# The physics of that gravity-defying chain of metal beads

Here’s a pretty mind blowing video. It was made by Steve Mould, who’s a science presenter and comic.

I was totally baffled when I first saw that. It’s so surprising that many sources covering this video assumed that the beads were actually magnets, presumably because that would make this strange phenomenon easier to swallow. But they aren’t magnets – what you’re seeing is just a boring old chain of metal beads, the kind that you might have at home hanging from blinds or from ceiling fans. (You can buy it here.) Which makes it even stranger.

So what’s going on in this incredible video? How does a seemingly unremarkable chain of metal beads somehow appear to defy gravity? The physics nerd in me had to find out. Fortunately, there’s an even more stunning slow-motion video where Steve offers us an explanation.

Look at it as a sort of tug-of-war. You can see the outer chain is going to be travelling really quickly as it falls, which means the inner chain is going to be travelling really quickly, as well. And if you’ve got something traveling really quickly, it’s got momentum… So you’ve got the inner chain traveling up, but it wants to change so it’s traveling down, but it can’t do that in an instant, because that would require infinite force.

Instead what it does is it changes direction slowly over the course of a loop, so that’s why it almost has to be a loop, because it needs that time and it needs that space to change directions.

That’s a nice explanation, but can we take it further. I’m in a particularly empirically zeal-ous mood, so in the spirit of this blog, let’s do a calculation and work out if this explanation fits the data.

The first thing we need to know is what momentum is. Momentum is a measure of how much stuff an object has, and how fast that stuff is moving. Something that’s fast and heavy has a lot of momentum. Something that’s light and slow has very little momentum. Mathematically, momentum is just the product of an object’s mass and its velocity.

momentum = mass x velocity

Or, as physicists like to write it:

Here, p is just shorthand for momentum.. don’t ask me why. m is mass, and v is velocity.

The second thing we need to know is that whenever an object changes its momentum, it experiences a force. If you think about it, this is pretty intuitive. For example, if you throw a tennis ball at a wall, its momentum changed from positive (forward) to negative (backward) when it bounces off the wall, because the wall slammed in to it (that’s the force). If you were to slap the table in front of you, the momentum of your hand would go from something to nothing. The pain you’d feel is the direct result of the force that brought about this change in momentum.

Now, in this bead-chain, just as Steve described, the inner part of chain is traveling upwards, and then suddenly, at the top, its being pulled downwards. As each beads turn the bend, its change in momentum causes a little upwards kick of a force. There are many beads in the chain, and so there’s a constant stream of little upward kicks, as the beads go around the bend. And it turns out that these kicks provide just enough of an upwards force to balance out the weight of the suspended part of the chain. That’s why the chain seems to hover in mid-air – it’s because the changing momentum of the chain provides a force that keeps it up.

This isn’t magic. It’s the same physics that’s behind this crazy water-powered jetpack. In the jetpack, a constant stream of water suddenly changes direction at a bend in the pipe, providing a steady stream of upwards kicks. This is what keeps Derek hovering in the air in that video, and it’s the same physics that keeps this chain afloat.

So how can we make this quantitative? Well, Newton figured out exactly how force is related to a change in momentum. What he taught us is that the force on an object equals the change in momentum divided by the time over which the momentum changes. Or,

force = change in momentum/duration

So if we know the change in momentum of the chain, we can work out the force.

Now, picture a tiny piece of chain that’s moving up. Its length is delta x, and it’s moving up at a speed v. If we call m the mass per unit length of chain, then the momentum of this upwards moving chunk of the chain is:

A moment later, this piece of chain goes around the bend, and is moving downwards. Its new momentum is now

(the negative sign is there because its velocity is now negative)

So now, we can work out the change in momentum, and how much force this results in.

We just worked out that the change in momentum of the chain provides an upwards kick of force, and the strength of that force is

But for this chain to be suspended in air, this force has to exactly counteract the weight of the suspended part of the chain. So let’s just set those two things equal to each other. (Remember, the weight of an object is just its mass multiplied by a conversion factor of g = 9.8 Newtons/kilogram)

And so, we’ve arrived at an equation relating the speed of the chain to the height of the beaker (H) and the length of the hump (L).

