingroup$ @eph, this answer simply (well, kind of, considering its length ;) ) should dispel the notion of two objects having the "exact" same dimensions. I think I got carried away a bit trying to simplify stuff to try and bring a general idea across. I agree with some of the comments where I made obvious errors (i.e. with naming the H atoms, which obviously would not behave like described here whatsoever). I'm dithering whether to improve or delete this answer at the moment. $\endgroup$
    – AnoE
    Commented Sep 26 at 12:44
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Mathematically it depends on whether the outer diameter of the ball and/or the inner diameter of the pipe include the points that correspond exactly to the ball radius. I.e., if the density of the ball is $$ \rho=\begin{cases} \rho_0, \text{ if } r< R,\\ 0, \text{ if } r \geq R \end{cases}, $$ whereas the pipe fills the space: $$ \rho=\begin{cases} 0, \text{ if } r\leq R,\\ \rho_1, \text{ if } r > R \end{cases}, $$ then the ball passes. But if we interchange the strict and non-strict inequalities above, then the points for $r=R$ belong to both the ball and the pipe, and the ball passes not. (However, if we interchange inequalities only in one of the equations above, the ball still passes.)

Physically the ball and the pipe are made from atoms, so their surface is necessarily not perfect. We will have to deal with the ruggedness of these surfaces, the atomic motion, the interactions between the atoms of the ball and the pipe, and other effects - all this renders the question ambiguous (if not meaningless.)

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Let me add a note in exterior influences that the other answers have not already mention.

If you let the perfectly fitting ball (by perfectly fitting, I mean that we idealize the situation such that the ball fits in the pipe with no air gap around it) fall into an open pipe then it might move through, under all the idealized assumptions of perfect roundness et. But it would not fall into a closed pipe. The air pressure would stop it as the air cannot escape around it as it falls. Thus, the open-pipe scenario is in fact not a good model for the bullet-in-gun-barrel, where the escaping air stream plays a large factor.

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    $\begingroup$ Perfect pipe and perfect ball will develop adhesion forces of some kind and in fact they will become one body. $\endgroup$
    – user13964273
    Commented Sep 23 at 11:57
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Talking mathematically, it won't fit. This is because the infinitely thin boundary of the sphere overlaps with the infinitely thin cylinder.

Talking physically, we cannot measure absolute precision and also QM also intervenes in terms of the position of electrons of the atoms making up the sphere and the cylinder. Since we cannot measure with absolute precision,the question doesn't make sense here.

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  • $\begingroup$ I'm not sure if the math part is right. It "sounds" right...and yet, I can trivially draw it. (Or at least the 2d analogy of it, a circle inscribed into a plane.) That said, that's more of a math thing than a physics thing, and this is the physics site, so less important perhaps. $\endgroup$
    – Beska
    Commented Sep 23 at 13:13
  • $\begingroup$ Mathematically, this is a definitional question. This answer is correct under its own definition of overlap. I'm unsure if that definition is well-accepted by the mathematics community (though in either case, this answer would be constrained to its own definition of overlap/fit). $\endgroup$
    – Brian
    Commented Sep 23 at 13:49
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    $\begingroup$ Both the line segment (0, L) and [0, L] have length L. Mathematically, both the ball and pipe having diameter L does not actually require that both occupy the same points. It's not necessary that both occupy the boundary, it's possible that neither occupy the boundary yet have identical diameter. $\endgroup$
    – Nuclear Hoagie
    Commented Sep 23 at 14:52
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Actually, in the real world, statements like "A's length is exactly the same as B's" is not well-defined. Take a simplest example, you can't even say "This particle is right at point P", only that "it's in the neighborhood of this point". Idealized physical models are called "idealized" as we aren't able to realize them in our real world.

When we talk about "the diameter of the ball is exactly the same as the inside pipe diameter", we are actually talking about "the atoms on the edge of the ball are exactly by the side of the atoms on the edge of the inside pipe". As we discussed above, you can't well define this statement even in classical situation, let alone we have to consider electron cloud model if quantum effects involved.

In conclusion, any idealized physics model is actually not well-defined in real world, so the word "exactly" can only exist in our imagination and mathematical models. Hence there's no way to talk about two objects of exactly the same diameter.

If you really want to know if the ball drop will roll down in this idealized model, I'll have to say that it totally depends on your own will and can't be solved in real terms of physics.

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A smoothbore gun does not lack in accuracy from the ball bouncing around inside the barrel, but rather from the aerodynamics of the ball after exiting the barrel.

The cartridge that wraps around the ball is typically much more compressible than the ball or the barrel, and in it's uncompressed state has a larger diameter than the barrel. However, it's close enough that it may be pushed into the barrel, compressing the cartridge so that it fills the space between the barrel and the ball. This prevents any rattling.

In general what happens when you stick a metal ball into a metal pipe of similar size depends greatly on what's in between. In the case of a musket you have the cartridge. If you had a nice lubricating oil you could likely get the ball to slide nicely down the pipe even if the ball is slightly larger than the hole. This is what's known as an interference fit. The greater the interference the more force would be required to force the ball through the pipe.

With nothing, not even air, between the ball and the pipe, it's possible that the ball and pipe will cold weld to each other effectively becoming one solid. However, cold welding is usually prevented by debris or oxide layers so it would be hard to actually achieve this.

All of these details are true regardless of the fact that both the ball and pipe typically will have surface roughness of at least tens of microns, which would be peaks and valleys hundreds of thousands of atoms tall.

To learn more about interference fits you can also look at contact mechanics which helps determine what will happen when two things start pressing against one another.

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  • $\begingroup$ Tens of microns is actually pretty rough for a metal surface. It's fairly easy to polish a surface down to submicrometer roughness. The valleys will still be hundreds of atoms deep, but not many thousands (not that this makes much of a difference for the point you're making). - The engineering tolerance of a surface is of course a different story, there 10 microns is a pretty good figure. $\endgroup$
    – leftaroundabout
    Commented Sep 26 at 13:19
  • $\begingroup$ @leftaroundabout Admittedly, I was using the high end of the surface roughness reference chart I have at my desk, which corresponds to mostly unpolished surfaces. I'm not sure how much polishing mucket balls get, but I was mostly just contrasting with the answers that were talking about the atoms as if the pipe and ball would be anywhere close to atom smooth. $\endgroup$
    – Eph
    Commented Sep 26 at 13:35
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This is a very practical problem faced by engineers and technicians daily, what size shaft and bearing to specify for a particular behaviour when they are mated.

When you buy components for a particular diameter, you can specify a class of tolerances such that they can slip fit with a gap, through press fit, all the way to fit when the bearing is electrically heated and the shaft cooled by LN2, then seize together permanently once they have cooled. Look up 'bearing tolerances' for the ranges available.

If you are not interested in real tolerances and materials, but a spherical cow in a vacuum, then they will behave as ideally as you can argue. You might argue mathematically, is the radius dimension included or excluded from the range that the part occupies. Or perhaps from physics, where is the 'surface' in terms of electron cloud density, or in terms of forces as surfaces approach each other.

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It will fit. Cut a circle in a piece of paper. Measure the diameter of the round piece of paper. Measure the diameter of the hole in the paper. They will be the same.

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    $\begingroup$ I like this intuitionist approach. In the limit of an arbitrarily fine blade, there is no gap between the circle and the hole (meaning they are of identical size), yet the circle obviously fits in the hole. $\endgroup$
    – Nuclear Hoagie
    Commented Sep 24 at 18:19

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