In words, it says that if you take the square root of the length of the suspended part of the chain (in meters), and multiply it times 2.2, you’ll get the speed of the chain (in meters/second). So the speed depends on the square root of the suspended length of chain.

Now we’re ready to play. Let’s plug in numbers into this equation from the video, and the see if our predicted speed of the chain matches the real, observed speed of the chain in the video.

In the first video, Steve tells us that his chain of beads is 50 meters long. I timed how long it took to fall, and it was about 13 seconds. Dividing the two, we get an average speed of 3.8 meters/second. What about the prediction? To find out, we need the height of the beaker and the height of the bump in the chain. To do this, I just imitated his pose and used a tape measure to get the lengths, correcting for the fact that Steve is a couple of inches taller than me.

I get that the height of the beaker is H = 1.36 meters and the height of the hump L = 0.3 meters. Plugging this in gives a predicted speed of 3.1 meters/second. That’s within 20% of the measured speed. Not too shabby, but we can do better.

For the next step, I used the open source physics software package Tracker to analyze the slow motion videos. First up is this bit of video:

I tracked the motion of a single bead falling in the chain, and plotted its position over time. To calibrate the length scale in the video, I assumed that Steve’s eyebrows are 14 cm away from his chin (for the simple reason that this is true for my face :)).

By fitting the trajectory of this motion to a straight line, I can get the speed of the chain from the slope of that line. That speed turns out to be 2.83 m/s.

That’s the measured speed. How about the prediction? To get this, I need to find H (height of the beaker) and L (height of the hump) once again, which I did just like before. I found H = 1.18 meters and L = 0.25 meters. The kind folks at BBC’s Earth Unplugged were nice enough to tell me the frame rate in that clip is 2000 FPS.

Putting this all together, here’s what I found:

The observed speed of the chain from the video was 2.83 meters/second

The calculated speed of the chain was 2.87 meters/second

That’s within 1.5% of the measured speed! Way closer than we have any right to expect.

Maybe I just got lucky? Let’s try another segment of video. This time, Steve holds the beaker up higher.

Here, again is the trajectory of a bead on the chain. It fits nicely to a straight line. (I didn’t use data from the beginning of this curve, because I think that the chain is moving towards the camera then, and so we’re not able to capture its entire speed. Later on, it seems like the chain is falling in the plane of the camera, so we get its true speed.)

The slope of that line tells me that the observed speed of the chain was 3.23 meters/second

How about the calculated speed? Like before, I need to find H and L. I estimated that H = 1.83 meters and L = 0.145 meters (To calibrate the length scale in that video, I used the fact that the length of a 1 liter glass beaker is 14.5 cm tall).

Plugging these numbers is, we get that the calculated speed of the chain was 3.22 meters/second.

That’s within a half of a percent of the observed value. To be honest, that accuracy is a bit coincidental. If I changed which part of the data I fit to a line, and assume that the camera’s perspective distorts the length L, this discrepancy increases. But it still stays within five percent or so. So I’d conclude that this model holds up admirably to the test of experiment.

What did I learn from this analysis?

First, science works! You can take a mind-bending video and analyze it using physics, and if your model works, you should be able to predict the outcome with reasonable accuracy. It took  three or four failed attempts before getting to the equation in this blog post, and so science also has a way of smacking you in the head and telling you that you’re wrong. 🙂

Second, this falling chain isn’t accelerating. I think it moves at a constant speed. At least, this is what I assumed in the calculation, and it fits the data quite well. I don’t think I can explain why this has to be true. If it is true, however, it would make the video seem all the more eerie and add to the strange hovering effect. (Normal falling objects don’t fall at a constant speed, they accelerate downwards.) Update (July 2): I added a bit to the technical physicsy discussion below to prove this point. The main idea is that as this chain  accelerates, it approaches its equilibrum speed, and once it gets there, all the forces on it are perfectly balanced, so it stops accelerating.

Third, the mass of the beads don’t matter at all. If you look at the equation predicting the speed of the chain, you’ll see that the mass doesn’t show up anywhere. This is because a chain of beads that’s twice as heavy would also have twice the momentum kick. These two effects cancel each other out, and the motion is independent of the chain’s weight. This is a prediction of this model, and we could easily test it out with a chain of plastic beads.

And so, in conclusion, Steve’s explanation of the video is remarkably spot-on, and the model based on this idea is able to explain the speed of the chain to within a few percent of accuracy. So thanks, Steve, for blowing my mind and for making my world a little more interesting!

Hat-tip to Kyle Hill for putting me on to this video, whose twitter feed is an endless source of fascinating links. And I owe a thanks to the creator of this video Steve Mould, and the slow-mo video team at BBC’s Earth Unplugged for providing me some of the numbers I needed for this.

### Geeky Physicsy Afterword (here be dragons):

I played a little fast and loose in my derivation of the velocity in this blog post. For those of you with a background in physics, here’s a more careful derivation that gives the same answer.

Break the chain into three parts: an upwards moving part of length L, a downwards moving part of length H+L, and a small chunk at the very top. Let m (or mu) be the linear mass density of the chain.

For the upwards moving part, there are three forces acting on it. The tension at the top (T1), the weight of the chain (m L g), and the contact force at the bottom of the chain. At the bottom, where the chain start moving, a chunk of chain goes from having zero momentum to having momentum m dx v in a time dt. The force it takes to do this is dp/dt = m dx v / dt = mu v^2 (where we’ve used dt = dx/v). So this contact force is m v^2. This contact force should be pointed downwards, because it pulls the chain down.

For the downwards moving part of the chain, the three forces are the tension (T2), the weight of the chain (m (L+H) g) and the contact force, which like before is m v^2, except this time it’s pointed upwards.

Lastly, let’s look at the topmost chunk of the chain. The forces on this chunk are T1 and T2, and they are matched by the force caused by the change in momentum, which is 2 m v^2.

So we’ve got a set of three equations:

Solving these three equations gives us the speed in terms of L and H:

The reason that this is the same answer as the simpler derivation in the blog post is that the two contact forces on the chain (m v^2) are equal and opposite, so they cancel out. In the end all that matters is the weight of the chain, and the force at the top.

Update (July 2):

OK, so we’ve solved for the speed at which the chain will be perfectly balance. But how does the chain know to arrive at that exact equilibrium speed, where the forces will cancel out and there’s no acceleration?

To work out how this happens, we need to study the general case where the chain is accelerating. Unlike the equations above, where I assumed zero acceleration, now F = m a.

So the three equations become modified into:

Whose solution is now:

We can rewrite this solution as a differential equation:

And the solution to this differential equation is (up to an overall minus sign, which I ignored):

Notice that as time passes, the speed of the chain approaches the same equilibrium speed from before!

Let’s make a plot of this equation. This will tell us what happens to the chain’s speed over time. Let’s say we start it from rest at various initial speeds, and that the equilibrium speed is 3 m/s, and H = 1 meter (not too far from the values in this videos).

Here’s what the plot of the speed equation looks like, for different values of the starting speed:

Isn’t that neat? No matter what speed you start the chain at, it will converge to its equilibrium speed in under a second!

• Mitch

Amazing as always, I look forward to the next post and thanks once again for the Twitter correction.

• Will Slaton

Nicely done! I’ll share your analysis with my physics club.

There are several “falling chain” classical mechanics problems. The bungee jumper being one, see here:

http://seniorphysics.com/physics/bungee_physics.pdf

Note: that article has a really nice demo of a chained block and an identical block falling together. The chained block falls at a greater acceleration! Something that could be measured with video analysis.

That said, I wonder if you can analyze the video of the chain’s motion before it hits the ground? This should also be a variable-mass problem and result in a vertical acceleration that is different that g. I agree that once the chain hits the ground it becomes a constant speed problem. I have a hunch that the acceleration is less than g in this case.

Aside: All this business of chains hitting the ground results in some interesting physics. See here: http://ruina.tam.cornell.edu/research/topics/fallingchains/ Reference and link from that page to an excellent AJP article on this.

Lastly, I am still uncertain as to why the loop at the top gets started in the first place. Ideas? I suspect that, as the slo-mo video points out, that collisions at the point of impact propagate back up the chain (and collisions of the chain with the rim). I think the tension in the chain will also depend on height and hence, the transverse wave speed will depend on height. A small pulse at the bottom will propagate from a low to a high wave speed as it ascends the falling chain. I haven’t worked it out, but if it is similar to reflection/transmission phenomena for two attached strings with different wave speeds, then the amplitude of the pulse should increase as it goes up the chain. (Also, if the chain was stationary the standing traverse waves on it would be Bessel functions – maybe in the moving case too, especially if its moving with constant velocity.) So, this means the top-loop shouldn’t be formed until the chain starts to hit the ground – not sure if the videos show this due to how they framed the scene.

Anyway, lots to explore with this mechanical system! Thanks again!

• Will Slaton

After thinking about it some more I am not convinced that the pulses traveling up the chain result in the vertical loop – they probably contribute the the wiggles of the chain.

It would be very interesting to see how the vertical loop develops when the chain is pulled from a flat table, a low dish, from a slightly steeper bowl, and then an almost vertical container. I think now think the initial vertical loop gets going because the chain has to go UP to get out of the beaker.

• Right – I feel the same way. I couldn’t come up with a way to predict the height of the hump L, which leads me to believe that you might be able to vary it. Say you just pull the hump higher as the chain falls — will it stay in this new position? It’s possible for this to happen, as long as the chain speeds up to compensate. Is there a limit to how large L can be? Much to play with here.

• Jesse Crossen

I think L has to be variable, just like it is in a hydraulic siphon, and theoretically it should be able to approach H up to the point where the friction of bending the chain overwhelms the energy produced by converting mgH into kinetic energy. Intuitively I thought that increasing H would increase L as well, but your analysis shows L going from about 1/5H to 1/12H even as H is increased. I think the easiest way to raise the loop might be to slowly lower the beaker while the chain is moving, just like you might do with a hydraulic siphon. I’m really tempted to buy some chain now and try this.

• pablortega

Momentum is derived from impetus and impetus from in+petere. that’s why p is used 🙂
Etymology
From Latin impetus (“a rushing upon, an attack, assault, onset”), from impetere (“to rush upon, attack”), from in (“upon”) + petere (“to seek, fall upon”).

• Thanks for that. I always assumed it was because m was already taken for mass. 🙂

Then again, scientists pretty much seem to assume it was chosen because m was taken and p was left over at the time. Argonne Nat Lab’s “Ask A Scientist”: http://www.newton.dep.anl.gov/askasci/phy00/phy00725.htm

Originally, apparently it was M (from “motu/motus” probably):

“The first correct statement of the law of conservation of momentum was by English mathematician John Wallis in his 1670 work, Mechanica sive De Motu, Tractatus Geometricus: “the initial state of the body, either of rest or of motion, will persist” and “If the force is greater than the resistance, motion will result”.[52] Wallis uses momentum and vis for force. Newton’s Philosophiæ Naturalis Principia Mathematica, when it was first published in 1687, showed a similar casting around for words to use for the mathematical momentum. His Definition II defines quantitas motus, “quantity of motion”, as “arising from the velocity and quantity of matter conjointly”, which identifies it as momentum.[53] Thus when in Law II he refers to mutatio motus, “change of motion”, being proportional to the force impressed, he is generally taken to mean momentum and not motion.[54] It remained only to assign a standard term to the quantity of motion. The first use of “momentum” in its proper mathematical sense is not clear but by the time of Jenning’s Miscellanea in 1721, four years before the final edition of Newton’s Principia Mathematica, momentum M or “quantity of motion” was being defined for students as “a rectangle”, the product of Q and V, where Q is “quantity of material” and V is “velocity”, s/t.[55]” [ http://en.wikipedia.org/wiki/Momentum#History_of_the_concept ]

Also, perusing that Wikipedia article, it seems to me the physicists at the time of choosing “momentum” as term was deliberately trying to separate the physics of motion (la: motus) from the flawed philosophic concept of “impetus”.

• andre

“Second, this falling chain isn’t accelerating.”
I only find this intuitive. Gravity is pulling down on the chain, and the ‘turning’ part is pulling up the chain, these forces balance out, so there’s no net force on the chain so there’s no acceleration.

• I agree that if the forces are balanced, then the chain won’t accelerate (F=ma where F is zero). But the thing that puzzles me is why are the forces exactly balanced? It seems like a strangely fine tuned situation. Maybe whenever there’s a non-zero acceleration, this causes the speed of the chain to increase or decrease until it reaches the speed that makes the force balance.

• Jesse Crossen

I think it’s more intuitive if you think about it from an energy standpoint instead of a force standpoint. If the free length of chain is considered as the “falling object”, you can see that it isn’t actually falling, since its center of mass stays at a constant height. So, no gravitational potential energy can be converted to kinetic energy. I guess that begs the question of why the system becomes stable in the first place, but note that the chain does accelerate as the leading length drops toward the floor, then holds at a steady velocity, and will accelerate again once the tail length comes out of the beaker. So it might simply accelerate until the energy generated by a falling beads equals the energy to lift a bead out of the beaker. Once no more beads are being lifted out of the beaker, then it’s actually in free fall. I think the same thing might happen if the beads were passing through some kind of friction brake, in which case they would only accelerate until the energy released by a falling bead was the same as the heat generated by the brake. The height and friction of the brake would then control the final velocity, but in that case the mass term would not be cancelled out because the resistance is not based on lifting beads, and the “perfect balance” would seem more intuitive.

• Ben

I think this is right. In the steady state, the work done by gravity on each bead as it falls a height H (from the beaker to the ground) goes into heat at the brake and at the ground. If the bead chain is moving at a constant speed v, then heat is generated at the ground at a rate of 1/2 (mu)v^2 and at the brake at a rate Fv, where F is the average force of friction on the beads as they are pulled out of the beaker pile.

Then (mu)gH = 1/2(mu)v^2 + Fv.

• Ben

(Oops, forgot a factor of v in those rates! Should be

(mu)vgH = 1/2(mu)v^3 + Fv

and so

(mu)gH = 1/2(mu)v^2 + F

• Hey, I updated the post with more details on this.

• Deepak

Regarding the latest update, I feel like it might be necessary to allow for the length L to vary. By keeping L fixed, you’re essentially keeping the weight that needs to be supported by the momentum change (and therefore velocity) fixed, so it kinda makes sense that the velocity will always saturate to this value: if its slower, the “kick force” is less, gravity takes over and it speeds up, etc. But if you let the length L change, then a slower speed can be accommodated by a shorter length. Though its not nearly obvious how I should relate L(t) with anything else in the problem.

• Deepak

Ok, so if your top piece was a length \$ds\$, then you’d have \$2mu v^2 – T_1-T_2 = 2mu ds frac{d^2 L}{dt^2}\$. But in your model the \$ds=0\$. In any case, we’ll need one more equation if we let L vary… I’m trying to work the thing out assuming it makes a semicircular arc of radius L at the top – but something looks funny… Will get back if I have any luck..

• Deepak

Actually, after some thought, when the speeds haven’t hit steady state yet, there is no reason why the speed of the up piece and the down piece need to be the same. And their difference will make the “loop” grow. Just like if you whip a rope and hold one end fixed, the other end still whips way up sort of piling on the fixed end.

So now we have two tensions, two speeds, and L, and four equations. Gotta search for the fifth one…

It is a good idea, but I think you may be overdoing the analysis. Perturbing the derived v(t), we see that dL can be ignored compared to da, as long as H >> L.

I would think that condition needs to apply to set the bead chain in motion anyway, seeing the other comments on the loop.

• Will Slaton

• Ben

Could you include a diagram with vectors that shows what each of the tensions and the “contact force” correspond to? I’m having trouble understanding this derivation.

For example, why are T1 and T2 in the third equation the same as the T1 and T2 in the first two equations? The tensions in each equation describe forces acting on different chunks of the chain, so there’s no reason to assume that T1 in the first equation should be the same as T1 in the third equation, and so on.

• Deepak

They’re on different chunks, but are acting at the same interface, and so have to have the same value, but of course opposite directions. Newton’s third law.

• Ben

After reading the paragraph describing the system a few times I see that that’s true, cool. I’m still confused what this “contact force” refers to. It seems to be just the mass flow term in dp/dt = v(dm/dt) + m(dv/dt), but why it’s referred to as a contact force is confusing, nor do I understand how the signs of these terms are derived in each equation. :

• Hi Ben, I apologize if I was being overly cryptic – that physicsy section was written in a rush. I’m attaching an image that better explains the setup.

The contact force are the forces on the vertical part of the chain either due to the ground or due to the part of the chain that’s at rest. I found it intuitive that the ground provides an upwards force to the chain (pushing it upwards) whereas the chain that’s at rest provides a downwards force (‘pulling down’ on the chain as it moves up).

However, you could also derive the signs without relying on intuition – you just have to consider a segment of chain and look at dp = p_final – p_initial. In the case of the chain hitting the ground, p_final = 0, p_initial = -mu dx v, so dp = +mu dx v, so the force is positive and = dp/dt = mu v^2.

Hope that is a little clearer, and let me know if anything is still confusing.

• It might also help to look at the solution for the weight of falling chain as measured on a scale. The key point is you can solve for the normal force on the vertical part of the chain (which I call the contact force) using F = dp/dt

http://www.feynmanlectures.info/solutions/falling_chain_sol_1.pdf

• gordon judd

I am always skeptical of intuitive concepts that involve pushing on string. As given in the arXiv:1005.2887 paper the curl on the end of a falling chain will produce added tension in the chain that will increase its acceleration not decrease it.

To get a better idea of the tangential velocity profile of the chain I decide it would make sense to just measure it. I just painted markers on the chain and the measured the time it took for those markers to cross a horizontal reference line. That gives a distance vs time curve and the derivative of the resulting distance vs time curve will give the velocity. The video used to make that measurement is at

From this plot you can see that for a chain falling from a container with H=1.53 and a nominal loop height of L=.22 the steady state velocity was around 4.3 m/s which is 38% higher than your value of 3.1 m/s.

Thus I think it will take a 2 dimensional ODE to solve this problem and it will involve the concepts in the slack dynamics paper to get real values for the force involved in producing the complicated force over time function to produce the momentum change in pulling the chain off the pile.

• Will Slaton

Colleague of mind found an American Journal of Physics article that talks about the physics of how the loop at the top develops. “The dynamics of a falling chain: II,” M.G. Calkin, AJP 57 (2), Feb 1989, pgs 157-159.

• septerr

Impressed

• Great post Aatish, it has inspired me to scribble mechanics again!
A small query: In the video, when the glass is held higher, they say “As you can see it goes much higher”, and it looks as if it does. Yet, in your video analyses, in the first case L=0.25 m, and in the second case L=0.145 m, i.e. you are saying that it goes much lower in the second case… It seems like it gains height after a while in the video. Could it be that the video-segment you used for getting L is not really representative, or is it really so counter intuitive that L becomes smaller when H becomes larger?

• falconphysics

I’m a bit confused by the constant velocity. When I look at the graphs you have from tracker they seem to show a parabolic curve, which I would expect for accelerating masses.

Additionally, I have a video (I’ll upload to YouTube later) that I took of 12 feet of beaded chain falling captured at 500 fps. In the beginning the beads are each crisp and clear but the motion blur increases as the beads leave the container and by the end the beads are a blur.

I was wondering if this might better be modeled as an Atwood’s Machine?

• natson wang

It is not necessary to talk about the force. Let’s talk about speed. The arch only form when the beads in the cup going up faster than beads falling down. But how come this happened? It happened when the dropping bead were interrupted by cup edge. The beads kept touching cup edge and arch sustained. This video is not real because the beads fly out without touching the cup. The real video is here:

• gordon judd

“This is a prediction of this model, and we could easily test it out with a chain of plastic beads.”

In doing that you find the linear mass density of the chain appears to have an effect on the height of the chain above the pile of chain. Also at least for a plastic chain the constraint imposed by the height of the container also has an effect. Taller containers will produce higher loop heights above the pile of chain in the jar.

Can you give instructions on how to include a .jpg picture in a post? I will show some pictures that compare the height observed for a falling plastic chain compared to a chain made of brass.

• Hi – when you type in a comment here, there is a small photo button in the bottom left of the box. You can click on it to upload a photo.

• gordon judd

Thank you. Those photos are shown above, As you can see from those frame grabs, the linear mass density makes quite a difference in the height of the loop produced for the same height (1.5 m) of the container.

• gordon judd

“It would be very interesting to see how the vertical loop develops when the chain is pulled from a flat table,”

A horizontal pulling on a chain laid out in a serpentine fashion will produce a vertical loop as well as described here:

Slack Dynamics on an Unfurling String:http://arxiv.org/abs/1202.0795

That analysis is so complicated that I hope someone here can explain its concepts in a more understandable form. I understand the premise of using the wave equation and finding that:

“The shape X itself is simply one of two waves, namely that which travels at speed

• gordon judd

Here is screen grab taken from a high speed video of the motion for a falling piece of plastic bead chain having a linear mass density of around .011 kg/m. The bottom of the 9 cm tall glass was 1.5 above the ground. It appears the velocity of the end of the chain was around .5 m/s as it started its fall so that would be in the ball park of the velocity going around the loop prior to straightening as well.

The second photo shows the loop observed for a #6 brass chain having a rho_l of around .049 kg.m. It also had a faster end velocity of around 1.1 m/s.

• hughhunt

This is all pretty good – well done – but I don’t think the explanation here is complete. Take a look at http://www.youtube.com/watch?v=FmyZkHFYcNc . I don’t think the theory proposed here can explain why the chain lifts off a flat table.

• hughhunt

In fact, look closely at these vids, and this one too http://youtu.be/6_KdvJfImXw and you’ll see that the chain gets pushed to the left. This
means that there is a force acting on the chain to the right (Newton
III) – so we’re nearing an explanation for what holds the self-siphoning
chain up in the air.

• hughhunt

The other thing that needs to be explained is what equation you would use to determine L given H and g ? If you take moments of all forces about the centre of the semicircular loop you get that L=0, which is not observed. There is something else going on, and it is certainly to do with coriolis forces that are generated as waves go around the corner of a translating chain. You can see here http://www.youtube.com/watch?v=Tf0_r7wl6_A when the chain is released through the neck of a bottle that the loop doesn’t occur. Here wave motion in the chain is constrained.

• Manuel

When you consider the the topmost chunk of the chain you say 3 forces act on it, T1, T2 and the change in momentum. Why is the weight not considered?

• ArcusTangens

Not being a physicist, I would like to apologize beforehand if this question is boring, obvious, or stupid.
In short: would the observed effects be found if the beaded chain was replaced by a rope? I.e., is the effect dependent on the granularity or segmentation of a chain?

I was very intrigued by the paper illustrating that a chained block in free fall falls more rapidly than a similar, untethered block. However, neither in the self-siphoning chain or in the chained block experiments have I been able to find examples where ropes, strings, or other non-segmented chain-like objects have been substituted for the chain. The tentative explanation for the upward loop in the self-siphoning chain given in this video (http://www.youtube.com/watch?v=6ukMId5fIi0), that the momentum of the upward moving part of the chain will translate into an upward curve, seems not to rely on the chain being, well, a chain, as opposed to a rope. I have this nagging feeling that this may be wrong, but given my lack of expertize, I don’t trust my intuition. Anyone around who can put this to rest?